"Showing your work" is why most of the potential answers would fail. Drawing that square in arbitrary position is not at all straightforward. The centered diamond, the centered square, or the square in 1 corner can all be done easily with compass and straightedge. The diamond can be done with straightedge only. Anything else is going to be more pain than it's worth.
Back in the early 70's, I was in EBP / MGM programs. I was in the 3rd and 4th grades doing up to 7th grade level work. Education was beginning to go through its defunding period and my programs were the first to get cut. So, in the 5th grade, doing 5th grade level work, my teacher would give me a failing grade because my work to solve problems were more advanced, not what she taught in the class. I got very bored in school when those programs got cut. BTW in the 4th grade, we were working on dodecahedrons, not squares lol.
What I was hoping was an explanation for a way to construct a square in a corner, using a square root of 2 as the radius of a circle to draw the intersections with the existing lines to draw the shape of the shaded area. The other solutions aren't as simple to construct as the first answer and a square in a corner answer.
Which usually isn't the case. If your answer is different than what's given in the teacher's edition of the book, it's wrong. Most teachers lack the ability to analyze different solutions.
True, but schools teach you to always "show how you reached your conclusion" so by that overarching rule you'd know that's not enough. Technically, you could answer every math question in your final year correctly, but by only writing down the final numbers and thereby failing anyway
@@migssdz7287 I occasionally posed problems where the idea is to check an equality. If a student tries the problem and botches it, he gets partial credit. Or if the problem has a derivative and the student has trouble with derivatives, he also gets partial credit. The way to get zero is simply not to take the derivative or otherwise check the equality, but instead do something fancy that shows himself clueless. (If the question is to check if x=2 satisfies x+2 = 5, all too often, it doesn't occur to the student to add 2+2 and answer "no" based on the result. In this case, a "no" by itself is good.)
That was my first inclination. But then, knowing that the task is possible but realizing that I could never actually accomplish, I changed my mind to "no."
My solution was to draw a crappy square, declare the shaded and unshaded areas equal, put out a warning that drawing is not to scale, and call it a day
@@haroldhayes4824 simplifying by taking parts of an equation and declaring it as something is often helpful. My default was always X, Y, Z, A, B, C as needed. I got so locked into this pattern that I once redefined every variable in a problem like this and my prof left me a note telling me I should break the habit lol
@@VocalMabiMapleI remember problems in college where I ran out of Latin and Greek letters and ended up using some hiragana. I suck at making simplifying assumptions.
I did the algebra to find that if you shaded a border ~0.29289 blocks thick, you would shade half and get a perfect square, then realized triangles exist...
my immediate thought when i saw the thumbnail was, just shade half of each four squares. mind you i flunked math, im an art person. i think its funny how different our brains can work. theres no way i could do the math and figure that out. the only math thing i remember, and i like to say to people sometimes, is a few digits of pi. off the top of my head i remember 3.14159265358979323846264338327. i want to learn more, it would be easy to memorise since its already segmented in my mind. once i completely memorise the second segment, i could keep building on it. theoretically. im actually really lazy and dont want to put in the work. thanks school, i remembered one thing but you burnt me out too much!
What is happening, is that WE are assuming limitations that were not stated. I did it to myself for about the first 45 seconds of this video ( a little longer because I paused it to look over the Tweet. ) And the "textbook answer" was the easy answer, as it can be done without any of the actual math, and even without a ruler if you are good at straight lines.( I am not... ) And as James said, the way the question was asked, the technically correct answer, is YES. As the question does not state that we are supposed to show or demonstrate the a solution.
@@janelle9998being able to problem solve, use logical thinking, and project geometric shapes is extremely useful in life. I’m not sure why one would assume the opposite.
I completely missed the simple version of drawing a diamond and went right for the smaller square in the middle, which is harder to accurately measure.
You just draw a unit circle in the center, then construct the diagonals, and connect the intersection points and you have your square. My solution would have been to set the compass to sqrt(2) via a diagonal of a small square, then use it to measure what I need to chop off on two edges, putting the half area square in a corner
We shouldn't call a rotated square "a diamond". A diamond is an actual shape where sides are equal, but pairs of angles differ. If all 4 sides and angles are equal, it's a square, regardless of the orientation. So unless the teacher explains that it's not how you draw it, but the properties that define a shape, many kids will call a square rotated 45º "a diamond", and nobody will correct them, even as adults.
How do you make 1.4142 in geometry? This was my fist question. And I made separate triangle and copied to the squares. And the 1st solution never hit me, regardless I've been so close. I guess my mind is inside so many boxes...
@@YourNickIsTaken it only needs to be root 2 if you each smaller squares are 1x1, which was chosen arbitrarily by the video. We don't have access to the actual paper but if I did I'd've grabbed a ruler.
I thought way outside the box at first and then I realized you'd just need to bisect each square diagonally so that the lines create a new square that covers half of each square
My thought was to draw a diagonal of sqrt(2) length and rotate that to form a square of area 2. I like your solution a lot better - going to watch now, I bet that's Presh's technique.
@@Gideon_Judges6 Actually, my first thought was to use the standard tool of geometry problem authors. Cut the square straight across. You could say that's a rectangle, but you could also say it's a square - not drawn to scale! 🤪
As a 4th grade teacher, questions like this drive me crazy. I also hated Language Arts questions asking students how the words an author used made them feel, since anything they wrote down was a valid answer.
It’s weird because the question says “Can you shade half of it so the unshaded part is *also* a square?” It’s just bad question writing like this that trips up people on tests. It makes you think the shaded part also has to be a square.
Took me a while, gone as far as to calculate the square root of 2. But then eventually, was looking at different ways to shade precisely the half, until realization has come. Very nice riddle, I am so glad I didn’t continue watching the video before resolving it on my own.
The second solution is what came to my mind, along with the thinking of "people probably aren't getting this because they're probably assuming that the shaded area also has to be square somehow". The first solution with the rotated square is actually so incredibly obvious once you see it that I'm ashamed I didn't think of it instead.
The first solution is the easiest to draw and the easiest to understand. Like you, I came up with the second solution, but it's not as easy to draw to scale, but it's easy to provide the math to explain where the square sits. Once I saw that, I realized that there are an infinite number of solutions that just need the math shown to prove they are correct.
I did one of the solutions where the square is on one of the four corners of the bigger square. So I also kind of made the obvious solution into a little bit more complicated one for no reason lol.
this is what i thought too lol im 20. now idk if that’s the answer the creators of this puzzle were looking for but it is AN answer, and its a correct one so 🤷🏻♀️
Okay but what about the cross in the middle? Cause that’s I thought too but there’s that division in the middle. So no, you won’t have a square, you’ll have 4 triangles.
Mine was the first answer (the one with triangles), I got it right after examining the problem in two seconds, I don’t get how people think it’s hard, it’s one of the easiest things I ever solved
The L-shaped shading was the first thing that came to my mind after I studied the diagram and what the question was and was not asking for. I'm reminded of the classic test with "Read all instructions before marking your paper" at the beginning, progressing through increasingly nightmarishly conplex equations you haven't been taught yet and the last instruction is "Disregard all previous instructions, sign your name and hand in the otherwise-blank test."
yeah i remember getting one of those, i think it was like 3rd grade for me? never liked them much, i always had the mentality of focusing on what i currently had to do instead of worrying about future questions and stressing myself out. of course it bit me in the ass that time though as i was one of 24 idiots who were all looking for someone who's name started with L who was born in November and had pancakes for breakfast.
That kind of happened to me, but in a more normal way. It was an algebra 1 test, and had normal algebra 1 stuff. It told you to read directions carefully, and one of the questions said at the end to draw a smiley face. Not many people did it. I got a bit of extra credit.
The fact that so many adults have trouble with this type of creative problem solving is exactly why we should be giving these types of problems to our children.
to be fair, to really PROVE this solution is valid, you need geometry level education (pythagorean theorem, 45-45-90 triangles, and others more based on how rigorous you wanna be with the proofs), which i dont think 4th graders have
@@henry_3300You don't, if you draw a diagonal inside a square it's pretty damn obvious it's cutting it in half. Kindergartners can see that Of course you need higher education to prove it right, but that's not what the question is asking
@@henry_3300 Here's your super advanced proof then: Assume a square of side length 1 Cut it in its diagonal This will generate two congruent right triangles with sides 1 and hypotenuse √2 By that, it's correct to say both have equal areas as the triangles are equal
I solved it almost immediately. I also failed every single math class I’ve ever taken and I only did 2 yrs of High School. This just proves to me that all that extra math you did would just slow me down 😅
I don't think that's the confusion, I think it's the fact that people assume they have to make it symmetrical, while also making it a non-rotated square
I understood it to mean that the unshaded portion should be a square, half the size of the full square. The division into four smaller squares is only there to help me visually gauge the size of 'one half'.
