It's very simple bro... No need to do so much calculations.. I didn't watch ur full video...... Divide diameter in ratio... 2:3...because all figures are square... So right side square 3x3=9,left side square divide its height in 2:1 ratio... So left side lower square will be 2x2=4,and upper one small square 1x1=1..now add all these figures.. 9+4+1=14
@@VaradMahashabde mathematics is just a practical thing.... According to figure it can be done... If you take 1:4 it will be foolishness... You have to try as practically....
Khang Nguyễn Quốc I learned addition at 2 and 11 years later I know calculus 2. But man do they teach us slow as hell in my school, I just started exploring other topics in math and it’s pretty awesome I enjoy it.
I started by mentally enlarging square h and shrinking square a and see what this would do to the whole diagram when square h and t became equal. Bingo 2 squares of 6.25 each.
Yes kinda, but you can view it as a point. To visualize it, I recommend the "Limit-Approach" having a getting smaller and smaller till it reaches 0, like Captain Oates described
@@NStripleseven Not sure if you're trolling, but no. A circle is the _set_ of points from a fixed distance r>0 from another fixed point. A point is simply just (x1, x2... xn) in n-dimensional space.
The solution is really simple if you use vectors: You have 3 unknows: t, h, a so you need 3 equations: 1. 5=t+h 2. t=a+h 3. The vector from the centre of the circle to the edge of the circle where the 2 smaller squares meet must have length 5/2: r = (5/2-t-a;t-a). Length of r must equal 5/2 and you have your 3rd equation. This gives then 2 solutions: one where t=3 and one where t=5/2 so the areas are 14 and 12.5 respectively.
Comparing my solution to yours, here's what occurs to me: There's a very similar big picture: given n unknowns find n equations by interpreting the figure, then isolate variables, substitute and simplify. You name b-a and b+c-a and thus have more variables and equations, no biggie. We still extract the same information from the figure, which is unsurprising since that set of information is sufficient but no proper subset is. I think the biggest take-away difference is that my solution requires approximately the first math I learned in second ed (school year 11) and a lot of algebra. Your solution requires a theorem I've either forgotten or never learned, and in exchange for using fancy theorems you need simpler algebra and less of it. That's an interesting trade-off. Now I'm wondering: I'm solving a quadratic polynomial. I've heard the phrase "completing the square". You're using some trigonometry. If you copy a right-angle triangle and glue the two copies along the hypotenuse, you get a rectangle. A square is a kind of rectangle. Is a rectangle an incomplete square in some sense? Is there some deeper level at which we're attacking the same quadratic phenomenon but via different paths? In programming there's often a trade-off between simplicity and generality; am I doing something that's more general and less simple? Am I seeing the same trade-off in math? One more difference between our solutions: in my head at least, I was putting the three variables of interest front and center, noting the two obvious (linear) constraints and searching for just that one last constraint. Your solution looked like you were throwing more names and more equations into the mix until you had a sufficient amount. That is, my solution feels top-down goal-oriented where yours looks bottom-up and exploratory. This is mainly a matter of organization, presentation and perception, though, but I wonder whether there's a difference in mindset here, and what their relative strengths and weaknesses are.
In Vietnam, we had learned Calculus in Grade 6 and some students learn Matrix, Integral, Derivative, Inequality in Grade 10. If you want to study Advanced High School, you will always deal with these questions like this.
Papa, I was so upset when you said that you wouldn’t solve it analytically that I decided to give it a try. I solved the problem rather quickly by setting the origin on the left of the circumference. Then I labeled the sides a>b>c, and wrote down the equations tying them, i.e. a+b=2r=5, b+c=a. Finally, I obtained the last relation by finding the x such that the height of the circumference is b, called that x0 and imposed that x0+c=b. x0 depends only on r and b. Done! 😃
1:09 It's pretty easy to do using coordinates. Putting the center of the circle (radius r) at the origin, and calling the square side lengths a (top left), b (bottom left), and c (right), you have b+c=2r and a+b=c, so c=a+b and a+2b=2r. The point where the squares with side lengths a and b coincide with the circle is (-r+b-a, b), so plugging this in to x^2+y^2=r^2 and simplifying gives 2b^2+a^2-2ab=2r(b-a). Replacing 2r by a+2b and simplifying further gives a(2a-b)=0, so a=0 or b=2a. We don't want a=0, so b=2a. Substituting this into a+2b=2r gives a=2/5*r, b=4/5*r, so c=a+b=6/5*r. Thus, the area is A=a^2+b^2+c^2=(4+16+36)/25*r^2=56/25*r^2. r=5/2 gives A=14.
An alternative solution can be: 1) Draw a line parallel to the diameter of circle. 2) Divide the semicircle into 2 quadrants. 3) Take the part left out of 5/2 × 5/2 square as x. 4) start calculating the sides of the squares. 5) They will be (5/2-x), (5/2+x) and 2x 6) Total area is 25/4 + 6x^2. 7) Some how we need to get x so that we can calculate the total area. 8) Draw a line from centre of semicircle to the point where the first square is cut by semi circle. 9) Hurrrah!!!!, Now we can see a right triangle of sides (5/2-x), 3x and 5/2 10) Use Pythagoras theorem to solve x. 11) The quadratic equation is 2x^2 -x =0 12) So,x has 2 values 0.5 or 1 13) The total area is either 14 or 12.5 If you liked this solution, send your love :)
We can think about the problem as a circle with diameter 5 inscribed in it's square. We draw lines horizontally and vertically through the center of the square and remove the lower half. Now we have a half circle inscribed in two squares with the same size, sharing a common point P on the circle. The center line will be translated to the left while maintaining both square's square property. Moving the vertical center line x units left will move point P x units up and the other P x units down while both P stay on the translated line. Reaching the initial problem picture with the translation, 2x will yield the side length of square one. The square side lengths in terms of x and circle radius R are: (2x)^2, (R-x)^2 and (R+x)^2 Solving for x: R^2 = (3x)^2+(R-x)^2 gives x=R/5 So the sum is: (2R/5)^2 + (4R/5)^2 + (6R/5)^2 where 2R/5=1 so the result is 1^2+2^2+3^2
I solved it a bit simplier using Pythagorian theorem only. There it is: Let "a" be the lenght of right square, "b" the lenght of the left square, "c" the lenght of the smallest square and "r" the perimeter of circle. Then we know this: a+b=2r b+c=a and the tricky one: r+sqrt(r*r-b*b)=a+c (Pythagorian theorem in triangle constructed using center of the circle) Then it is just solving 3 equations of 3 variables to get that a*a+b*b+c*c=(56/25)r*r a=(6/5)r b=(4/5)r c=(2/5)r So nice puzzle, thanks for the video.
