I saw the triangel, but didn't really think much of it, but then this comment was highlighted under the video, so I guess I had a clue going forward to solve the problem... So I was robbed of the feeling of solving it all by myself... (darn it!) 😄
We Iranians have a proverb that says: "Easy is a puzzle after it is solved!" Me after seeing the solution to the question that I failed to answer: "That was easy." :))))
I noticed that triangle had to be the key to solving this, but didn't realize you could use it like that to make the problem trivial, cool solution indeed
i noticed that too so i started making contructions and wrote evrything in terms of the side of the square and the angle between the side of the square and the base after using a lot of trignometry i ended up with 2 equations which gave me the answer that the side of the square is 5
I solved it using rotations. I considered the coordinates of certain points before and after the same rotation on diffetent coordinate grids: (2x;3x)->(?;17) (5x;2x)->(?;26) Where x is the side length of the squares. This results in essentially the same system of equations of the video: 3cos(a)x+2sin(a)x=17 2cos(a)x+5sin(a)x=26 Where a is the angle that the squares have to be rotated in the clockwise direction, so that they line up with the rectangle. This yields: sin(a)x=4 cos(a)x=3 And using the pythagorean trigonometric identity, I got: x^2=25 In short, instead of counting triangles, I counted squares with my head tilted 60°.
Being a Vienamese, I love how you solve this beautiful geometry puzzle in the easy way for everyone to understand( also I have encountered this puzzle when I was 13 , it"s pretty tough for a 13 years old boy :)) )
I thought I was going to find an even easier way of the solution. I got this equation: sqrt(2x^2)/2 + 5x = 17, I solved it, and found the area was 8.872, which is wrong. After sometime, I realized that I was missing + in my Left hand side of my equation. And finding that will be the same method as shown in the video. And I am 13. I got pretty close to solving it through my own method, didn't I?
I solved it!! I really did! and what "pushed" me to find the solution no matter what is that sentence that i found quite demotivational "a version of this problem was given to students who were 12 to 13 year old in Vietnam" thank you Presh for this "achievement dose" i'm having right now ^^
Likewise, I eventually got there; Now I'll be strutting about feeling all 'big brain' for the rest of the day (and will conveniently forget the aforementioned sentence)!
@@Smiththehammer you would also need to know that the corners of the squares are touching the big square at 4 locations, 2 on the length and 2 on the width, and that a small triangle is formed at one of the corners... without this information you can’t solve it with words... jus saying
@@Smiththehammer Uhm, actually, that isn't exactly the case. You can't know how many triangles would be placed inside the square without the scale. You might think they are, let's say, 5 triangles, but the problem wasn't to scale and it actually was 5.25 triangles.
I was told never to use a ruler to measure the diagram, use only the numbers that are given to you. If you followed that rule, you would not be able to use the triangles method that was used to solve this puzzle
I used projections, which led to basically the same solution. Nice problem. Before I came up with the projections idea, I was thinking of putting additional copies of the same puzzle next to each other and to try to find how many rectangles I can construct to fill it in. This was a kind of a dead-end, but it made me think about the projections, so it helped me solve the problem.
In Vietnam, 13 years old, 7th class, children know how to solve pair of linear equations in two variables that is taught in 10th class in India!! (CBSE)
Pfft, believe me, don't. The vietnam education system is so outdated and badly designed that the only thing keep it alive are outstanding teachers. Source: vietnamese myself.
I have admit these simple problems are so much damn fun to watch, it makes me so happy to watch this, everytime I laugh uncontrollably. very clever indeed. Keep it up #MindYourDecisions
Is there any pure algebraic solution for this problem? Because I think , these type of clever trick solutions will not come in mind during the exam competition.
Note that the sum of the semi-diagonals of the small squares spanning the horizontal and vertical dimensions are in a 7/5 ratio. This is incommensurate with the dimensions of outer rectangle which are in a 26/17 ratio. Thus the diagram is not possible.
I had a 30 minutes in pullman to spend doing something. After thinking very deeply for half an hour i got it. I'm so proud of myself. Now I just wish I was 12
To be honest whenever I attempted this.... I felt an urge to extend the squares' sides such that they meet the sides of the rectangle.... And somehow be able to see something.... I just couldn't see this.....
Iam Vietnamese too and im 14 ,when I see this video I was very surprised because I have done this math before so I very happy when we have a Math problem from Vietnam on your Video
I fixed it pretty closely to this method. However, I worked out expressions for the change in height and horizontal distance across the diagonals of the squares. If you are going down or going across it's a+b in both cases.
I started with the same small triangle . Instead of labeling them as a,b,c . I labeled the hypotenuse as 5a I.e the side of the square..and the other sides as 4a and 3a. Solving for the vertical edge 17a =17, a=1. The square is 5ax5a= 25
Wow! ... Very nice problem! I saw the triangle first, but I was very curious to find out how to solve it and did not think any further! :-D Big LIKE! 👍 Best regards from Romania! 🤩 _Gabriel_
That's amazing i'm new here before watching this video I don't like maths and now it got interest on maths. I have share this video with my maths teacher.
If one side of the square is a-bi, the other side rotated by 90 degrees is b+ai, so the real axis direction is 5a+2b=26, the imaginary axis direction is 2a+3b=17 ∴a=4 b=3 ∴Area= a^2+b^2=16+9=25
I did in a much more convoluted way but i worked. First I found the distance between the left touch point and the top touch point using the grid to be s * sqrt(13) where s is the side of the little square. The same way for the distance between the left and right touch points to be s * sqrt (29). Then using the triangles formed by these two lines and the sides of the rectangle I came up with a single equation involving s^2 = A, namely A = (9 - sqrt(13 * A - 289))^2 + (17 - sqrt(13 * A - 289) - sqrt(29 * A - 676))^2. The solution to this implicit equation is 25, which I found by checking the values with a calculator while getting a feel for what A should be before attempting to isolate A, which I ended up not doing since it was 25 exactly.
I have found another way to solve this. First create all possible boxes inside the rectangle by extending the lines. The rest left are traingles and some quad. If you look carefully each left traingle completes another triangle on opposite side making a full sq. Total sq inside is 18. Now divided area by total no of sq inside. You will get area of each small squares.
