Well it will apply in shapes with a 90 degree triangle ofc. Squares and rectangles sure, but not trapezoids or kites and such (at least not all of them
I can't remember ever learning that an inscribed triangle along the diameter is a right triangle, but it makes sense. That's the conceptual step I was missing.
@@Zieki99 Not that I can recall, but highschool was more than a decade ago and I don't use geometry in my day-to-day job. Again, it's just something I can't remember ever being covered, not that we didn't cover it.
The last time I needed to solve problems like this was around 15 years ago, but i still come here and try to solve these once in a while. The way you teach and explain is so good! Kudos for not only keeping students challenged, but people like me as well!
I never comment. Never subscribe. But you are crushing it. I save all of the problems that involve basic algebra geometry and algebra 2 concepts for my high school students. Andy Math out here differentiating instruction for me. God bless you and your family
For all those saying, "its obvious" or "i could just visualise it" etc.... That means nothing in an exam as its not a proof. You will get next to 0 points just giving an answer or an explanation like that. You need to PROVE the answer. Yes you can guess but it doesnt mean its right. What if y isnt 2x? Thats why we have proofs.
what if y isn't 2x? (x + 2y) / 5 = 1 assuming 5x = 5 5x - 2y =x y=2x and x=1 let's fill this in x + 2y =5 1 + 2(2) = 5 hrmmmmmm sir I have a suspicion it y may be 2x
Yeah, it can bite you back Especially shapes where they LOOK like they're touching each other so you can visualize and make comparison....but turns out they werent when try to proof it and end up not getting an answer on all 5 options
I’d also assume the other unstated (but visually implied) given was that the arc intersected exactly at the lower left corner of the pink square. There’s nothing that says it is drawn to scale, but I don’t see a way to solve it without that implied corner contact variable. Good stuff.
Tried solving it but without that it's impossible. I saw that it was probably what I was missing but given it wasn't stated in the question you can't assume it's a fact.
This is also how I would approach this problem to begin with, how do you scale purple and blue square so that the third, pink square, will touch both the semicircle and align with top of the blue one? Purple and blue depend on each other (otherwise they either aren't squares, or don't add up to 5 base), so there is only one degree of freedom, then for each pair the pink square is implied and either accepts or rejects the solution.
Ah I see. The entire time I was wondering how it even makes sense to find a unique solution, given that you can draw the other 2 squares for any blue square, but with that restriction that doesn't really hold.
@@chaoticsquidno it's Def solveable without the circle. We know it's Fibonacci sequence. Which means we can run that sequence with a base dummy variable adding up each time and dividing that by five. The dummy variable represents the edge of orange square, and fibonnaci sequence dictates it'll repeat five times by now.
Good assumption, I just used the blue square is a little over 2.5 so nearest while number fitting scale being 3 and working through 😅yours is a much better assumption and much cooler too
I solved it a different way. The radius of the semicircle is half the diameter, so r=5/2. Drawing a line from the center of the semicircle's base to the point at which the two smaller squares and the circle meet results in a line of length 5/2. Draw a line straight down from that point. You now have a right triangle with hypotenuse of length 5/2 and a height of y. Thanks to the Pythagorean Theorum, the base of that triangle, which goes from where that vertical line of length y touches the base to the midpoint of the semicircle's base, is the square root of the difference between (5/2)^2 and y^2. So the triangle's sides are sqrt((25/4)-y^2), y and 5/2. The length of the line extending from that triangle's right angle to the left edge of the semicircle is y-x. Therefore, the sum of y-x and sqrt((25/4)-y^2) is equal to the radius, which is 5/2. That's (5/2)=y-x+sqrt((25/4)-y^2) Solve for x, and you get x=y-(5/2)+sqrt((25/4)-y^2). Replace x with that expression in 5=2y+x, and you get 5=3y-(5/2)+sqrt((25/4)-y^2). Isolate the radical to get (15/2)-3y=sqrt((25/4)-y^2). Square both sides to get (225/4)-45y+9y^2=(25/4)-y^2. Move everything over to one side and combine like terms to get the quadratic function 10y^2-45y+50=0. Simplify to 2y^2-9y+10=0. Utilize our old friend, the quadratic formula, to get y=(9+-sqrt((-9)^2-4(2)(10)))/2(2). Simplify: y=(9+-sqrt(81-80))/4. Simplify: y=(9+-sqrt(1))/4. Simplify: y=(9+-1)/4. Conduct plus/minus operation: y=10/4, 8/4. Simplify: y=5/2, 2. y must be less than the radius, which is 5/2. Therefore, y can only be 2. Plug it in to 5=2y+x, solve for x, and x=1. Plug x and y into x^2+y^2+(x+y)^2 to get 14.
Was the triangle necessary? I could tell immediately that a lemgth of pink was ½ a length of purple; which means if diameter is 5, blue must be 3 and purple 2, thus pink 1. So it's 3×3+2×2+1×1=14 without the triangles... It's also a Fibionacci sequence, but I wanted to show my work. 😋
An alternative way to solve this problem:trace a line beetwen the center of the semicircle and the point where the semicircle intersect the lower left corner of the smaalest square and apply the pitagorean theorem în the right triangle.
Very nice! However there exists another solution just using the assumptions you made. The equation at 2:13 assumes y is not 0, but for y=0 you do acually get another solution for the total area, being 12.5 square units.