The question did state half which means 50% of that must be shaded no more no less and that each portion must be a square the thing people are getting confused on is the fact that they look at the whole thing and assume that it is a big square which it is not it is for individual squares which means the simple answer is just to shade two squares which means you're left with a square two but still a square it did not say that you cannot have two it did insinuate that they're at least must be one. Such a simple problem it's just people inject information about the problem that isn't present and that isn't needed like it being a big square and that if you shade half you're left with a rectangle which you are not because they are two individual squares.
@@MrChrisRP My guess was that as well, but I say "guess" because I don't know exactly where to put the sides of the smaller square. (If you mean the one at 5:22). I would GUESS the sides go 1/3 of the way from the big square's edges to its center, but that's a total guess. Idk enough geometry for that.
I started to think that if I wanted an uprigth square, then if one square is 1x1, then the large square's area is 4, then half of it is 2, so the sides of the half is sqrt2, and that is side to diagonal ratio of a square and then the tilted square hit me. You just diagonally cut the corners, and those would make up two squares, so the middle is a half sized square. Maybe this was the aim of the problem. But it did take me five minutes
@frankmerrill2366 Exactly this. Sure, if the 4th grader actually had the knowledge and the tools to somehow make it geometrically accurate, then it could be valid. But otherwise, there is just one solution and it's the first one.
Divide each square into four squares, divide each of those into four squares, shade as required, specifically the outside ring of squares. The inside is a square, the number of shaded and unshaded squares are identical.
It’s weird because the question says “Can you shade half of it so the unshaded part is *also* a square?” It’s just bad question writing like this that trips up people on tests. It makes you think the shaded part also has to be a square.
The "square" below is made up of four small "squares" can you shade half so that the unshaded part is also a "square" The question is right they already talked about the existence of 5 other squares and thats why it says "also"
@@discussions. coping? about what? understanding basic grammar? or is it you that is coping and whats with the snarky remark if we are talking about this thing sensibly continue to do so or dont continue this conversation as people like that wont learn anything from it anyway
The unshaded part is still a square. it just has another square inside of it. That doesn't make it stop being a square. A square is defined as a rectangle with equal length sides. It doesn't matter if you draw another square INSIDE it. Heck, you can draw a giraffe with a top-hat riding a bicycle inside of it, it's still a square.
I ran across this image several times, a few of those i tried in vain to find a solution, and i see this video now, my mind immediately snapped to what you named the first solution. Felt soo good that i had an eureka moment honestly
It's not. If thats not stated, shape created using their diagonals can only be assumed to be a square. That statement removes the need for any assumptions.
I think of this as well. I think using parentheses or air quotes should be more common to avoid misunderstandings. Whether "shade the small squares" in full or "shade" the small squares was the difference between me getting paid and an angry customer IRL.
There was nothing in the language that implied you need to completely shade any of the small squares. The only requirement was that half of the large square be shaded.
First answer is big brain. I'm not sure I would've gotten it as a 4th grader. My first thought was measure it, calculate the area, halve it, sqrt it, measure that in a corner of the diagram and shade around it.
The thing is, the "intuitive" solution does require some knowledge to prove it's a square. Still doable by 4th graders I'd say, just not in terms of degrees but by simply saying "this is exactly half of a right angle and this is another half so this is a whole right angle" and "these four squares are the same so these diagonal lines are also the same". I once solved a similar question that the teacher couldn't solve, dividing a triangle in 4. Luckily I was a Zelda fan already
No, the problem had everything that you needed to solve it in the description. Took me less than a minute to think up the same solution that was described in the video.
Yeah if I think about how a 4th grader me would've approached it (assuming I found a solution), I would've thought of origami, because the first solution where you draw straight lines between middle points of each side is a step in some origami patterns. Adult me did however think of the sqaure root of 2 thing, but I was thinking about drawing it vertically/horizontally, either in the center or aligned to the top left.
If some adults can't solve it then I believe it's reasonable to overthink the question a bit. Plus it's a fun thought exercise and demonstration of geometric knowledge to think of multiple solutions :D
6:15 I thought of this instantly, but then thought "no, it can't be that easy." I'm now apparently smarter than someone with a PHD And yet I'm failing precalculus
I got the first 2 solutions by the time I read the question. The other solutions you can get by just moving the unshaded square around were interesting. I like that.
I loved this because I got the second and third solutions but missed the "intended" solution. That's why I love this channel. I always learn something new!
My first thought was to leave a square in the middle by shading a box around it, and then realized you could essentially place that new box in any orientation or area within the area. I don't understand any of the math formulas that one could use to figure that out, but figured I could divide the small squares into whatever number of smaller squares needed to make visually determining the size easier sort of like real world measurement. 1/2, 1/4, 1/8, 1/16, and so on.
Logical approach: Connect diagonal lines to form the inner square. Mathematical approach: Half of the square is 2a*a If a=1 this means that x*x=2*1*1->x*x=2->x=square root of 2=1.414 I would draw the square of 1.414*1.414 anywhere inside the square on the picture and shade the remaining area. In this way, half of the entire area would be inside a square, and the other half would be shaded.
@@ezura4760 No, the trap is starting the question with "Can you..." as as the question is asking about your ability to do whatever. As such, the only correct answers possible are "Yes" and "No".
(Before watching the video) I'd calculate the size of a square half the area of the big one, draw it anywhere inside the big square and shade the rest. (After watching the video) The "textbook" solution is so clean. Love this kind of problems.
I haven't watched it yet, but it was to segment the squares down further, to get enough segments to shade around the right and bottom edge, to leave exactly half unshaded in the top left? I would have assumed that, though it's a dork move not to have segmented it properly to begin with, making the student fully provide the rest of the work as well as the solution.
@@ImMimicute yeah my initial thought was square in center so i just labeled the extension into the square from the sides as x and then setup the quadratic as 4x^2+4(2-2x)x=2 which solves to x=1-sqrt(2)/2. then i used that to discover that the area of the unshaded region was sqrt(2) and was like, oh, obviously, the unshaded square has half the area of the original square. so then i realized that you could put a square of side length sqrt(2) anywhere within the original square and still get an answer. the solution with the diamond square in the center looks nice, but as long as you label the square within the larger square as having side length sqrt(2), your answer is correct.
5:55 I came up with this instantly but thought it was going to be wrong because math questions never tell you how to do it and they usually have only one correct answer labeling the outside the box solutions as incorrect because Schools don't always allow you to think outside the box and would rather you get the answer they want you to get.
my first thought was to OCD it and overcomplicate it by calling the side length 2 (indicated by it having 2 units), find a square that takes up exactly half the area, find that side length, and then I realized I could just make a diamond because dividing each square into a triangle halves it...
you don't have to be exactly correct in your drawing, just show that the inner square is 1/2 the area of the outer square. marking the length of each side would probably suffice.
That's what I'm thinking. I appreciate the other solutions and they're correct in some sense, but the question asks whether I can shade half of it, and during the test with limited time and tools the answer is no for all but the first solution, which I can simply use a ruler for
I would agree. The other ones are just free handing and it's hard to tell unless you specify each side is sqrt(2). This is more of a geometry problem at a concept level. Not a numbers kind of math problem.
It took me 1 min to solve that and the best thing is I didn't use any hectic math , I knew half the shaded portion of the large square would also be the half shaded portion in the smaller squares and the diagonal of the smaller gave me an unshaded square in middle. That's it
Shade 3/5 of top right square vertically (shade right) Shade 3/5 of bottom left square (shade bottom) Shade the bottom right square leaving a smaller unshaded square in its top left corner than corresponds with the new vertical line from top right square and new horizontal line from bottom left square Do this in a way to make a new unshaded square that takes up slightly more room than the top left sqare ending up with a sqaure exactly half of the total area
Two possible solutions: 1. Shade over the words "half of it", leaving an unshaded square. 2. Shade over either the I or the t in the word "it", leaving an unshaded square.
A thing that I learned while working as a tutor is that everybody, and I mean EVERYBODY, has gaps in their knowledge. PhD doesn't mean "smart" or "genius" or anything like that. It means that they were dedicated and found something that interested them enough to become, in some sense, THE expert on. PhDs are about advancing our knowledge as a species and that takes dedication, but not always math knowledge.
@@gildedbear5355 It either takes dedication or money and P-hacking, apparently. Academia and the journal system are kinda wrecked tbh. I didn't realize how bad until I started training AIs with the nonsensical publish-or-perish papers
I'd find the area of the entire square and draw an outline of a square equal to half of that in the very middle and shade the outside part. I don't expect a 4th grader to do this, maybe a 7th grader (I'm not a teacher this is just based on the complexity of the stuff taught at schools).
nah, you just gotta know the curriculum. this question is most likely testing the properties of a square (all sides are equal, can be split in half diagonally). from there, you just gotta apply a bit of creative thinkin', and boom! you got your answer.
IDK, could be. Only took me a few seconds thinking about it to find a solution, but I was in the gifted program and solve problems (comp programmer) for a living. I would expect most people to have a harder time with it, or not get it at all.
No, because you don't have to do any math to solve it. This video overcomplicates the issue. You don't need to know that the diagonal of a square with side length 1 has length sqrt(2); you only need to know that if you draw a diagonal line though a square, half of the area will be on either side of that line, which is obvious.
