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 \(APD\) and \(BQC\) both equal \(120^\circ\). What is the ratio of the area of \(ABQCDP\) to the remaining area inside \(ABCD\)?

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

Hint:

Take the side of the square as 2. The line \(L\) is the horizontal midline, so \(P\) and \(Q\) are found from the angle \(120^\circ\). The hexagon \(ABQCDP\) is the square with two small triangles cut off.

Answer 1

The Correct Option is C

Step 1: Set up coordinates:
Let the side of the square be \(2\): \(A(0,0)\), \(B(2,0)\), \(C(2,2)\), \(D(0,2)\). Then \(H(0,1)\) and \(F(2,1)\), so \(L\) is the line \(y = 1\).

Step 2: Locate P:
Let \(P = (p, 1)\) with \(p > 0\). The line \(L\) is the perpendicular bisector of \(AD\), so \(\angle APD = 120^\circ\) means \(\angle APH = 60^\circ\), where \(AH = 1\). So \(\tan 60^\circ = \dfrac{AH}{PH} = \dfrac{1}{p}\), giving \(p = \dfrac{1}{\sqrt{3}}\).
By symmetry, \(Q = \left(2 - \dfrac{1}{\sqrt{3}}, 1\right)\).

Step 3: Identify the regions:
The hexagon \(ABQCDP\) is the square from which triangles \(APD\) and \(BQC\) have been removed. (\(P\) and \(Q\) lie inside the square, so these two triangles are the "remaining area".)

Step 4: Compute the areas:
Area of triangle \(APD = \tfrac{1}{2} \times AD \times PH = \tfrac{1}{2} \times 2 \times \dfrac{1}{\sqrt{3}} = \dfrac{1}{\sqrt{3}}\). The triangle \(BQC\) has the same area.
Remaining area \(= \dfrac{2}{\sqrt{3}}\). Area of \(ABQCDP = 4 - \dfrac{2}{\sqrt{3}}\).

Step 5: Ratio:
\(\dfrac{4 - 2/\sqrt{3}}{2/\sqrt{3}} = \dfrac{4\sqrt{3}}{2} - 1 = 2\sqrt{3} - 1\).

Final Answer:
The correct option is (3): \(2\sqrt{3} - 1\).