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Points $E$, $F$, $G$, and $H$ lie inside a rhombus $ABCD$, such that the triangles $\triangle AEB$, $\triangle BHC$, $\triangle CGD$, and $\triangle DFA$ are isosceles right triangles with hypotenuses $AB$, $BC$, $CD$, and $DA$. The sum of areas of $ABCD$ and $EFGH$ is $S$. Find, with the proof, the length of $CD$. Express your answer in terms of $S$ only.

I know the area of the rhombus $\left( \frac12 d_1d_2 \right)$ and I know the four triangles are equivalent and can find their areas to subtract from $ABCD$. I'm unsure how to go about finding the remaining area (four triangles) to get the area of $EFGH$. My approach could be wrong though.

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  • $\begingroup$ is it possible to include a sketch? $\endgroup$ – Narasimham Apr 15 '15 at 7:25
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WARNING: NOT RIGOROUS (intended to walk someone through how a proof would be constructed in an environment more conducive to diagrams).

First, we note that if $ABCD$ is a square, then $EFGH$ collapses to a single point at the center. Let $\theta$ be the the smaller angle of the rhombus $ABCD$ and $x=CD$ be its side length. We see that $EFGH$ is a square since $$|EF|=|FG|=|GH|=|HE|=2\frac{x}{\sqrt{2}}\sin\left(\frac{\theta-\frac{\pi}{2}} {2}\right)$$ and the perpendicular bisectors of $EF$ and $GH$ are themselves perpendicular. Hence $$EFGH=2x^2\sin^2\left(\frac{\theta-\frac{\pi}{2}}{2}\right)=x^2(1-\sin(\theta))$$ $ABCD$ can similarly be expressed in terms of $\theta$ and $x$. Let the longer side of the rhombus be $d_1$, the shorter $d_2$. Then $d_1=2x\cos\left(\frac{\theta}{2}\right)$ and $d_2=2x\sin\left(\frac{\theta}{2}\right)$. Therefore $$ABCD=\frac{1}{2}\left(4x^2\left[\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)\right]\right)=x^2\sin(\theta)$$ and $S=x^2$.

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  • $\begingroup$ I like how you derived the other way to calculate the area of a rhombus. $\endgroup$ – jxh Apr 17 '15 at 18:58
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This question is much easier to deal with if you have a picture.

Let $\theta_{XYZ} = |\angle XYZ|$. That is, $\theta_{XYZ}$ denotes the measured size of $\angle XYZ$.

Without loss of generality, choose point labeling so that $\theta_{ABC}\leq\theta_{BCD}$.

enter image description here

Let $L$ denote the length of $\overline{CD}$, noting that all the sides of rhombus $ABCD$ have the same length. Note that the area of $ABCD$ is $L^2\sin\theta_{ABC}$.

Since the $4$ interior right triangles are all identical, the $4$ interior isosceles triangles ($\triangle AEF$, $\triangle BEH$, $\triangle CGH$, and $\triangle DFG$) are identical as well. Since the diagonals of $ABCD$ are perpendicular bisectors of the bases of the isosceles triangles that they cross, it follows that $EFGH$ is a square. Each side of the square has length $L\sqrt2\sin\frac{\theta_{EBH}}2$, therefore its area is $L^22\sin^2\frac{\theta_{EBH}}2 = L^2(1-\cos\theta_{EBH})$. Since $\theta_{EBH} = \frac\pi2 - \theta_{ABC}$, the area can be rewritten as $L^2 - L^2\sin\theta_{ABC}$.

As $S$ is the sum of the areas of $ABCD$ and $EFGH$, then $S = L^2$, thus $L = \sqrt S$.


A hint to the final answer could be obtained by considering the two degenerative cases: $\theta_{ABC} = \frac\pi2$ and $\theta_{ABC} = \frac\pi4$ (we can discount $\theta_{ABC} < \frac\pi4$, because then the points of $EFGH$ would lie outside the rhombus, although the result remains valid so long as the relative positions of where the points of $EFGH$ go are specified differently than just "inside").

For $\theta_{ABC} = \frac\pi2$, it is clear $ABCD$ is a square, and $EFGH$ collapses to a point, so $L = \sqrt S$.

For $\theta_{ABC} = \frac\pi4$, it is easier to see that $EFGH$ is a square and how to compute the length of the sides, since the sides of the interior right triangles are colinear with the sides of the outer rhombus.

enter image description here

A rigorous explanation of why $EFGH$ must be a square is left as an exercise.

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  • $\begingroup$ I pretty much get everything, but where does the length of the side of the square come from? $\endgroup$ – bigbenmetalhead Apr 16 '15 at 21:54
  • $\begingroup$ Note that $2 BE^2 = L^2$. With $BD$ as the perpendicular bisector of $EH$, we have $\sin\frac{\theta}2 = \frac{EH/2}{BE}$, which means $EH = 2 BE \sin\frac{\theta}2$. $\endgroup$ – jxh Apr 29 '15 at 22:13

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