2015 AIME II Problem 7

Question

Triangle

ABC

has side lengths

AB = 12,

BC = 25,

and

CA = 17.

Rectangle

PQRS

has vertex

P

on

\overline{AB},

vertex

Q

on

\overline{AC},

and vertices

R

and

S

on

\overline{BC}.

In terms of the side length

PQ = w,

the area of

PQRS

can be expressed as the quadratic polynomial

\text{Area}(PQRS) = \alpha w - \beta \cdot w^2.

Then the coefficient

\beta = \frac{m}{n},

where

m

and

n

are relatively prime positive integers. Find

m + n.

Answer

Answer: 161

Concepts:similarity Heron’s Formula area

Difficulty rating: 2470

Small hint:

Heron’s formula gives area

90

for

ABC,

so the altitude from

A

to

\overline{BC}

is

\frac{36}{5}

Big hint:

Triangle

APQ

is similar to

ABC

with ratio

\frac{w}{25},

so the rectangle’s height is

\frac{36}{5}\left(1 - \frac{w}{25}\right)

Solution:

By Heron’s formula with

s = 27,

the area of

ABC

is

\sqrt{27 \cdot 2 \cdot 10 \cdot 15} = \sqrt{8100} = 90,

so the altitude from

A

to

\overline{BC}

has length

h = \frac{2 \cdot 90}{25} = \frac{36}{5}.

Since

\overline{PQ} \parallel \overline{BC},

triangle

APQ

is similar to

ABC

with ratio

\frac{w}{25},

so the distance from

A

down to line

PQ

is

\frac{w}{25}h,

and the rectangle’s height is

PS = h - \frac{w}{25}h.

The area is

\begin{aligned} &w \cdot h\left(1 - \frac{w}{25}\right) \\ &= \frac{36}{5}\,w - \frac{36}{125}\,w^2. \end{aligned}

Thus

\beta = \frac{36}{125},

and

m + n = 36 + 125 = 161.