How To Solve For The Area - Viral Math Problem
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- Опубліковано 10 лют 2025
- Thanks to Reio in Romania for emailing me this fun problem! What is the area? This puzzle was shared with the tagline "you should be able to solve this."
Solution to 5th grade Chinese challenge problem
• Genius student solved ...
My blog post for this video
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Reference
Math problem from anime Steins;Gate
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Happy anniversary! Thanks for nearly 2 million views in one year! I really thought it's obvious why this equation is true (a + b) + (c + d) = (b + c) + (a + d).
Please tell us, I can't figure out why
MindYourDecisions You could've just said "by the associative and commutative laws of addition". That would've been about as fast as saying "notice that" or "it's obvious that" while still giving viewers enough information to research their missing puzzle piece.
Luke Johnson : Ah, yes, the plight of so, so many proofs in textbooks! Just spell out all the steps that are so "obvious" or "clear" -- it doesn't even take that many more words usually!
@@HeracIeid This is the first real world application of commutative and associative laws I have encountered!
Other way to see it is because the whole square is 2a + 2b + 2c + 2d. Let's know the area of each triangle, their angles and length of their legs, just for fun.
I contemplated the thumbnail of this video for 20 minutes before finally deciding to watch it.
SAME.. I stared for like 10 min.. I think the answer is 28... (16+32=48) and (20-48=28)...
Let's see what's the answer now
I just guessed and said 28 and I was right
Given triangle ABC, a = 3, b = 4, sin A = 0.6 . Find c and draw it geometrically. It's not so easy as you think!
@Moonlight io use trigonometry and algebra, and you will be surprised!
@Moonlight io well, then it's time to start to learn!
28? Aw man, so close! My answer was "Argentina"
Presh Talwalker: You will really need to change your name, at least the first name. I have never seen a name that I really don't like.
I got 25
Ha! I guessed what i subtract and add and and I got 28 😝
I seriously got 28....but i did not use the exact same method !
So were my grandpa’s
When you can solve complex analysis problems but didn't even know where to start solving this
Who don't understand why the sum of opposite areas are equal, just try to count what triangles these areas consist of.
Yellow area - a, b
Red one - c, d
Green one - b, c
Blue one - a, d
So
Yellow and Red areas together consist of a, b, c, d.
Green and blue areas together cosist of a, b, c, d too.
So, it's why they are equal
Actually, it's not the "opposite areas" which are equal, because that would mean a+b = c+d. I also at first somehow read and heard "opposite areas are equal" which is just wrong. It's really "the *sum* of the opposite areas is equal". I guess more ppl just misheard/misread it and missed that critical word "sum".
TheFeldhamster you're right, I corrected
Excellent explanation. Was wondering where he pulled this magical statement from.
nice bro...
Another way to look at it would be to consider the interior point as a locus of a kind.
Another method: Designate the side of the square as 2a. Run a diagonal line between the center points of each side of the square to the adjacent sides. The square is now divided into 8 pieces. The outer 4 pieces are 45 degree right triangles with area (a^2)/2 . The other four pieces are triangles each of a different size. One triangle goes with each of the four areas the square is divided into. We denote the areas of these triangles T16, T20, T32, and T?, with the number corresponding to the identified area of the original square. Now rotate this cluster of four triangles by 45 degrees clockwise. You see that you have a square whose side is sqrt(2)*a. Now notice that based on the base and height formula for a triangle that T16+T32 = a^2. Now add the two 45 degree triangles that go with the area that's 16 and the area that's 32. You get (a^2)/2 + (a^2)/2 + T16+T32 = 16+32. Hence 2a^2=48 and a^2=24. Thus a=6sqrt(2). Now we know that T20= 20-(a^2)/2= 20-12=8. We also know from the square containing the four triangles, using the area of a triangle formula that T20+T? = a^2. Hence T?= a^2-T20 = 24-8=16. The area the problem asks for is T? plus one of the 45 degree corner triangles whose area is (a^/2)/2 which is 12. So the area the problem asks for is 16+12=28.
how does T16+T32 = a^2
Because the area of a triangle is a^2/2 and since they are two triangles with equal sides, then it's a^2/2*2, so a^2 @@account5223
"I don't know" is also a correct answer
I put that answer for my exams and I got an F! I got Fantastic!
@@-Chicken_ I think in this case F stands for "Fair"
I think, i‘m a genius. Just continue watching and don’t even try solving, because he‘ll tell you the solution.
wow，intelligent！
Thats cheating!
By the way MY method was way faster...
Solution:
Take the square root of 76 and then scroll through the comments...
The first step I left out
Genius answer , pal , total 101% genius
What praise should I give you for such an genius answer
Einstien's brain is the mere dust of your feet , pal , genius
Benhard Wiesner, You cannot apply the square root like that......can You?
I solved it from the first try 😊
“The sum of opposite areas are equal” , you lost the majority of us there mate.
Yes but it is not essential to resolve the problem !
If the sum of "opposite areas" (definition, please?) are equal, why even bother to subdivide into triangles?
Aha! Don't think of this as "opposite areas are equal," think of this:
a + b + c + d = a + b + c + d, so ...
(a + b) + (c + d) = (b + c) + (a + d) merely by rearranging, and the rest follows.
@@josepherhardt164 Thanks Bro.
How does the side length matter?
If the whole square area equals to 100%, and other areas' sum is 16%+20%+32% = 68%, so the rest would be 32%, in current measures is 32 cm^2. Can this be correct?
@@MrMeGaSeNt we don't know if its 16%, just that it's 16 cm^2.
I solved it in a different way with matrices.
I first thought about a square cut into 4 even pieces (like how a window looks like) so the dot would be right in the middle
I labeled each area z
Then I thought about what would happen if u moved the intersection (the dot) directly left or right. If I moved the dot to the right, then the left 2 squares would increase the same amount the right 2 decreased , since you can't magically gain or lose more area by partitioning them differently.
So I made x the amount a square gains/loses when the dot moves left or right.
The same reasoning for the dot moving up or down. Made the change y
So upper right was z + x + y = 32
Upper left was z - x + y = 20
Lower right was z - x - y = 16
3 equations 3 unknowns. z = 24, x = 6, y = 2
The equation of the lower left square was z + x - y. Plug in and you get 28
nice
Damn, great solution.
