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).

I contemplated the thumbnail of this video for 20 minutes before finally deciding to watch it.

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

28? Aw man, so close! My answer was "Argentina"

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.

I got negative twelve.

"I don't know" is also a correct answer

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.

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.

When you can solve complex analysis problems but didn't even know where to start solving this

“The sum of opposite areas are equal” , you lost the majority of us there mate.

I really appreciate that you talk through the answer slowly so I can watch a bit and then have a go :)

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 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

I think, i‘m a genius. Just continue watching and don’t even try solving, because he‘ll tell you the solution.

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!!!

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!

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.

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 :')

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

My day job is modeling/drafting with Revit, but I enjoy learning what I should already know. Thanks!

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 ?

A Chinese fifth grader solved the more difficult problem in less than a minute... *cries myself to sleep*

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.

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)

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.

i am a simple romanian: i see the world romania i get my heart warmed and like the video

I got Jamaica on the 2nd question, is it correct?

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.

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

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.

your videos stimulate my mind. thank you so much

the Chinese kid who solved the end problem in under one minute answered with "Red Triangle is Supreme over all. Long live Red Triangle".

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.

you can inscribe a different square connecting the the middles of the sides of former square and then u see 32-y+16-y=20-y+x-y where y is the rectangular 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.

I finally solved one of these! Although my way of solving was far less elegant.

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.

Sir,I have seen your all videos .Your method of explanation is very good.(From India)

sooo glad I came across this channel.

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

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

Wowww man you are amazing I have nothing to do with math or trigonometry. But your puzzles are interesting. Thank you very much for your content

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.

3:52 You can do that also with system of equations

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

One of the best puzzles ever!

Assigning the point coordinates (a,b) and writing expressions for the original areas worked fine too. //=h, ha/2+hb/2=16, h(2h-a)/2+h(2h-b)/2=32 , insert first in second and solve for h. Then you get the total area of the square 4h^2=96.

hi presh, I solved this problem in another way, using the same principle but without any algebra, it is a much more visual solution. To explain, Let me label the upper right region as A and the upper left as B and so forth moving counter clockwise. (so D is our unknown area) Start by drawing in the lines between the centers of adjacent sides of the square. We have now decomposed each region into two triangles, like in your method. But here one is in the corner of the square (and has fixed area) and the other has a base of that diagonal and a height dependant on the position of the point. let's call these the variable triangles. Now draw a line through the point and parallel to the base of the variable triangle from region D (call this line m). Notice that this line is also parallel the base of the variable triangle from region B. What we notice as we drag the point around on line m is that the areas of regions B and D stay the same, and consequently area is just shifted back and forth between regions A and C. If we drag the point to the middle of line m, the figure is symmetric and by that symmetry, regions A and C have the same area. Originally they had 16+32=48 area between them, and they still must, so each has an area of 24. Now draw the main diagonal that is also the symmetry line, and passes through our point, and is parallel to the bases of the variable triangles of regions A and C. (Call this line d) If we drag our point around on line d we notice that regions A and C don't change area and the area shifts between regions B and D. And! we notice that we can put the point in the middle of the entire square. When we do this all areas are the same, so all the areas in this position must be 24. Shifting the point back so that region B only has an area of 20 means that 4 units of area mush have shifted from it into region D. 24 +4 = 28 done.

Sovled within ten seconds, with the correct assumption that opposite segments equals half the total area. It's easy to see this is true when we have the intersection point in the exact centre (all segments equal sized squares), but if we visualise the intersection point moving from the centre to the corner, we see one segment get larger, while its opposite corner becomes smaller. Meanwhile, the other two segments change shape equally, but they do not change area relative to one another. If they never changed area relative to one another, they they must have remained the same area as they were when the intersection point was in the exact centre. Thus, opposite segments are equal to half the total area, regardless of the location of the intersection point.

I am a biggest fan of you. I always watch ur videos and i improve my math skills. I want to meet u once in my life

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.

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...

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.

One should make the difference between to guess and to demonstrate. To guess right is good, to prove is better and is real maths. I couldn't solve without pen and paper. Finally found a method different with the same result.

I just Love your channel 😍

ahh nice problem!!

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.

Wonderful explaination sir.. I am ur fan now

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.

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

Well thank God I solved it as fast as a fifth grader in China. I can be proud of that

You can show the diagonally opposite areas are equal in a slightly different way: Start by connecting the midpoints of the square's sides. Opposite triangles in that new diamond must have total area equal to half the diamond bc their heights sum to a side length (so 0.5bh +0.5bh' = 0.5b(h + h') = 0.5b^2)

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

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`m Japanese! I love this channel!

If you draw lines from the interior point to the corners of the square, you end up with 8 triangles. Pairs of triangles which share the same side of the square have equal area as they have the same base and height. So you can set up the following equations (i'll use compass directions to denote the triangles of equal area). W+N = 20. N+E = 32. S+E = x. S+W = 16. (W+N)+(S+E) = 20+x. (N+E)+(S+W) = 32+16. 20+x = 48. Therefore x = 28.

Thanks for the video!!

I don't think dividing the shapes to triangles helped in solving ... only the fact that the sum of the opposites is equal needed.

