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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 APD and BQC both equal 120$^o$. What is the ratio of the area of ABQCDP to the remaining area inside ABCD?

1. $\frac{4 \sqrt{2} } {3}$
2. $2 + \sqrt{3}$
3. $\frac{10-3 \sqrt{3} } {9}$
4. $1+ \frac{1}{\sqrt{3} }$
5. $2 \sqrt{3} -1$
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+1 vote

$5^{th}$ option must be the right one.

Here is the diagram of square $ABCD$

Consider the side length of the square be $a$ units.

Then $AH = BF = \frac{a}{2}$

From the rectangle $ABFH$,

$\angle AHP = \angle BFQ = 90^{\circ}.$

Now since $HF$ is the perpendicular bisector of $DA$ and $BC$,

$\angle APH = \frac{1}{2}\angle APD = 60^{\circ}$ and

$\angle BQF = \frac{1}{2}\angle BQC = 60^{\circ}$.

Hence $\angle PAH = \angle QBF = 30^{\circ}.$

Area of $\Delta APD =$ Area of $\Delta BQC = 2\left ( \frac{1}{2}\times \frac{a}{2} \times \frac{a}{2} \tan\left ( 30^{\circ} \right ) \right )$

The area of region $ABQCDP$ = Area of square $ABCD -$ (Area of triangle $APD$ $+$ Area of triangle $BQC$).

so area of the region  $ABQCDP$

$= a^{2} - \left ( 2\left ( \frac{1}{2}\times \frac{a}{2} \times \frac{a}{2} \tan\left ( 30^{\circ} \right ) \right ) + 2\left ( \frac{1}{2}\times \frac{a}{2} \times \frac{a}{2} \tan\left ( 30^{\circ} \right ) \right ) \right )$

$= a^{2} - \left ( \frac{a^{2}}{2\sqrt3}\right ).$

$= a^{2}\left ( 1 - \frac{1}{2\sqrt3} \right )$

Area of the region left after removing the area of $ABQCDP$

$= a^{2} - a^{2}\left ( 1 - \frac{1}{2\sqrt3} \right ) =\frac{a^{2}}{2\sqrt3}$.

Hence the required ratio $= \frac{a^{2}\left ( 1 - \frac{1}{2\sqrt3} \right )}{\frac{a^{2}}{2\sqrt3}}.$

On simplifying, this will give $\left ( 2\sqrt3 - 1 \right ).$

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