Liked ur approach but something is a miss here.. You got ur ans coz of square or a parallelogram which have this as a property .. But according to your method(area can never be created) this can be applied to any of the figure like quadrilaterals.. N there it fails.. Try it urself.
Alex Xela
This is so much better
I really appreciate that you talk through the answer slowly so I can watch a bit and then have a go :)
There is a much simpler way to arrive to the same answer. Starting with the upper left and working clockwise, let's label each area A, B, C, and D respectively. Since each line segment bisects the outer wall, then (A + C) = (B + D), giving us (20 + C) = (32 + 16), or (20 + C) = 48. Subtract 20 from both sides: C = 28.
This is way easier
Can you explain why areas A+C will be equal to B+D? Didn't understand that part..
@@girishshivshankar6339+
@@girishshivshankar6339 If you join the mid points of the sides of the square then you get a smaller inner square. Now each of A,B,C,D is composed of a triangle outside the inner square and a triangle inside the inner square. The outside triangles are all the same size. Because it's a square the sum of the heights of opposite triangles in the inner square are equal. Because their bases are also equal, the sum of their areas is equal.
@@girishshivshankar6339 Divide the square into two rectangles vertically at intersecting point. 2b is half the area of left rectangle, 2d is half the area of right rectangle. The remaining area is 2a+2c.
Hell yeah ! When I finally decided to watch the video I was like, "ok might as well try to find a solution before watching" and I found 28cm2 in my head. God it's nice to get a boost in math confidence for once :')
A Chinese fifth grader solved the more difficult problem in less than a minute... *cries myself to sleep*
it is so easy for a Chinese like me.
And i‘m lerning integration now, although i‘m just a ninth grader :)
For the bonus: Call the point where the red area and the 72 area meet M, the point where the 10 area meets side BC N, and the point where the 8 area meets side CD P. Area(triangle ADM) + area(triangle BPM) = 1/2 area(parallelogram ABCD) = area(triangle ADN). The missing spots are the same in each side, so call the sum of their areas y and the red area x. Then x + y + 72 +8 = y +79 + 10 -> x = 9.
Thus the red area is 9.
Nice
Who told those are of same size
@@KM-om1hm It's been a couple years and I've only taken a glance at the problem but I'm gonna go with probably some combination of the definition of a parallelogram and my brain. I'll look more closely after work and try to get a more accurate answer.
@@OmniLiquid okay. Thank you
@@OmniLiquid The reason is the triangle area formula mentioned in the video. If we call the height of the parallelogram 'h', and represent the length of line segment AM as [AM], then the area of ADM is 0.5h[AM] and the area of BPM is 0.5h[MB]. Add these and you get 0.5h[AB]. Since the area of the parallelogram is defined as h[AB], the sum of the two triangles is half the total area. You can show in a similar way that the area of ADN is half the area of the parallelogram.
I got negative twelve.
I got 76 :c
I got 28...
Good ol' negative area.
how could yoy do that? great!
Negative 12 would work too
Hey. A very similar question was asked in Turkey's university entrance exam yesterday. I don't know if I could solve it if I didn't watch this video. I haven't got the results yet but you made me make one question more and closer to the university I want. Thanks a lot.
I hope you've made it!
Hey, I'm curious, what was the exam that you took? Was it SAT? YOS?
Hope you can make it to uni, btw.
@@abshariadamIt was YKS I think.
I think your problem perfectly encapsulates the beauty of math, in the sense that, although it's a really hard problem, it's solution is also quite simple, but not intuitive!
Beautiful problem!
i am a simple romanian: i see the world romania i get my heart warmed and like the video
I just simply assumed that the sum of both "diagonal sections" should be equal, that SW + NE = NW + SE, specifically NW20 + SEx = SW16 + NE32, so 20 + x = 48, so x = 28.....no need to divide every section in half or consider as triangles. Or was it mere coincidence that this worked in this case ?
It wasn't coincidence because your assumption was correct in this case, but in maths you should always prove your assumptions before using them in your calculations (as Presh did in this video).
it wouldn't be coincidence since with a point anywhere in the square coming from the midpoints of each side, you can split it up in the same way as he did with a, b, c and d without knowing the exact area. since (a + b) + (c + d) will always be equal to (b +c) + (d + a) this means that the sum of both opposite sections in any square split up that way will be equal. I hope this made sense
I Believe your approach works in every instance regardless of where that point is.
They are always the same, and it is not a coincidence. As you can see, both diagonal areas are (a+b+c+d).
You can also get to this conclusion by noting that any deviation from the center point creates a equal but opposite change to the areas of the diagonally opposite areas, so they must stay constant.
How did you come up with your assumption?
Imagine the point is coincident with one of the corners. Then it becomes more clear why your approach works.
"these videos build confidence" not when you have no idea how to solve something
I made it with a harder way through figuring out the side length of square = 4√6. After watching the video, I found it's actually a simple question.
How did you calculate the side ? ! !
He didn't
Fooling himself only
@@critisizerr245 I did the same thing, so no, he wasn't fooling himself.
中文解答：Please have a look at MY SOLUTION: ua-cam.com/video/h01cab2ngM0/v-deo.html
I solved it by just looking at the thumbnail and randomly:
b+c=20
b+a=16
Difference is 4.
Then a+d=32-4=28
Pretty cool considered i didnt use anything lol
same way...but by suming up opposites areas...
For everyone confused why he took so long and it only took you all second, hes doing the same thing. Just proving the method you probably used is true. The sum of opposite areas are equal, so he can justify his answer, which is usally the hardest part.
a+b= 16, b+c=20, c+d=32 => a+b+b+c+c+d=16+20+32=> a+d+2*(b+c)=68 => a+d=68-2*(b+c)=68-2*20=28. No need to perplex things by means of mentioning this about the opposite areas. Cute problem thought. Thanks for sharing it!!!
Who else checked the comments for 'clues' but realized they were so early that all the comments aren't very helpful yet... Well I did...