Let a square's edge length be x and t be the angle of rotation of the squares (with an angle of 0 meaning the squares' edges are parallel to the rectangle's). The vector that points from a square's bottom vertex to its rightward vertex is (I'll refer to this direction as forward), and the vector pointing to its leftward vertex is (I'll refer to this direction as up). To get from the vertex along the bottom edge of the rectangle to the vertex along the top edge, you need to go up 3 and forward 2 square edges. That combined vector is 3 + 2 = . The y-component of this vector must have a magnitude of 17. To get from the left edge to the right you need to go forward 2 and down 5, or 2 - 5 = . The x-component of this vector must have a magnitude of 26. From this, the following equations are made: 3xcos(t) + 2xsin(t) = 17 2xcos(t) + 5xsin(t) = 26 Solve the first equation for xcos(t): xcos(t) = (17 - 2xsin(t))/3 Plug into the second equation, solve for t: 5xsin(t) + 2(17 - 2xsin(t))/3 = 26 --> t = arcsin(4/x) Plug back into the first equation: xcos(arcsin(4/x)) = (17 - 2xsin(arcsin(4/x)))/3 --> xsqrt(1 - (4/x)^2) = 3 --> sqrt(x^2 - 16) = 3 --> x^2 = 25 = a square's area
The right triangle you saw was exactly what i saw just after that diagram was fully drawn 😀😀😃😃 I also knew from the very beginning that the triangle would be a special triangle or most probably a 3...4...5 triangle This is just amazing that there is a hidden unknown right triangle and that too that its a special 3...4...5 triangle..WOW !!! Math is just amazing 😍😍
Easy question just by rotating anti clockwise the interior pattern such that it becomes mirror image of L and then you can simply imagine that 9 blocks covers the half rectangle so 9x area of one block = (17x26)/2. So area of one block 24.5 approx 25.
I used vectors . The side of the square pointing down being vector (a b). Pointing up is then (-b a). 3(a b)+2(-b a)+2(a b)=(26 x). Similar for (y 17). You get a=4 an b=-3. So side=5. Qed.
Ok, that is/was smart. Small suggestion: When you "expand" the equations (2:47 - ) and you double the values (to ultimately get two with 6b) explain that before just doing it. I was kind of thrown by what you were doing. Explain that we have 3b and 2b and we are going to make them both 6b and to do that we will do this....... That way people may not be as thrown by wondering what you were doing.
If we look vertically right side. There are two squares joint by there diagonals and a square by half of the diagonal. So, if we assume its diagonal as a(2)^1/2 then a(2)^1/2 + a(2)^1/2 + a/(2)^1/2 = 17. But after calculating a, area is coming 23.12 . Why?
My solution came up like what if the squares were rotated so that the left 3 squares placed at the bottom of the rectangle. Then we have, 5s < 26 and 3s < 17, therefore the most likely answer would be 5 if it's an integer. Of course there are many assumptions for this approach
I did essentially the same thing, but without using the little triangle. Instead of a and b, I had xcosθ and xsinθ, and equivalently solved that these were 3 and 4.
i didnt saw that but when i got a question of why are there many squares got me thinking and i realized after the video that those are the way to get the area of square
I thought at using the diagonals of the squares. That got me the fact that 3.5 diagonals are 26 and 2.5 diagonals are 17. Not really nice numbers... And they dont really match at all. As ofc the squares aren't placed at exactly 45° for me to count like that :< oof.
my method is putting the square inside a bigger and non-tilted square then find the triangles (pretty much the same but it's the outside triangle instead) it's really cool that the numbers ended up not in decimals lol
I'd propose another solution which instantly came to my mind, although not as precise Take the square touching the top line and bottom line, give their hypotenuse a value of 1. Now the middle square that connects the top and bottom square also has a hypotenuse of value 1 but the distance between top square and bottom square is only half hypotenuse i.e. vertical side of rectangle = 2 hypotenuse+ 1/2 hypotenuse of middle square Vertical side of rectangle = 2.5 hypotenuse 17 = 2.5 hypotenuse Hypotenuse = 6.8 But (hypotenuse)^2 = 2* (square side)^2 Then (side of square) = (23)^(1/2) Area of square = (23)^(1/2) * (23)^(1/2) Area of square = 23
I don't quite understand what you mean by "is touching the top line and the bottom line", but if your answer is 23 then you obviously made a mistake somewhere along the way. Nice try though...
ouch is right, glad I learned something new, but it kills me when I hear that it was given to 13-year-olds, gives new meaning to thinking outside the box. I never saw the triangle piece, I thought it was going to have to do with noticing that 6 of the squares would fold to make a box and the 7th one overlapped, but from there I had no idea how I was going to figure it out.
I did find a geometric solution by noting a right angled isosceles triangle insid the figure. But I didnt feel fully satisfied. I knew it wasnt the easiest and intended solution. So I feel like solving it is not the absolute goal, the goal is to grasp the problem and find the best appropriate solution (but I do feel better about myself if I solve it before watching the nice solution)
I can't believe I got so close with my estimate. Even though the squares are not rotated at exactly 45 degrees, I just treated them like they were, and counted the number of diagonals that made up the length and width, to solve for the diagonal and use pythagoras to get the side. Ended up estimating the diagonal at 7, and the area of the square as 24.5.