Your way is definitely prettier (I never would have come up with all that), but I thought using analytical geometry was a bit easier: If the sides of the squares from big to small are a, b, c, and the lower left is the origin: c = a - b (comparing sides of the squares) R=(a+b)/2 (radius of the semicircle is half the diameter) (x - R)^2 + y^2 = R^2 (the equation of the circle is shifter R in the horizontal direction) And the key point is that intersection of the squares and circle: (b-c,b) Plug that point into the equation of the circle, make the substitutions for c and R, and the equation melts down to a smooth: 3b = 2a. So then if a+b = 5, we know a, b, and c. The answer follows quickly. I consider your way more proper, but maybe the way above can help. Thanks for the video!
Another approach of solution: With standard x and y axes; Eqn of semi circle will be y = sqr(r2 - x2) for a given radius r. From which the abscissa of the vertex of right angled triangle which is p will be given by p^2 = r^2-(3r-3p)^2. by resolving the Quadratic Eqn: p = 4r/5 or r; from which p = 4r/5, a = 2r/5 and t = 6r/5 for any given value of r.
How I did it: See that the medium square side + the big square side = 5, but also the big square side is the medium square side + the small square side, so 2 * x (the medium square side) + y (the small square side) = 5. Guess and check x = 2, y = 1, 2x + 1 = 5, we see in the top left corner that y has to be half of x, 1 is half of 2, so everything checks out. So the small square is 1x1, medium is 2x2, large is 3x3
Try software Inkscape which is a *FREE* open-source software. With Inkscape, you can neatly draw your squares, circles, etc. Every serious mathematician should have Inkscape on PC. Not only for graphic designs and visual presentations, Inkscape is useful also for drawing geometric shapes, measuring lines & angles, technical drawing, etc. I regularly use Inkscape to solve geometry puzzles.
I solved it by constructing inverse and mirrored drawing at the bottom - completing the circle and drawing another 3 squares in different order. Now I could construct right triangle with sides: (2h), (t - h + 2a) and 5 (as circle diameter between connection points of both smaller squares) Later I noticed that I can write side: (t - h + 2a) as (15 - 6h) Now, from Pythagorean theorem: (15 - 6h)^2 + (2h)^2 = 5^2 Gave me: h = 2, from that, t = 3, a = 1 and the answer 14. (also h could be 2.5 but that's broken)
I found an other solution to the problem: h = p+a, therefore it is one side length of the "middle sized square", because it's given that the rectangle, of which one sidelength is h, is a square => t = h + a => t = 2a + p => 5 = 3a + 2p p = (5-3a)/2 You can construct a right triangle with the radius of the circle with d = 5, h, and 2.5-p as its sidelengths. ( 2.5 - p because the distance from the center of the circle to the start of p is 2.5, so the sidelength of the triangle that is necessary is 2.5 - p). Pythagoras => (2a + (5 - 3a)/2)^2 + (2.5 - (5 - 3a)/2 )^2 = 2.5^2 a = 1 => p = (5 - 3*1)/2 = 1 A = 5t - p*a => 5 * (2*1 + 1) - 1 = 14
So, this problem has a pretty simple solution if you draw yourself a nice right-angled triangle. Draw a vertical radius and then draw a radius to the bottom left-hand corner of the smallest square. Now from that corner, take a horizontal line across to form a right-angle with the vertical radius. If we set the side length of the large square to (r+x), then the medium square has side length (r-x) and the small square has side length 2x. Now, our nice right-angled triangle has a hypotenuse of r, and right-angled sides of (r-x) and 3x. Pythagorus gives us r^2 = (r-x)^2 + (3x)^2 = r^2 -2xr + 10x^2 --> 2xr = 10x^2 --> r = 5x or x=r/5. The solution follows from plugging the radius measurement into the area formula for the squares = (r+x)^2 + (r-x)^2 + (2x)^2. For r=5/2, x=1/5: A = 3^2 + 2^2 + 1^2 = 9+4+1 = 14.
One can also solve completely without the triangle. Let s be the side length of the middle sized square. Then 5-s is the is the side length of the large square and s/2 the side length of smallest square. Also we know that 5 times s+s/2 equals the area of the the big square plus the middle square plus two times the small square. One equation, one variable. Solve for s. Done. (Maybe/probably Someone noted before didn’t check)
Call the length of biggest square = L. Length of medium square M = 5-x. Length of small square S = x - (5-x) = 2x-5. The corner of the small square that is on the semicircle is M = 5-L above and (L-2.5 + S) = (L-2.5)+(2L-5) = 3L-7.5 to the left of the center of the circle. Pythagorean Theorem: (5-L)^2 + (3L-7.5)^2 = 25. Solve for L. L=2.5 or L=3. So the squares have length of [L=3, M= 2, S=1] or [L=2.5, M=2.5, S=0]. Assuming we use the value of L that makes S non-zero, the areas of the squares are 9, 4 and 1, for a total of 14.
I did it another way : Let 'a' be the increment of the big square relatively to the radius length (= 5/2). It is quite easy to convince oneself that the little square has side 2a. The two other squares have side 5+a and 5-a. One can find a by looking for the intersection point of x^2 + y^2 = (5/2)^2 and y = 7.5 - 3x. 'a' will be 2.5-x. I'm sorry not to be more explicative than that but this is something one can quite easily see geometrically too. It leads to a simple quadratic equation which leads to x = 2 or 2.5, thus x = 2 because the little square is not a point. Thus a=0.5 and the area is (5/2-a)^2 + (5/2+a)^2 + (2a)^2 = 14
I solved it a bit different. Let the side length of the medium square equal “a” and the length of the small square equal “b”. The large square would have side length a + b. Therefore, one equation is a + a + b = 5. To get another equation, draw a line segment from the radius to the point where the small square intersects the circle. We can construct a right triangle from this with side lengths a and 1.5b and hypotenuse 2.5. Therefore, a^2 + (1.5b)^2 = 2.5^2 Or 4a^2 + 9b^2 = 25 Solving the system of equations yields a = 2 and b = 1. Thus the sides of the squares are 1, 2, and 3
Hey Flammable Maths. I'm new to the channel. I woke up at 04:00AM tonight and went on a marathon of your videos. Just wanted to say I really enjoyed your content over the last hours and subbed after just a few videos. It's really great stuff keep going :) My first videos were the factorial derivative and product integral ones... I'll stick around
I clicked for the title 😂 Also recognised the Fibonacci spiral in there so I figured it would be 1+4+9 but tried to prove it also and started with a^2 + 5c^2 = b^2. Nice problem 👍
This problem could also be solved pretty fast applying the compass-and-straightedge construction of a real number's square root. For the application of this construction, p must be 1 which give us the height h is equal to sqrt(5-p), so h=2 and so the medium square's area is h^2=4. The side of the larger square is 5 - 2 = 3, so its area is 3^2=9. The small square's side is 3 - 2 = 1, so its area is 1. So the total area is 4 + 9 + 1 = 14 Basically, this problem just solves itself. Anyway, interesting problem. Keep going.