To be really accurate, maybe the only solution that can be done by hand and precision, is the first one with the diagonals. The others are okay but you can’t tell if they are the half
I love home excited you get when you start revealing the answer. Like you don't sound mean or harsh at all, you're just happy that you can help people literally think outside the box.
If you’ve never seen the puzzle, I felt that the first solution was more creative than the second. The second and third solutions were the first to come to my mind. I forgot that an angled square is still a square. Moreover, the first solution is more efficient because it doesn’t require a ruler to scale √2, which somehow makes it seems like a "better/smart" answer to the kids in class.
You'd just have to label your sides. Assuming you label one of the original small lengths as X, you could freehand the shading quickly and sloppily then label an unshaded side as √2X. No need to spend the time carefully measuring. That's why I'd put it in a corner.
How come I only see the first solution as one square with four triangles on its edges rather than two squares? I don't know why I assumed you can't have the shapes overlap.
@@Schmeethe88 In grade 4 you would identify diagonals of equal length with "tics" across the line segment. The students would also know about 90⁰ and 45⁰ angles in squares and diagonals. I would expect more of a verbal explanation than proper notation on a grade 4 math creativity question.
I thought of the second solution you gave almost immediately. As a former 4th grade teacher, I would also accept all of the answers you gave because I believe that true inquiry-based, student-centered education is the only way to go. thanks for the video! 😃
@the_nikster1 then you are failing your students, because your aren't teaching them how to develop a solution that is rigorous. A major issue in our world today is that people don't understand the concept of proving something with logic, they just make assertions and try to bully others into accepting their derangement. Math is about rigorously proving your assertions are valid based on fundamental axiomatic logic.
That's not really true as you can't directly measure sqrt(2) as it is non-rational. You can construct it as follows: 1. Draw both diagonals in one of the squares 2. Take a compass and measure from one corner to the intersection of the diagonals 3. Draw a circle on every middle of the outer edges of the big square 4. Connect the intersections with lines parallel to the outer edges. 5. Shade the area outside of the now created square
No You can make square root 2 geometrically. So the second is also possible, or any of the corner fitted one. Actually, with copying the distance, you can make all the solutions with just a ruler and a pair of compasses
@@m.a.6478cant you just say that since every sub square is half shaded (since diagonal divides a square in diagonal), their combined total is half the entire shape? Much simpler
Shade in each square at an angle. Cutting each square will provide a square in the middle witch is a diamond shape. You shaded half and a square is left.
Thank you for sharing this, I've always had doubts about my problem solving skills but this video confirmed that I need to give myself more credit some times. ( I am aware that 4th graders are definitely capable of this but I I got the answer within about a minute or so, which is a lot quicker than I expected)
Here is how you do it: get a pair of calipers and measure the exact length and width of the square. Then you find the volume. Afterwards take the volume and divide it in half and take the square root. Now CAD a square with the dimensions and then 3d print an outline of it(make sure that the outline’s width is expanding outwards and not inwards). Take the outline, put it on any location in the square and shade inside of the outline. Presto - its now a perfect shading of half of it.
@@KingBobbitoeasy, it's 0 thickness by definition so the volume is 0, so now you just gotta do 0/2 = 0 then do sqrt(0), & your CAD becomes really easy too 😝 Ok we're being jerks, we know they meant area 😄
@@KingBobbito You find the thickness of the paper, and make sure you use a tattoo machine or something to shade extra layers to ensure that it is half.
Exactly so, but I think you mean the unshaded area moved around inside the shaded area! Once you have proved and defined the unshaded area as half of the overall area of the large square, you can move that shaded area about laterally and vertically for an infinite number of positions within the large square. Also, if you take the first solution he showed, with the diagonally arranged unshaded square, that square itself is centred on a rotation axis so the square can also be rotated about this axis, again with an infinite number of positions.
Draw a quarter circle connecting the right and bottom edges. Then draw a diagonal line through the square. Then find the intersection and draw lines down and right. Then shade that part of the square. You shaded half of the square and you can easily check that the unshaded part of the square is a smaller square.
Yes. Draw diagonal lines: top left square: draw from top right to bottom left, shade outside triangle; top right square: draw from top left to bottom right, shade outside triangle; bottom left square: draw from top left to bottom right, shade outside triangle; bottom right square: draw from top right to bottom left, shade outside triangle. Inside triangles form an unshaded square and you've used half of the large square to do it. 😮
You can also construct the corner squares by drawing a circular arc from the corner and the diagonal (radius of sq2), then extend two perpendicular lines from the intersection with the outer square lines, and shade the remaining area.
That's the solution I thought of, as well! Also, it's one that is valid even if you assume that the grader: 1) Considers a tilted square a rombhus instead of a square (which would invalidate the first solution the video proposes). 2) Means for you to come up with a solution that doesn't involve the use of a ruler with marks, but rather a classic "Euclidean" geometrical construction (which would invalidate all other solutions proposed in the video).
The corner solution can be constructed if you allow a compass and straight edge. Draw a quarter circle centered on a corner, through the center of the large square. For each point where that circle intersects the large square draw a perpendicular line. The smaller square is determined by the original corner we chose, the two intersection points between the quarter circle and the large square, and the intersection of the two perpendiculars. The proof that this is the correct size square is exactly the same as the one presented for the first solution presented here.
But remember, this Is fourth grade math. They're testing If you know how to do something correctly, not your drawing skills. You'd probably pass with some wobbly lines as long as It's clear what you tried to do Is right
Draw a diagonal in each quarter, so that those diagonals form a square (losange) inside at 45º to basis. That square is 4x45º triangles, and the outer 4 x45º triangles can be shaded.
I never noticed this before, but you do outros. In this video, I particularly liked your concluded comments. They are quite motivational. Been watching for a year. Keep it coming brother!
Im glad I clicked on this, as I was thinking the answer was easy in that you'd just shade two contiguous squares. Somewhere along the way (one of the "benefits" of being old!) I'd reversed the quote, "All squares are rectangles, not all rectangles are squares." Once my "obvious" answer was shown wrong, i refreshed myself on the quote and the rotated square answer shoje through. Thats why it's always good to brush up on things you're sure you know, as you might be surprised what youve forgotten/misremembered!
"You could also shade all of the large square at 50% transparency" - If you shade all of the large square, even at only 0.00000001% transparency, then you have still shaded 100% of the square, leaving nothing unshaded.
I'm pretty sure thats what the answer was supposed to be based on the fourth grade thing, I'm pretty sure they teach that rectangles are squares at that grade level, at least where I live
One of the defining features of a square is its 4 equal sides, that is more defining features than a rectangle so the rectangle is the larger pool of shapes that squares fit under. Squares are rectangles because they have parallel lines and 90 degree corners, but a rectangle that has a 2 to 1 ratio in side length is not a square @@thumtak_
I absolutely adore math problems that truly speaking aren't the most complicated, they just evoke deeper thought. One day i hope for the opportunity to be a math teacher, especially for the opportunity to see how different minds come to different but correct answers, as that is my favorite part about math.
2 simple ways is to shade 1/2 of the outside edges towards center leaving a perfect square, or connect the diagonals leaving a perfect square. Was this too easy, or were they looking for something more complicated ?
Before watching the video (@0:06) my guess is you draw your own lines from the bottom middle to left middle to top middle to right middle back to bottom middle. Then you shade in the outside edges in order to have the middle diamond be square.
This is great really. I made the mistake of thinking that the shaded and the unshaded parts both bath to be squared and could frankly think of nothing at all. I like how you explained it so well!
Yes. Draw diagonal lines that join the midpoints of each edge. Each of these lines will bisect each of the smaller squares diagonally. Then, shade the outer portion, leaving a central unshaded square lying diagonally with exactly half of the volume of the large square.
It's about the thought process really, so the teacher would probably decide whether to give them hints or just let them tackle it with whatever tools they have. Probably depends on whatever stuff they've been learning too - if they've been playing around with shapes and working stuff by manipulating them (like experimenting with finding the area of a triangle without giving them the formula derived from the basic approach) then this question might fit neatly into that, without anyone even thinking about measuring or drawing circles
Yeah there's a bit of ambiguity here, sure you can measure everything and compute the needed size of a square with half the area, and then place it anywhere within the original. You could also compute the radius of a circle that has the correct area etc. The only real answere, given the way the question is phrased, is the rotated 45 degree square.
@@21palica Also, since the square root of 2 is an irrational number you cannot draw a line that is EXACTLY as long as the square root of 2, however you could draw a line that exactly crosses the intersections of the original diagram.
I remember solving a logic puzzle like that during counseling in I don't remember what grade. There comes a point where you realize you're being misdirected. The solution involved using the pieces to draw the correct shape, but after numerous tries it became clear there weren't enough pieces and negative space had to be used. I"ll admit that after trying a few I suspected wouldn't work, I settled on the second solution shown. But the first one is simpler and easier to get precise given the guidelines present. Realistically questions like this are less about the solution than about the attempt.