Wouldn't that mean that the point where the circle intersects the corner of the squares is on the bottom left? This is a problem since the triangle we used to make that would have 0 angles.
we could consider the case x=0 still a solution that represents the case when area of the little square vanish or gets smaller and smaller leaving only two equal symmetric squares, in this configuration the area is 12.5
Love your videos, doing these things with you are one of my favorite activities. Please take care of your own health and don't overdue with the videos and or any other job. Love you.
yeah.. forget the semicircle. like no. specially with how the question is stated just go in with a ruler and determine the other sides from the bottom line
I did it a different way, label lengths of the squares from largest to smallest as a,b,c. Then we can create a set of equations a+b = 5 (1) b+c = a (2) To get a third equation we can take the point where the corner a is on the semicircle and use pythagoras, noting that the radius is r = 2.5: b^2 + (2.5 - c-a)^2 = 2.5^2 (3) add together (1) and (2) to get a+c = 5 + a - 2b = 5 + (a+b) - 3b = 10 - 3b (4) substitute (4) into (3) to get b^2 +(3b-7.5)^2 = 2.5^2, ==> 10b^2 - 45b + 50 = 0, ==> (2b - 5)(b - 2) = 0 ==> b = 2, 2.5 if b is 2.5 then a = 2.5 and c = 0 so total area is 12.5 (trivial solution) , and for b = 2, we get a = 3 and c = 1 so total area is 14.
Thank you so much for the great content! As one who works as a math teacher, your content has been a huge inspiration on how to make challenging and fun puzzles!
x and y are the coords of the bottom-left corner of the pink square: Pink Square Length = 5 - 2y Pink Square Length = y - x y = (5+x)/3 But that same corner point also falls on the semicircle, which can be expressed as: (x-5/2)(x-5/2) + yy = 25/4 Substituting the first expression into the second and simplifying yields the quadratic: 2xx - 7x + 5 = 0 Which has solutions: x = 5/2 , x = 1
@@armandocarrion479 The x=5/2 solution corresponds to the pink square having no size. The consequence of this is that purple and blue both have side length equal to the radius (5/2). It's a valid, albeit trivial solution.
i did this using the Pythagoras formula. you know the center of the circle is 2.5 from the edge. you can draw a triangle that goes from the center of the circle to intersection of the circle and the 2 smaller boxes, then down perpendicular to the base of the semi circle. this has sides 2.5 (hypotenuse is a radius of the circle), y and 2.5-y+x. we can substitute x = 5-2y into that last side then the side lengths into Pythagoras formula to give us a quadratic in y. solve that to get y = 2 or y = 2.5 giving x = 1 or x = 0 (which we can discard) and finish up getting the areas.
Bro, i used to love maths and geometry up until highschool, but i had a shitty teacher that made me hate these subjects. Now thanks to Andy im relearning how cool this is
I just figured they must be Fibonnaci squares, pink+purple=blue. And 5 is a number in the Fibonnaci sequence. So 2+3=5, and 1 comes before 2 in the sequence. So the squares' areas are 1^2 + 2^2 + 3^2=14 Though if the diameter was 6 or 4 then they'd still be Fibonnaci squares, but with non integer lengths, so you'd be better off with the method in the video
The only way I see that working is if the side lengths of the squares have to be integers. Otherwise, there's ways I can manipulate the squares where pink is ALMOST the size of purple. That gives us an area of approximately 2x by 3x where 3x = 5. Final equation for the area gives us 5 × 5 × 2/3 = 50/3 square units, which is more than 14. Other way around, make pink approach zero. Area is now 2y^2 where y equals 5/2, the answer is 25/2 square units which is less than 14. Any other answers will lie between those extreme values.
@tomdekler9280 I suppose I did also assume that. Good point. I came to the point of, 5=2x+y where x is purple and y is pink. From there I said 2x must be an even number less than 5. this is where I assumed an integer value, it's also where I said pink must be smaller because otherwise purple could be 1.
Fibonacci and Generalization - Building on your process, paying close attention to the triangle which is the sum of two smaller ones, with specific ratio: We have x. Next, we have y=x+x=2x. Next, we have x+y=2x+x=3x. Next, we have x+2y=3x+2x=5x. This is the Fibonacci sequence where each term is multiplied by x. The first, same value as second, term is missing (we have 1, 2, 3, and 5, instead of 1, 1, 2, 3, and 5 - which correspond to the "bite" missing from the "complete" rectangle). using s for side lengths and a for areas, each followed by 1-3 for smallest to largest squares. Let's call the 4th term: z. specifically: z=5x. so: s1=z/5, s2=2z/5, s3=3z/5. Squaring for area: a1=z²/25, a2=4z²/25, a3=9z²/25. Summing: Total area = 14z²/25. In your example, z=5, so the total area = 14*5²/25=14. For z=6, for example, total area would be 14*6²/25=20.16. For z=10, twice the 5, the result should be quadrupled: 14*10²/25=56 - and it is.
Approached this slightly differently. Made a right triangle from the center of the circle to where it intersects the pink and purple squares. One side is y, hypotenuse is 2.5 and the other side is the radius of the circle (2.5) minus the portion of purple square that sticks out past the pink square (y - x). So right triangle with sides y and 2.5 - (y - x) and hypotenuse 2.5. Plugged in 5-2y for x and solved the resulting quadratic form Pythagoras.Not quite as elegant but still worked.