Gifted Guy all you can i find in this section are comments like "fresh dog walker"
Hello there. I always loved math. I loved it so much that I became an actuary. I saw this problem a different way. I figured, if you pick any point inside of a square, and draw a line from that point to the center of every side of the square, then the sum of opposite areas has to equal half of the area of the square. I used that principle and came up with the same answer of 28.
Please have a look at MY SOLUTION: ua-cam.com/video/h01cab2ngM0/v-deo.html
Let a be the half-length of a side. Total area is T = 4a². Also, T = SW + NW + NE + SE = 68 + SE. So SE = 4a² - 68.
Let x and y be the coordinates of the point. Use the area of a triangle: A = 1/2 × base × height. For SW: 16 = 1/2 × a × x + 1/2 × a × y = a/2 × (x + y). For NE: 32 = 1/2 × a × (2a - x) + 1/2 × a × (2a - y) = a/2 × (4a - x - y). Sum these two, the x and y cancel out: 48 = 2a².
Replace in the first equation: SE = 48 × 2 - 68 = 28.
This is how I solved it as well. It turns into 3 equations with 3 unknowns
Or just 16+32=20+x
Well thank God I solved it as fast as a fifth grader in China. I can be proud of that
The moment u started talking about triangles I was convinced I could do it and I actually did it! (with a long detour, but I got there none the less)
Meanwhile, Autocaptions: “ Hey, this is pressed Tell Walker” 😹😹😹😂🤣
ua-cam.com/video/Ik7beS1YbnM/v-deo.html
I couldn't think of a simpler way to do it, so I set it up as 4 equations and 4 unknowns, using nothing but areas of rectangles and right triangles, and was able to grind through the algebra to solve for the total area, from which the resulting area is 28. I'm embarrassed that it took me over 1/2 hour.
I went through 6 eqns and 6 unknowns ! FML brute forcing my way through everything XD
Awesome
My first idea was 16 equations and 16 unknowns...
Bă,Raio din România, mi-am spart creierul cu problema asta vreo 2 ore,apoi cand am vazut cum a fost rezolvată m-am luminat:făină de tot problema cu o rezolvare elegantă.Bravo!
The discussion about the smaller triangles being equal was interesting and potentially useful for something else but that had nothing to do with the actual solution. Knowing that the sum of the areas of the opposing polygons being equal is all you needed to solve the problem. I didn't know that so I learned something new today.
But the smaller triangles is how he showed you that was true. He divided it into smaller triangles, with pairs of the same area, by noting they had the same base and height. Then he took the original polygons and noted what their areas were in terms of those smaller triangles. The SW polygon has an area of a+b, NW is b+c, NE is c+d, and SE is d+a. Then you can note that the combined area of the SW and NE polygons is a+b+c+d, and the combined area of the NW and SE polygons is b+c+d+a. Then you can see that those sums are the same; they're the sums of the same four equivalent areas.
The point of the video is showing that that's true, not merely telling you that that's true.
I had a similar problem on my algebra test a few years ago, but I didn't know the thing about opposite areas being equal, so I had to go the long away around.
In this case, I did the same thing by dividing everything into triangles and labeling the equal areas a, b, c, and d. Then I did some algebra. You know a+b=16, b+c=20, and c+d=32, and what you're looking for is a+d. So based on that you know that d=b+12 and a=c-4.
Then to finish, you do (a+d)=(b+12)+(c-4) and then simplify it to =b+c+8. And since we already know that b+c=20, we know that (a+d)=20+8=28
I spent more than 1 hr to this problem.... played with x to the half of square side, used the equation of area of quadrilateral, and a multiple of other ideas. But i didn't got any answer 😅 you are amazing...
My explanation to why the the sum of a pair of opposing areas= Sum of the other pair:
Consider a square inscribed inside the original square, with its vertices at the midpoints of the original square.
Within this inscribed square are four triangles.
Drop altitudes(heights) from the interior point to the bases of these triangles (sides of the inscribed square)
Sum of one pair of opposing heights= Side length of inscribed square= Sum of the other pair of opposing heights.
Multiplying the heights by the inscribed square side length and you will get their areas.
Hence
Sum of one pair of opposing triangles in inscribed square= Sum of the other pair
Adding the areas of the triangles generated in the space between the original square and the inscribed square, and since the four of them are equal, you get
Sum of one pair of opposing areas= Sum of the other pair
Hope that helps.
I got what you said, but for people who don't try it this way, it's hard!
I had the same approach at the first time, and it shows the true nature of the problem... but I think that Presh's solution was more elegant...
Marcos Guilherme To be frank if I were in a time-limited exam, I would have gone first for Presh's solution. It was more methodological and alelgebraic. It isn't as intuitive and requires you to write down the algebra or form them in your mind to realize the relation. Whereas I think my method aids more the geometrical understanding of the relation. It took me a while to think of my method. In an exam I definitely would recommend Presh's.
I got Jamaica on the 2nd question, is it correct?
Hahaha emre bravo
Did you divide by two?
Huh, I got Jamaica/sqrt(Cuba) on that question.
Nah dude. U forgot to multiply by Canada and divide the remainder with Caribbean
Chaos Master26 u gotta subtract the USSR first though...
I am a retired Aerospace Structural Design Engineer with too much time on my hands. I solved the problem graphically by building a Catia model and measuring the 4 areas. Then I adjusted the size of the square and the position of the center point until I came up with the correct areas for 20, 32, and 16 square cm. At that point, the fourth area was 28 sq cm and the square was exactly 9.797959 cm on a side, which is the square root of 96. Brute force method exercises the brain too.
I can actually solve this initial problem way easier. I did it in under a minute. It is simple! due to the fact that all lines connect to the center of one of the lines of the square, it does not matter where you put the connection point.
in these circumstances diagonally opposing area's together are ALWAYS half of the square. So the equation to solve this is simply
16+32-20=28
Sasha Shadowhive Wow, I didn’t know that! Thanks for telling me this!