I came up with the triangle method immediately. However, to restate what you said in my own words, the original area was 4. So we needed to draw a square with 0.5 of that area, or 2. A square with an area of 2 will always have sides of sqrt(2). So any square with sides sqrt(2) that fits completely inside the original squares qualifies.
Since the diagonal of the smaller squares is of length root 2. I thought of using a compass to measure out the needed side-length of the square which is half the total area. Placing it in the upper right, you would have a shaded square and a non-shaded L shape which are both half the total area.😊
With a compass you can inscribe a circle inside the big box, draw the diagonals,of the big square, then connect the intersection of the circle and the diagonals.
Pedant here! 🙋♂️ I am not intending to denigrate your correct logic, but you need a pair of compasses to do this. With a compass you could then allign your sides to face North, East, South ans West. Now excuse me whilst I put on my smarty pant.
ez, shade each quadrant diagonally by half and have it where the shaded part is towards the center. repeat it 4 times, and you get a diamond shape, which is a diagonal square.
Not gonna lie, I solved this pretty easily 😅 I am very far from smart when it comes to maths The way I did it was by dividing all squares into 4 squares. So 16 by 16 grid. Just shaded the outside 8 blocks
If you're a 4th grader submitting one of those "creative" solutions, you'd better hope the teacher is smart enough to recognise it as valid.
"Showing your work" is why most of the potential answers would fail. Drawing that square in arbitrary position is not at all straightforward. The centered diamond, the centered square, or the square in 1 corner can all be done easily with compass and straightedge. The diamond can be done with straightedge only. Anything else is going to be more pain than it's worth.
3:23
Back in the early 70's, I was in EBP / MGM programs. I was in the 3rd and 4th grades doing up to 7th grade level work. Education was beginning to go through its defunding period and my programs were the first to get cut. So, in the 5th grade, doing 5th grade level work, my teacher would give me a failing grade because my work to solve problems were more advanced, not what she taught in the class. I got very bored in school when those programs got cut. BTW in the 4th grade, we were working on dodecahedrons, not squares lol.
What I was hoping was an explanation for a way to construct a square in a corner, using a square root of 2 as the radius of a circle to draw the intersections with the existing lines to draw the shape of the shaded area. The other solutions aren't as simple to construct as the first answer and a square in a corner answer.
Which usually isn't the case. If your answer is different than what's given in the teacher's edition of the book, it's wrong. Most teachers lack the ability to analyze different solutions.
Technically, the answer to the question that was asked is "yes".
True, but schools teach you to always "show how you reached your conclusion" so by that overarching rule you'd know that's not enough. Technically, you could answer every math question in your final year correctly, but by only writing down the final numbers and thereby failing anyway
And technically if you don't find how to draw it, answering "no" would also be correct
Once you shade it, the shaded part is no longer a square. So the unshaded part is not also a square, it is the only square. So the answer is "no". lol
@@migssdz7287 I occasionally posed problems where the idea is to check an equality. If a student tries the problem and botches it, he gets partial credit. Or if the problem has a derivative and the student has trouble with derivatives, he also gets partial credit. The way to get zero is simply not to take the derivative or otherwise check the equality, but instead do something fancy that shows himself clueless. (If the question is to check if x=2 satisfies x+2 = 5, all too often, it doesn't occur to the student to add 2+2 and answer "no" based on the result. In this case, a "no" by itself is good.)
That was my first inclination. But then, knowing that the task is possible but realizing that I could never actually accomplish, I changed my mind to "no."
My solution was to draw a crappy square, declare the shaded and unshaded areas equal, put out a warning that drawing is not to scale, and call it a day
I passed a calc final exam in my undergrad like this. I just defined something as “a” and skipped some steps.
@@haroldhayes4824 simplifying by taking parts of an equation and declaring it as something is often helpful.
My default was always X, Y, Z, A, B, C as needed. I got so locked into this pattern that I once redefined every variable in a problem like this and my prof left me a note telling me I should break the habit lol
@@VocalMabiMaple lmao
Engineer?
@@VocalMabiMapleI remember problems in college where I ran out of Latin and Greek letters and ended up using some hiragana. I suck at making simplifying assumptions.
I did the algebra to find that if you shaded a border ~0.29289 blocks thick, you would shade half and get a perfect square, then realized triangles exist...
I was on that same train, lol
that was the train i was on too, lol
We Motherfuckin Choo Choo over here.
Good call, I just drew a smaller square in the middle but yours is much more accurate
my immediate thought when i saw the thumbnail was, just shade half of each four squares. mind you i flunked math, im an art person. i think its funny how different our brains can work. theres no way i could do the math and figure that out. the only math thing i remember, and i like to say to people sometimes, is a few digits of pi. off the top of my head i remember 3.14159265358979323846264338327. i want to learn more, it would be easy to memorise since its already segmented in my mind. once i completely memorise the second segment, i could keep building on it. theoretically. im actually really lazy and dont want to put in the work. thanks school, i remembered one thing but you burnt me out too much!
Sometimes the hard part about math is not the problem itself but rather the lack of details provided in the problem
Honestly yeah like can you shade it so the unshaded part has a hole
Yeah, if they had said half the area of the square it would've been easier
What is happening, is that WE are assuming limitations that were not stated. I did it to myself for about the first 45 seconds of this video ( a little longer because I paused it to look over the Tweet. ) And the "textbook answer" was the easy answer, as it can be done without any of the actual math, and even without a ruler if you are good at straight lines.( I am not... ) And as James said, the way the question was asked, the technically correct answer, is YES. As the question does not state that we are supposed to show or demonstrate the a solution.
And the funny thing is knowing the answer to this question has no purpose later in life for those students
@@janelle9998being able to problem solve, use logical thinking, and project geometric shapes is extremely useful in life. I’m not sure why one would assume the opposite.
I completely missed the simple version of drawing a diamond and went right for the smaller square in the middle, which is harder to accurately measure.
You just draw a unit circle in the center, then construct the diagonals, and connect the intersection points and you have your square.
My solution would have been to set the compass to sqrt(2) via a diagonal of a small square, then use it to measure what I need to chop off on two edges, putting the half area square in a corner
@@gernottiefenbrunner172 You assume I own a compass.
I thought of both, but liked the outer area more.
We shouldn't call a rotated square "a diamond". A diamond is an actual shape where sides are equal, but pairs of angles differ. If all 4 sides and angles are equal, it's a square, regardless of the orientation. So unless the teacher explains that it's not how you draw it, but the properties that define a shape, many kids will call a square rotated 45º "a diamond", and nobody will correct them, even as adults.
Got the “diamond” (rotated square) option right away 😜
The second solution you showed "square in the center" was my first thought.
Same, and then trying to work out exactly where to put the sides of the smaller square made my 4:30am brain error out.
It was my first thought as well
How do you make 1.4142 in geometry? This was my fist question. And I made separate triangle and copied to the squares. And the 1st solution never hit me, regardless I've been so close. I guess my mind is inside so many boxes...
@@YourNickIsTaken it only needs to be root 2 if you each smaller squares are 1x1, which was chosen arbitrarily by the video. We don't have access to the actual paper but if I did I'd've grabbed a ruler.
Yeah theres no way figuring out an irrational number is part of the intented solution @@blakdeth
I thought way outside the box at first and then I realized you'd just need to bisect each square diagonally so that the lines create a new square that covers half of each square
If each Line = L
Area= 4L^2
If we bisect each square diagonally, through Pythagoras
Each side = √2L
Area = 2L^2
First time I got an answer right, fast and think: nah it can't be this easy.
Just made my day.
the way school often makes you doubt yourself just cause you found an easy solution
Same
samee
Yeah i was like just fit a square inside a square, how? Half each quadrants and then boom a square inside a square or that's how i did it
Well, it’s still 4th grade. „It can’t be that easy“ fits more to 9th or 10th grade.
Easy. Connect the diagonals to make a square tilted 45 degrees.
My thought was to draw a diagonal of sqrt(2) length and rotate that to form a square of area 2. I like your solution a lot better - going to watch now, I bet that's Presh's technique.
Right? I was wondering why this is 9+ minutes, but I will continue to watch.
@@Gideon_Judges6 Actually, my first thought was to use the standard tool of geometry problem authors. Cut the square straight across. You could say that's a rectangle, but you could also say it's a square - not drawn to scale! 🤪
😅
here I am thinking about b*sqrt(2)=a lol
Both "yes, I can" or "no, I can't " would be valid answers
Absolutely! If the question is "Can you?" and you can't, then to say so would be your correct answer :D
@@ruthholbrook Would I? No I would not, but others might..
As a 4th grade teacher, questions like this drive me crazy. I also hated Language Arts questions asking students how the words an author used made them feel, since anything they wrote down was a valid answer.
It’s weird because the question says “Can you shade half of it so the unshaded part is *also* a square?”
It’s just bad question writing like this that trips up people on tests. It makes you think the shaded part also has to be a square.
