Question

Consider a square $ABCD$ with midpoints $E, F, G, H$ of $AB, BC, CD$ and $DA$ respectively. Let $L$ denote the line passing through $F$ and $H$. Consider points $P$ and $Q$, on $L$ and inside $ABCD$, such that the angles $\angle APD$ and $\angle BQC$ both equal $120^\circ$. What is the ratio of the area of $ABQCDP$ to the remaining area inside $ABCD$?

Accepted Answer

E. Let the side of square $ABCD$ be $a$. Area of square $= a^2$. In $	riangle APD$, $AP = PD = x$. $\angle APD = 120^\circ$. Using Sine Rule in $	riangle APD$: $$\frac{a}{\sin 120^\circ} = \frac{x}{\sin 30^\circ} \implies x = \frac{a}{\sqrt{3}}$$ $$	ext{Area}(	riangle APD) = \frac{1}{2} x^2 \sin 120^\circ = \frac{1}{2} \left(\frac{a^2}{3}ight) \frac{\sqrt{3}}{2} = \frac{a^2}{4\sqrt{3}}$$ By symmetry, $	ext{Area}(	riangle BQC) = 	ext{Area}(	riangle APD) = \frac{a^2}{4\sqrt{3}}$. Area of region $ABQCDP = a^2 - 2 	imes 	ext{Area}(	riangle APD) = a^2 - \frac{a^2}{2\sqrt{3}} = a^2\left(1 - \frac{1}{2\sqrt{3}}ight)$. Remaining area inside $ABCD = 2 	imes 	ext{Area}(	riangle APD) = \frac{a^2}{2\sqrt{3}}$. $$	ext{Ratio} = \frac{a^2\left(1 - \frac{1}{2\sqrt{3}}ight)}{\frac{a^2}{2\sqrt{3}}} = 2\sqrt{3} - 1$$.

Area Ratio in Square with Intersecting Angles

Explanation

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