Hey Brilliant, I found this interesting problem in Catalonia's Olympiad.

We have five squares stowed in this way: (See the photo)

Prove that ABCD sqare has the same area as AEF triangle.

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If \(ABCD\) has side \(x\), then \(BX \,=\, x\cos\theta\) and \(DY = x\sin\theta\). Since \(\angle ABX = 180^\circ-\theta\) and \(\angle ADY = 90^\circ+\theta\), we deduce from the Cosine Rule that \[ \begin{array}{rcl} y^2 & = & x^2 + x^2\cos^2\theta - 2x^2\cos\theta\cos(180^\circ-\theta) \; = \; x^2(1 + 3\cos^2\theta) \\ z^2 & = & x^2 + x^2\sin^2\theta - 2x^2\sin\theta\cos(90^\circ+\theta) \; = \; x^2(1 + 3\sin^2\theta) \end{array} \] Using the Sine Rule on \(ABX\) and \(ADY\), \[\begin{array}{rcl} \sin\alpha & = & \frac{x\cos\theta \times \sin(180^\circ-\theta)}{y} \; = \; \frac{\sin\theta\cos\theta}{\sqrt{1+3\cos^2\theta}} \\ \sin\beta & = & \frac{x\sin\theta \times \sin(90^\circ+\theta)}{z} \; = \; \frac{\sin\theta\cos\theta}{\sqrt{1+3\sin^2\theta}} \end{array} \] and hence \[ \cos\alpha \; = \; \frac{1 + \cos^2\theta}{\sqrt{1 + 3\cos^2\theta}} \qquad \cos\beta \; = \; \frac{1 + \sin^2\theta}{\sqrt{1 + 3\sin^2\theta}} \] Hence \[ \cos(\alpha+\beta) \; = \; \frac{(1+\cos^2\theta)(1+\sin^2\theta) - \sin^2\theta\cos^2\theta}{\sqrt{(1 + 3\cos^2\theta)(1 + 3\sin^2\theta)}} \; = \; \frac{2}{\sqrt{(1 + 3\cos^2\theta)(1 + 3\sin^2\theta)}} \] Now \(\gamma = 90^\circ - \alpha-\beta\) and so \(\sin\gamma = \cos(\alpha+\beta)\). Thus \[ |AEF| \; = \; \tfrac12yz\sin\gamma \; = \; \frac{yz}{\sqrt{(1 + 3\cos^2\theta)(1 + 3\sin^2\theta)}} \; = \; x^2 \; = \; |ABCD| \]

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Alt text

There's no need to bust out the complicated trigonometry. ;)

We note that \(X'A=XA\), \(AB=BC=CD=DA\) and \(Y'A=YA\). Also \(|BCV|=|XBA|=|ADY|\), which can be easily proved by trigonometry--noting that \(|BCV| = \frac{1}{2} \times BV \times BC = \frac{1}{2} \times BX \times BA =|XBA|\) and the same can be done for triangle \(ADY\).

We construct \(X'B'A\) by rotating \(XBA\) \(90\) degrees clockwise about \(A\) and \(X_1'B'C\) by rotating \(XBA\) \(90\) degrees clockwise about \(B\). Similarly we construct \(Y'D'A\) by rotating \(YDA\) \(90\) degrees anticlockwise about \(A\) and \(Y_1'B'C\) by rotating \(YDC\) \(90\) degrees anticlockwise about \(D\).

Rotation preserves area and congruency, and we see that by angle chasing, \(X'B'\) is equal in length and parallel to \(X_1'B\) and the same with the pairs \(B'D'\) and \(BD\), \(Y'D'\) and \(Y_1'D\). Therefore we see that \(X'Y'D'AB'\) is congruent to \(X_1'Y_1'DCB\).

More importantly we also see that \(|BDHV|=|X'Y'D'B'|=|X_1'Y_1'DB|\) as all three share the same height \(VH\) and lengths \(BV\) and \(DH\).

Therefore

\(|X'Y'A|=|X'Y'D'AB'|-|X'B'A|-|Y'D'A|=|X_1'Y_1'DCB|-|X_1'BC|-|CDY_1'|\)

\(=|X_1'Y_1'DB|+|BCD|-|BVC|-|DCH|=|BDHV|+|BCD|-|BVC|-|DCH|\)

\(=|BCD|+|BVC|+|DCH|+|BCD|-|BVC|-|DCH|=2|BCD|=|ABCD|\)

And we are done! ;)

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I forgot to add in \(\sin \angle{BVC} = \sin \angle{XBA}\) as \(\angle{BVC} = 180 - \angle{XBA}\). Oops. ;)

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This is obvious, but the area of ABCD equals the sum of the areas of the two smallest squares. I don't know if that would help though.

EDIT: I played with the problem a little bit. Label the side of the smallest square to the side of the biggest square a-e, respectively. After tinkering with the Pythagorean Theorem, I have come to the conclusion that \(5c^2=d^2+e^2\), so if you can prove that \(\dfrac{d^2+e^2}{5}\) is the area of \(\Delta AEF\) then you are done.

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teach me your ways master

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wow so complex

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Do the two lower squares have coincident bottom sides, and does C lie on the extension of both bottom lines?

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I think we can assume that. If it wasn't the case, the figure would probably have been drawn differently.

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