@@comershaw7610 no problem. I just don't understand why he uses such a roundabout way
Keep in mind this only works if the connection points on the lines are turn symmetrical. 2 easy examples are, like in this problem, the middle of each line or in each corner. But if the connection point is, for example, on 2cm from the left corner (if you look at it from the center of the square) and 5 cm from the right corner, this theory will still work if ALL 4 POINTS are 2 cm from their respective left corner and 5 from their right corner. If the points are no longer the same this will not work
@@sashashadowhive6128 Yeah, but try to prove that it actually is correct solution (I know it is, but I wouldn't say the proof is obvious, meanwhile, in his solution, the proof is quite obvious, it's pretty much the solution itself).
3:52 You can do that also with system of equations
Yep I ended up doing that. A bit longer to do AND it doesn't really use the method the sender wanted us to use but it gets the job done !
@@exlzyor2006 Yes, I'm more comfortable with algebra than geometry. I used x = base of all of those triangles (i.e. half the length of each side of the square), y = how far below the midpoint was from centre, and z = how for to the left of the midpoint from centre. So for the "a + b" shape, my formula was x(x-y)/2 + x(x-z)/2 = 16. And so on. Fortunately, there were ways of simplifying the algebra, so it didn't take all that long to get the right answer. But still not nearly as elegant as Presh's approach.
@@aquamarine99911 elegant or not, I solved it in head very fast using different division and sum of heights. Inscribed square is a half. Areas b+d=a+c=2S because of sum of heights. S is ⅛ of the whole square. 4S = 16+32 => S=12 and square is 96.
Unknown area will be 96-16-20-32=28
Bro mine is coming 25.5cm² . Please give a response
Let equals side be x and other according to their type like y,z,k,z,l
Area of first =20cm²=x*x*y*z
Area of second=16cm²=x*x*y*k
Area of third=32cm²=x*x*z*l
Area of fourth=?=x*x*k*l
Area of 4th= Area of 2nd * Area of 3rd / . Area of 4th
(x*x*y*k)*(x*x*z*l)/x*x*y*z=x*x*k*l(which is 4th )
16*32/20=25.6 cm²
@@shadowblock5814 Look, connect middle points of the square - you'll get 4 equal isosceles right triangles (note them s) and inscribed square with side equal L√2/2, and area being a half of initial square. Inscribed square is split into 4 triangles, let's note them clockwise a,b,c,d. And correspondingly we'll note initial shapes A,B,C,D. a and c have base equal to the side of inscribed square, and the sum of their heights is equal to the side, too. The same to b and d. Thus, sum of their areas are equal a+c=b+d. Equality will stand if you add the same to both sides.
2s+a+c=2s+b+d => (s+a)+(s+c)=(s+b)+(s+d) => A+C=B+D => D=A+C-B= 16+32-20=28.
Thanks for the fun problem. The last assumption is making it complicated. I come up with a solution without assuming that. Here is what I come up with:
a + b = 16 . . (i)
b + c = 20 . . (ii)
c + d = 32 . . (iii)
a + d = ?
(ii) - (i)
c - a = 4
c = a + 4
replace c in (iii)
a + 4 + d = 32
a + d = 28
Simple I had 28 in a minute!!
Where the point in the inner square is, is indifferent for the answer!!
The inner square (connect half way points) is half the size.
Now each of the two opposing triangles of the 4 triangles of the point to the edges of the inner square take half the size.
So half the size is 16+32 because half the inner square plus half the outer parts of the larger square equals half the larger square.
Now you know that
16+32=20+x and x = 28
Guessing 28 is not good enough.
TheReactor8 your explanation makes no sense sir. Your language is so confusing. Without explaining what you mean by inner square. Reread your first sentence starting with “now each” and think if anyone would be able to understand such a long sentence without label or a reference picture. Impossible.
The key to solve this problem is proving that the "sum of the opposite areas of triangles are equal", which is not obvious. At @4:05, he just mentions "notice" and moves forward. That begs some explanation as it confuses lots of people.
Proof : Denote the midpoints along the sides of the square as EFGH, starting from top moving clockwise. Denote the vertices as ABCD, starting from top-left, moving clockwise. Let the intersection in the middle of the square be O. Consider the square EFGH.
Within this square -
Area of triangle EOH plus triangle FOG = (1/2*HE*h1) + (1/2*GF*h2)= HE(h1+h2)=HE*HE.
This is because in the square EFGH, h1+h2 equals one of the side, HE. And also, as it is a square HE = GF.
Similarly,
Area of triangle EOF plus triangle GOH = (1/2*EF*l1) + (1/2*GH*12)= EF(11+l2)=EF*EF.
11+l2 equals one of the side, EF. And also, as it is a square EF= GH.
Q.E.D
How do we solve otherwise ?
I agree he just glosses over it and does not explain why clearly. It is a problem with these type of videos. It's mostly a math teacher showing off and does not help the students get better.
I looked at the thumbnail and had a guess at 28, looked like a valid guess so I checked the video, and sure enough.
I finally solved one of these! Although my way of solving was far less elegant.
the Chinese kid who solved the end problem in under one minute answered with "Red Triangle is Supreme over all. Long live Red Triangle".
4:06 the sum of opposite areas are equal. This is a beautiful and (to me) surprising general result that i find even more interesting that the solution to this particular problem.
Once I saw the equal triangle, I was able to solve it. I was trying to do it algebraically.
For the first question, let's call the centre point "P" which links to the centre point on the four square edges to divides the square into 4 areas.
The first question then can also be solved by dividing the 4 areas in the other way: Link the middle point of each square edges to form a 45 degree square (call it the centre square). This centre square divides the 4 areas into 8 triangles. The sum of the top-right + bottom-left area, could be calculated as the sum of their four triangles. These four triangles has the same base length: the edge length of the center square! The total height of the 4 triangles is exactly the diagonal of the original square. Thus the sum 4 triangles is 1/2 x Centre square edge length x Outer square diagonal length. The same argument can be applied to the top-left + bottom-right area, and results in the same number (since the area sum will not depend on the position of point "P".)
Now we know the two diagonal area sums are the same, and the problem can be solved as 32 + 16 - 20 = 28
你好。請看我的：Please have a look at MY SOLUTION: ua-cam.com/video/h01cab2ngM0/v-deo.html
Nice man. Neat trick.