Here is another simple solution: move the inner point at an angle of 45degrees until you hit the vertical line that divides the square in half. The areas 20 and x will remain constant. Therefore also the sum of the others will be constant (48). Now we have a symmetrical situation on the left and on the right half of the square, so 20 + x = 48.

I used the "area of the triangle with same bottom and same height" too, but in a different way, first, move the connecting point to the diagonal line (the top left - bottom right one), the area of the blue part and 20-part stay the same, so sum of the other parts still be 32+16=48, and now they are equal so each is 24, Next move the connecting point to the other diagonal line, so two 24-part-s area stay the same, and now each of them is exactly 1/4 of the whole square, so the area of the square is 24x4=96 Problem solved I find it easier in my brain than typing it out, may be it need a visual demonstration to understand it easily

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

Alternatively: Draw the inner diamond of the square, where the 4 cornerpoints of the diamonds are the middles of the edges of the square. All edges of this diamond are equal by the pythagorean theorem and the fact that all half-lines of the square are the same size. (1) The angles of the inner diamond are all 90 degrees, proven by the 1/1/sqrt(2) triangles having 45 degree angles. Focussing on the diamond. draw a line with an angle of 90 degrees from all edges of the diamond through the dot. The lines from opposite sides coincide, leaving 2 new lines. Now the base of the triangle is the edge of the diamond, which is the same for all sides. See (1). The height of blue + the height of green is also equal to this. The area of the blue triangle + area green triangle = 0.5 *e dgeLength * height(blue) + 0.5 * edgeLength * height(green) =0.5* edgeLength * (height(blue)+height(green)) = 0.5 * edgeLength * edgeLength Repeat this for the red and yellow area's. The area of all 1/1/sqrt(2) triangles are all the same, therefore, green area+blue area = red area + yellow area I hope it's somewhat understandable

Once I saw the equal triangle, I was able to solve it. I was trying to do it algebraically.

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...

I constructed an in-square with the outer square's mid points. If the side of outer square = 2x Side of in-square = Sqrt(2) of x Area of the portion other than in-square = 4 equal areas = (x^2)/2 = 12 sq cm(each) (This is obtained by considering the portions with areas 32 sq cm & 16 sq cm) Now , by considering the portion 20 sq cm , wanted area & In-square, The area of wanted portion inside in-square = 16 sq cm Hence, Area of wanted portion = 12 sq cm + 16 sq cm = 28 sq cm

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

"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.

I got it right without all this just in my head. My reasoning behind this is that the 16 and 20 section has a difference of 4, so I subtracted 32 by 4 and got 28. I don't know if this can apply to every other situation, but it applied here.

You don't even have to go with associative property shenanigans, you can separately prove that 2a + 2c = 2b + 2d. All you need to do is mentally draw in the height lines for the 2n triangles. The combined height lines of opposing triangles are the same length as the sides of the square, and the base is also the side of the square.

I found this way. If the sides are equally cut then 16+32=20+x And x becomes 28 easy man easy

I finally got one of these right!

Merci beaucoup à vos extraordinaires raisonnements. Logiques et pédagogiques. From Morocco.

This looks like Vivienne's theorem. Very interesting, thank you so much !

This was literally in a math olympiad I participated. I did not finish it, but later I did🙂

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.

Very great exercise! I didn't make it unfortunately. Thank you for your explanations.

"these videos build confidence" not when you have no idea how to solve something

Please make a video on trisecting an angle!!

I don't get why the sum of opposite areas are equal. Somebody pls explain?

Bravo Reio!

You're a lifesaver, I already would sell my soul for the solve :). How about Fermat's theorem? :) Thanks to both of you!

Meanwhile, Autocaptions: “ Hey, this is pressed Tell Walker” 😹😹😹😂🤣

28 was my first guess after 15 seconds. Then I let myself get into the mood and my eyes deceived me into saying 20

Thank you for these brain treats.

I got the correct total area (96 cm^2) by applying principles but directly knowing the geometric rules. This was a good refresher. Usually these videos introduce me to completely new concepts and techniques.

The only person you should try to be better than, is the person you were yesterday -Shazistic

Não vi o vídeo ainda. Espero que minha solução esteja correta. Tome o quadrado ABCD e seja *O* o ponto interno tal que *[AQOM]=20*, *[MONB]=32*, *[OPDQ]=16* e *[OPCN]=x* é a área a ser descoberta, onde *M*, *N*, *P*, *Q* são os pontos médios de *AB*, *BC*, *CD* e *DA*, respectivamente. Seja *2a* o lado do quadrado. Veja que os triângulos AQM, MBN, NCP e PDQ são congruentes (ALA) e retângulos. Daí, suas áreas são equivalentes e iguais a *a²/2*. Isso nos leva ao fato de que *MQO=20-a²/2* *MNO=32-a²/2* *NPO=x-a²/2* *PQO=16-a²/2.* Veja que MQPN é quadrado de lado *a√2* (prova deixada a cargo do leitor). Isso leva ao fato de que *[MQO]+[PNO]=[QPO]+[MNO]=a²* (também fica a cargo do leitor). Daí *i) 20-a²/2+x-a²/2=a²* *→x=2a²-20* *ii)16-a²/2+32-a²/2=a²* *→2a²=48→a²=24* De i) e ii): *x=2•24-20=28*. Logo, *[PCNO]=28*

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

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