"no i can't" show proof or your answer is incorrect
Took me a while, gone as far as to calculate the square root of 2. But then eventually, was looking at different ways to shade precisely the half, until realization has come. Very nice riddle, I am so glad I didn’t continue watching the video before resolving it on my own.
The second solution is what came to my mind, along with the thinking of "people probably aren't getting this because they're probably assuming that the shaded area also has to be square somehow".
The first solution with the rotated square is actually so incredibly obvious once you see it that I'm ashamed I didn't think of it instead.
The first solution is the easiest to draw and the easiest to understand. Like you, I came up with the second solution, but it's not as easy to draw to scale, but it's easy to provide the math to explain where the square sits. Once I saw that, I realized that there are an infinite number of solutions that just need the math shown to prove they are correct.
I also did the exact same thing
I did one of the solutions where the square is on one of the four corners of the bigger square. So I also kind of made the obvious solution into a little bit more complicated one for no reason lol.
The second solution (and all but the first solution) requires using the square root of 2, though. So somewhat outside the scope of 4th grade math.
the only solution i thought is to shade an L but i have no idea how thick the L is
You have to think inside the box with this one!
Eeey that's a good one.
🤓☝️ errmmm a box is 3 dimensional and needs and opening or else it's just a square or cube
nice joke
@@lav-kittyYou know what? Flats you, Uncubes your cube!
Take my like and give me a big hug for that joke
@@KillerKatz12 NOOO NOT MY CUBE
Easy peasy and I am 76. Just color the four outer triangles of each block.
W
I'm 75 found it easy, thinking outside the boxes.
this is what i thought too lol im 20. now idk if that’s the answer the creators of this puzzle were looking for but it is AN answer, and its a correct one so 🤷🏻♀️
It’s worded in a way that made me think you could only shade in two full squares, not half of each square.
Okay but what about the cross in the middle? Cause that’s I thought too but there’s that division in the middle. So no, you won’t have a square, you’ll have 4 triangles.
5:55 is the solution that i came up with.
Same...ig it's easier to think of that just by looking if one's familiar with shapes...
Ditto
Same
Mine was the first answer (the one with triangles), I got it right after examining the problem in two seconds, I don’t get how people think it’s hard, it’s one of the easiest things I ever solved
Same
The L-shaped shading was the first thing that came to my mind after I studied the diagram and what the question was and was not asking for. I'm reminded of the classic test with "Read all instructions before marking your paper" at the beginning, progressing through increasingly nightmarishly conplex equations you haven't been taught yet and the last instruction is "Disregard all previous instructions, sign your name and hand in the otherwise-blank test."
and from that moment forth, i started all tests at the last question...
i loved that test, i and my friend were the only ones in a class of 30 to just put our name on it 😅
yeah i remember getting one of those, i think it was like 3rd grade for me? never liked them much, i always had the mentality of focusing on what i currently had to do instead of worrying about future questions and stressing myself out. of course it bit me in the ass that time though as i was one of 24 idiots who were all looking for someone who's name started with L who was born in November and had pancakes for breakfast.
That kind of happened to me, but in a more normal way. It was an algebra 1 test, and had normal algebra 1 stuff. It told you to read directions carefully, and one of the questions said at the end to draw a smiley face. Not many people did it. I got a bit of extra credit.
I'm glad that reading "also" isn't part of the instructions
I drew a square in the center and added a “not to scale” warning
Bruh!! 💀
Born to be an engineer.
stolen
@@kray3883I'm a software engineer and it was exactly what I was thinking - just without the "not to scale" warning.
I thought "Make a square inside of it. But then I would have to calculate the proportions... I can't be bothered. Oh yeah the diagonals, simple"
The fact that so many adults have trouble with this type of creative problem solving is exactly why we should be giving these types of problems to our children.
to be fair, to really PROVE this solution is valid, you need geometry level education (pythagorean theorem, 45-45-90 triangles, and others more based on how rigorous you wanna be with the proofs), which i dont think 4th graders have
@@henry_3300You don't, if you draw a diagonal inside a square it's pretty damn obvious it's cutting it in half. Kindergartners can see that
Of course you need higher education to prove it right, but that's not what the question is asking
@@nullieee Exactly!
@@nullieee i did say prove, so. doesnt matter if intuition says so, because intuition isnt enough in proofs.
@@henry_3300 Here's your super advanced proof then:
Assume a square of side length 1
Cut it in its diagonal
This will generate two congruent right triangles with sides 1 and hypotenuse √2
By that, it's correct to say both have equal areas as the triangles are equal
I solved it almost immediately. I also failed every single math class I’ve ever taken and I only did 2 yrs of High School. This just proves to me that all that extra math you did would just slow me down 😅
elementary be enough
No normal person would go through those calculations to solve this, it's primarily just a logic problem.
He lost me with the extra math.
The phrasing of the question is the confusion -- most readers would take the "also" to refer back to the shaded portion (also being a square).
Maybe. But that would be a poor assumption. We humans do that quite often.
I don't think that's the confusion, I think it's the fact that people assume they have to make it symmetrical, while also making it a non-rotated square
I understood it to mean that the unshaded portion should be a square, half the size of the full square. The division into four smaller squares is only there to help me visually gauge the size of 'one half'.
The question did state half which means 50% of that must be shaded no more no less and that each portion must be a square the thing people are getting confused on is the fact that they look at the whole thing and assume that it is a big square which it is not it is for individual squares which means the simple answer is just to shade two squares which means you're left with a square two but still a square it did not say that you cannot have two it did insinuate that they're at least must be one. Such a simple problem it's just people inject information about the problem that isn't present and that isn't needed like it being a big square and that if you shade half you're left with a rectangle which you are not because they are two individual squares.
@@grimmspectrum1547 no, that's not a plausible interpretation. The entire unshaded part has to be "a" square (a as in one).
"Yes. Next question."
It's actually quite easy if you think about it for a moment.
That lady should give up her PhD.
@@shauryaaher1579 I figured it out quickly then saw this video is over 9 minutes long, so I left this comment then left the video.
“No” is also an answer, “No, next question!”
@@martinconnelly1473dang ur smart
The tilted square was also my first solution.
Other solutions are more difficult to "construct".
But with compass and ruler, it is possible.
My first was a square in direct center with same orientation as largest square. It's where my mind went first. Nice!
@@MrChrisRP My guess was that as well, but I say "guess" because I don't know exactly where to put the sides of the smaller square. (If you mean the one at 5:22).
I would GUESS the sides go 1/3 of the way from the big square's edges to its center, but that's a total guess. Idk enough geometry for that.
All the other solutions are approximations, because the length of the sides is an irrational number. The first solution is precise.
I started to think that if I wanted an uprigth square, then if one square is 1x1, then the large square's area is 4, then half of it is 2, so the sides of the half is sqrt2, and that is side to diagonal ratio of a square and then the tilted square hit me. You just diagonally cut the corners, and those would make up two squares, so the middle is a half sized square. Maybe this was the aim of the problem. But it did take me five minutes
@frankmerrill2366 Exactly this.
Sure, if the 4th grader actually had the knowledge and the tools to somehow make it geometrically accurate, then it could be valid.
But otherwise, there is just one solution and it's the first one.
Divide each square into four squares, divide each of those into four squares, shade as required, specifically the outside ring of squares. The inside is a square, the number of shaded and unshaded squares are identical.
Shade the whole thing. Look at the back, that was left unshaded. It's exactly half
Thats what I thought
theres no square in the back unless the print ink bleeds through
You are thinking in three dimensions!
That's fine, I agree with your solution. There's a counter surface of a perfect plane with no depth.
I love it but no square there thats double the size of the shading. Wiuld pass you though, thats to good of an answer to fail
There is no front or back in 2D planes though
As a little old lady quilter, I came up with the first solution fairly easily, and the other solutions have given me new ideas for quilt blocks.
Awww, one day I'd like to get into quilting. I didn't even think of it from that angle 😂
As a fellow quilter, I agree 😊
3:41 answer
Same here. Squares don't need to be set on one edge, they can be set on point.
Me too!
It’s weird because the question says “Can you shade half of it so the unshaded part is *also* a square?”
It’s just bad question writing like this that trips up people on tests. It makes you think the shaded part also has to be a square.
th e word "also" implies both are squares
The "square" below is made up of four small "squares" can you shade half so that the unshaded part is also a "square"
The question is right they already talked about the existence of 5 other squares and thats why it says "also"
@@mastergamer-zq8xc bro is coping
@@discussions. coping? about what? understanding basic grammar? or is it you that is coping and whats with the snarky remark if we are talking about this thing sensibly continue to do so or dont continue this conversation as people like that wont learn anything from it anyway
The unshaded part is still a square. it just has another square inside of it. That doesn't make it stop being a square. A square is defined as a rectangle with equal length sides. It doesn't matter if you draw another square INSIDE it. Heck, you can draw a giraffe with a top-hat riding a bicycle inside of it, it's still a square.
I ran across this image several times, a few of those i tried in vain to find a solution, and i see this video now, my mind immediately snapped to what you named the first solution. Felt soo good that i had an eureka moment honestly
The "four small" squares is the diversion. Makes us think that we must fully shade small squares in order to achieve the result.