The solution by MYD is elegant, I have to admit. However, such mental agility is not possessed by all of us. Instead, I solved it using brute force, i.e. some vector algebra / trig and end up with the exact same equations with just a little bit more work. There are only 4 points where the seven boxes touch the outside rectangle. My first step was to rotate the seven boxes so they are upright. The figure I imagined resembles a toilet seat. In that upright configuration, I assign coordinates to the 4 points. Make one of them the origin, for simplicity. Here are coordinates for my 4 points in the upright configuration (yours can be different, the end result is the same) P1 = [0,0] origin P2 = [d,0] where d is the length of a side of the box P3 = [2d,3d] P4 = [-d,5d] When that upright figure is rotated by an angle theta, you get the configuration that we started with. Using linear algebra [new coordinates (x',y')] = rotation matrix times [old coordinates (x,y)] which when written out looks like the following x' = x cos (theta) - y sin (theta) y' = x sin (theta) + x cos (theta) So the coordinates in the rotated configuration that we find at the beginning of the problem are P1' = [0,0] no change P2' = [d*cos(theta), d*sin(theta) ] P3' = [2d*cos(theta) -3d*sin(theta), 2d*sin(theta)+3d*cos(theta) ] P4' = [-d*cos(theta) - 5d*sin(theta), -d*sin(theta)+5d*cos(theta) ] For my set of points, I note that the height (17) is the difference between the y-coordinates between P3' and P1'. After simplifying, I get eq [1]: 3d*cos(theta) +2d*sin(theta) = 17 Also, the width (26) is the difference between the x-coordinates between P2' and P4'. After simplifying and rearranging the cosine term first to match eq [1], I get eq [2]: 2d*cos(theta) + 5d*sin(theta)=26 As you can see, equations [1] and [2] resemble the ones MYD came up with after using his mathemagics. I manipulated these two equations to eliminate cos(theta). I get the following: d*sin(theta) = 4 I then manipulated the two equations to eliminate sin(theta), which results in the following: d*cos(theta) = 3 I could laboriously solve for d and theta precisely by substitution, but at this point, I can see readily that only hypotenuse d=5 has lengths 3 and 4 as cosines and sines, i.e. we are looking at a 3,4,5 triangle. Since d = 5, area = 25. Again, keep in mind that how you defined coordinates for your 4 points could be different than mine. The principle is the same. Derive coordinates for the 4 points where the 7 boxes touch the outer rectangle. Use the length = 26 and height = 17 to come up with a system of two equations, from which you get the same answer regardless of how you defined your 4 points.
That is straightforward enough, assuming you are fluent in rotated coordinate systems. I would have had to look it up, so instead I did it another way. As you noted, there are 4 points where the little squares intersect with the big rectangle, one on each edge. All you have to do is draw a line between intersection points on opposite edges of the big rectangle. The "horizontal" line has a length of c*sqrt(5^2 + 2^2) = c*sqrt(29) and the "vertical" line has a length of c*sqrt(13) where c is the edge of the small squares. Then all you need is the angle between the "horizontal" line and the horizontal edge of the rectangle, and the "vertical" line and the vertical edge of the rectangle. You can construct a right triangle with each of those lines as the hypotenuse, and then use the definition of sine (opposite edge over hypotenuse) to compute sin(h)=26/(c*sqrt(29) and sin(v)=17/(c*sqrt(13)). I consider that all straightforward so far. The only trick is to compute the angles h & v. To do that I first define the angle tan(t)=b/a using the notation in the video (I do not actually use b or a, I am just mentioning them to specify the angle t that I am defining). Then the horizontal angle can be computed as h=90-t+y where tan(y)=2/5 (from the little squares, 5 over and 2 up) and similarly the vertical angle v=t+x where tan(x)=2/3 (3 over and 2 up). It is somewhat difficult to describe without a diagram, but if you draw it I think it will be clear. Then we have 17 / c sqrt(13) = sin(t+x) and 26 / c sqrt(29) = sin(90 - t + y) which is two equations in two unknowns c & t since x & y are known: previously defined with tan(x) and tan(y). Then divide the two equations by each other to cancel out c and solve for t using the trig formula sin(p+q)=sin(p)cos(q)+cos(p)sin(q) as needed, and substituting in for terms like sin(x) which we can get from tan(x)=2/3 so sin(x)=2/sqrt(13), then plug t back in to the previous equations to solve for c=5 and so area=25.
A different way to visualize these triangles would be to use the same orientation as much as possible. Across the length, you can lay the shorter triangle 3 items at the bottom, then 2 taller triangles, and then 2 shorter triangles. For the height, you would have 2 tall triangles from the top and then 3 short triangles. My solution was effectively based on that picturization, but I used larger triangles instead of many of the same smaller triangle. If x is the side of the square and a is the angle between the square grid and the base of the rectangle, then the equations were 2x sin(a) + 3x cos(a) = 17 3x sin(a) + 2x cos(a) + 2x sin(a) = 26 effectively the same equations. just a little different visualization
The reason we don't use this method for solving areas is because most of our diagrams are not drawn accurately. This only works on diagrams that are accurate.
The image doesn't really need to be accurate - the only necessary factors are that the squares are touching the rectangle at the same points in the image, with the right triangle in the bottom right. Everything will still work the same due to geometric principles, even if it doesn't look right when drawing it.
I assumed 6 square side equal to diagonal of rectangle Diagonal will be sq root of 17sq + 26sq = 31 Thus side 5.1 Close to answer of yours 5 N area will be also close
@MindYourDecisions, I tried a slightly more intuitive/visual approach to solving this and definitely not mathematically rigorous. Inspecting the picture of the rectangle, the maximum number of squares that could be placed along the width(17) of the rectangle was 3. The largest integer (I made this assumption) multiple of 3 that is less than 17 = 15 => the side of the square was 5 => area = 25.
I found that the corner right triangle can be copied to the left 3 times to create a block that fits below the 2 squares on the right and to the right of the center square, while also continuing the diagonal line. The block is made of one copy that is double the scale, one that identical, and one that is a scale factor of *b÷a.* I then found that *5a + 2b = 26* and *2b - a = b^2 ÷ a.* I then found that the *2b - a = b^2 ÷ a* can be simplified to *(a - b)^2 = 0.* I then tried to solve for *a* and *b* when *a - b = 0* and *5a + 2b = 26,* but it was inaccurate. I checked my Diagram and nothing seemed wrong with it. What got me something inaccurate?
I constructed 4 similar/identical triangles to the left. The first 3 triangle copies from the right create a block that is like those first 2 squares from the right, except shifted down by *a* and left by *b,* while also cut off by the 26×17 rectangle's boundary. The left triangle copy is triple the scale and is adjacent to the left 3 squares.