There's a much easier way I only realised after solving it myself using Pythagoras: Let sides of small, medium and large square = s, m, and l respectively m = 2s l = s + m = s + 2s = 3s Diameter d = m + l = 2s + 3s = 5s = 5 Therefore s = 1, m = 2 and l = 3 etc.
Let x be the side length of the middle sized square (h in the video) and r the radius. Then the big square has side length 2r - x and the small one 2r-2x. Joining the point on the semicircle (vertex of the small square) with the center of the semicircle gives you a right triangle with side lengths 3(r-x), x and r. Pythagorean theorem gives you 9(r-x)^2 + x^2 = r^2 and solving it for x gives x = r or x = 4/5 r. The first solution is nonsense since the small square wouldn't exist in that case, so x = 4/5 r. Plugging in gives the total area is 56/25 r^2. Finally, letting r = 5/2 gives 14. Cheers!
need just pitagora’s theorem call x the side of the medium square you find quadratic equation in x with two solutions, but one is not acceptable because implies the smallest square has zero area, and that is not consistent with the problem, so total area = 14
more easily......look at the triangle OPP´ ( O = center of the circle , P =intersection point between circle and square, P´ = projection of P on the diameter ) . So ( r-p)^2+(2p)^2 = r^2 ( r= circle radius).. then p=2/5 r . A=(2p)^2+p^2+(3p)^2= 14 p^2 = 56/25 r^2
hum... I'm confused... Consider this as "a" gets approaches zero, "h" approaches "t". In the limit where a=0, h=t. In this case we basically have two squares side by side and a semi-circle touching the top where the squares meet. so surely h = 2a is not necessarily true? e.g. when "a" is a small positive what am I missing here? As far as I can tell, "h" can be anything between 0 to t
I understand his solution is more general, but in this case (where the sum of the sides at the base is 5) I noticed that the triangle has sides of 3,4,5 (pythagorean triple) and proceeded with the other triangle (that with hyphotenuse 3 has sides 1 and 2) to obtain the two sides of the little squares. After that also the side of the bigger square is immediate.
no need for any theorems. the one for right triangle would be : an inscribed angle is half the central angle and since the diameter is a straight line and there are 180 degrees that makes a straight line. The result would be a right triangle subtended by the diameter.
I just used pythagoras and the diameter in function of the squars sides. Making a system with two variables and two equations. It's right??? Because it was pretty easy to do this..... edit1: ''used'', ''to do'' edit2: edit1
Side of the medium square =h Side of max square = 5-h Side of smal square = 5-2h The possible values positive integers(for elementary school only positive integers number) for h are 1 or 2 Atot25 impossible for h=2 Atot=14 Atot=14
Papa Flammy, not sure if that's interesting enough but could you maybe do something oh the virial theorem? It pops up occassionally in astrophysics and I think it's kind of awesome how general it is ^^
I can't spot a flaw in the reasoning shown but I'm not quite getting it. To me the point where squares h and a touch the circumference of the semicircle can be anywhere from the base line up to the mid point. At the extreme points some of the squares go to zero dimension. The result could therefore be anything between 12.5 and 25.
Nice problem! If I am not mistaken this is actually a Fibonacci related problem because a^2 =1^2, h^2=2^2, t^2 = 3^2 and the missing square in de corner is ap which is the same as a^2= 1^2.
@@PapaFlammy69 Thanks! I initially(mistakenly) thought that any size would do as long as the widths of the two bottom rectangles add up to 5. Your approach is very nice in that the final equation for a and h is simple. Here is mine using triangle similarity: www.overleaf.com/read/pksjxtdrxsqt
I used a method where I just used the pythagorean theorem, system of equations, and the formula to calculate the area of quadrilaterals to solve the problem
The equations I used are (1) 5(d+r) - r(d-r) = d^2 + r^2 + (d + r)^2 (2) (5/2)^2 = d^2 + (r+(5/2-d))^2 The first equation is the equation computing the area of the three squares, and the second equation is the equation to calculate pythagorean theorem. d = side of medium square, r = side of small square d + r = side of big square.
Couldn’t you have skipped the pq equation because as it’s drawn, if they are all squares, you know a would be .5h because h is a bisector of the bottom left square and the top left square lined up with the middle of the bottom left square?
So first I solved this problem using GeoGebra Classic 6, and found that the length of the side of the first square to the left is 2, therefore the total area of the squares is 2²+3²+1²=14. Then I solved this problem myself and the most difficult thing I did was solving a second degree polynomial and calculating the distance between 2 points in a Cartesian plane. I don't know what age is an elementary school student in Vietnam is, but I'd say a 12-13 year old student could have done all I did.
So first I defined the bottom left corner of the left square as the origin (0,0) of my axis. Let's say the length of the side of the left square is x and the diameter of the semicircle is D. Therefore, the right square has sides D-x. If the height of the right square is D-x and the height of the left square is x, therefore the top square has height (D-x)-x=D-2x. The left corner of the top square is a point in the semicircle and has coordinates (x-(D-2x), x)=(3x-D, x) and the center of the semicircle is at (D/2, 0), so the distance between those two points should be half the diameter D of the semicircle (its radius). => (3x - D - D/2)² + (x - 0)² = (D/2)² => 10x² - 9*D*x + 2D² = 0 => x = (9D ± √(81D²-80D²) )/20 => x = (1/2)D or x = (2/5)D If x = (1/2)D then the top square would have side length equal to zero, therefore x = (2/5)D. In our problem, it was given that D = 5 so the left square has side length 2, the right square has side length 3 and the top square has side length 1 and the total area of the squares is 2²+3²+1²=14.