Though, if anyone could explain 4:00 in more detail, like, how to get that identity, that would be very kind.
there is another way for this example wich is figuring out the middle worth of every section(in this case its 24) wich makes the square 96 cm² bc there are 4 sections so you just subtract 68 from 96 and you get 28
But if the three known areas have a total surface of 68, how do you get to an average of 24?
@@NilfNilf1972 i went from twenty to the number wich is in the middle of 16 and 32 wich is 24
24 is not the average of 16, 20 and 32...
@@timonburkard3481 i know but in this case the average from the biggest and smallest section is the overall average
@@jolioding_2253 Thanks, i understand now what you mean. Anyway i do not understand why in general (a+b) + (c+d) = (b+c) + (a+d) ?
Me: Wow, I have no idea. It probably involves math I never learned of something.
Him: So this problem can be solved using the equation A=.5bh
Me: Goddamnit
Neat problem and neat solution. How about the obvious follow on? What are the coordinates of the interior point? The areas of the triangles a,b,c,d are proportional to the apex heights as the base lines are uniform; mark the lengths as A,B,C,D with A+C=B+D= square’s edge. You should be able to deduce A,B,C,D proportional to 10,6,14,18 and the square edge 24. The total area was 96 sq.cm so the edge is the sq.root, about 9.8cm. The coordinates x,y (B,A) are 6/24, 10/24 of the edge length, or about 2.45, 4.08.
28 was my first guess after 15 seconds. Then I let myself get into the mood and my eyes deceived me into saying 20
Me too. I was like: 'hmmm.. let's go with 28' and it worked (this is btw my way of surviving math lessons)
Im from China and that was exactly how I failed my grade 5 math yet got 4.0 gpa in calculus at one of the best universities in Canada
5th grade, damm Asian
go back to china...
@@trinkCOKEorDIE bruh
I got it after he said how triangles can have the same area if they have the same height and base, and him making number sections. This is very interesting.
I predicted 28 because of the relationship of the shapes but I didn't know that it was the answer
20-16=4
32-4=28
But this thing doesn't apply to everything
I took the Relation between 16 and 32. The middle is 24 what describes the size of all squares If the Point were in the middle. And than Just 24-20=4 so the opposide has to be 24+4=28
... I think that Always works
Same
LOOL SAME! I THOUGHT IT WOULDNT WORK BUT I DID IT ANYWAYS
same I did it the same way
@@GabbaGandalf-fo7cg I feel like that is mathematically the same as the "sum of opposites are equal" method, just executed in a different way :P
This was literally in a math olympiad I participated. I did not finish it, but later I did🙂
A similar problem was in an olympiad I participated in as well. Are you Chilean?
I solved by finding the coordinates of the point in terms of the length of each side of the square. I did this by shifting the point where the lines meet vertically to the horizontal bisector and horizontally to the vertical bisector and then used the fact that the area of each triangle stays the same to express the point in terms of the length of the each side of the square. I then solved for the unknown area using the coordinates and the equation 68+unknown area=length of side of square ^2. Simultaneous equations and banged out the correct answer.
I`m Japanese!
I love this channel!
I don't think dividing the shapes to triangles helped in solving ... only the fact that the sum of the opposites is equal needed.
that fact is derived from observing the division into triangles
I agree that it is not needed, but it can be helpful for those who don't know/remember that the sum of opposite areas are equal.
Here, you have a+b+c+d=a+b+c+d : the equality is demonstrated.
You can't proove it otherwise.
@@sanpedro5018 > thanks, I was wondering where this "sum of the opposites are equal" was coming from, but this is a lot more clear :)
Ya you are right
I looked at it a minute and assumed the opposing areas should be equal area since the segments are drawn from the midpoints of the sides and all share a point. Then I did that last line in my head and came out with 28. I was so happy to see my assumption was correct. Not sure if I was lucky or logical, but I feel smart.
Or
d+a=??
d=32-c
a=16-b
So
32-c+16-b=??
Because b+c=20, then -b-c=-20
Therefore 48-20=28
I did something kind of similar. We know that b=16-a. c is then 4+a because they are the green triangles adding up to 20. c is also 32-d. So 4+a=32-d, or a=28-d. a+d=(28-d)+d=28.
32-c+16-b=?
32+16=b+c
48=b+c
B+C=20 from ..fig .🙄
@@kiranshetti7991 i think this is what i did, the area equality of a square divided from a central point/equally on the sides meant i just intuitively knew that each half was the same area as 16 + 32 = 48, so 20 + ? = answer, or 28
@@kiranshetti7991 you cant actually take b and c to right hand side because r.h.s is not equal to 0.
It is actually equal to ? Which is actually a+d
@@arshjeetsingh7113 Exactly he forgot ? = a+d. What he did was add b+c to a+d, which is 48.
ahh nice problem!!
Problematic(puzzle channel) ????
No. It's NOT a nice puzzle because it requires the knowledge of a geometric formula.
A puzzle that requires me to look up a math formula is not a puzzle at all - it's a MATH problem.
Disliked this video.
edmis90 I disagree with you. He doesn't simply produce a formula but explains showing different triangles.
edmis90 Your point: All question with solving geometry problem are not nice puzzle
Make a square and attach an elastic band across the square from the centre point of 2 opposite sides and attach to a pole in the centre. Wherever you move the pole away from the centre, one side gets larger and the other side gets smaller by the same amount. The area of the square remains the same. Now attach another elastic band across the square from the centre point of the other 2 sides and attach to the pole. This time wherever you move the pole from the centre, the 2 elastic bands which created 4 equal squares now make 4 different shapes and areas but still the opposite corners increase and decrease by the same amount. The area of the square remains the same. As we are dealing in square centimetres, we can use the sum 16 + 32 = 20 + x which of course = 28 No need to make any triangles.
I finally got one of these right!
Me too ! :)
Lol I took a completely different approach but got the right result. My approach wasn’t that clever, i divided all 4 areas into rectangular triangles and rectangles. I described those mathematically by using one variable for the (half) side length of the big square, one for the horizontal offset of the „point“ and one for the vertical offset. I ended up with 3 variables and equations (each for every known area) which turned out to be very easy to solve after some terms cancelled out.