Hey, I am researching on Riemann hypothesis for the last 3 years and successfully I have found a formula for 'a' in Zeta(a+ib) in terms of 'b' and some hard looking definite integrals which are in terms of 'a' and 'b'which I was not able to solve so I am challenging you to solve those integrals . I used Riemann xi function and Riemann functional equation for Zeta function to find the formula for 'a'. Please make a video on those integrals. Integrals are:- 1)integral from 1 to infinity of x^((a-2)/2)*f(x)*cos(ln(x)*b/2) + x^(-a-1)*f(x)*cos(ln(x)*b/2) 2)integral from 1 to infinity of x^((a-2)/2)*f(x)*sin(ln(x)*b/2) - x^(-a-1)*f(x)*sin(ln(x)*b/2) Where f(x) = summation from n=1 to infinity of e^(-(n^2)*π*x)
I hit that method myself, getting the same equations, the same solutions, and the same final area. (I just happened to call the sides "ℓ cos θ" and "ℓ sin θ".)
hahaha, I did the same, started with lower triangle and worked for 17cm length and extending the square lines to the sides. once I wrote the equation for 17cm in terms of a sin(theta) and a cos(theta) length it was obvious what's happening here. Solved for theta=37 and a.
I got the answer .... only different thing I did was assume an angle x and use lcosx and lsinx instead of what you used as a and b ( where l is side of square......beautiful problem indeed
I solved it almost in the same way, but I used sines and cosines instead of just a and b, not realizing I didn't need the Pythagorean identity. The procedure was quite similar, though a bit longer.
How can you be sure that(other than observations) by placing the triangles over the squares, the height and base of triangles will be parallel to sides of the rectangle?
@@sutapadey5274 It is not an answer, because by rotating to 90 degree it does not mean it will matches the square lines. I also struggle to prove it is parallel
I see it now. By rotating triangle in 90 degree it has to match the square lines because the hypotenuse of triangle before rotation and hypotenuse after rotation has to make 90 degree and thus it is matching the square. But it is not trivial and should be mentioned in the video as explanation
Another of those problems that make us question why we never saw that method before. That's quite intuitive and simple!
Why would you not see it?
@@leif1075 right, I'm certain you saw it right away
@@nathangarrett4771 i did honestly so hope you weren't being sRcastic.
Another one of those problems I never needed to figure out in my life.
Meanwhile, my instinct was to try messing with basis vectors, but I gave up almost immediately
That's just genius, I saw the triangle but couldn't find anything to do with I
Same here... couldn't make much with triangle
th-cam.com/video/CrFgKpSy2YQ/w-d-xo.html
Same
Agreed
I saw the triangel, but didn't really think much of it, but then this comment was highlighted under the video, so I guess I had a clue going forward to solve the problem... So I was robbed of the feeling of solving it all by myself... (darn it!) 😄
I really didn't see THAT coming. Clever Solution indeed.
We Iranians have a proverb that says: "Easy is a puzzle after it is solved!"
Me after seeing the solution to the question that I failed to answer: "That was easy." :))))
👍
😄
Lol In Persian it’s
معما چو حل گشت آسان شود
Or like: a solved puzzle is an easy puzzle
Line which disappointes
" A version of this problem was given to 13 year old kids in Vietnam "
Indeed
It's actually very common for us who studying in special school or " trường chuyên" in Vietnam
In China, we say: it doesn’t hurt you too much but really humiliating.
Ouch
th-cam.com/video/CrFgKpSy2YQ/w-d-xo.html
I noticed that triangle had to be the key to solving this, but didn't realize you could use it like that to make the problem trivial, cool solution indeed
i noticed that too so i started making contructions and wrote evrything in terms of the side of the square and the angle between the side of the square and the base
after using a lot of trignometry i ended up with 2 equations which gave me the answer that the side of the square is 5
Even though I'm a high schooler, I couldn't solve it....
It was a very interesting solution. Thank you for bringing such amazing Math puzzles....
Hi mate, I got a natural sol'n using purely co-geom. To see it please order comments by time.
Can't stop crying after the legendary line...............
0:19
Bro can u tell me that how do you share that time in comment section
Plz...
@@VivekSharma-vv3dd just notice the time in video. Then write it and post.
Like this 0:19
@@mohitzen ok thnx
@@VivekSharma-vv3dd Just write the time stamp of the exact time without any spacing
I also want to say as others did, this was one of the best geometry ones I've seen. How intuitive, goes to prove we always continue to learn in life.
I'm from Vietnam. Thanks so much for this interesting problem and an outstanding solution
me too
th-cam.com/video/CrFgKpSy2YQ/w-d-xo.html
I solved it using rotations. I considered the coordinates of certain points before and after the same rotation on diffetent coordinate grids:
(2x;3x)->(?;17)
(5x;2x)->(?;26)
Where x is the side length of the squares. This results in essentially the same system of equations of the video:
3cos(a)x+2sin(a)x=17
2cos(a)x+5sin(a)x=26
Where a is the angle that the squares have to be rotated in the clockwise direction, so that they line up with the rectangle. This yields:
sin(a)x=4
cos(a)x=3
And using the pythagorean trigonometric identity, I got:
x^2=25
In short, instead of counting triangles, I counted squares with my head tilted 60°.
Being a Vienamese, I love how you solve this beautiful geometry puzzle in the easy way for everyone to understand( also I have encountered this puzzle when I was 13 , it"s pretty tough for a 13 years old boy :)) )
Yeah :))
I thought I was going to find an even easier way of the solution. I got this equation:
sqrt(2x^2)/2 + 5x = 17, I solved it, and found the area was 8.872, which is wrong. After sometime, I realized that I was missing + in my Left hand side of my equation. And finding that will be the same method as shown in the video. And I am 13. I got pretty close to solving it through my own method, didn't I?
Yup :3
@@shambhav9534 You are not 13 you are 31 😂
way to go, and you keep studying young student. Nothing makes me more proud as an educator to see youth working hard on education.
I solved it!! I really did! and what "pushed" me to find the solution no matter what is that sentence that i found quite demotivational "a version of this problem was given to students who were 12 to 13 year old in Vietnam" thank you Presh for this "achievement dose" i'm having right now ^^
Congratulations 🎊
😑😑😑😑But I am not..