It's not. If thats not stated, shape created using their diagonals can only be assumed to be a square. That statement removes the need for any assumptions.
@@marciusnhasty You're contradicting yourself lol
I think of this as well.
I think using parentheses or air quotes should be more common to avoid misunderstandings.
Whether "shade the small squares" in full or "shade" the small squares was the difference between me getting paid and an angry customer IRL.
Color half of each
There was nothing in the language that implied you need to completely shade any of the small squares. The only requirement was that half of the large square be shaded.
This is the first time I've gotten the solution on my own, and I feel so happy!
4th grade homework. 💀💀💀
@@trucid2 I have a 3rd grade brain, lmao
First answer is big brain. I'm not sure I would've gotten it as a 4th grader. My first thought was measure it, calculate the area, halve it, sqrt it, measure that in a corner of the diagram and shade around it.
you probably only think of these other ways because you know more maths now and know there are other ways to do it.
First answer is pretty basic to most fourth graders because they're learning better math concepts nowadays.
The thing is, the "intuitive" solution does require some knowledge to prove it's a square. Still doable by 4th graders I'd say, just not in terms of degrees but by simply saying "this is exactly half of a right angle and this is another half so this is a whole right angle" and "these four squares are the same so these diagonal lines are also the same".
I once solved a similar question that the teacher couldn't solve, dividing a triangle in 4. Luckily I was a Zelda fan already
@@tomdekler9280I really love how the Triforce was able to help you in math!
@@tomdekler9280 I hope that wasn't a math teacher.
6:51 amogus
The reason these problems are confusing is ALWAYS because the question was written extremely poorly and in a way that is hard to understand.
No, the problem had everything that you needed to solve it in the description. Took me less than a minute to think up the same solution that was described in the video.
Sorry but it just means one thing, you are child like and innocent but we are not @@TroySavary
@@trilokyamohanchakra6351 No, it means I can actually solve an easy problem and you can't.
@@TroySavary coz your thinking isn't maligned like ours into making simple things complex
I don’t have a PhD either, but I am a BSEE + MBA. Took me all of 11 seconds to see the solution.
I played this at 1.25 speed, and it solved my hardest issue with these videos. Bless Presh, sometimes he drags on.
Or maybe you have a small attention span😅
Maybe time to uninstall tiktok
you a gold fish
Hahaha, I'm so intellectually superior, because I watched it at regular speed. I'm *very* smart, you're probably just a TikTok kid 🤓
@@VAVORiAL u built like a gold fish
Pov: overthinking the question
Remember the this is 4th grade "lets think this like adults" is your first mistake
Yeah if I think about how a 4th grader me would've approached it (assuming I found a solution), I would've thought of origami, because the first solution where you draw straight lines between middle points of each side is a step in some origami patterns.
Adult me did however think of the sqaure root of 2 thing, but I was thinking about drawing it vertically/horizontally, either in the center or aligned to the top left.
I bogo sorted this
I bogo bogo sorted this
If some adults can't solve it then I believe it's reasonable to overthink the question a bit.
Plus it's a fun thought exercise and demonstration of geometric knowledge to think of multiple solutions :D
@@beckettthirion3147The reason they couldn't solve it is *because* they were overthinking it
3:40 i instantly thought of that one to be honest
Same... My only problem was the word : "Shade". Forgot what it meant for a brief moment. ( English ain't my native language ) 🤣
@@sirjazz1268 Me too, english ain't my native language and i had to rethink "Shade" twice ahah.
Same here
Ok, Big boy good job, now divide it by 0.
6:15
I thought of this instantly, but then thought "no, it can't be that easy."
I'm now apparently smarter than someone with a PHD
And yet I'm failing precalculus
My sis who’s six said the ans lol
Precalculus is precalculus. Don't be ashamed. Everything above trigonometry is unreasonable for anyone under the age of 35 to suffer through.
same.
@samirdoncic6395her dp says that she has a phd in making false claim
@@ModernAegis
How so?
I got the first 2 solutions by the time I read the question. The other solutions you can get by just moving the unshaded square around were interesting. I like that.
Isn’t it great how our brains work? I saw the third solution right away but not the first two.
I loved this because I got the second and third solutions but missed the "intended" solution. That's why I love this channel. I always learn something new!
I did the exact same thing
Same here. Odds are we'd have gotten marked down still because it didn't match the solution in the textbook and we "did math wrong."
Same. I actually immediately thought of the third one.
My first thought was to leave a square in the middle by shading a box around it, and then realized you could essentially place that new box in any orientation or area within the area. I don't understand any of the math formulas that one could use to figure that out, but figured I could divide the small squares into whatever number of smaller squares needed to make visually determining the size easier sort of like real world measurement. 1/2, 1/4, 1/8, 1/16, and so on.
Logical approach:
Connect diagonal lines to form the inner square.
Mathematical approach:
Half of the square is 2a*a
If a=1 this means that x*x=2*1*1->x*x=2->x=square root of 2=1.414
I would draw the square of 1.414*1.414 anywhere inside the square on the picture and shade the remaining area.
In this way, half of the entire area would be inside a square, and the other half would be shaded.
I got the answer but was convinced it was wrong because in my mind you have to colour inside the already present lines, because… rules….
exactly!
That’s the trap! LOL
@@ezura4760
No, the trap is starting the question with "Can you..." as as the question is asking about your ability to do whatever. As such, the only correct answers possible are "Yes" and "No".
A common error, reading more limitations into the question than were stated.
@@cigmorfil4101no one's keeping me from answering with "maybe"
(Before watching the video)
I'd calculate the size of a square half the area of the big one, draw it anywhere inside the big square and shade the rest.
(After watching the video)
The "textbook" solution is so clean. Love this kind of problems.
I haven't watched it yet, but it was to segment the squares down further, to get enough segments to shade around the right and bottom edge, to leave exactly half unshaded in the top left?
I would have assumed that, though it's a dork move not to have segmented it properly to begin with, making the student fully provide the rest of the work as well as the solution.
I was sat there for like 5 minutes doing quadratics trying to figure out how much I had to cut off
Felt like such a goof when he bisected the squares
@@ImMimicute yeah my initial thought was square in center so i just labeled the extension into the square from the sides as x and then setup the quadratic as 4x^2+4(2-2x)x=2 which solves to x=1-sqrt(2)/2. then i used that to discover that the area of the unshaded region was sqrt(2) and was like, oh, obviously, the unshaded square has half the area of the original square. so then i realized that you could put a square of side length sqrt(2) anywhere within the original square and still get an answer. the solution with the diamond square in the center looks nice, but as long as you label the square within the larger square as having side length sqrt(2), your answer is correct.
That's exactly what I would've done too😭
As a 7th grader, I got the answer at the thumbnail
5:55 I came up with this instantly but thought it was going to be wrong because math questions never tell you how to do it and they usually have only one correct answer labeling the outside the box solutions as incorrect because Schools don't always allow you to think outside the box and would rather you get the answer they want you to get.
Actually the solution was a bunch of triangles inside of the box 😊
my first thought was to OCD it and overcomplicate it by calling the side length 2 (indicated by it having 2 units), find a square that takes up exactly half the area, find that side length, and then I realized I could just make a diamond because dividing each square into a triangle halves it...
The "wanted" solution is the only realistic one though, as shading those other ones by hand is very difficult
you don't have to be exactly correct in your drawing, just show that the inner square is 1/2 the area of the outer square. marking the length of each side would probably suffice.
That's what I'm thinking. I appreciate the other solutions and they're correct in some sense, but the question asks whether I can shade half of it, and during the test with limited time and tools the answer is no for all but the first solution, which I can simply use a ruler for
I would agree. The other ones are just free handing and it's hard to tell unless you specify each side is sqrt(2). This is more of a geometry problem at a concept level. Not a numbers kind of math problem.
In fact it is very easy. Cut out the realistic solution and put it onto another 2 by 2 square and shade it with a pencil.
The only correct solution is the one listed in the teachers guide, because they don't understand it either.
0:37 diagonally shade each square… it’s a diamond but also a square
Yep
Not even close
@@SeptemberStudioswdym?? It is close
@@DevMan2342 no, it doesn’t even look like a square either way
@@SeptemberStudios you misunderstood. Shade the diagonal half of each small square on the outside and you have an unshaded square in the middle. Wdym?
It took me 1 min to solve that and the best thing is I didn't use any hectic math , I knew half the shaded portion of the large square would also be the half shaded portion in the smaller squares and the diagonal of the smaller gave me an unshaded square in middle. That's it
I did exactly the same thing. No math skill at all in this feeble brain, just logic.
This isn't exactly hectic math though, but yeah I did the same as you
That sounded like a Professor Layton puzzle for me 😂
"diamonds" are squares too, is it what it's called? Diamond?
