A diagonal divides a large, outer square into two equal parts. A smaller square is inscribed in each part. Let the area of the blue square be $A,$ and let the area of the orange square be $B.$

What is $\frac AB?$

$\dfrac{A}{B}=\dfrac{8}{9}$
$\dfrac{A}{B}=1$
$\dfrac{A}{B}=\dfrac{9}{8}$

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As pointed out in the comments, it's not entirely proven that the upper portion can be broken into 9 congruent triangles.

The premise gives enough information to say the marked triangle marked with a star above is a 45-45-90 one. This means it is isosceles and the two legs are congruent. This can then be used symmetrically on the other side of the square B to establish the length of the square of B is 1/3 of the diagonal of the entire square.

This implies a 3 by 3 grid can be laid down which intersects the points of B, leading to the 9 congruent triangle result.

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I suggest to you that it can be seen that the triangles must have the same dimensions and the same angles of 45 degrees, 45 degrees and 90 degrees, and so are congruent. What a fantastic solution this is! I got the same result but by calling the length of a side of square A as L and then finding the length of a side of square B. But this solution is extremely elegant!! Regards, David

David Fairer
- 3 years, 2 months ago

The really briliant

Mohd O
- 3 years, 2 months ago

How do you know which to cut in 4 and 9?

Nathan Davies
- 3 years, 2 months ago

I don't follow why it is not entirely proven that the upper portion can be broken into 9 congruent triangles. The large square diagonal is at 45deg to the sides, therefore so are the sides of B (2 sides parallel and 2 perpendicular). If Lb is the length of each side of B, then the hypotenuses of the triangles adjacent to B are all Lb long, and all right-angle triangles with the other 2 angles of 45deg. The same can be said about the 4 triangles inside B. The remaining 2 triangles are also 90-45-45deg with the same length side. Therefore all 9 are congruent.

Mark Billingham
- 3 years, 2 months ago

Total Down triangle is cut into two (½) not the small square

Tamil Tamil
- 3 years, 2 months ago

"A diagonal divides a large, outer square into two equal parts.." the diagonal does not necessarily have to be at 45 deg. However, if diagonal is limited in definition to 45 deg., then, as per the diagram above, the upper triangle can clearly (just visually) be divided into 9 congruent triangles. Given the discrete choice of answers, one must assume diagonal to mean 45 deg. Lay the 3x3 grid over the whole square and the upper triangle, along with the existing lines should make it pretty clear - see Thomas' post above.

El Dal
- 3 years, 2 months ago

Given the limited choices, I can tell what the answer is by looking at the sizes.

Dennis Rodman
- 2 years, 6 months ago

For the larger square we can say it has a side length of s.

For square A to fit and still be a square, it will need to have half the length of the original, or a side length of $\frac{1}{2}$ s

so the area of square A is $\frac{1}{4}$ $s^{2}$

For square B to fit and still be a square, we can split the diagonal of the big square into 3 equal pieces. To find the length of the diagonal we can use Pythagorean theorem or the 45-45-90 special right triangle rule, we can say that the diagonal has a length $\sqrt{2}$ s so the side length of square B is $\frac{\sqrt{2}}{3}$ s

This means the area of square B is $\frac{2}{9}$ $s^{2}$

so the ratio of the areas $\frac{A}{B}$ is $\frac{\frac{1}{4}s^{2}}{\frac{2}{9}s^{2}}$ which simplifies to $\boxed{\frac{9}{8}}$

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You can remove s by taking s=1

Pierre Thierry
- 3 years, 2 months ago

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That would only prove it for s=1. By having s cancel itself out you prove it for any value of s.

Thor - Magne Saevareid
- 3 years, 2 months ago

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Length is always relative. You can choose the unit of length to be the size of the square.

Pierre Thierry
- 3 years, 2 months ago

For your areas you should have s^2 "so the area of square A is 1/4 s^2" and the area of "B is 2/9 s^2. They still cancel in the ratio so the answer is the same, but s squared is required for the values as presented to be correct for the areas.