Likewise, I eventually got there; Now I'll be strutting about feeling all 'big brain' for the rest of the day (and will conveniently forget the aforementioned sentence)!
I had been try this for about an hour, but you did in a matter of seconds. Hands of to you. Awesome solution. I am a huge fan of your daily problems.
Back when I was in school 11 years ago things were never drawn accurately to scale or angle, so that clever solution would never have been possible.
@@Smiththehammer you would also need to know that the corners of the squares are touching the big square at 4 locations, 2 on the length and 2 on the width, and that a small triangle is formed at one of the corners... without this information you can’t solve it with words... jus saying
Why do you think that?
@@Smiththehammer Uhm, actually, that isn't exactly the case. You can't know how many triangles would be placed inside the square without the scale. You might think they are, let's say, 5 triangles, but the problem wasn't to scale and it actually was 5.25 triangles.
Exactly my thoughts, unless the exam States it's drawn to scale this method would be damn near impossible to get right.
I was told never to use a ruler to measure the diagram, use only the numbers that are given to you. If you followed that rule, you would not be able to use the triangles method that was used to solve this puzzle
I did it just this way... I said aloud "Is that seriously a 3-4-5 triangle?"
I used projections, which led to basically the same solution. Nice problem.
Before I came up with the projections idea, I was thinking of putting additional copies of the same puzzle next to each other and to try to find how many rectangles I can construct to fill it in. This was a kind of a dead-end, but it made me think about the projections, so it helped me solve the problem.
In Vietnam, 13 years old, 7th class, children know how to solve pair of linear equations in two variables that is taught in 10th class in India!! (CBSE)
Bro so what, how you can compare the basic knowledge with Deep analysis knowledge..
But still India's education is seen tough by the world lol....
@@bhuvanaharibhs8453 yes obviously.
yay another vietnamese here
But i was taught in 6th class in school.
I like the way each time you say "and thats the answer!"
If you like solving puzzles and riddles, do checkout Puzzle Adda where you find best puzzles, riddles and their solutions.
th-cam.com/users/puzzleadda
From Vietnam wish love
*Excellent* !!! Such a *Cool* *Mathematical Problem* !!! *Thanks* for posting 😊😊
Love from *Bangladesh* 🇧🇩 ♥️
Nomoshkar apnar naam ki ?
These videos make me _really_ jealous. I wish our education was as good as it is in Vietnam
@Lee Ruan cringe.
They’re probably more strict and make you stay in school longer.
Pfft, believe me, don't.
The vietnam education system is so outdated and badly designed that the only thing keep it alive are outstanding teachers.
Source: vietnamese myself.
@@elysiummaybee Believe me, I would much rather have your education system than ours. Source: am Russian, our education is fucked up beyond repair
*laughs in Brazilian*
I have admit these simple problems are so much damn fun to watch, it makes me so happy to watch this, everytime I laugh uncontrollably. very clever indeed. Keep it up #MindYourDecisions
Is there any pure algebraic solution for this problem? Because I think , these type of clever trick solutions will not come in mind during the exam competition.
Yes I was also thinking the same.How these tricks will strike in exam ? 🤔🤔
Don't worry I don't think so such questions come in exam 😁
Idk how kids of Vietnam have solved this question.
@@Anonymous-kw7ls they had probably done it using pieces of paper 😁
@@108_adityakumar6 how can i contact presh talwalkar regarding this query?
Just another example on why I love geometry, beautiful and simple. Thanks for showing this to all of us!
Note that the sum of the semi-diagonals of the small squares spanning the horizontal and vertical dimensions are in a 7/5 ratio. This is incommensurate with the dimensions of outer rectangle which are in a 26/17 ratio. Thus the diagram is not possible.
Nice solution. For 12-13 year olds it is important to include dimensions in the solution. The area of a single square is 25 [(units of length)^2]
I had a 30 minutes in pullman to spend doing something. After thinking very deeply for half an hour i got it. I'm so proud of myself.
Now I just wish I was 12
Good job anyway
Hmm.... This is something.... I couldn't figure it out even after it's about a week now..... More problems like this Presh.... Love from India... ♥️
To be honest whenever I attempted this.... I felt an urge to extend the squares' sides such that they meet the sides of the rectangle.... And somehow be able to see something.... I just couldn't see this.....
I think this is the most interesting geometry problem I've seen in quite a while.
Thanks for sharing.
The solution was already beautiful enough, but the presentation not only does the solution justice, it makes the beauty of the solution even better.
Nicely explained.Your mathematics problems are always awesome and tooo much different.😊😊
th-cam.com/video/CrFgKpSy2YQ/w-d-xo.html
Iam Vietnamese too and im 14 ,when I see this video I was very surprised because I have done this math before so I very happy when we have a Math problem from Vietnam on your Video
The triangle trick is mind-blowing, awesome solution.
I fixed it pretty closely to this method. However, I worked out expressions for the change in height and horizontal distance across the diagonals of the squares. If you are going down or going across it's a+b in both cases.
I did almost the same
I just took a as x sin theta and b as x cos theta and followed like you
Anyways, Presh's methods are always awesome!!!
Same
Alternate Solution: 😉😉
17 = 2.5( diagonal length) (from figure; approx)
17= 2.5* √2* a
a= 5 (approx)
Area= a^2 = 25
😉
Alternately,
You can also use 26= 3.5* diagonal
(Approx; from figure)
26= 3.5* √2*a
a= 5 (approx)
Area: 5^2 = 25
I started with the same small triangle . Instead of labeling them as a,b,c . I labeled the hypotenuse as 5a I.e the side of the square..and the other sides as 4a and 3a. Solving for the vertical edge 17a =17, a=1. The square is 5ax5a= 25
This is like guessing the answer!
Wow! ... Very nice problem! I saw the triangle first, but I was very curious to find out how to solve it and did not think any further! :-D
Big LIKE! 👍 Best regards from Romania! 🤩
_Gabriel_
The pitagorian triangle 3,4,5! Elegance at it’s finest.
I thought it's called the Egyptian triangle
A very beautiful solution indeed!
th-cam.com/video/CrFgKpSy2YQ/w-d-xo.html
That's amazing i'm new here before watching this video I don't like maths and now it got interest on maths. I have share this video with my maths teacher.