This is what I did. Its not elegant, and it takes a lot more work, but it gives the correct answer, and requires no brilliance. More interestingly, it gives you the coordinates of that point in the middle. The crazy thing is just how simple it gets when things start cancelling, and you know there was a way that didn't involve manipulating all these terms around.
The point is SQRT(6) over and SQRT(6)/3 up. The entire square has a side length of 4*SQRT(6).
That's exactly what I did: I expressed 16=(stuff) and 20=(stuff) in terms of the offsets and half-sidelength, s, and was able to easily get the offsets in terms of s. Then I spent a few hours down a total rat hole failing to make any headway with 32=(similar stuff). It gave me a quartic for s which turned out not to have easy to find factors (indeed, it turns out no "easy" real roots), so I bailed out on that.
Eventually (a good night's sleep probably helped), I ignored the 32= and focused on the missing area, equating its s- and offsets-based formula with 4s² -(16+20+32) hence solving for s.
I enjoyed it, but was very grateful to come out of the other side with the right answer! I guess I'll never make a mathematician (too late to carve out another career anyway, now) :-)
Glad to see I was not alone! I used the same method too. The equations were indeed easy, and I managed to get the coordinates of the point and then 28 without using pen and paper.
+1, me too. This is the approach that simple exploration leads you inexorably down, without requiring spotting a special"trick".
You did not need the "sum of opposite areas are equal" equation. It could be solved for 28 just by knowing that a+b=16,b+c=20,c+d=32,a+d=x. You use basic substitution across the first three equations, and you eventually get that a=28-d. As such, if a+d=x, then x must equal 28.
I found this way.
If the sides are equally cut then 16+32=20+x
And x becomes 28 easy man easy
Weeb- Desu yeah I did the same say too
How it works
Weeb- Desu me too
That is literally what the video proves, but instead you just said it.
His vids are good but they are all longer than 10 minutes and i just dont watch a proof of a questions answer for 10 mins
Don’t know if this is luck but I looked at the three areas and said that 16 +32 is 48 and 48 minus 20 is 28 and that’s how I got it
Mag Arts Not luck, it’s logical maths
Oh ok thanks. That actually makes me feel more confident
Wow presh I knew your solution simpler but I used this solution:
I have drawn height from the middle point into each side of square, for little area I said it h1 and h2, and for other ones 2a-h1 and 2a-h2
Then from triangles formula one by one 4 different equations I found:
a/2 (h1 +h2) = 16
a/2 (h1+(2a-h2)) = 20
a/2 ( (2a-h2) + (2a-h1) ) = 32
From all above we find
a=24^1/2 root square of 24
Then h1 = 6/(6^1/2)
h2 = 10/(10^1/2)
:)
Unnecessarily found all those triangle areas
im 8th grade in turkey and i found the second one 9 under a minute: the half of the shape is [(72+8+a)+(b+x)] and also the half of the shape is (79+10+a+b) so x = (79+10)-(72+8) = 9. Is it correct?
Wow, I had been staring at it for several minutes and not getting anywhere, but you're absolutely right! For those who don't get it: the triangle with base AD and tip between B and C (containing 79, 10 and two white areas) is exactly half the area of the parallellogram. But the combination of two triangles with their bases on the two parts of AB and tips in D and on DC, together, are also half the size of the parallellogram. Those consist of the same two white areas, 72, 8 and the red area. So the red area has to be 9. Brilliant, well done!
The way I found out was that the answer must be times by 4.
Sameee
Me, too, but my first off-the-cuff guess was 24 :D
The funny thing is, that you don't even need to know the 3rd area to find the area of the rectangle
we can combine the square's sides' midpoints with lines. Those lines will divide each area to 2 triangles. Another triangle is the same in each area. (the right triangle that is located in the square's corners.). those lines will also form a smaller square inside our main square.
We can mark half of the main square's side (the side length of those right triangles) with x. We can also by x measure the length of our right triangle's hypothenuse. (which is also the side length of the smaller square). the hypothenuse of a right triangle is also the base of the another triangle of an area.
we can mark height of one area's another triangle with h. now we can have 1 equation for 2 variables (because we can by h and x tell the area). we can also see, that the triangle's (which's height is h) opposite area's triangle's height is the width of the smaller square - h. This is how we can also find that area with the same variables.
And so we get 2 equations for 2 variables.
Hi Presh, I actually had a second way to go from the point where you labeled the areas
since we know that A+B =16 and B+C =20, we can subtract them from each other to get A-C = -4 or A=C-4, then the same for B+C = 20 and C+D=32 this gives us B-D= -12, then -D = -B - 12 or D=B+12
So now we can add those together and get A+D = C+B+12-4, or A+D=C+B+8, we also know that C=20-B from the statement of B+C=20. So lets substitute that in for C, so now we have A+D=(20-B)+B+8, the B's cancel out and you're left with A+D=20+8 or 28. But I could only do this once you set up the triangles of a, b, c and d, which allowed for using algebra afterwards
So I solved it right in the most wrong way, I saw they were all multiples of 4 and similar in size. 16 was a little less than 20, so I just did a little less than 32 but a multiple of 4 and got 28 lol
I literally did the same
Wow So brilliant...its funny😂
Same
MadLad
Same here. Just visual guess.
An area of a triangle is not (base x height) / 2 but (base x height x sin(a)) / 2 where a is the angle between base and height. The reason the triangle pairs have equal areas is that they have the same base and height but also the same sine value of the angle between them. This is because the angles are x and 180-x and sin(x)=sin(180-x)
When you said 28 because 32 plus 16 is 48 minus 20 is 28 and its right
Lul
Pressure tell walker
Jonas Daverio gahhahhahhha
Jonas Daverio Lol I was going to write it
I usually never put auto-caption but now, with these videos, I always do
Jonas Daverio pressure rice cooker
But actually, auto-caption is really not to blame, I would never have could write his name properly if he hadn't written it on each videos
1/ Given any point inside a square and connect this point to the 4 corners; the sum of the areas of the 2 opposite triangles are equal.