@@giaiaspiritit's just a rotated square
Shade 3/5 of top right square vertically (shade right)
Shade 3/5 of bottom left square (shade bottom)
Shade the bottom right square leaving a smaller unshaded square in its top left corner than corresponds with the new vertical line from top right square and new horizontal line from bottom left square
Do this in a way to make a new unshaded square that takes up slightly more room than the top left sqare ending up with a sqaure exactly half of the total area
Two possible solutions:
1. Shade over the words "half of it", leaving an unshaded square.
2. Shade over either the I or the t in the word "it", leaving an unshaded square.
well them you are involving the entire page into the problem, where the unshaded portion isnt a square
Solving a puzzle that a person with a PhD couldn't as a minor has been the biggest confidence boost I've recieved in a while.
A thing that I learned while working as a tutor is that everybody, and I mean EVERYBODY, has gaps in their knowledge. PhD doesn't mean "smart" or "genius" or anything like that. It means that they were dedicated and found something that interested them enough to become, in some sense, THE expert on. PhDs are about advancing our knowledge as a species and that takes dedication, but not always math knowledge.
That's not to lessen the effort that a PhD requires, just pointing out a disconnect.
@@gildedbear5355 It either takes dedication or money and P-hacking, apparently. Academia and the journal system are kinda wrecked tbh. I didn't realize how bad until I started training AIs with the nonsensical publish-or-perish papers
@@gildedbear5355 it also doesn't determine that one with a PhD knows more in said subject than one without a PhD.
"A Ph.D. in wha(?)" is an important question. Art history?
I came up with your second two solutions. The first one you proposed with the 45 degree lines seem the most elegant though.
Yeah exactly 💯 😂
I'd find the area of the entire square and draw an outline of a square equal to half of that in the very middle and shade the outside part. I don't expect a 4th grader to do this, maybe a 7th grader (I'm not a teacher this is just based on the complexity of the stuff taught at schools).
Just saw a comment of someone saying that you can just draw a diamond. Probably the best way to do this lol 😂
This is one of those gifted and talented kids’ questions.
nah, you just gotta know the curriculum. this question is most likely testing the properties of a square (all sides are equal, can be split in half diagonally). from there, you just gotta apply a bit of creative thinkin', and boom! you got your answer.
@@wyawardclown exactly, it's a solution the curriculum has to come up with a question for but never used during your life
IDK, could be. Only took me a few seconds thinking about it to find a solution, but I was in the gifted program and solve problems (comp programmer) for a living. I would expect most people to have a harder time with it, or not get it at all.
No, because you don't have to do any math to solve it. This video overcomplicates the issue. You don't need to know that the diagonal of a square with side length 1 has length sqrt(2); you only need to know that if you draw a diagonal line though a square, half of the area will be on either side of that line, which is obvious.
To be really accurate, maybe the only solution that can be done by hand and precision, is the first one with the diagonals. The others are okay but you can’t tell if they are the half
I figured it out, Shade a triangle into every corner and it will turn into a diagonal square
bro watched the video then commented like he didnt
@@perezoso2 yeah
I love home excited you get when you start revealing the answer. Like you don't sound mean or harsh at all, you're just happy that you can help people literally think outside the box.
6:08 This was my solution, but now I'm too invested to not watch the others lol
If you’ve never seen the puzzle, I felt that the first solution was more creative than the second. The second and third solutions were the first to come to my mind. I forgot that an angled square is still a square. Moreover, the first solution is more efficient because it doesn’t require a ruler to scale √2, which somehow makes it seems like a "better/smart" answer to the kids in class.
You'd just have to label your sides. Assuming you label one of the original small lengths as X, you could freehand the shading quickly and sloppily then label an unshaded side as √2X. No need to spend the time carefully measuring. That's why I'd put it in a corner.
I guessed the solution in my head just from only seeing the thumbnail, I think the second and third is more creative....
I thought of no. 3
How come I only see the first solution as one square with four triangles on its edges rather than two squares? I don't know why I assumed you can't have the shapes overlap.
@@Schmeethe88 In grade 4 you would identify diagonals of equal length with "tics" across the line segment. The students would also know about 90⁰ and 45⁰ angles in squares and diagonals. I would expect more of a verbal explanation than proper notation on a grade 4 math creativity question.
I thought of the second solution you gave almost immediately. As a former 4th grade teacher, I would also accept all of the answers you gave because I believe that true inquiry-based, student-centered education is the only way to go. thanks for the video! 😃
would you have accepted just the two right squares being shaded because in the first grade we learnt that all rectangles are also technically squares
@@Jayden-hl2qdNo I believe it's "all squares are rectangles but not all rectangles are squares"
@@imperfectly_meganoh, yeah, you are right. Whoops!
@the_nikster1 then you are failing your students, because your aren't teaching them how to develop a solution that is rigorous. A major issue in our world today is that people don't understand the concept of proving something with logic, they just make assertions and try to bully others into accepting their derangement. Math is about rigorously proving your assertions are valid based on fundamental axiomatic logic.
The first solution is the only one you can generate a proof for without direct measurement.
One you have proof for one single solution it is trivial to come up with a proof for all other solutions (as done in this video near the end).
That's not really true as you can't directly measure sqrt(2) as it is non-rational. You can construct it as follows:
1. Draw both diagonals in one of the squares
2. Take a compass and measure from one corner to the intersection of the diagonals
3. Draw a circle on every middle of the outer edges of the big square
4. Connect the intersections with lines parallel to the outer edges.
5. Shade the area outside of the now created square
because presh didnt showed them it doesnt mean you cant draw other solutions with a ruler and compass
No
You can make square root 2 geometrically. So the second is also possible, or any of the corner fitted one. Actually, with copying the distance, you can make all the solutions with just a ruler and a pair of compasses
@@m.a.6478cant you just say that since every sub square is half shaded (since diagonal divides a square in diagonal), their combined total is half the entire shape? Much simpler
Shade in each square at an angle. Cutting each square will provide a square in the middle witch is a diamond shape. You shaded half and a square is left.
Thank you for sharing this, I've always had doubts about my problem solving skills but this video confirmed that I need to give myself more credit some times. ( I am aware that 4th graders are definitely capable of this but I I got the answer within about a minute or so, which is a lot quicker than I expected)
Here is how you do it: get a pair of calipers and measure the exact length and width of the square. Then you find the volume. Afterwards take the volume and divide it in half and take the square root. Now CAD a square with the dimensions and then 3d print an outline of it(make sure that the outline’s width is expanding outwards and not inwards). Take the outline, put it on any location in the square and shade inside of the outline.
Presto - its now a perfect shading of half of it.
How do you find the volume of a square?
@@KingBobbitoeasy, it's 0 thickness by definition so the volume is 0, so now you just gotta do 0/2 = 0 then do sqrt(0), & your CAD becomes really easy too 😝
Ok we're being jerks, we know they meant area 😄
@@hokayson6518 lol
@@KingBobbito You find the thickness of the paper, and make sure you use a tattoo machine or something to shade extra layers to ensure that it is half.
7:20, since we can move the shaded area anywhere, there are an infinite number of answers.
Yep, that's what I was thinking! Infinite possibilities for the placement of the square
Exactly so, but I think you mean the unshaded area moved around inside the shaded area! Once you have proved and defined the unshaded area as half of the overall area of the large square, you can move that shaded area about laterally and vertically for an infinite number of positions within the large square. Also, if you take the first solution he showed, with the diagonally arranged unshaded square, that square itself is centred on a rotation axis so the square can also be rotated about this axis, again with an infinite number of positions.
Clicked cause I need to know the answer 0:32
The beauty of this "problem" is that there are numerous and creative ways to solve and answer the question.
Do you think having quarter of each square in the middle shaded would work, like it is a square but it has a hole in it
"Can you shade half of it so that the shaded part is also a square?"
- "No. I am not a smart man..."
At least you're not trying to showcase how smart you are in the comment section! 😂
Draw a quarter circle connecting the right and bottom edges. Then draw a diagonal line through the square. Then find the intersection and draw lines down and right. Then shade that part of the square. You shaded half of the square and you can easily check that the unshaded part of the square is a smaller square.
Technically you're wrong. Now you know how to do it you can do it. It asked if you can do it, not if you can figure out how to do it on your own.
That is correct!
Short answer:
Yes I can.
Explanation of long answer:
some folding is required.
Yes. Draw diagonal lines: top left square: draw from top right to bottom left, shade outside triangle; top right square: draw from top left to bottom right, shade outside triangle; bottom left square: draw from top left to bottom right, shade outside triangle; bottom right square: draw from top right to bottom left, shade outside triangle. Inside triangles form an unshaded square and you've used half of the large square to do it. 😮
You can also construct the corner squares by drawing a circular arc from the corner and the diagonal (radius of sq2), then extend two perpendicular lines from the intersection with the outer square lines, and shade the remaining area.
That's the solution I thought of, as well! Also, it's one that is valid even if you assume that the grader:
1) Considers a tilted square a rombhus instead of a square (which would invalidate the first solution the video proposes).