Please what is the right channel to send you suggested question.
I feel amazing for finding this one out!
If one side of the square is a-bi, the other side rotated by 90 degrees is b+ai, so the real axis direction is 5a+2b=26, the imaginary axis direction is 2a+3b=17 ∴a=4 b=3 ∴Area= a^2+b^2=16+9=25
I did in a much more convoluted way but i worked. First I found the distance between the left touch point and the top touch point using the grid to be s * sqrt(13) where s is the side of the little square. The same way for the distance between the left and right touch points to be s * sqrt (29). Then using the triangles formed by these two lines and the sides of the rectangle I came up with a single equation involving s^2 = A, namely A = (9 - sqrt(13 * A - 289))^2 + (17 - sqrt(13 * A - 289) - sqrt(29 * A - 676))^2. The solution to this implicit equation is 25, which I found by checking the values with a calculator while getting a feel for what A should be before attempting to isolate A, which I ended up not doing since it was 25 exactly.
man, this was so intuitive. I didn't even know drawing extra triangle in Vietnamese math contest was allowed (yes, I'm a Vietnamese)
You don't have to draw them on the paper; just imagine them in your head.
@@psychohist I know, but in Vietnamese contest we have to explain everything we just did, not just intuitive stuff in the video
me2
@Lee Ruan we don't say anything about learning by heart here. Did you misread anything?
Do checkout Puzzle Adda TH-cam Channel for more puzzles and riddles.
That solution is actually so cool, so simple yet most people like me don't see it
_"This will become very useful in the future so pay attention"_
I have found another way to solve this.
First create all possible boxes inside the rectangle by extending the lines. The rest left are traingles and some quad. If you look carefully each left traingle completes another triangle on opposite side making a full sq.
Total sq inside is 18. Now divided area by total no of sq inside. You will get area of each small squares.
Amazing. So simple, so elegant.
I am Vietnamese, and I love this!
(I must admit that i see the figure a bit familiar)
First it looked hard, but after the way to the solution, it was actually pretty easy. Great video! 👍👍👍
Let a square's edge length be x and t be the angle of rotation of the squares (with an angle of 0 meaning the squares' edges are parallel to the rectangle's).
The vector that points from a square's bottom vertex to its rightward vertex is (I'll refer to this direction as forward), and the vector pointing to its leftward vertex is (I'll refer to this direction as up).
To get from the vertex along the bottom edge of the rectangle to the vertex along the top edge, you need to go up 3 and forward 2 square edges. That combined vector is 3 + 2 = . The y-component of this vector must have a magnitude of 17.
To get from the left edge to the right you need to go forward 2 and down 5, or 2 - 5 = . The x-component of this vector must have a magnitude of 26.
From this, the following equations are made:
3xcos(t) + 2xsin(t) = 17
2xcos(t) + 5xsin(t) = 26
Solve the first equation for xcos(t):
xcos(t) = (17 - 2xsin(t))/3
Plug into the second equation, solve for t:
5xsin(t) + 2(17 - 2xsin(t))/3 = 26
--> t = arcsin(4/x)
Plug back into the first equation:
xcos(arcsin(4/x)) = (17 - 2xsin(arcsin(4/x)))/3
--> xsqrt(1 - (4/x)^2) = 3
--> sqrt(x^2 - 16) = 3
--> x^2 = 25 = a square's area
Same solution as the video, but reframed a little.
this question is also used as olympic question in my country Azerbaijan . i didnt expected find solution in this channel lol
I knew the triangle was gonna do something but didnt know what, very interesting solution
They got us in the first half, not gonna lie.
Highly recommended. Easy to understand. Thank you so much for saving my time.
The right triangle you saw was exactly what i saw just after that diagram was fully drawn 😀😀😃😃
I also knew from the very beginning that the triangle would be a special triangle or most probably a 3...4...5 triangle
This is just amazing that there is a hidden unknown right triangle and that too that its a special 3...4...5 triangle..WOW !!!
Math is just amazing 😍😍
Easy question just by rotating anti clockwise the interior pattern such that it becomes mirror image of L and then you can simply imagine that 9 blocks covers the half rectangle so 9x area of one block = (17x26)/2. So area of one block 24.5 approx 25.
Really, It is the best explanation which I have ever seen.
hello MindYourDecision , I'm from Vietnam , i like your video
I used vectors . The side of the square pointing down being vector (a b). Pointing up is then (-b a). 3(a b)+2(-b a)+2(a b)=(26 x). Similar for (y 17). You get a=4 an b=-3. So side=5. Qed.
Ok, that is/was smart.
Small suggestion: When you "expand" the equations (2:47 - ) and you double the values (to ultimately get two with 6b) explain that before just doing it.
I was kind of thrown by what you were doing.
Explain that we have 3b and 2b and we are going to make them both 6b and to do that we will do this.......
That way people may not be as thrown by wondering what you were doing.
If we look vertically right side. There are two squares joint by there diagonals and a square by half of the diagonal. So, if we assume its diagonal as a(2)^1/2 then a(2)^1/2 + a(2)^1/2 + a/(2)^1/2 = 17. But after calculating a, area is coming 23.12 . Why?
The diagonal is not perpendicular to the bottom, so you cant calculate like that.
My solution came up like what if the squares were rotated so that the left 3 squares placed at the bottom of the rectangle. Then we have, 5s < 26 and 3s < 17, therefore the most likely answer would be 5 if it's an integer. Of course there are many assumptions for this approach
Mind blown. Once he introduced the triangles I paused again to try use it but never saw that solution!
Very nice. They’re not all this entertaining and enlightening, but I enjoyed this one.
I did essentially the same thing, but without using the little triangle. Instead of a and b, I had xcosθ and xsinθ, and equivalently solved that these were 3 and 4.
this one had me stumped, peeked your startup then I got it, love the challenge, great job
i didnt saw that but when i got a question of why are there many squares got me thinking and i realized after the video that those are the way to get the area of square
Thats very amazing, keep rocking Talealker😎
I thought at using the diagonals of the squares. That got me the fact that 3.5 diagonals are 26 and 2.5 diagonals are 17. Not really nice numbers... And they dont really match at all. As ofc the squares aren't placed at exactly 45° for me to count like that :< oof.