Uhm you can actally solve it much easier by just noticing that the small and the middle squares side lenghts are together the side lengh of the larger and we know that the big and the mediums side lenghts are 5 together. And its really easy to see that they are 2 and 3
i did: say the small square is x^2, so the square under is probalby (2x)^2, 2+1=3 so the large square is (3x)^2. Wait the medium and large square is 2x+3x=5x, which is the lenght given, so area is x^2+(2x)^2+(3x)^2=14? and somehow was correct
Anybody else guess the right answer by looking at the perfect squares in the thumbnail? Was nice to get it right, but honestly my thought process made no use of the constraint imposed by the circle, it just looked like the square on top had half the side length of the square on bottom. Lucky guess
Before trying to actually use rigorous math: guessing 12.5 because I just moved the point at the middle, basically making 2 squares with side length of 2.5, so basically 2*2.5^2 = 12.5
I didn't even need any formulas to see that it was three squares with sidelength 1,2,3... couldn't you've chosen any other n than 5 for the diameter of the semicircle... Although if it had been 6 it would've been as easy, you would just have had to multiply each sidelength by 1.2 for 10 you would just double each sidelength... who needs any math when the answer is obvious and intuitive. It would've been a lot more interesting if you had made the diameter 1. Then a would have sidelength .2 b would have sidelength .4 and c would have sidelength .6 Oh and that triangles area is more fun to calculate... how many of you would know that it's 2²? Well it was still a nice video. It's a shame it was such a basic problem.
It took 1 minute to think and 10 seconds to solve The smallest square has a length of x It is half of the lower square Of length 2x The biggest square will be of length x +2x = 3x On the diameter 3x +2x is given as 5 So x is 1 Smallest square 1 Lower square 4 Biggest square 9 Total 14 Good problem
HERE IS THE EASIEST SOLUTION. Let the length of sides of the medium square be x. ----(i) Hence the length of sides of the largest square will be 5-x. -----(ii) Now, observe that the length of smallest square + the length of medium square = 5-x The length of smallest square + x = 5-x The length of smallest square = 5-2x. -----(iii) But we know that the smallest square's length is just half of the length of the medium square. It appears from the picture. Since, each of the 3 figures is just a square. Length of smallest square = x/2. -----(iv) Equate the equations iii and iv, we get, x/2 = 5- 2x x= 2 units = length of medium square. Now, from ii and iii, we get, Length of biggest square = 5-x = 3 units. Length of smallest square = 5-2x = 1 unit. Total area = 1x1 + 2x2 + 3x3 = 14 units squared. It was just a simple linear equation to be solved.
![Can You Figure It Out? The *Almost* Impossible Geometry Puzzle Solved. [Not Clickbait]](http://i.ytimg.com/vi/X0l83mkzkX8/mqdefault.jpg)
Support the channel by trying out Brilliant for free mah sons and Daughters
i already paid for brilliant this year lol
It's very simple bro... No need to do so much calculations.. I didn't watch ur full video...... Divide diameter in ratio... 2:3...because all figures are square... So right side square 3x3=9,left side square divide its height in 2:1 ratio... So left side lower square will be 2x2=4,and upper one small square 1x1=1..now add all these figures.. 9+4+1=14
I exist
@@babababablacksheep9452 It's wrong bro
Also why would you divide the diameter in 2:3
@@VaradMahashabde mathematics is just a practical thing.... According to figure it can be done... If you take 1:4 it will be foolishness... You have to try as practically....
3-year-olds in Vietnam even have multivariable calculus as a part of the curriculum, very challenging indeed
Isn't that a prerequisite for kindergarten?
I am an apple, expert in analytic number theory and I learned calculus at 0.5 seconds old. And I am a fckng fruit
You're bananas! (I'll get out, pardon me)
Mauricio Mondragon but, HOw DId YoU TyPe ThAt iF You’Re aN aPpLe?
Khang Nguyễn Quốc I learned addition at 2 and 11 years later I know calculus 2. But man do they teach us slow as hell in my school, I just started exploring other topics in math and it’s pretty awesome I enjoy it.
I did this when I was an embryo, can’t believe you couldn’t do it in less than a second.
Poor you in china they do this when they were sperm.
:'D
Sad Chicken In Japan, they solved this before the Big Bang occurred.
Fresh toadwalker is it you?
Hahahah thats not Presh Talwaker
Oily Macaroni Hello, Mr. oily macaroni constant!
I litterally didn't click at first thinking it was presh trashtalker
yes.
@@PapaFlammy69 Strange, I didn't hear "Gougu" anywhere in the video.
9:10
I think a=0 would work out,
it would conclude that h=t=2.5
It would just look like 2 equalsized Squares
You're correct. In my approach, you actually end up with a quadratic that gives you both solutions h=2 and h=2.5 ;)
And the semicircle would hit the middle top of the figure.
I started by mentally enlarging square h and shrinking square a and see what this would do to the whole diagram when square h and t became equal. Bingo 2 squares of 6.25 each.
You can’t .If a=0 ,then the square just doesn’t exist
Yes kinda, but you can view it as a point.
To visualize it, I recommend the "Limit-Approach" having a getting smaller and smaller till it reaches 0, like Captain Oates described
"Let's use elementary geometry instead of analytical geometry." Goes on to solve everything using algebra anyway.
Well, algebra geometry
Check the comment by Our math channel, solved by basic geometry concepts. You may like it.
"this point"
> draws circle
Ruinenlust Well, Yeah, isn't a point the same as a circle?
It's a very small circle
@@NStripleseven Not sure if you're trolling, but no. A circle is the _set_ of points from a fixed distance r>0 from another fixed point. A point is simply just (x1, x2... xn) in n-dimensional space.
Ruinenlust Circle with radius 0
And yeah, I saw the >0.
Ruinenlust Actually, a point is a 0-circle, so they are technically correct, this fact is just not useful.
Challenge: Every time he says “meaning overall” solve the integral of sqrt(tanx) and then check by differentiation
Drinking game for mathematician
I'd have a better shot of surviving actual shots
The heck?
:'D
This is evil.
The solution is really simple if you use vectors: You have 3 unknows: t, h, a so you need 3 equations:
1. 5=t+h
2. t=a+h
3. The vector from the centre of the circle to the edge of the circle where the 2 smaller squares meet must have length 5/2:
r = (5/2-t-a;t-a). Length of r must equal 5/2 and you have your 3rd equation.
