Question

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 ABQCDP to the remaining area of ABCD?

• A. \(\frac {4\sqrt 2}{3}\)
• B. \(2\sqrt 3-1\)
• C. \(2+\sqrt 3\)
• D. \(\frac {10-3\sqrt 3}{9}\)

Answer 1

The Correct Option is B

A square \(ABCD\) has midpoints \(E, F, G, H\) on sides \(AB, BC, CD, DA\) respectively. Line \(L\) passes through \(F\) and \(H\). Points \(P\) and \(Q\) lie on \(L\) inside the square with \(\angle APD = \angle BQC = 120^\circ\). We need the ratio of area \(ABQCDP\) to the remaining area of \(ABCD\). 

Since \(L\) is the perpendicular bisector of \(BC\) and \(DA\), \(L\) divides \(ABCD\) into two equal halves. Thus, \(L\) is parallel to diagonal \(AC\). Consider \(P\) and \(Q\) such that triangles \(APD\) and \(BQC\) have \(\angle APD = 120^\circ\) and \(\angle BQC = 120^\circ\), making both triangles equilateral due to \(\angle DAP = \angle BQC = 60^\circ\).

To find the ratio of area \(ABQCDP\) to the remaining area, first find the area of triangle \(APD\) and triangle \(BQC\). If the side of \(ABCD\) is \(s\), the height of equilateral triangle \(APD\) from \(P\) to line \(AD\) is \(\frac{s\sqrt{3}}{2}\).

The area of triangle \(APD\) is:

\[\frac{\sqrt{3}}{4}s^2\]

The same calculation applies to triangle \(BQC\), as it is congruent to \(APD\).

The total area of both triangles is:

\[\frac{\sqrt{3}}{4}s^2 + \frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{2}s^2\]

The area of the square \(ABCD\) is:

\[s^2\]

Thus, the remaining area is:

\[s^2 - \frac{\sqrt{3}}{2}s^2 = \left(1-\frac{\sqrt{3}}{2}\right)s^2\]

The ratio \(R\) of the area \(ABQCDP\) to the remaining area is:

\[R = \frac{\frac{\sqrt{3}}{2}s^2}{\left(1-\frac{\sqrt{3}}{2}\right)s^2}\]

Simplifying \(R\):

\[R = \frac{\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}} = 2\sqrt{3} - 1\]