I fell into the same trap :(
I watch videos like this to keep adding methods to my mind's toolbox.
my method is putting the square inside a bigger and non-tilted square then find the triangles (pretty much the same but it's the outside triangle instead) it's really cool that the numbers ended up not in decimals lol
I'd propose another solution which instantly came to my mind, although not as precise
Take the square touching the top line and bottom line, give their hypotenuse a value of 1. Now the middle square that connects the top and bottom square also has a hypotenuse of value 1 but the distance between top square and bottom square is only half hypotenuse i.e.
vertical side of rectangle = 2 hypotenuse+ 1/2 hypotenuse of middle square
Vertical side of rectangle = 2.5 hypotenuse
17 = 2.5 hypotenuse
Hypotenuse = 6.8
But (hypotenuse)^2 = 2* (square side)^2
Then (side of square) = (23)^(1/2)
Area of square = (23)^(1/2) * (23)^(1/2)
Area of square = 23
I don't quite understand what you mean by "is touching the top line and the bottom line", but if your answer is 23 then you obviously made a mistake somewhere along the way. Nice try though...
ouch is right, glad I learned something new, but it kills me when I hear that it was given to 13-year-olds, gives new meaning to thinking outside the box. I never saw the triangle piece, I thought it was going to have to do with noticing that 6 of the squares would fold to make a box and the 7th one overlapped, but from there I had no idea how I was going to figure it out.
I saw triangle but didn't think it was important and couldn't solve problem but when you told focus on this right triangle I saw how to do it
I did find a geometric solution by noting a right angled isosceles triangle insid the figure. But I didnt feel fully satisfied. I knew it wasnt the easiest and intended solution. So I feel like solving it is not the absolute goal, the goal is to grasp the problem and find the best appropriate solution (but I do feel better about myself if I solve it before watching the nice solution)
I can't believe I got so close with my estimate. Even though the squares are not rotated at exactly 45 degrees, I just treated them like they were, and counted the number of diagonals that made up the length and width, to solve for the diagonal and use pythagoras to get the side.
Ended up estimating the diagonal at 7, and the area of the square as 24.5.
Most of the time i managed to solve the problems you upload, but i never solve them in your clever way!
The solution by MYD is elegant, I have to admit. However, such mental agility is not possessed by all of us. Instead, I solved it using brute force, i.e. some vector algebra / trig and end up with the exact same equations with just a little bit more work. There are only 4 points where the seven boxes touch the outside rectangle. My first step was to rotate the seven boxes so they are upright. The figure I imagined resembles a toilet seat. In that upright configuration, I assign coordinates to the 4 points. Make one of them the origin, for simplicity. Here are coordinates for my 4 points in the upright configuration (yours can be different, the end result is the same)
P1 = [0,0] origin
P2 = [d,0] where d is the length of a side of the box
P3 = [2d,3d]
P4 = [-d,5d]
When that upright figure is rotated by an angle theta, you get the configuration that we started with. Using linear algebra
[new coordinates (x',y')] = rotation matrix times [old coordinates (x,y)] which when written out looks like the following
x' = x cos (theta) - y sin (theta)
y' = x sin (theta) + x cos (theta)
So the coordinates in the rotated configuration that we find at the beginning of the problem are
P1' = [0,0] no change
P2' = [d*cos(theta), d*sin(theta) ]
P3' = [2d*cos(theta) -3d*sin(theta), 2d*sin(theta)+3d*cos(theta) ]
P4' = [-d*cos(theta) - 5d*sin(theta), -d*sin(theta)+5d*cos(theta) ]
For my set of points, I note that the height (17) is the difference between the y-coordinates between P3' and P1'. After simplifying, I get
eq [1]: 3d*cos(theta) +2d*sin(theta) = 17
Also, the width (26) is the difference between the x-coordinates between P2' and P4'. After simplifying and rearranging the cosine term first to match eq [1], I get
eq [2]: 2d*cos(theta) + 5d*sin(theta)=26
As you can see, equations [1] and [2] resemble the ones MYD came up with after using his mathemagics.
I manipulated these two equations to eliminate cos(theta). I get the following:
d*sin(theta) = 4
I then manipulated the two equations to eliminate sin(theta), which results in the following:
d*cos(theta) = 3
I could laboriously solve for d and theta precisely by substitution, but at this point, I can see readily that only hypotenuse d=5 has lengths 3 and 4 as cosines and sines, i.e. we are looking at a 3,4,5 triangle. Since d = 5, area = 25.
Again, keep in mind that how you defined coordinates for your 4 points could be different than mine. The principle is the same. Derive coordinates for the 4 points where the 7 boxes touch the outer rectangle. Use the length = 26 and height = 17 to come up with a system of two equations, from which you get the same answer regardless of how you defined your 4 points.
That is straightforward enough, assuming you are fluent in rotated coordinate systems. I would have had to look it up, so instead I did it another way. As you noted, there are 4 points where the little squares intersect with the big rectangle, one on each edge. All you have to do is draw a line between intersection points on opposite edges of the big rectangle. The "horizontal" line has a length of c*sqrt(5^2 + 2^2) = c*sqrt(29) and the "vertical" line has a length of c*sqrt(13) where c is the edge of the small squares.
Then all you need is the angle between the "horizontal" line and the horizontal edge of the rectangle, and the "vertical" line and the vertical edge of the rectangle. You can construct a right triangle with each of those lines as the hypotenuse, and then use the definition of sine (opposite edge over hypotenuse) to compute sin(h)=26/(c*sqrt(29) and sin(v)=17/(c*sqrt(13)). I consider that all straightforward so far. The only trick is to compute the angles h & v.
To do that I first define the angle tan(t)=b/a using the notation in the video (I do not actually use b or a, I am just mentioning them to specify the angle t that I am defining). Then the horizontal angle can be computed as h=90-t+y where tan(y)=2/5 (from the little squares, 5 over and 2 up) and similarly the vertical angle v=t+x where tan(x)=2/3 (3 over and 2 up). It is somewhat difficult to describe without a diagram, but if you draw it I think it will be clear.