Cool! Here's how I solved it instead: - Grabbed a screenshot - Cropped the figure - Resized it so that the long side equals 500 pixels - Measured all sides, assuming that 1 unit = 100 pixels - Obtained measurements of 2.97, 2.03, and 0.95 units - Rounded up the measurements to the nearest integers, assuming that the author of the problem used integers - Obtained measurements of 3, 2 and 1 - Calculated the area of each square - Summed the areas
HALT! Not so fast! There is no stated reason to discard the x=0 case! Yes, just because it is on the diagram, does not mean the square actually exists! If x=0, then x+2y=5 => y=5/2. We then must add x^2 + y^2 + (x+y)^2 to find the areas. 0^2 + (5/2)^2 + (5/2)^2 = (5/2)*(5/2)*2=25/2=12.5 square units. This "extraneous" solution is the same a building two squares, one purple and one blue, with side length equal to the radius, which means the "intersection point" between the circle and the corner of the "pink square", which is actually a point by this construction, is exactly perpendicular from the center of the circle, one radius length away along both squares' side. How exciting.
if x were to be 0 it would break all squares containing it to be triangles, which is not a valid solution, as the image is in real scale, and all squares don't have x as 0.
I agree you can construct a valid geometric solution with x=0. I think it depends on the interpretation of the problem, which states there are 3 squares. If x=0, then the area of the pink square is zero. Is that still a square?
@@themathhatter5290How do you distinguish it from a degenerate circle though? Or a degenerate triangle? Or a degenerate regular polygon with n sides? They are all just points.
Had to fill in an abandoned septic tank years ago. The inside dimensions of the tank was 5' deep, 5'wide and 6' long with rounded ends. Use pi and radius formulas to determine the amount of controlled density fill that was needed. Stuff is order in complete cubic yards.
Another way to solve this is to form right triangle, hypotenuse center to intersection circle & red & purple square, drop perpendicular down as presented. If large square has side X, middle square Y, & small square Z. X + Y = 5, Z = X - Y. hypotenuse + 2.5, vertical leg = Y, horizontal leg = 7.5 - 3Y. Solve for Y, get y = 5/2 (rejected), Y = 2. Square dimensions are 1, 2, 3, area = 14.
With math problems like this a lot of the time they're not to scale, so it's important to check, but it's satisfying that the solution to this one actually is what it looks like.
Well, if you want to really overcomplicate it, sure. But the fact that small and medium fit into one side of big, and big plus medium equal 5, and they are all true squares... means medium is 2, small is 1 and big is three. No equations required.
For me it's purely visual, I saw that the length was 5 then found that the smallest square is one by one long with an area of 1 and simply counted how many small squares would be present in the shape we are given. Granted this doesn't work in every scenario but I got to the answer a lot quicker than the video does and got a good chuckle when me and the video arrived to the same conclusion.
When i was studying maths i was taught to always tey different approach to what seemed like a difficult problem. But when I started working in difficult problems, I learned quickly that most difficult prblems were not as difficult as they would seem at first glance.
Completely forgot about that right angle theorem; there are just so many from Geometry to remember .. I kept thinking about trying to use Pythagorean theorem on the x and y blocks to find the radius of that semi-circle..
I defined x as the length of a side of the smallest square like you did. Then I found ways to express the sides of the other squares with using x as the only variable: pink edge = x purple edge = 2.5-0.5x blue edge = 2.5+0.5x Then I put a right triangle between the bottom left corner of the smallest sqaure, the center point of the half circle and somewhere on the base line directly underneath that bottom left corner of the smallest square. The hypothenuse of that triangle would be identical to the radius of the half circle which is 2.5 and the other sides would be 1.5x and 2.5-0.5x (the purple edge). Using the pythagorean theorem I found x (the pink edge) to be 1 and substituting that into the above I found the purple and blue edges to be 2 and 3 respectively. 1²+2²+3²=1+4+9=14
You say x can’t equal 0, but I don’t see why not. You get the pink square with an area of 0, and the purple and blue squares each with an area of 7.75, for a total area of 12.5, making it a rectangle.
I support your comment. x = 0 is a totally legit solution. It shouldn' t be a surprise that a problem that implies a quadratic equation has two solutions.
I understand the process as soon as i see the problem. Math is very interesting, exciting and challenge. I love to go back in time and wanna challenge these math problems again😢
I have an exam today, and they ask a lot of area type geometry questions, I have been following you for a long time, if I get atleast one question that have concepts that you used, all Credit goes to you❤
I solved this exercise by comparing the area of the entire rectangle 5*(x+y) minus the white area missing (x-y)^2 with the summatory of the 3 squares 5(x+y) - (x-y)^2 = x^2 + y^2 + (x+y)^2 then i replaced y = 5-2x and solved the cuadratic equation I have so much fun with your videos 💯💯💯
Doesn't the fact that the semicircle intersects the corner of the small square and terminate at the corner of the medium square necessarily imply that the medium square is twice the length of the smaller square? If so, that saves a lot of time.
I don't think it does. Draw a quarter circle, and draw a square where the ends of the quarter circle touch two opposing corners. Then, you can shrink the square any amount you want, and put a smaller square so that you have an arrangement similar to the problem. You can make the ratio of the side lengths as big as you like.
Learned something new with your way of solving it. I managed to solve it but I assumed the smaller triangle intersect was right at the middle - Which I think it’s not a given so I’m considering my process luck 😅
Seeing this video reminded of my college days and learning of a square with negative dimensions. I think veritisium made a video about real life shapes that exist as the number i. Great content! Keep it up brother.
I got the right answer just by eyeballing the bases of the squares. Though, your method is probably way more reliable, and I doubt my calculus teacher would be happy if I just estimated everything on the next test, lol.
I'm pretty sure x=0 is a valid solution. The purple and blue boxes then end up being the same size, each having a side of 2.5 units. Total in this case is 0 + 2.5^2 + 2.5^2 = 12.5.
I was wondering what the relevance of the semi-circle was, and now I see it - it lets you create the inscribed triangle which gives you a 90 degree angle and produces the similar right triangles which then allows a way to figure out the length of x and y.