2/Just connect the 4 midpoints we have a smaller square of which the sum of the opposite parts have the equal area.
moreover every 4 parts has added up 4 equal triangles
so ? + 20= 32+16 ----> ?= 28 sq cm
is 28 right? i think the opposite parts should always be same size. So 16+32= 48 and 20+x=48 --> 28
6man12 EXACTLY WHAT I DID
If you would answer that on your exam in my course, you wouldn't get any credit for the 28 unless you could explain WHY the sum of the opposite areas must be equal. THINKING something is true never counts.
I'm soo bad at Math but I solved like this in a minute;
16+32=20+X
X=28
Well... If the sums of the "opposite" areas are equal, then it is really this simple. No need for all that other stuff w/ the triangles. He uses that in his solution... but not so directly.
I'd never considered that equality before. It certainly was not pointed out in my Math classes.
I got George Washington
I think I got it wrong again. Damn
Pretty smart solution. What I'd do after splitting the areas up into the triangles with areas a, b, c and d is: total area of the square T=2*a+2*b+2*c+2*d and rewrite to T=2*(a+b)+2*(c+d). We know a+b=16 and c+d=32 so you get T=2*16+2*32=96. We also know T=16+32+20+(a+d) so (a+d)= 96-68=28cm2. But stating that a+b+c+d=a+b+c+d (which is obviously true) and rewriting that as (a+b)+(c+d)=(b+c)+(a+d) to get to (a+d)=28 is pretty neat too.
I went about this in a different way: If you connect the midpoints of the sides of the square you get another square inside the square we started with, the inscribed square. Let's call the length of the sides of the big square 2a so that all the lines marked ‖ (equal-sign) have a length a. The sides of the inner square then become √2a^2. You'll see that each of the 4 areas we created has the same triangle in them, with two legs of length a and one of √2a^2. Since all 4 areas have this same triangle in them this triangle can be discarded when looking at how the areas relate to each other. This I'll use in a bit.
You'll also see we now have 4 triangles in the inner square, each with one of the sides of the inner square as their base and the other legs are the lines we drew to make the 4 areas. Let's call the top-left triangle B, the top-right one C, the bottom-left one D, the bottom-right one E and the intersection point S. Now drop a line from the intersection point S perpendicular to the base of each of the triangles B, C, D and E. These lines are the heights of those 4 triangles in the inner square and you'll see that the height of triangle B (b) plus the height of the triangle E (e) equals the length of their base. The same goes for the combined heights of triangles C (c) and D(d). So b+e=c+d=√2a^2.
As we know the area of a triangle is 1/2*base*height. These 4 triangles all have the same base so the relation between their areas depends on the relation of their heights. I've also determined that the relation between the 4 areas we created at the beginning depends on the relation between the area of these 4 triangles. Since their area depends upon their heights (as they all have the same base) I can state that the relation between the heights of these triangles and the areas we created at the beginning is b+e=c+d. So 20+e=32+16 gives e=28 which is the area we were looking for. Total area of the square then is 16+20+32+28=96 and its sides are √96=4√6 long.
i'm a engineering student and i couldn't figure out whats the solution lol
LeifGrethe xD
At least you're not an architect or a civil engineer
Same. And I'm a 1st year chemical engineer. Guess i should have followed what my mom asked me to study; Shakespearean
i forgot that 1/2 baseXheight is the surface of a triangle and i couldnt solve it.
watches bridge collapse...yeah I can do that
I don't get why the sum of opposite areas are equal. Somebody pls explain?
OhSaft The explanation is in the video, get a paper and a pen and try to replicate it
Vitor Nathan Gonçalves how do you proove it?
Once all of the a, b, c, and d areas are labelled, you can see that the sum of the areas of opposite regions is always a+b+c+d.
You could try cutting them up into triangles, and since the area of a triangle is 1/2*b*h and they share a common base, opposing triangles, when added up, would have the area 1/2*1/2(Length of cube)*length of cube. Do this for the adjacent triangle in the same quadrilateral and you’d find out that the area for them is half the area of the entire square
way too complicated but this was also my first thought to be honest :D
5th grader problem
First let us name some stuff.
Going clockwise, let's call the intersections between AB, BC and CD: E, F, and G, respectively. Also going clockwise, let's name the polygons of unknown area a, b, c, d, e and ?, with the one that's in the middle called f.
What do we know?
From the equation for triangles, we know that ADE + BEG = DEG + BCG = 1/2 * ABCD.
We also know that ADF = ABF + CDF = 1/2 * ABCD
From that we know that ADF = ADE + BEG. This equation will be easy to work with because it only refers to e and f as unknowns, which will simplify out.
Working out the math:
? + e + 72 + f + 8 = e + 79 + f + 10
? = 79 + 10 - 72 - 8
? = 9
This question is a bit difficult for 5th graders.
I literally scratched to get rid that dot ( 0:14 ) inside the square. 😂😂
Once you know the sum of the opposite areas is the same, you don't really need the triangles anymore.
MatrixQ yes, but you need the triangles to prove that
David Cox - Not really. I figured out that the sum of opposite areas were the same, but my approach was completely different, and didn't use those particular triangles. (Mind you, Presh's approach was a lot simpler.)
Let's say the large square has sides 2s. Start by putting the interior point (let's call it C) in the center, so that each quadrant is a square of area s^2. Now, we'll move the interior point horizontally by x and vertically by y, to a new position C'.
It's hard to demonstrate this without pictures, but moving from C to C' makes the two leftmost quadrants both gain an area of sx/2, the two upper quadrants both gain an area of sy/2, and the opposite quadrants lose the same areas. This works because it's possible to construct each quadrant by starting with the original square with side s, and then adding or removing one triangle with base s and height |x|, and another triangle with base s and height |y|.
For example, let's take the case in the thumbnail, and look at the 20 cm^2 quadrilateral in the upper left. Start with a square quadrant with side s. Now add a triangle formed by the bottom of the square and point C', forming a pentagon with area s^2 +sy/2. Now remove the triangle formed by the right side of the square and point C', and we're left with the quadrilateral in the thumbnail with area s^2 + sx/2 + sy/2. By a similar process, the opposite quadrilateral has area s^2 - sx/2 - sy/2. In this particular case, s = 2sqrt(6), x = -sqrt(6), and y = sqrt(6)/3.
