Consider a square ABCD with midpoints E, F, G and H of sides AB, BC, CD and DA. Let L denote the line passing through F and H. Consider points P and Q on the line L inside the square such that the angles APD and BQC are both equal 120 degrees. What is the ratio of area ABCQPD (i.e. area of hexagon A-B-C-Q-P-D) to the remaining area of ABCD?

Question

Adhir Mishra has three children: Urmila, Raghu and Sumit. Sumit married Roma, the eldest daughter of Mr. and Mrs. Mohan. The Mohans married their youngest daughter to the eldest son of Mr. and Mrs. Sharma and they had two children Sandeep and Shaifali. The Mohans have two more children, Roshan and Bimla, both elder to Sheila. Sohan and Shivendar are sons of Sumit and Roma. Leela is the daughter of Sandeep.

Answer 1

Take the square to have side length $a$ (the ratio will not depend on $a$, so choosing $a=1$ earlier was only a simplification, not a restriction). Let $M$ be the midpoint of $AD$ and $N$ the midpoint of $BC$; the line $L = MN$ passes through the center of the square and is parallel to $AB$ and $CD$.

Because $P$ lies on line $L$, and $L$ is the perpendicular bisector of $AD$, triangle $APD$ is isosceles with $PA = PD$. Call this common length $r$. The apex angle at $P$ is given as $120^\circ$, so the two base angles at $A$ and $D$ are each $\frac{180^\circ - 120^\circ}{2} = 30^\circ$.

Drop a perpendicular from $P$ to $AD$; since the triangle is isosceles, this perpendicular bisects $AD$ and bisects the apex angle into two $60^\circ$ halves. In the resulting right triangle, the half-base is $\frac{a}{2}$ and the angle at $P$ is $60^\circ$, so:

\[\tan 60^\circ = \frac{a/2}{h} \quad\Rightarrow\quad h = \frac{a/2}{\sqrt3} = \frac{a}{2\sqrt3}\]

where $h$ is the perpendicular distance from $P$ to side $AD$, which is exactly how deep triangle $APD$ cuts into the square. The area of triangle $APD$ is then:

\[[APD] = \frac12 \times a \times h = \frac12 \times a \times \frac{a}{2\sqrt3} = \frac{a^2}{4\sqrt3}\]

By the mirror symmetry of the construction, triangle $BQC$ has exactly the same area $\frac{a^2}{4\sqrt3}$. The two removed triangles together have area $\frac{a^2}{2\sqrt3} = \frac{a^2\sqrt3}{6}$.

The whole square has area $a^2$, so the hexagon $ABCQPD$, being the square minus these two triangles, has area $a^2 - \frac{a^2\sqrt3}{6} = a^2\cdot\frac{6-\sqrt3}{6}$.

The required ratio of the hexagon's area to the remaining (removed) area is:

\[\frac{a^2(6-\sqrt3)/6}{a^2\sqrt3/6} = \frac{6-\sqrt3}{\sqrt3} = \frac{6}{\sqrt3} - \frac{\sqrt3}{\sqrt3} = 2\sqrt3 - 1\]

The $a^2$ cancels out, confirming the ratio does not depend on the size of the square, exactly as it should not. This matches option (2), so the ratio is $\boxed{2\sqrt3 - 1}$.

Answer 2