David Richner
- 3 years, 2 months ago

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Yes that would be correct, I carelessly overlooked that since it cancels in the end.

Callum M
- 3 years, 2 months ago

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How can you prove by your answer that remaining part of both halves isn't equal??

Mr. India
- 3 years, 2 months ago

Square A's half does not leave any remaining part. For clarity: 2A=.5 2B+c=.5, where c is the remaining triangular part. Then, 2B+c=2A» B+(c/2)=A. A>B. I hope it's clear.

A Former Brilliant Member
- 3 years, 2 months ago

I don't think this is sufficient proof. The triangles around B are all smaller than the ones around A. So if B were sufficiently larger than A, it might compensate for the third tiny leftover-triangle you mentioned.

Alessandro Simovic
- 3 years, 2 months ago

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No, bro. It can be easily seen from the equations I wrote some comments below that A is larger than B.

A Former Brilliant Member
- 3 years, 2 months ago

Here is a little visualization: $2.25B=2A\Rightarrow \boldsymbol{\frac{A}{B}=\frac{2.25}{2}=\frac{9}{8}}$

Because the upper and lower sum represent both a half of the triangle, it follows:
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From the figure we can just observe that A is larger than B

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let L=the length of the largest square,

A=( $\frac{L}{2}$ )^2= $\frac{L^2}{4}$

We can cut the top left side of the diagonal into 9 identical triangles to show that sqrt(B) is $\frac{diagonal}{3}$

B=( $\frac{diagonal}{3}$ )^2=( $\frac{sqrt(2L^2)}{3}$ )^2= $\frac{2L^2}{9}$

A* $\frac{1}{B}$ = $\frac{L^2*9}{4*2L^2}$ = $\frac{9}{8}$

:-)

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The whole square (let's call it $S$ ) is split into two equal halves. Taking the half with the square $A$ , we notice that it is comprised of $A$ as well as two other sections, $t_{1}$ and $t_{2}$ . $A$ is a square extending from the corner of $S$ to its diagonal, and therefore it is a quarter of $S$ , and thus half of the right triangle made by the diagonal. Triangles $t_{1}$ and $t_{2}$ , then, also together form half the right triangle. So we can say $t_{1}+t_{2}=\frac{1}{4}S=A$ . Together, the left triangle half can be written as $A+t_{1}+t_{2}=A+A$ , or $2A$ .

As for the right half, it is comprised of square $B$ and three triangles: $t_{3}$ and $t_{4}$ , which are identical, and $t_{5}$ in the corner. Noticing that $t_{5}$ can be folded down onto the intersection of the diagonals of $B$ informs us that $t_{5}=\frac{1}{4}B$ , because it can only cover a fourth of the area. Likewise, triangles $t_{3}$ and $t_{4}$ can be folded across $B$ , and each take up half of the area of $B$ . So $t_{3}=t_{4}=\frac{1}{2}B$ . The right half of square $S$ , then, is written $B+t_{3}+t_{4}+t_{5}=B+\frac{1}{2}B+\frac{1}{2}B+\frac{1}{4}B$ , which simplifies to $\frac{9}{4}B$ .

Finally, since each half of $S$ is equivalent, we can set the left half, $2A$ , equal to the right half, $\frac{9}{4}B$ . So, $2A=\frac{9}{4}B$ , and rearranging gives us $\frac{A}{B}=\frac{9}{8}$ .

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A can be formed using A1 and A2, and B can be formed using B1 and B2. Since the two halfs of the square are equal, we have

$\begin{aligned} A + A1 + A2 &= B + B1 + B2 + C\\ 2A &= 2B + C\\ A &= B + \frac{C}{2} > B\\ A/B &> 1 \end{aligned}$

(where the labels represent the areas of the geometric object)

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Cut the lower half and the upper one into respectively 4 and 9 triangles of equal areas, as below.

Then $A=\frac{1}{2}$ and $B=\frac{4}{9}$ , so $\frac{A}{B}=\frac{1}{2}\frac{9}{4}=\frac{9}{8}$