I solved using the equation of the circle for 3 points like one of the other video you showed, you can declare 2 variables hx and Xx and you can write a system with 4 incognita and 4 equations solving for hx and Xx you get the same result but in a less elegant fashion
The construction is not fully constrained, so x could absolutely be equal to 0, leaving the other answer as 25/2. The circle still intersects the meeting between the BL of the pink square and the top edge of the people square.
Wow, amazing how these can be solved, when at a first glance it seems impossible! Love watching these to brush up on my math skills👍 Trigonometry is my favourite!👍
Add a square in the top left corner 3x + y = 5 2y + x = 5 y = 5 - 3x 2(5 -3x) + x = 5 10 - 6x + x = 5 -5x = -5 x = 1 3(1) + y = 5 3 + y = 5 y = 2 I suppose the only way this doesn’t hold up is if we don’t know that the pink square’s side is not half of the purple square’s side.
Nice explanation, but i think you could solve it quicker by comparing the pink square to the purple square. You can see on the top that the purple square's side is double the one of the pink square one, so x = 0.5y, or y= 2x The rest is simple. y+x= 3x, 3x + 2x = 5, 5x = 5, x = 1. Pink = 1^2, Purple = (2*1)^2, Blue = (1*1 + 2*1) ^2
So, I am not sure I would choose to entirely neglect the x=0 solution, given the information provided. It’s not too hard to solve that though, and depending on who’s marking it, may be required to get full credit
Think there is another way here without circle theorems - using the relation x+2y=5 and also noticing that 5(x+y) = x^2 + y^2 +(x+y)^2 + (y-x)^2, which is just saying the area of the large rectangle equals the sum of all the interior squares. Using the first relation and eliminating x leads to a quadratic equation in y with solution either y=5/3 or y=2. You can rule out the first of these since it would imply y=x which is clearly false, and this leads to the solution. Great problem!
In the first 5 seconds i said "that's easy, that one's 9, that one's 4, that one's 1." The big square was taking 3/5 of the bottom so 3x3, the medium square was taking 2/5, so 2x2, and the small square's side looked like half the medium one's, so 1x1. I was right without math
There's a simpler solution I used: You know the diameter is x + 2y = 5, but you can also conclude that the diameter is 3x+y When put into a equationsystem you define y = 5 - 3x and then substitute y in the other equation and you get x + 2(5 - 3x) = 5 Then you simplify into x + 10 - 6x = 5 and 10 - 5x = 5 and 5x = 5 and x = 1 Now you substitute x in any equation from before and we get x + 2y = 5 = 1 + 2y = 5 and 2y = 4 and y =2 Then you work out the areas as before
HOW EXCITING🔥🔥🔥🔥🔥
Me and all my homies when Andy Math say HOW EXCITING🎉🎉🎉🎉🎉🎉🎉🎉🎉
🗣️🔥
Keep it at 777 likes
Put a box around that comment
I’ve never seen someone so excited to solve the area of shapes
"how exciting" 3:55 love his constant excitement every video hahahah
Am I John Cena ni-
-ce friend of mine 😁😁
Well it will apply in shapes with a 90 degree triangle ofc. Squares and rectangles sure, but not trapezoids or kites and such (at least not all of them
This guy is such a G. He's genuinely excited to just do geometry all day, and in such a simple and zen way we can follow.
I can't remember ever learning that an inscribed triangle along the diameter is a right triangle, but it makes sense. That's the conceptual step I was missing.
Never heard of Thales theorem?
@@Zieki99 Not that I can recall, but highschool was more than a decade ago and I don't use geometry in my day-to-day job. Again, it's just something I can't remember ever being covered, not that we didn't cover it.
Separate it to 2 isosceles triangles from the center, and apply all angles sum to 180 you get α+β=90
Shut up nerd
@@Zieki99I thought it was circle theorems
The last time I needed to solve problems like this was around 15 years ago, but i still come here and try to solve these once in a while. The way you teach and explain is so good! Kudos for not only keeping students challenged, but people like me as well!
Pretty much same here. Im gonna start pausing at the start to try an do em myself. Hoping it'll keep my mind sharp!
I never comment. Never subscribe. But you are crushing it. I save all of the problems that involve basic algebra geometry and algebra 2 concepts for my high school students.
Andy Math out here differentiating instruction for me. God bless you and your family
For all those saying, "its obvious" or "i could just visualise it" etc.... That means nothing in an exam as its not a proof. You will get next to 0 points just giving an answer or an explanation like that. You need to PROVE the answer. Yes you can guess but it doesnt mean its right. What if y isnt 2x? Thats why we have proofs.
what if y isn't 2x?
(x + 2y) / 5 = 1
assuming 5x = 5
5x - 2y =x
y=2x and x=1
let's fill this in
x + 2y =5
1 + 2(2) = 5
hrmmmmmm
sir I have a suspicion it y may be 2x
if y=2x then what exactly IS y supposed to be
also I can just divide into fifths
my proof is 5/5 is 1
Yeah, it can bite you back
Especially shapes where they LOOK like they're touching each other so you can visualize and make comparison....but turns out they werent when try to proof it and end up not getting an answer on all 5 options
I’d also assume the other unstated (but visually implied) given was that the arc intersected exactly at the lower left corner of the pink square. There’s nothing that says it is drawn to scale, but I don’t see a way to solve it without that implied corner contact variable.
Good stuff.
Tried solving it but without that it's impossible. I saw that it was probably what I was missing but given it wasn't stated in the question you can't assume it's a fact.
This is also how I would approach this problem to begin with, how do you scale purple and blue square so that the third, pink square, will touch both the semicircle and align with top of the blue one? Purple and blue depend on each other (otherwise they either aren't squares, or don't add up to 5 base), so there is only one degree of freedom, then for each pair the pink square is implied and either accepts or rejects the solution.