This gives then 2 solutions: one where t=3 and one where t=5/2 so the areas are 14 and 12.5 respectively.
Who needs vectors when we have Pythagoras theorem! 😉
So it does work out if a=0.
but technically you don't have 3 squares anymore in the case of 12.5
diD yOu FiGuRe It oUT??/?/? now for 10 seconds of monetized silence!
You mean minutes?
*beeping noises intensify*
Presh presh presh
From the figure at 0:14, before our host starts drawing over it, calling the side lengths a, b and c with a
Comparing my solution to yours, here's what occurs to me:
There's a very similar big picture: given n unknowns find n equations by interpreting the figure, then isolate variables, substitute and simplify. You name b-a and b+c-a and thus have more variables and equations, no biggie. We still extract the same information from the figure, which is unsurprising since that set of information is sufficient but no proper subset is.
I think the biggest take-away difference is that my solution requires approximately the first math I learned in second ed (school year 11) and a lot of algebra. Your solution requires a theorem I've either forgotten or never learned, and in exchange for using fancy theorems you need simpler algebra and less of it. That's an interesting trade-off.
Now I'm wondering: I'm solving a quadratic polynomial. I've heard the phrase "completing the square". You're using some trigonometry. If you copy a right-angle triangle and glue the two copies along the hypotenuse, you get a rectangle. A square is a kind of rectangle. Is a rectangle an incomplete square in some sense? Is there some deeper level at which we're attacking the same quadratic phenomenon but via different paths?
In programming there's often a trade-off between simplicity and generality; am I doing something that's more general and less simple? Am I seeing the same trade-off in math?
One more difference between our solutions: in my head at least, I was putting the three variables of interest front and center, noting the two obvious (linear) constraints and searching for just that one last constraint. Your solution looked like you were throwing more names and more equations into the mix until you had a sufficient amount. That is, my solution feels top-down goal-oriented where yours looks bottom-up and exploratory. This is mainly a matter of organization, presentation and perception, though, but I wonder whether there's a difference in mindset here, and what their relative strengths and weaknesses are.
In Vietnam, we had learned Calculus in Grade 6 and some students learn Matrix, Integral, Derivative, Inequality in Grade 10. If you want to study Advanced High School, you will always deal with these questions like this.
I feel dumb
@@leant6487 Don't worry. I study 14 hours/day in High School so you don't feel dumb (Study from 6:00 AM to 10 PM (2 hours relax)
Calculus in grade 6??????!?!??
In my school we were taught decimal multiplication and the most advanced topic was the basics of sets?
Papa, I was so upset when you said that you wouldn’t solve it analytically that I decided to give it a try. I solved the problem rather quickly by setting the origin on the left of the circumference. Then I labeled the sides a>b>c, and wrote down the equations tying them, i.e. a+b=2r=5, b+c=a. Finally, I obtained the last relation by finding the x such that the height of the circumference is b, called that x0 and imposed that x0+c=b. x0 depends only on r and b. Done! 😃
1:09 It's pretty easy to do using coordinates. Putting the center of the circle (radius r) at the origin, and calling the square side lengths a (top left), b (bottom left), and c (right), you have b+c=2r and a+b=c, so c=a+b and a+2b=2r. The point where the squares with side lengths a and b coincide with the circle is (-r+b-a, b), so plugging this in to x^2+y^2=r^2 and simplifying gives 2b^2+a^2-2ab=2r(b-a). Replacing 2r by a+2b and simplifying further gives a(2a-b)=0, so a=0 or b=2a. We don't want a=0, so b=2a. Substituting this into a+2b=2r gives a=2/5*r, b=4/5*r, so c=a+b=6/5*r. Thus, the area is A=a^2+b^2+c^2=(4+16+36)/25*r^2=56/25*r^2. r=5/2 gives A=14.
I did exactly the same
An alternative solution can be:
1) Draw a line parallel to the diameter of circle.
2) Divide the semicircle into 2 quadrants.
3) Take the part left out of 5/2 × 5/2 square as x.
4) start calculating the sides of the squares.
5) They will be (5/2-x), (5/2+x) and 2x
6) Total area is 25/4 + 6x^2.
7) Some how we need to get x so that we can calculate the total area.
8) Draw a line from centre of semicircle to the point where the first square is cut by semi circle.
9) Hurrrah!!!!, Now we can see a right triangle of sides
(5/2-x), 3x and 5/2
10) Use Pythagoras theorem to solve x.
11) The quadratic equation is
2x^2 -x =0
12) So,x has 2 values 0.5 or 1
13) The total area is either 14 or 12.5
If you liked this solution, send your love :)
Wow!!! Looks so simple.
Thank you.
We can think about the problem as a circle with diameter 5 inscribed in it's square.
We draw lines horizontally and vertically through the center of the square and remove the lower half.
Now we have a half circle inscribed in two squares with the same size, sharing a common point P on the
circle.
The center line will be translated to the left while maintaining both square's square property.
Moving the vertical center line x units left will move point P x units up and the other P x units down
while both P stay on the translated line.
Reaching the initial problem picture with the translation, 2x will yield the side length of square one.
The square side lengths in terms of x and circle radius R are: (2x)^2, (R-x)^2 and (R+x)^2
Solving for x: R^2 = (3x)^2+(R-x)^2 gives x=R/5
So the sum is: (2R/5)^2 + (4R/5)^2 + (6R/5)^2 where 2R/5=1 so the result is 1^2+2^2+3^2
I solved it a bit simplier using Pythagorian theorem only. There it is:
Let "a" be the lenght of right square, "b" the lenght of the left square, "c" the lenght of the smallest square and "r" the perimeter of circle.
Then we know this:
a+b=2r
b+c=a
and the tricky one:
r+sqrt(r*r-b*b)=a+c (Pythagorian theorem in triangle constructed using center of the circle)
Then it is just solving 3 equations of 3 variables to get that a*a+b*b+c*c=(56/25)r*r
a=(6/5)r
b=(4/5)r
c=(2/5)r
So nice puzzle, thanks for the video.
Your way is definitely prettier (I never would have come up with all that), but I thought using analytical geometry was a bit easier: If the sides of the squares from big to small are a, b, c, and the lower left is the origin:
c = a - b (comparing sides of the squares)
R=(a+b)/2 (radius of the semicircle is half the diameter)
(x - R)^2 + y^2 = R^2 (the equation of the circle is shifter R in the horizontal direction)
And the key point is that intersection of the squares and circle: (b-c,b)
Plug that point into the equation of the circle, make the substitutions for c and R, and the equation melts down to a smooth: 3b = 2a. So then if a+b = 5, we know a, b, and c. The answer follows quickly.
