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# Interesting Problem

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.

Note by Jordi Bosch
4 years, 2 months ago

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alt text

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 $|AEF| \; = \; \tfrac12yz\sin\gamma \; = \; \frac{yz}{\sqrt{(1 + 3\cos^2\theta)(1 + 3\sin^2\theta)}} \; = \; x^2 \; = \; |ABCD|$

- 4 years, 2 months ago

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! ;)

- 4 years, 2 months ago

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

- 4 years, 2 months ago

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.

- 4 years, 2 months ago

- 3 years, 7 months ago

wow so complex

- 3 years, 7 months ago

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

- 4 years, 2 months ago

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

- 4 years, 2 months ago