Ah I see. The entire time I was wondering how it even makes sense to find a unique solution, given that you can draw the other 2 squares for any blue square, but with that restriction that doesn't really hold.
@@chaoticsquidno it's Def solveable without the circle.
We know it's Fibonacci sequence. Which means we can run that sequence with a base dummy variable adding up each time and dividing that by five. The dummy variable represents the edge of orange square, and fibonnaci sequence dictates it'll repeat five times by now.
Gonna be honest I just assumed they had areas of 1,4 and 9 because of the Fibonacci sequence
same
Good assumption, I just used the blue square is a little over 2.5 so nearest while number fitting scale being 3 and working through 😅yours is a much better assumption and much cooler too
Correct but you still need to prove it...
its an exponential sequece, not the fibonacci sequence... the fibonnaci sequece goes 1,1,2,3,5,8,13....
Such an elegant solution with whole numbers.
I love how well made the puzzle is; so simple yet so many straightforward steps to solve it. Would buy a book full of these
I solved it a different way.
The radius of the semicircle is half the diameter, so r=5/2. Drawing a line from the center of the semicircle's base to the point at which the two smaller squares and the circle meet results in a line of length 5/2. Draw a line straight down from that point. You now have a right triangle with hypotenuse of length 5/2 and a height of y.
Thanks to the Pythagorean Theorum, the base of that triangle, which goes from where that vertical line of length y touches the base to the midpoint of the semicircle's base, is the square root of the difference between (5/2)^2 and y^2. So the triangle's sides are sqrt((25/4)-y^2), y and 5/2.
The length of the line extending from that triangle's right angle to the left edge of the semicircle is y-x. Therefore, the sum of y-x and sqrt((25/4)-y^2) is equal to the radius, which is 5/2. That's (5/2)=y-x+sqrt((25/4)-y^2)
Solve for x, and you get x=y-(5/2)+sqrt((25/4)-y^2).
Replace x with that expression in 5=2y+x, and you get 5=3y-(5/2)+sqrt((25/4)-y^2).
Isolate the radical to get (15/2)-3y=sqrt((25/4)-y^2).
Square both sides to get (225/4)-45y+9y^2=(25/4)-y^2.
Move everything over to one side and combine like terms to get the quadratic function 10y^2-45y+50=0. Simplify to 2y^2-9y+10=0.
Utilize our old friend, the quadratic formula, to get y=(9+-sqrt((-9)^2-4(2)(10)))/2(2).
Simplify: y=(9+-sqrt(81-80))/4.
Simplify: y=(9+-sqrt(1))/4.
Simplify: y=(9+-1)/4.
Conduct plus/minus operation: y=10/4, 8/4.
Simplify: y=5/2, 2.
y must be less than the radius, which is 5/2. Therefore, y can only be 2.
Plug it in to 5=2y+x, solve for x, and x=1.
Plug x and y into x^2+y^2+(x+y)^2 to get 14.
I like that you always go for geometry problems that can be solved by high school level mathematics yet are still challenging.
Was the triangle necessary? I could tell immediately that a lemgth of pink was ½ a length of purple; which means if diameter is 5, blue must be 3 and purple 2, thus pink 1. So it's 3×3+2×2+1×1=14 without the triangles... It's also a Fibionacci sequence, but I wanted to show my work. 😋
An alternative way to solve this problem:trace a line beetwen the center of the semicircle and the point where the semicircle intersect the lower left corner of the smaalest square and apply the pitagorean theorem în the right triangle.
Very nice! However there exists another solution just using the assumptions you made. The equation at 2:13 assumes y is not 0, but for y=0 you do acually get another solution for the total area, being 12.5 square units.
i got to your solution in my head before clicking the video (i assumed x=0, same effect), so are there finite solutions?
Wouldn't that mean that the point where the circle intersects the corner of the squares is on the bottom left? This is a problem since the triangle we used to make that would have 0 angles.
I’m a healthcare professional and your videos fill the void of math in my field.
I like that you explain how to solve the problem very succinctly and clearly.
we could consider the case x=0 still a solution that represents the case when area of the little square vanish or gets smaller and smaller leaving only two equal symmetric squares, in this configuration the area is 12.5
Who made bro so high and mighty in mathematics 😭
Fun fact, although is only a small sample, those squares apear in representations of the Fibonacci sequence.
Your videos really helping me for my Olympiads =)
I can only be gay for Andy
Wtf man?!? He is just doing maths.
@@ВалепЙикстуер some people cannot control themselves I guess 😂 desperate peeps really
@@ВалепЙикстуерhaven’t you ever heard people say math is gay? I would know, I loved math in school, and now I’m gay.
Love your videos, doing these things with you are one of my favorite activities. Please take care of your own health and don't overdue with the videos and or any other job. Love you.
yeah.. forget the semicircle. like no. specially with how the question is stated just go in with a ruler and determine the other sides from the bottom line
I did it a different way, label lengths of the squares from largest to smallest as a,b,c. Then we can create a set of equations
a+b = 5 (1)
b+c = a (2)
To get a third equation we can take the point where the corner a is on the semicircle and use pythagoras, noting that the radius is r = 2.5:
b^2 + (2.5 - c-a)^2 = 2.5^2 (3)
add together (1) and (2) to get
a+c = 5 + a - 2b = 5 + (a+b) - 3b = 10 - 3b (4)
substitute (4) into (3) to get
b^2 +(3b-7.5)^2 = 2.5^2,
==> 10b^2 - 45b + 50 = 0,
==> (2b - 5)(b - 2) = 0
==> b = 2, 2.5
if b is 2.5 then a = 2.5 and c = 0 so total area is 12.5 (trivial solution) , and for b = 2, we get a = 3 and c = 1 so total area is 14.