I think you can even show that this is true without resorting to formulas (though those are the actual proof, of course). If you put the interior point in the middle and move it left to right, the two quadrants on the side that get bigger get bigger by the same amount that the ones on the other side get smaller. Same goes for up and down. If you mix this, the opposing quadrants correlate to each other, because they gain/lose from the left/right motion and the up/down motion. If you choose one quadrant, the opposite quadrant correlates on both the up/down and the left/right movement.
All the cut lines for the rectangles start at the mid point all that moves is the centre cross point. So they have to be equal. Move centre point left rectangles on left shrink proportionally to the right rectangles growth.
MatrixQ - While that works on an intuitive level, it doesn't really prove anything. Yes, it's obvious that moving the interior point left makes the left two quadrilaterals shrink and the right two grow, but it's not obvious that the opposite quadrilaterals shrink & grow by the same amount. Without the various proofs given by Presh, me, and others, it would have been plausible to claim something like, "when the interior point is at such-and-such location and you move it 1 cm left, the top two quadrilaterals shrink/grow by 1cm^2 and the bottom two shrink/grow by 2 cm^2". That claim would have obeyed your intuitive idea that the left/right sides shrink/grow by the same amount, while violating the stronger claim that the opposite quadrilaterals shrink/grow by the same amount.
Easier way:
Draw a line from each middle of each side to the middle of it's adjacent sides to create a (square) diamond inside the square.
Draw two perpendicular lines paralell to the sides of the diamond that pass through the point and end at the sides of the diamond.
Since the combined height of the oposite triangles inside the diamond is the same for the sw->ne and nw->se region, the sum of the area of the sw&ne regions is equal to the sum of the ne&sw region, hence 32+16=20+x
x=28
OMG THIS IS HOW I DID IT :D
I'm going to guess 25.6cm^2
Now I'll watch.
Edit - realized my method was flawed, paused at 1:14. Redone, area is 20cm^2
[Let a = top left, b = bottom left, c = top right, d = bottom right, L = length to midpoint, W = width to midpoint
a + b + c + d = A
2L * 2W = A
a = 20
b = 16
c = 32
68 + d = A
L = W
32 = LW + X + Y
20 = LW + X - Y
16 = LW - X - Y
12 = 2Y, Y = 6
4 = 2X, X = 2
32 = LW + 6 + 2, LW = 24
3LW + X - Y + d = 4LW
3(24) + 2 - 6 + d = 4(24)
72 - 4 + d = 96
68 + d = 96
d = 28]
Edit again, I realized in my last equation I mixed up X and Y when I plugged everything in, corrected. Got 28, though not before seeing the whole video. Hooray!
👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏👏
The sum of opposite areas are equal, I doubt recall this, may I ask the name or where you got that part because I dont seem to be able to solve it without that and I missed where it came from
Because they're triangles leaning on each other forming a circle so to say hence their sum is always 360 no matter how many times you slice them.
Also they share same base length and same height if you mean a and a, b and b etc
I subtracted b+c from a+b to get a-c=(-4), then added c+d to get a+d=28. The b’s & c’s cancel out this way which confirms Presh’s math that (in a different order) 16-20+32=28
Did you figure it out?
加藤みずっち took about 5 secs to know how to solve it...
加藤みずっち: Yes, I did. It's very easy.
加藤みずっち 1 second to figure out how to do it, and to solve it near 1 minute
If it took you 1 sec to figure it out it should not take you 1 min to solve it. 5 sec to figure it out and 15 sec to solve it makes more sense.
Øystein Well I was slow minded to calculate 😂
"The sum of opposite areas are equal."
Then there is no need to show that the area of a triangle is 1/2 of the base times height.
16 +32 = 20 + ?
? = 28
Unless you are trying to show that this is how it is derived.
Not sure if someone already mentioned it, but provided that sum of opposite areas are euqal, there is no need to split areas into two. All you have to do is state that A+B = C+D, therefore 16+32 = 20+D, which equals 28.
so the second ? is 9 although it looks bigger imo. there are triangles that are half area of the paralelligran sharing areas. so i got 72 + 8 + ? + sharedarea = 79 + 10 + sharedarea
Paul Googol I get the same answer: 9 ^^
Yes, it's 9, and the big white triangle below is 41.
Can’t you just say that opposite quadrants always add to half the square? So quadrant 1 + 3 = 2 + 4
16 + 32 = 20 + n
n = 28
True, but prove it.
I just looked at it and had that thought about the sum of the diagonal areas being equal, but just couldn't recall how on earth I'd prove it, years of not trying Maths problems.
Ya thats what i thought at first
As long as the bases are separated evenly, it's true, and it's proved in the video tho
Sure. That's because the area of one of the quadrants is = ½ (distance from side midpoint to neighbour side midpoint) (length of projection of the quadrant onto the diagonal that starts in the quadrant's corners). So, the sum of the area of two opposite quadrants is ½ (distance from side midpoint to neighbour side midpoint) (length of the diagonal). Since both diagonals have the same length, quadrant 1 + quadrant 3 = quadrant 2 + quadrant 4.
I solved this problem in less than 1 minute, although it was only an approximation because I did not know that sums of opposite areas in a parallelogram are equal. My solution was simple: I added all the 3 given areas 20+32+16=68. Now, since the entire figure is a square with 4 units of length on each side, I reckoned that each unit is 2.5 cm long, so the total area should be 10x10=100 sq cm. Subtract from this the given areas of 68 sq cm and we get 32 sq cm, approximately. Since it is obvious from the figure that the sum of the upper 2 areas =20+32=52 sq cm, they should occupy more than half (50) the area of the square, the other half should be less than 50 sq cm, that is, 16+32=48 sq cm. If the total area is actually 96 and not 100, the sides of the square would have unreal values (4sqrt 6) which is improbable in real life, although sqrt 6 is approx. equal to 2.45 cm, just a little less than my assumption of 2.5 cm.
I got to 28, but i didn't use that method nor did i use a couple methods found further down in the comments.
I just saw the left 2 sections where the bottom one was 4 less than the top one. I just did the same thing for the right side 32 - 4 = 28.
that's pure luck
This is the n. 1 best solution. It even uses the principles used in the poster’s solution.
Same here:)))