I consider your way more proper, but maybe the way above can help. Thanks for the video!
Another approach of solution:
With standard x and y axes; Eqn of semi circle will be y = sqr(r2 - x2) for a given radius r.
From which the abscissa of the vertex of right angled triangle which is p will be given by p^2 = r^2-(3r-3p)^2.
by resolving the Quadratic Eqn: p = 4r/5 or r; from which p = 4r/5, a = 2r/5 and t = 6r/5 for any given value of r.
3:00 In English i think it's called "geometric mean".
oh, really?
thx!
How I did it: See that the medium square side + the big square side = 5, but also the big square side is the medium square side + the small square side, so 2 * x (the medium square side) + y (the small square side) = 5. Guess and check x = 2, y = 1, 2x + 1 = 5, we see in the top left corner that y has to be half of x, 1 is half of 2, so everything checks out. So the small square is 1x1, medium is 2x2, large is 3x3
Papa flammy:
*I’m sorry, I can’t draw a square or a semi-circle😔😔😔*
Also Papa Falammy 2 minutes later:
*draws two perfect lines*
:'D
Try software Inkscape which is a *FREE* open-source software. With Inkscape, you can neatly draw your squares, circles, etc. Every serious mathematician should have Inkscape on PC.
Not only for graphic designs and visual presentations, Inkscape is useful also for drawing geometric shapes, measuring lines & angles, technical drawing, etc. I regularly use Inkscape to solve geometry puzzles.
@@pinklady7184 how much are they paying you and how do I get in on that
I solved it by constructing inverse and mirrored drawing at the bottom - completing the circle and drawing another 3 squares in different order.
Now I could construct right triangle with sides: (2h), (t - h + 2a) and 5 (as circle diameter between connection points of both smaller squares)
Later I noticed that I can write side: (t - h + 2a) as (15 - 6h)
Now, from Pythagorean theorem: (15 - 6h)^2 + (2h)^2 = 5^2
Gave me: h = 2, from that, t = 3, a = 1 and the answer 14. (also h could be 2.5 but that's broken)
I found an other solution to the problem:
h = p+a, therefore it is one side length of the "middle sized square", because it's given that the rectangle, of which one sidelength is h, is a square
=> t = h + a => t = 2a + p
=> 5 = 3a + 2p p = (5-3a)/2
You can construct a right triangle with the radius of the circle with d = 5, h, and 2.5-p as its sidelengths. ( 2.5 - p because the distance from the center of the circle to the start of p is 2.5, so the sidelength of the triangle that is necessary is 2.5 - p).
Pythagoras => (2a + (5 - 3a)/2)^2 + (2.5 - (5 - 3a)/2 )^2 = 2.5^2 a = 1
=> p = (5 - 3*1)/2 = 1
A = 5t - p*a => 5 * (2*1 + 1) - 1 = 14
just wanna say keep using and demonstrating new theorems in your vids they are really interesting
thanks
So, this problem has a pretty simple solution if you draw yourself a nice right-angled triangle.
Draw a vertical radius and then draw a radius to the bottom left-hand corner of the smallest square. Now from that corner, take a horizontal line across to form a right-angle with the vertical radius.
If we set the side length of the large square to (r+x), then the medium square has side length (r-x) and the small square has side length 2x. Now, our nice right-angled triangle has a hypotenuse of r, and right-angled sides of (r-x) and 3x. Pythagorus gives us r^2 = (r-x)^2 + (3x)^2 = r^2 -2xr + 10x^2 --> 2xr = 10x^2 --> r = 5x or x=r/5. The solution follows from plugging the radius measurement into the area formula for the squares = (r+x)^2 + (r-x)^2 + (2x)^2. For r=5/2, x=1/5: A = 3^2 + 2^2 + 1^2 = 9+4+1 = 14.
"Now a door closed.. magically.." that was incredible xD your reaction was so serious, like a eurika moment or something xD
One can also solve completely without the triangle. Let s be the side length of the middle sized square. Then 5-s is the is the side length of the large square and s/2 the side length of smallest square. Also we know that 5 times s+s/2 equals the area of the the big square plus the middle square plus two times the small square. One equation, one variable. Solve for s. Done. (Maybe/probably Someone noted before didn’t check)
Call the length of biggest square = L.
Length of medium square M = 5-x.
Length of small square S = x - (5-x) = 2x-5.
The corner of the small square that is on the semicircle is M = 5-L above and (L-2.5 + S) = (L-2.5)+(2L-5) = 3L-7.5 to the left of the center of the circle.
Pythagorean Theorem: (5-L)^2 + (3L-7.5)^2 = 25. Solve for L.
L=2.5 or L=3. So the squares have length of [L=3, M= 2, S=1] or [L=2.5, M=2.5, S=0].
Assuming we use the value of L that makes S non-zero, the areas of the squares are 9, 4 and 1, for a total of 14.
11:36 damn papa dat sum quicc math
I did it another way :
Let 'a' be the increment of the big square relatively to the radius length (= 5/2).
It is quite easy to convince oneself that the little square has side 2a. The two other squares have side 5+a and 5-a.
One can find a by looking for the intersection point of x^2 + y^2 = (5/2)^2 and y = 7.5 - 3x. 'a' will be 2.5-x. I'm sorry not to be more explicative than that but this is something one can quite easily see geometrically too.
It leads to a simple quadratic equation which leads to x = 2 or 2.5, thus x = 2 because the little square is not a point.
Thus a=0.5 and the area is (5/2-a)^2 + (5/2+a)^2 + (2a)^2 = 14
nice :)
Let the arc be the graph of the function sqrt(R^2 - x^2), let the left and right squares have side lengths s = R-d, S = R+d, s
I solved it a bit different. Let the side length of the medium square equal “a” and the length of the small square equal “b”. The large square would have side length a + b. Therefore, one equation is a + a + b = 5. To get another equation, draw a line segment from the radius to the point where the small square intersects the circle. We can construct a right triangle from this with side lengths a and 1.5b and hypotenuse 2.5.