I solved it with the same approach!
Thank you so much for the great content!
As one who works as a math teacher, your content has been a huge inspiration on how to make challenging and fun puzzles!
x and y are the coords of the bottom-left corner of the pink square:
Pink Square Length = 5 - 2y
Pink Square Length = y - x
y = (5+x)/3
But that same corner point also falls on the semicircle, which can be expressed as:
(x-5/2)(x-5/2) + yy = 25/4
Substituting the first expression into the second and simplifying yields the quadratic:
2xx - 7x + 5 = 0
Which has solutions:
x = 5/2 , x = 1
That'll make x bigger than y, which has no sense on the graphic, x=1 is the valid answer
@@armandocarrion479
The x=5/2 solution corresponds to the pink square having no size.
The consequence of this is that purple and blue both have side length equal to the radius (5/2).
It's a valid, albeit trivial solution.
i did this using the Pythagoras formula. you know the center of the circle is 2.5 from the edge. you can draw a triangle that goes from the center of the circle to intersection of the circle and the 2 smaller boxes, then down perpendicular to the base of the semi circle. this has sides 2.5 (hypotenuse is a radius of the circle), y and 2.5-y+x. we can substitute x = 5-2y into that last side then the side lengths into Pythagoras formula to give us a quadratic in y. solve that to get y = 2 or y = 2.5 giving x = 1 or x = 0 (which we can discard) and finish up getting the areas.
Bro, i used to love maths and geometry up until highschool, but i had a shitty teacher that made me hate these subjects. Now thanks to Andy im relearning how cool this is
Seeing someone solve things like this perfectly is so satisfying 😭
I like the way you solve problems. Quicker and much more exciting than the other youtubers.
Wish you will reach a million subscriber this year 😊
Interesting… at a glance, I wondered if those were the proportions, but assumed it wouldn't be so simple!
I love how clear his explanation is. No unnecessary talking so that even a non-native speaker who always sucked at math can follow easily!
I just figured they must be Fibonnaci squares, pink+purple=blue. And 5 is a number in the Fibonnaci sequence.
So 2+3=5, and 1 comes before 2 in the sequence. So the squares' areas are
1^2 + 2^2 + 3^2=14
Though if the diameter was 6 or 4 then they'd still be Fibonnaci squares, but with non integer lengths, so you'd be better off with the method in the video
it could've also been like 1.9 and 3.1 and then the pink one being 1.2, it didn't say they had to be whole numbers
You can do this without the semicircle or triangles. You do, however, have to infer that pink is smaller than purple.
The only way I see that working is if the side lengths of the squares have to be integers.
Otherwise, there's ways I can manipulate the squares where pink is ALMOST the size of purple.
That gives us an area of approximately 2x by 3x where 3x = 5.
Final equation for the area gives us 5 × 5 × 2/3 = 50/3 square units, which is more than 14.
Other way around, make pink approach zero.
Area is now 2y^2 where y equals 5/2, the answer is 25/2 square units which is less than 14.
Any other answers will lie between those extreme values.
@tomdekler9280 I suppose I did also assume that. Good point.
I came to the point of, 5=2x+y where x is purple and y is pink. From there I said 2x must be an even number less than 5. this is where I assumed an integer value, it's also where I said pink must be smaller because otherwise purple could be 1.
Fibonacci and Generalization -
Building on your process, paying close attention to the triangle which is the sum of two smaller ones, with specific ratio:
We have x. Next, we have y=x+x=2x. Next, we have x+y=2x+x=3x. Next, we have x+2y=3x+2x=5x.
This is the Fibonacci sequence where each term is multiplied by x.
The first, same value as second, term is missing (we have 1, 2, 3, and 5, instead of 1, 1, 2, 3, and 5 - which correspond to the "bite" missing from the "complete" rectangle).
using s for side lengths and a for areas, each followed by 1-3 for smallest to largest squares.
Let's call the 4th term: z. specifically:
z=5x. so:
s1=z/5, s2=2z/5, s3=3z/5. Squaring for area:
a1=z²/25, a2=4z²/25, a3=9z²/25. Summing:
Total area = 14z²/25.
In your example, z=5, so the total area = 14*5²/25=14.
For z=6, for example, total area would be 14*6²/25=20.16.
For z=10, twice the 5, the result should be quadrupled: 14*10²/25=56 - and it is.
Hola andy, me encantan tus videos, siempre me sorprende la forma tan sencilla en la que solucionas los problemas. Saludos desde Colombia ^^
¡Gracias!
I cant believe this channel does not have more subscribers! Im am so glad i found you in my algorithm
Approached this slightly differently. Made a right triangle from the center of the circle to where it intersects the pink and purple squares. One side is y, hypotenuse is 2.5 and the other side is the radius of the circle (2.5) minus the portion of purple square that sticks out past the pink square (y - x). So right triangle with sides y and 2.5 - (y - x) and hypotenuse 2.5. Plugged in 5-2y for x and solved the resulting quadratic form Pythagoras.Not quite as elegant but still worked.
Nice! I actually prefer this solution.
Cool! Here's how I solved it instead:
- Grabbed a screenshot
- Cropped the figure
- Resized it so that the long side equals 500 pixels
- Measured all sides, assuming that 1 unit = 100 pixels
- Obtained measurements of 2.97, 2.03, and 0.95 units
- Rounded up the measurements to the nearest integers, assuming that the author of the problem used integers
- Obtained measurements of 3, 2 and 1
- Calculated the area of each square
- Summed the areas
😂😂 out of the box thinker😂
HALT! Not so fast! There is no stated reason to discard the x=0 case! Yes, just because it is on the diagram, does not mean the square actually exists!
