2019 AIME I Problem 3

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

Since

15^2 + 20^2 = 25^2,

the triangle has a right angle at

R

Subtract the three corner triangles from

\triangle PQR;

each corner triangle has two sides of length

5

around a known angle, so its area is

\frac{1}{2} \cdot 25 \cdot \sin(\text{angle})

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.

Problem 3 in other years