Therefore, a^2 + (1.5b)^2 = 2.5^2
Or 4a^2 + 9b^2 = 25
Solving the system of equations yields a = 2 and b = 1. Thus the sides of the squares are 1, 2, and 3
Hey Flammable Maths. I'm new to the channel. I woke up at 04:00AM tonight and went on a marathon of your videos. Just wanted to say I really enjoyed your content over the last hours and subbed after just a few videos. It's really great stuff keep going :)
My first videos were the factorial derivative and product integral ones... I'll stick around
awwww :))
The good old HÖHENSATZ
Papa seriously going through puberty now? Those (completely unintentional) voice cracks
I know I’m really late but didn’t he divide by a at 9:18 so couldn’t there be another solution
When you told us on discord that brilliant was sponsoring you, I was a bit nervous lol. This was great! I'm so proud of you. :)
I clicked for the title 😂
Also recognised the Fibonacci spiral in there so I figured it would be 1+4+9 but tried to prove it also and started with a^2 + 5c^2 = b^2. Nice problem 👍
This problem could also be solved pretty fast applying the
compass-and-straightedge construction of a real number's square root.
For the application of this construction, p must be 1 which give us the height h is equal to sqrt(5-p), so h=2 and so the medium square's area is h^2=4.
The side of the larger square is 5 - 2 = 3, so its area is 3^2=9.
The small square's side is 3 - 2 = 1, so its area is 1.
So the total area is 4 + 9 + 1 = 14
Basically, this problem just solves itself.
Anyway, interesting problem. Keep going.
There's a much easier way I only realised after solving it myself using Pythagoras:
Let sides of small, medium and large square = s, m, and l respectively
m = 2s
l = s + m = s + 2s = 3s
Diameter d = m + l = 2s + 3s = 5s = 5
Therefore s = 1, m = 2 and l = 3
etc.
Let x be the side length of the middle sized square (h in the video) and r the radius. Then the big square has side length 2r - x and the small one 2r-2x. Joining the point on the semicircle (vertex of the small square) with the center of the semicircle gives you a right triangle with side lengths 3(r-x), x and r. Pythagorean theorem gives you 9(r-x)^2 + x^2 = r^2 and solving it for x gives x = r or x = 4/5 r. The first solution is nonsense since the small square wouldn't exist in that case, so x = 4/5 r. Plugging in gives the total area is 56/25 r^2. Finally, letting r = 5/2 gives 14. Cheers!
need just pitagora’s theorem call x the side of the medium square you find quadratic equation in x with two solutions, but one is not acceptable because implies the smallest square has zero area, and that is not consistent with the problem, so total area = 14
1:28 why you do this?
more easily......look at the triangle OPP´ ( O = center of the circle , P =intersection point between circle and square, P´ = projection of P on the diameter ) . So ( r-p)^2+(2p)^2 = r^2 ( r= circle radius).. then p=2/5 r . A=(2p)^2+p^2+(3p)^2= 14 p^2 = 56/25 r^2
hum... I'm confused...
Consider this
as "a" gets approaches zero, "h" approaches "t".
In the limit where a=0, h=t. In this case we basically have two squares side by side and a semi-circle touching the top where the squares meet.
so surely h = 2a is not necessarily true? e.g. when "a" is a small positive
what am I missing here? As far as I can tell, "h" can be anything between 0 to t
I just didn't see it when filming the video ^^
9:15
What's the problem with a=0 ?
It does seem like a solution, just that in it there are two identical squares.
I rather derived for the triangle that h^2= p^2+q^2+pq. What theorem is that?
I understand his solution is more general, but in this case (where the sum of the sides at the base is 5) I noticed that the triangle has sides of 3,4,5 (pythagorean triple) and proceeded with the other triangle (that with hyphotenuse 3 has sides 1 and 2) to obtain the two sides of the little squares. After that also the side of the bigger square is immediate.
I got 14 just looking at the thumbnail, but I just assumed the 5 was split into 2 and 3, and the small square was 1. But I was right lol
bruv
Bruh I did that using coordinate geometry and you just guessed that :(
My disappointment is immeasurable and my day is ruined!
no need for any theorems. the one for right triangle would be : an inscribed angle is half the central angle and since the diameter is a straight line and there are 180 degrees that makes a straight line. The result would be a right triangle subtended by the diameter.
What an amazing video! It was well explained. Nice solution :0
I learned a lot 😄
Brilliant is awesone!
where can i buy this blackboard? I want one of these. thx
You could probably get a sponsorship with then if you try
I just used pythagoras and the diameter in function of the squars sides. Making a system with two variables and two equations. It's right??? Because it was pretty easy to do this.....
edit1: ''used'', ''to do''
edit2: edit1
0:10 Engineer's Semicircle
Cries in approximation.
A mathematician is more likely to get away with it, if an engineer produces that circle somethings gonna blow up :D
9:25 a can absolutely be zero, if and only if the right triangle is also an isosceles one.
Side of the medium square =h
Side of max square = 5-h
Side of smal square = 5-2h
The possible values positive integers(for elementary school only positive integers number) for h are 1 or 2
Atot25 impossible
for h=2 Atot=14 Atot=14
1:34 "it is what it is" more like
"that's just how it is, wap bap"
Papa Flammy, not sure if that's interesting enough but could you maybe do something oh the virial theorem? It pops up occassionally in astrophysics and I think it's kind of awesome how general it is ^^
I can't spot a flaw in the reasoning shown but I'm not quite getting it. To me the point where squares h and a touch the circumference of the semicircle can be anywhere from the base line up to the mid point. At the extreme points some of the squares go to zero dimension. The result could therefore be anything between 12.5 and 25.
Is there some way this can be generalised to n squares such that the total area is \sum_{k=0}^n k^2 (= 1/6 n(n+1)(2n+1) I think)?
Nice problem! If I am not mistaken this is actually a Fibonacci related problem because a^2 =1^2, h^2=2^2, t^2 = 3^2 and the missing square in de corner is ap which is the same as a^2= 1^2.
What are the givens? Only diameter of the circle being 5?
yes
@@PapaFlammy69 Thanks! I initially(mistakenly) thought that any size would do as long as the widths of the two bottom rectangles add up to 5. Your approach is very nice in that the final equation for a and h is simple. Here is mine using triangle similarity: www.overleaf.com/read/pksjxtdrxsqt
You should press the play button to start a video :) Love the vids. Keep on making them Papa!
I used a method where I just used the pythagorean theorem, system of equations, and the formula to calculate the area of quadrilaterals to solve the problem
The equations I used are
(1) 5(d+r) - r(d-r) = d^2 + r^2 + (d + r)^2
(2) (5/2)^2 = d^2 + (r+(5/2-d))^2
The first equation is the equation computing the area of the three squares, and the second equation is the equation to calculate pythagorean theorem. d = side of medium square, r = side of small square d + r = side of big square.