Then we have 17 / c sqrt(13) = sin(t+x) and 26 / c sqrt(29) = sin(90 - t + y) which is two equations in two unknowns c & t since x & y are known: previously defined with tan(x) and tan(y). Then divide the two equations by each other to cancel out c and solve for t using the trig formula sin(p+q)=sin(p)cos(q)+cos(p)sin(q) as needed, and substituting in for terms like sin(x) which we can get from tan(x)=2/3 so sin(x)=2/sqrt(13), then plug t back in to the previous equations to solve for c=5 and so area=25.
I was like "this seems hard, ill just skip" and then I saw the triangle and instantly know the solution, really should have tried
A different way to visualize these triangles would be to use the same orientation as much as possible. Across the length, you can lay the shorter triangle 3 items at the bottom, then 2 taller triangles, and then 2 shorter triangles. For the height, you would have 2 tall triangles from the top and then 3 short triangles.
My solution was effectively based on that picturization, but I used larger triangles instead of many of the same smaller triangle. If x is the side of the square and a is the angle between the square grid and the base of the rectangle, then the equations were
2x sin(a) + 3x cos(a) = 17
3x sin(a) + 2x cos(a) + 2x sin(a) = 26
effectively the same equations. just a little different visualization
You are really perfect teacher 💪
The reason we don't use this method for solving areas is because most of our diagrams are not drawn accurately. This only works on diagrams that are accurate.
The image doesn't really need to be accurate - the only necessary factors are that the squares are touching the rectangle at the same points in the image, with the right triangle in the bottom right. Everything will still work the same due to geometric principles, even if it doesn't look right when drawing it.
Although I guess actually thinking to use the method given can be difficult if the image is drawn poorly.
I assumed 6 square side equal to diagonal of rectangle
Diagonal will be sq root of 17sq + 26sq = 31
Thus side 5.1
Close to answer of yours 5
N area will be also close
@MindYourDecisions, I tried a slightly more intuitive/visual approach to solving this and definitely not mathematically rigorous. Inspecting the picture of the rectangle, the maximum number of squares that could be placed along the width(17) of the rectangle was 3. The largest integer (I made this assumption) multiple of 3 that is less than 17 = 15 => the side of the square was 5 => area = 25.
I found that the corner right triangle can be copied to the left 3 times to create a block that fits below the 2 squares on the right and to the right of the center square, while also continuing the diagonal line. The block is made of one copy that is double the scale, one that identical, and one that is a scale factor of *b÷a.* I then found that *5a + 2b = 26* and *2b - a = b^2 ÷ a.* I then found that the *2b - a = b^2 ÷ a* can be simplified to *(a - b)^2 = 0.* I then tried to solve for *a* and *b* when *a - b = 0* and *5a + 2b = 26,* but it was inaccurate. I checked my Diagram and nothing seemed wrong with it. What got me something inaccurate?
I constructed 4 similar/identical triangles to the left. The first 3 triangle copies from the right create a block that is like those first 2 squares from the right, except shifted down by *a* and left by *b,* while also cut off by the 26×17 rectangle's boundary. The left triangle copy is triple the scale and is adjacent to the left 3 squares.
Proud of my country. We learned Math before we were born.
Hey, I am researching on Riemann hypothesis for the last 3 years and successfully I have found a formula for 'a' in Zeta(a+ib) in terms of 'b' and some hard looking definite integrals which are in terms of 'a' and 'b'which I was not able to solve so I am challenging you to solve those integrals . I used Riemann xi function and Riemann functional equation for Zeta function to find the formula for 'a'.
Please make a video on those integrals.
Integrals are:-
1)integral from 1 to infinity of x^((a-2)/2)*f(x)*cos(ln(x)*b/2) + x^(-a-1)*f(x)*cos(ln(x)*b/2)
2)integral from 1 to infinity of x^((a-2)/2)*f(x)*sin(ln(x)*b/2) - x^(-a-1)*f(x)*sin(ln(x)*b/2)
Where f(x) = summation from n=1 to infinity of e^(-(n^2)*π*x)
I hit that method myself, getting the same equations, the same solutions, and the same final area. (I just happened to call the sides "ℓ cos θ" and "ℓ sin θ".)
hahaha, I did the same, started with lower triangle and worked for 17cm length and extending the square lines to the sides. once I wrote the equation for 17cm in terms of a sin(theta) and a cos(theta) length it was obvious what's happening here. Solved for theta=37 and a.
Splendid, insightful and above all, fun. Thanks PT
I got the answer .... only different thing I did was assume an angle x and use lcosx and lsinx instead of what you used as a and b ( where l is side of square......beautiful problem indeed
I solved it almost in the same way, but I used sines and cosines instead of just a and b, not realizing I didn't need the Pythagorean identity. The procedure was quite similar, though a bit longer.
I just fell in love with the solution.
Yay this is the first problem I actually worked out on my own here. Same method pretty much.
I am in Vietnam but I really like a lot of homework. But it's very hard to solve.
Hi bro
If you like solving puzzles and riddles, do checkout Puzzle Adda where you find best puzzles, riddles and their solutions.
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I’m just happy I understood the answer
How can you be sure that(other than observations) by placing the triangles over the squares, the height and base of triangles will be parallel to sides of the rectangle?
because you just rotate it 90° as square corner is 90°
@@sutapadey5274 It is not an answer, because by rotating to 90 degree it does not mean it will matches the square lines. I also struggle to prove it is parallel
@@lato1813 it will. the triangles are similar and the have tge same length hypotenuse so tgey are congruent
I see it now. By rotating triangle in 90 degree it has to match the square lines because the hypotenuse of triangle before rotation and hypotenuse after rotation has to make 90 degree and thus it is matching the square. But it is not trivial and should be mentioned in the video as explanation
@@lato1813 it is assumed that everyone will realise. i don't think there are many little children watching this video who need an explanation
I saw the right angle triangle thought it was going to be key, but was completely stuck at that point.
Me too. I'm thinking too long and asking how am I suppose to use that triangle