If x=0, then x+2y=5 => y=5/2. We then must add x^2 + y^2 + (x+y)^2 to find the areas. 0^2 + (5/2)^2 + (5/2)^2 = (5/2)*(5/2)*2=25/2=12.5 square units.
This "extraneous" solution is the same a building two squares, one purple and one blue, with side length equal to the radius, which means the "intersection point" between the circle and the corner of the "pink square", which is actually a point by this construction, is exactly perpendicular from the center of the circle, one radius length away along both squares' side.
How exciting.
if x were to be 0 it would break all squares containing it to be triangles, which is not a valid solution, as the image is in real scale, and all squares don't have x as 0.
@@giovannicesaramorim9adigan961 A square of side lengths 0 is a point, not a triangle
I agree you can construct a valid geometric solution with x=0. I think it depends on the interpretation of the problem, which states there are 3 squares. If x=0, then the area of the pink square is zero. Is that still a square?
@@calvolito9535 It is a degenerate c, but It could successfully be argued that yes, a degenerate square is still a square.
@@themathhatter5290How do you distinguish it from a degenerate circle though? Or a degenerate triangle? Or a degenerate regular polygon with n sides? They are all just points.
Mind blowing! absolutely loved it.
Had to fill in an abandoned septic tank years ago. The inside dimensions of the tank was 5' deep, 5'wide and 6' long with rounded ends. Use pi and radius formulas to determine the amount of controlled density fill that was needed. Stuff is order in complete cubic yards.
Im gonna have test from plane geometry soon, this is actually gonns be pretty helpful (we do these kinds of exercises) 🔥🔥
Another way to solve this is to form right triangle, hypotenuse center to intersection circle & red & purple square, drop perpendicular down as presented. If large square has side X, middle square Y, & small square Z. X + Y = 5, Z = X - Y. hypotenuse + 2.5, vertical leg = Y, horizontal leg = 7.5 - 3Y. Solve for Y, get y = 5/2 (rejected), Y = 2. Square dimensions are 1, 2, 3, area = 14.
At each equation found you can feel that it makes he happier 😂
With math problems like this a lot of the time they're not to scale, so it's important to check, but it's satisfying that the solution to this one actually is what it looks like.
Well, if you want to really overcomplicate it, sure. But the fact that small and medium fit into one side of big, and big plus medium equal 5, and they are all true squares... means medium is 2, small is 1 and big is three. No equations required.
For me it's purely visual, I saw that the length was 5 then found that the smallest square is one by one long with an area of 1 and simply counted how many small squares would be present in the shape we are given. Granted this doesn't work in every scenario but I got to the answer a lot quicker than the video does and got a good chuckle when me and the video arrived to the same conclusion.
There was no guarantee that the squares' sides would be integers, or that the sizes were to scale.
When i was studying maths i was taught to always tey different approach to what seemed like a difficult problem.
But when I started working in difficult problems, I learned quickly that most difficult prblems were not as difficult as they would seem at first glance.
Completely forgot about that right angle theorem; there are just so many from Geometry to remember .. I kept thinking about trying to use Pythagorean theorem on the x and y blocks to find the radius of that semi-circle..
I defined x as the length of a side of the smallest square like you did.
Then I found ways to express the sides of the other squares with using x as the only variable:
pink edge = x
purple edge = 2.5-0.5x
blue edge = 2.5+0.5x
Then I put a right triangle between the bottom left corner of the smallest sqaure, the center point of the half circle and somewhere on the base line directly underneath that bottom left corner of the smallest square. The hypothenuse of that triangle would be identical to the radius of the half circle which is 2.5 and the other sides would be 1.5x and 2.5-0.5x (the purple edge). Using the pythagorean theorem I found x (the pink edge) to be 1 and substituting that into the above I found the purple and blue edges to be 2 and 3 respectively. 1²+2²+3²=1+4+9=14
You say x can’t equal 0, but I don’t see why not. You get the pink square with an area of 0, and the purple and blue squares each with an area of 7.75, for a total area of 12.5, making it a rectangle.
Because the question explicitly states that there are 3 squares, twice. Also, how did you get (5/2)² = 7.75?
I support your comment. x = 0 is a totally legit solution. It shouldn' t be a surprise that a problem that implies a quadratic equation has two solutions.
I understand the process as soon as i see the problem. Math is very interesting, exciting and challenge. I love to go back in time and wanna challenge these math problems again😢
It’s been 2 plus decades doing this and enjoyed. I had to pause to get my memory going 😂
I have an exam today, and they ask a lot of area type geometry questions, I have been following you for a long time, if I get atleast one question that have concepts that you used, all Credit goes to you❤
You are just brilliant at explaining this stuff!
This guy helps me with geo better than any tutor
I solved this exercise by comparing the area of the entire rectangle 5*(x+y) minus the white area missing (x-y)^2 with the summatory of the 3 squares
5(x+y) - (x-y)^2 = x^2 + y^2 + (x+y)^2
then i replaced y = 5-2x and solved the cuadratic equation
I have so much fun with your videos 💯💯💯
That’s how I did it too!
Beautiful! 🔥
Wish my math lessons were this intriguing when I was a youngster !
It’s fun when you can approximately do it by eye and assumption of whole numbers, but the correct algebraic method is interesting to follow along
Doesn't the fact that the semicircle intersects the corner of the small square and terminate at the corner of the medium square necessarily imply that the medium square is twice the length of the smaller square? If so, that saves a lot of time.