Always have to check if 5 is 0 before dividing both sides by 5.
Hi, ich bin Vietnamese und studiere jetzt in Deutschland. Ich folge deinem Kanal seit langem und bin ein Fan von dir
Dude that brilliant advertisement was brilliant, how on earth did you come up with it??
hehe :D
Couldn’t you have skipped the pq equation because as it’s drawn, if they are all squares, you know a would be .5h because h is a bisector of the bottom left square and the top left square lined up with the middle of the bottom left square?
1:45 what about diameters not passing through the center? Lol
Finally a question I can think of solving.
So first I solved this problem using GeoGebra Classic 6, and found that the length of the side of the first square to the left is 2, therefore the total area of the squares is 2²+3²+1²=14. Then I solved this problem myself and the most difficult thing I did was solving a second degree polynomial and calculating the distance between 2 points in a Cartesian plane. I don't know what age is an elementary school student in Vietnam is, but I'd say a 12-13 year old student could have done all I did.
So first I defined the bottom left corner of the left square as the origin (0,0) of my axis. Let's say the length of the side of the left square is x and the diameter of the semicircle is D. Therefore, the right square has sides D-x. If the height of the right square is D-x and the height of the left square is x, therefore the top square has height (D-x)-x=D-2x. The left corner of the top square is a point in the semicircle and has coordinates (x-(D-2x), x)=(3x-D, x) and the center of the semicircle is at (D/2, 0), so the distance between those two points should be half the diameter D of the semicircle (its radius).
=> (3x - D - D/2)² + (x - 0)² = (D/2)²
=> 10x² - 9*D*x + 2D² = 0
=> x = (9D ± √(81D²-80D²) )/20
=> x = (1/2)D or x = (2/5)D
If x = (1/2)D then the top square would have side length equal to zero, therefore x = (2/5)D. In our problem, it was given that D = 5 so the left square has side length 2, the right square has side length 3 and the top square has side length 1 and the total area of the squares is 2²+3²+1²=14.
Uhm you can actally solve it much easier by just noticing that the small and the middle squares side lenghts are together the side lengh of the larger and we know that the big and the mediums side lenghts are 5 together. And its really easy to see that they are 2 and 3
i did: say the small square is x^2, so the square under is probalby (2x)^2, 2+1=3 so the large square is (3x)^2. Wait the medium and large square is 2x+3x=5x, which is the lenght given, so area is x^2+(2x)^2+(3x)^2=14? and somehow was correct
Anybody else guess the right answer by looking at the perfect squares in the thumbnail? Was nice to get it right, but honestly my thought process made no use of the constraint imposed by the circle, it just looked like the square on top had half the side length of the square on bottom. Lucky guess
bruh
If you bring out the math-hitler one more time, I think I must sue you for smart money.
This time I hit my head on a box due to laughter :D
Before trying to actually use rigorous math: guessing 12.5 because I just moved the point at the middle, basically making 2 squares with side length of 2.5, so basically 2*2.5^2 = 12.5
Ah so that is the degenerate case when a = 0.
Does a contour integral around the weird shape and concludes it is zero vis the residue theorem 😆
bruv
Flammable maths: posts a video Me: IS THAT A NEW VIDEO I SEE?
Wow a papa geometry video, how (fresh) of you
After watching the vid it’s actually neat and nice 100k behind
SIR YOUR APPROACH WAS SO SCARY! WHY COULD YOU NOT DO IT EASILY? THE SOLUTION I USED WAS DAMN SIMPLE.
If A is nice and spicy then
Flemmy's wife name must start with ∆ ! I guessメ
So good solution for this exam .
I didn't even need any formulas to see that it was three squares with sidelength 1,2,3... couldn't you've chosen any other n than 5 for the diameter of the semicircle...
Although if it had been 6 it would've been as easy, you would just have had to multiply each sidelength by 1.2 for 10 you would just double each sidelength... who needs any math when the answer is obvious and intuitive. It would've been a lot more interesting if you had made the diameter 1. Then a would have sidelength .2 b would have sidelength .4 and c would have sidelength .6
Oh and that triangles area is more fun to calculate... how many of you would know that it's 2²?
Well it was still a nice video. It's a shame it was such a basic problem.
Yeah, the video was fun even though it's far more basic than most of his content
This is the best math video ever pleade upload more elementary geonometry so I can get ahead of the rest in my class lol
It took 1 minute to think and 10 seconds to solve
The smallest square has a length of x
It is half of the lower square
Of length 2x
The biggest square will be of length x +2x = 3x
On the diameter 3x +2x is given as 5
So x is 1
Smallest square 1
Lower square 4
Biggest square 9
Total 14
Good problem
The click-baity title makes this video even better.
ye :'D
love the vids
Or a=0, area = 25÷2 = 12.5
Excellent
Solves elementary schools problem: ....So with the use of the Peano axoms...
xD
HERE IS THE EASIEST SOLUTION.
Let the length of sides of the medium square be x. ----(i)
Hence the length of sides of the largest square will be 5-x. -----(ii)
Now, observe that
the length of smallest square + the length of medium square = 5-x
The length of smallest square + x = 5-x
The length of smallest square = 5-2x. -----(iii)
But we know that the smallest square's length is just half of the length of the medium square. It appears from the picture. Since, each of the 3 figures is just a square.
Length of smallest square = x/2. -----(iv)
Equate the equations iii and iv, we get,
x/2 = 5- 2x
x= 2 units = length of medium square.
Now, from ii and iii, we get,
Length of biggest square = 5-x = 3 units.
Length of smallest square = 5-2x = 1 unit.
Total area = 1x1 + 2x2 + 3x3 = 14 units squared.
It was just a simple linear equation to be solved.
Yes, it‘s easy to read the meme :)
my spermatozoides just started with your question, just wait a few months for the right answer!.
The weird noises you make are funny 😂😂
:D
so great , I try to know this exam .
"yes they are supposed to be squares"
- papa flamey
I think this can be solved using Fibonacci..
Edit: yes it can. Even if the base wasn't 5, you could use Fibonacci as a ratio.
8:30 Wait what? Kitchen? I was under the assumption that he was filming these at some school, because aren't you a teacher? Welp this is awkward.
Please do more high school level problems!