2) Means for you to come up with a solution that doesn't involve the use of a ruler with marks, but rather a classic "Euclidean" geometrical construction (which would invalidate all other solutions proposed in the video).
honestly i suck at maths but figured it out in 10 seconds
The obvious answer was always a no.
@@thatguy3000 fair
It's not a math problem it's an art problem but this channel always overcomplicates things
@@DIRTkat_ofcit would probably still be put in a maths test/paper though
@@Wasted117. definitely not. Well I guess it may depends on where you live, but from where I am, i'm pretty sure it wouldn't be put in math
All the alternate solutions are indeed possible, but the first one is the only one that can be done with a straight adage and pencil.
Edit: I think you're right.
-The second too. Needs a bit more construction, but it's not too hard. I posted how in another main comment.-
The corner solution can be constructed if you allow a compass and straight edge.
Draw a quarter circle centered on a corner, through the center of the large square. For each point where that circle intersects the large square draw a perpendicular line. The smaller square is determined by the original corner we chose, the two intersection points between the quarter circle and the large square, and the intersection of the two perpendiculars. The proof that this is the correct size square is exactly the same as the one presented for the first solution presented here.
But remember, this Is fourth grade math. They're testing If you know how to do something correctly, not your drawing skills. You'd probably pass with some wobbly lines as long as It's clear what you tried to do Is right
@@zaelgreen1670that’s exactly how I did it
No you can bisect the 45 degree diagonal to find a sqrt 2 landmark on the edges.
Draw a diagonal in each quarter, so that those diagonals form a square (losange) inside at 45º to basis. That square is 4x45º triangles, and the outer 4 x45º triangles can be shaded.
Geometric triangulation...simple dimple. Dissect each square diagonally to form the middle square and shade outer portion.
Figured it out straight away. Shade the 4 squares in hald diagonally to form a unshaded square in the centre
5:22 was my initial thought
Facts
Same!
I never noticed this before, but you do outros. In this video, I particularly liked your concluded comments. They are quite motivational. Been watching for a year. Keep it coming brother!
Could you please show more "brain teaser" type problems? They are loads of fun! I got this in five seconds.
Im glad I clicked on this, as I was thinking the answer was easy in that you'd just shade two contiguous squares. Somewhere along the way (one of the "benefits" of being old!) I'd reversed the quote, "All squares are rectangles, not all rectangles are squares." Once my "obvious" answer was shown wrong, i refreshed myself on the quote and the rotated square answer shoje through. Thats why it's always good to brush up on things you're sure you know, as you might be surprised what youve forgotten/misremembered!
The answer for the first one is to shade most of the corners so that leaves a white square in the middle while also passing as “half shaded.”
wym "the answer to the first one"? are you literally just repeating what is said in the video?
You could also shade all of the large square at 50% transparency.
This is an epic response!
@@FirstDarkAngel2001 epic failure
if everything is lightly shaded there is no unshaded area at all that could form a square
"You could also shade all of the large square at 50% transparency" - If you shade all of the large square, even at only 0.00000001% transparency, then you have still shaded 100% of the square, leaving nothing unshaded.
That's still shaded. Any amount of shading is shading. You're trying to hard to be celver and ending up wrong
the 5:21 was what came to my mind after like 20 seconds, no thoughts, just square 😀
Same!
Same!
I thought the same thing I'm in 6th
“I got it in less than a minute!”
My dumbass saying a rectangle is a square cause a square can be a rectangle
I'm pretty sure thats what the answer was supposed to be based on the fourth grade thing, I'm pretty sure they teach that rectangles are squares at that grade level, at least where I live
@@thumtak_ but... Every square is a rectangle, but not every rectangle is a square
@@piern1k336 every rectangle is a square, every square is not a rectangle
One of the defining features of a square is its 4 equal sides, that is more defining features than a rectangle so the rectangle is the larger pool of shapes that squares fit under. Squares are rectangles because they have parallel lines and 90 degree corners, but a rectangle that has a 2 to 1 ratio in side length is not a square @@thumtak_
@@piern1k336 you have it backwards
Even a 4th grade math question feels like a legal document 😂
I absolutely adore math problems that truly speaking aren't the most complicated, they just evoke deeper thought. One day i hope for the opportunity to be a math teacher, especially for the opportunity to see how different minds come to different but correct answers, as that is my favorite part about math.
2 simple ways is to shade 1/2 of the outside edges towards center leaving a perfect square, or connect the diagonals leaving a perfect square. Was this too easy, or were they looking for something more complicated ?
Before watching the video (@0:06) my guess is you draw your own lines from the bottom middle to left middle to top middle to right middle back to bottom middle. Then you shade in the outside edges in order to have the middle diamond be square.
This is great really. I made the mistake of thinking that the shaded and the unshaded parts both bath to be squared and could frankly think of nothing at all. I like how you explained it so well!
Yes. Draw diagonal lines that join the midpoints of each edge. Each of these lines will bisect each of the smaller squares diagonally. Then, shade the outer portion, leaving a central unshaded square lying diagonally with exactly half of the volume of the large square.
The second solution is the one I immediately thought of but the first is definitely the simplest and most elegant imo
Surely there’s an unspoken “with some accuracy without a ruler, protractor or pair of compasses” in that original question?
This is exactly what I thought. The answer was obvious, but I think I would call my teacher to ask if I am allowed to use the ruler.
It's about the thought process really, so the teacher would probably decide whether to give them hints or just let them tackle it with whatever tools they have. Probably depends on whatever stuff they've been learning too - if they've been playing around with shapes and working stuff by manipulating them (like experimenting with finding the area of a triangle without giving them the formula derived from the basic approach) then this question might fit neatly into that, without anyone even thinking about measuring or drawing circles
At a maths test, you are very likely to have a ruler and a pencil. Nowhere in the question does it say you are not allowed to use these
Yeah there's a bit of ambiguity here, sure you can measure everything and compute the needed size of a square with half the area, and then place it anywhere within the original. You could also compute the radius of a circle that has the correct area etc. The only real answere, given the way the question is phrased, is the rotated 45 degree square.
@@21palica Also, since the square root of 2 is an irrational number you cannot draw a line that is EXACTLY as long as the square root of 2, however you could draw a line that exactly crosses the intersections of the original diagram.
I remember solving a logic puzzle like that during counseling in I don't remember what grade.
There comes a point where you realize you're being misdirected.
The solution involved using the pieces to draw the correct shape, but after numerous tries it became clear there weren't enough pieces and negative space had to be used.
I"ll admit that after trying a few I suspected wouldn't work, I settled on the second solution shown.
But the first one is simpler and easier to get precise given the guidelines present.
Realistically questions like this are less about the solution than about the attempt.
I came up with the triangle method immediately. However, to restate what you said in my own words, the original area was 4. So we needed to draw a square with 0.5 of that area, or 2. A square with an area of 2 will always have sides of sqrt(2). So any square with sides sqrt(2) that fits completely inside the original squares qualifies.
It's so simple! Shade the two diagonally opposite small squares and the remaining two will also be squares exactly half of the area of the big square.
Right when it showed up I was like: 'Bruh, you just cut the perpendicular squares in half and then shade the remaining area in"
Yeah It's absolutely not deep enough to warrant mathematical equations, though it's cool that some people try regardless.
I know. The answer was obvious after a few seconds of thought.
Since the diagonal of the smaller squares is of length root 2. I thought of using a compass to measure out the needed side-length of the square which is half the total area. Placing it in the upper right, you would have a shaded square and a non-shaded L shape which are both half the total area.😊
With a compass you can inscribe a circle inside the big box, draw the diagonals,of the big square, then connect the intersection of the circle and the diagonals.
Pedant here! 🙋♂️ I am not intending to denigrate your correct logic, but you need a pair of compasses to do this. With a compass you could then allign your sides to face North, East, South ans West. Now excuse me whilst I put on my smarty pant.
That was, for me, one of the best of recent weeks. Thank you - it made me look 'outside the box' a bit.
ez, shade each quadrant diagonally by half and have it where the shaded part is towards the center. repeat it 4 times, and you get a diamond shape, which is a diagonal square.
I sat for around 10 minutes, thinking without watching the video, and I solved it, but it was a brain teaser
bro its litterally to just shade half of the square to make a rectangele which is also a square
took me around 1 minute by looking at the thumbnail, but then again, I was creative
@@sathvikchillapalli5977ugh. Everything about this comment is incorrect.
Square = rectangle. Not the other way around @@sathvikchillapalli5977
@@sathvikchillapalli5977 it's the otherway around, squares are rectangles but with equal sides
The square in a square was my thinking.
I was thinking the shade each inner corner
Yeah 5:42
I went with the second one, as well, but then thought of the first one. Took me nearly a minute with these edibles in me 😂
YES. It takes just a bit of "thinking outside the box(es)".
The solutions are all _inside_ the boxes...
@@nthgth No solution lies inside ONE single box, Einstein.
Not gonna lie, I solved this pretty easily 😅
I am very far from smart when it comes to maths
The way I did it was by dividing all squares into 4 squares. So 16 by 16 grid. Just shaded the outside 8 blocks