Question:
Square \(ABCD\) has midpoints \(E, F, G, H\) of sides \(AB, BC, CD, DA\) respectively. Let \(L\) be the line through \(F\) and \(H\). Points \(P, Q\) are on \(L\) inside \(ABCD\) such that \(\angle APD = \angle BQC = 120^\circ\). What is the ratio of area of \(ABQCDP\) to the remaining area?
Square \(ABCD\) has midpoints \(E, F, G, H\) of sides \(AB, BC, CD, DA\) respectively. Let \(L\) be the line through \(F\) and \(H\). Points \(P, Q\) are on \(L\) inside \(ABCD\) such that \(\angle APD = \angle BQC = 120^\circ\). What is the ratio of area of \(ABQCDP\) to the remaining area?
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For complex geometry ratio problems, coordinate geometry often simplifies calculation.
Updated On: Jul 30, 2025
- \(\frac{4\sqrt{2}}{3}\)
- \(2 + \sqrt{3}\)
- \(\frac{10 - 3\sqrt{3}}{9}\)
- \(1 + \frac1{\sqrt{3}}\)
- \(2\sqrt{3} - 1\)
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The Correct Option is C
Solution and Explanation
By coordinate geometry, place square of side \(2\), locate midpoints, find intersections.
Calculate polygon \(ABQCDP\) area and compare to remainder.
The ratio simplifies to \(\frac{10 - 3\sqrt{3}}{9}\).
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