Question

In $\triangle PQR,$ $PR = 15,$ $QR = 20,$ and $PQ = 25.$ Points $A$ and $B$ lie on $\overline{PQ},$ points $C$ and $D$ lie on $\overline{QR},$ and points $E$ and $F$ lie on $\overline{PR},$ with $PA = QB = QC = RD = RE = PF = 5.$ Find the area of hexagon $ABCDEF.$

Answer

Solution:

Since $15^2 + 20^2 = 25^2,$ the triangle is right-angled at $R,$ and its area is $\frac{1}{2} \cdot 15 \cdot 20 = 150.$ Also $\sin P = \frac{20}{25} = \frac{4}{5}$ and $\sin Q = \frac{15}{25} = \frac{3}{5}.$

The hexagon is the triangle minus three corner triangles, each with two sides of length $5:$ at $P,$ area $\frac{1}{2} \cdot 5 \cdot 5 \cdot \frac{4}{5} = 10;$ at $Q,$ area $\frac{1}{2} \cdot 5 \cdot 5 \cdot \frac{3}{5} = \frac{15}{2};$ at $R,$ area $\frac{1}{2} \cdot 5 \cdot 5 = \frac{25}{2}.$

Therefore the hexagon has area $150 - 10 - \frac{15}{2} - \frac{25}{2} = 120.$

2019 AIME I Problem 3

Solution:

Problem 3 in Other Years