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 of area ABCQPD (i.e. area of hexagon A-B-C-Q-P-D) to the remaining area of ABCD?

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

Hint:

P and Q lie on the perpendicular bisectors of AD and BC; use the 120 degree apex angle to find how far each one cuts into the square.

Answer 1

The Correct Option is B

Step 1: Set up coordinates for the square.
Let the square ABCD have side 1, with A = (0, 0), B = (1, 0), C = (1, 1) and D = (0, 1), so AB, BC, CD and DA are the four sides in order. Then H, the midpoint of DA, is (0, 0.5) and F, the midpoint of BC, is (1, 0.5). The line L through F and H is the horizontal line y = 0.5, which is also the perpendicular bisector of both AD and BC.
Step 2: Locate P using the 120 degree angle condition.
Let P = (p, 0.5) with 0 < p < 0.5, so P lies on segment HF closer to the A-D side. Because P sits on the perpendicular bisector of AD, PA = PD, so triangle APD is isosceles with base AD = 1. By the law of cosines applied to angle APD = 120 degrees: AD^2 = PA^2 + PD^2 - 2 PA PD cos120 = 3PA^2. Since AD = 1, this gives PA^2 = 1/3, so PA = 1/sqrt(3).
Step 3: Convert PA back into the coordinate p.
Since PA^2 = p^2 + 0.25, setting this equal to 1/3 gives p^2 = 1/3 - 1/4 = 1/12, so p = 1/(2sqrt3) = sqrt(3)/6. By the same argument applied to angle BQC = 120 degrees, Q = (1-p, 0.5) sits the mirror-image distance from the B-C side.
Step 4: Find the area of triangle APD, and by symmetry triangle BQC.
Triangle APD has base AD = 1 lying on the line x = 0, and its apex P is a horizontal distance p from that line, so its height is p. Its area is 1/2 x base x height = p/2 = sqrt(3)/12. Triangle BQC is congruent to triangle APD by the left-right symmetry of the figure, so it has the same area sqrt(3)/12.
Step 5: Compute the two areas and their ratio.
The hexagon ABCQPD is the square with these two corner triangles APD and BQC cut away, so its area is 1 - 2 x sqrt(3)/12 = 1 - sqrt(3)/6 = (6-sqrt3)/6. The remaining area of the square (the two removed triangles together) is 2 x sqrt(3)/12 = sqrt(3)/6. Dividing the two gives (6-sqrt3)/sqrt3 = 6/sqrt3 - 1 = 2sqrt3 - 1.
Step 6: Match with the options.
This equals option (2), 2sqrt3 - 1 ~ 2.46. None of the other three listed expressions equal this value, so option (2) is correct.