I don't think it does. Draw a quarter circle, and draw a square where the ends of the quarter circle touch two opposing corners. Then, you can shrink the square any amount you want, and put a smaller square so that you have an arrangement similar to the problem. You can make the ratio of the side lengths as big as you like.
Very simple and straightforward explanation 👌
Merci pour ces vidéos, je suis impressionné par la facilité dont résout ces problèmes
Learned something new with your way of solving it. I managed to solve it but I assumed the smaller triangle intersect was right at the middle - Which I think it’s not a given so I’m considering my process luck 😅
This shows how logical reasoning gets you the correct answer very quickly, but proving that it is correct is a long and confusing path.
Seeing this video reminded of my college days and learning of a square with negative dimensions.
I think veritisium made a video about real life shapes that exist as the number i.
Great content! Keep it up brother.
I forgot how satisfying it was to solve geometry equations! the same as like figuring out a puzzle
Excellent vídeo. I dont understand english well, but It IS Very "visible" resolution and easy to follow The math😊
WE GETTING OUT OF MATH CLASS WITH THIS ONE 🙏🙏🙏🙏😭😭😭🌛🔥🔥🔥
This is an exiting classic!
This is appeared in my recommendations and is the best recommendation that UA-cam gave me today.
I got the right answer just by eyeballing the bases of the squares. Though, your method is probably way more reliable, and I doubt my calculus teacher would be happy if I just estimated everything on the next test, lol.
I'm pretty sure x=0 is a valid solution. The purple and blue boxes then end up being the same size, each having a side of 2.5 units. Total in this case is 0 + 2.5^2 + 2.5^2 = 12.5.
I was wondering what the relevance of the semi-circle was, and now I see it - it lets you create the inscribed triangle which gives you a 90 degree angle and produces the similar right triangles which then allows a way to figure out the length of x and y.
I solved using the equation of the circle for 3 points like one of the other video you showed, you can declare 2 variables hx and Xx and you can write a system with 4 incognita and 4 equations solving for hx and Xx you get the same result but in a less elegant fashion
Another video from my favorite math teacher youtuber
I love watching these. I'm hoping to remember some of it when I need it later!
The construction is not fully constrained, so x could absolutely be equal to 0, leaving the other answer as 25/2. The circle still intersects the meeting between the BL of the pink square and the top edge of the people square.
I think x=0 is a valid solution, the problem doesn't specify that each square must have non-zero area
Clear, clean and elegant solution.
Although I would improve a little the graphic animation, still it's an excellent video.
Big like ❤👍
Cool problem! I ended up with 14 by setting (2.5-y+x)^2 + y^2 = 2.5^2 based on where the semicircle overlaps with the pink square’s corner.
I wouldn't have thought to use a similar triangle proportion. Obscure methods are exciting.
Wow, amazing how these can be solved, when at a first glance it seems impossible!
Love watching these to brush up on my math skills👍 Trigonometry is my favourite!👍
Add a square in the top left corner
3x + y = 5
2y + x = 5
y = 5 - 3x
2(5 -3x) + x = 5
10 - 6x + x = 5
-5x = -5
x = 1
3(1) + y = 5
3 + y = 5
y = 2
I suppose the only way this doesn’t hold up is if we don’t know that the pink square’s side is not half of the purple square’s side.
Nice explanation, but i think you could solve it quicker by comparing the pink square to the purple square. You can see on the top that the purple square's side is double the one of the pink square one, so x = 0.5y, or y= 2x
The rest is simple. y+x= 3x, 3x + 2x = 5, 5x = 5, x = 1. Pink = 1^2, Purple = (2*1)^2, Blue = (1*1 + 2*1) ^2
How do you know the purple square is twice the size of the pink one? There's no guarantee that the sides are of integer lengths.
@@charltonrodda I eyeballed it 😂
I like your funny words, magic man
You're right, that WAS a fun one :D
I remembered how to form that first right triangle but didn't figure out the step to form the similar similar smaller ones. Cool stuff!
love all you great videos Andy! thank you very much! :)
So, I am not sure I would choose to entirely neglect the x=0 solution, given the information provided. It’s not too hard to solve that though, and depending on who’s marking it, may be required to get full credit
Think there is another way here without circle theorems - using the relation x+2y=5 and also noticing that 5(x+y) = x^2 + y^2 +(x+y)^2 + (y-x)^2, which is just saying the area of the large rectangle equals the sum of all the interior squares. Using the first relation and eliminating x leads to a quadratic equation in y with solution either y=5/3 or y=2. You can rule out the first of these since it would imply y=x which is clearly false, and this leads to the solution. Great problem!
Thank you for revising all concepts sir❤
I might be thinking about this wrong, but wouldn’t x always be 1/2 y.
In the first 5 seconds i said "that's easy, that one's 9, that one's 4, that one's 1." The big square was taking 3/5 of the bottom so 3x3, the medium square was taking 2/5, so 2x2, and the small square's side looked like half the medium one's, so 1x1. I was right without math
For some reason, this problem was super easy to visualize and solve without doing any algebra
Its cool that i could just try guessing these squares and still get it right
There's a simpler solution I used:
You know the diameter is x + 2y = 5, but you can also conclude that the diameter is 3x+y
When put into a equationsystem you define y = 5 - 3x and then substitute y in the other equation and you get x + 2(5 - 3x) = 5
Then you simplify into x + 10 - 6x = 5 and 10 - 5x = 5 and 5x = 5 and x = 1
Now you substitute x in any equation from before and we get x + 2y = 5 = 1 + 2y = 5 and 2y = 4 and y =2
Then you work out the areas as before
I’m proud that I found you before you reached a million subs (which I know for sure will happen!)