1964 AMC 12 Problem 35

Question

The sides of a triangle are of lengths

13,

14,

and

15.

The altitudes of the triangle meet at point

H.

If

AD

is the altitude to the side of length

14,

what is the ratio

HD:HA?

3:11

5:11

1:2

2:3

25:33

The

13

-

14

-

15

triangle has area

84

, so the altitude to side

14

is

12

The altitude foot divides the side of length

14

into segments

5

and

9

; use coordinates to locate

H

Heron’s formula gives area

84,

so

AD=\frac{2(84)}{14}=12.

The adjacent

13

-side has projection

\sqrt{13^2-12^2}=5,

leaving

9

on the base. Put

D=(0,0),\ A=(0,12),

B=(-5,0),\ C=(9,0).

Line

AC

has slope

-\frac{4}{3},

so the altitude through

B

has slope

\frac{3}{4}

and meets

AD

at height

\frac{15}{4}.

Thus

\begin{gathered} HD=\frac{15}{4},\\ HA=12-\frac{15}{4}=\frac{33}{4}, \end{gathered}

giving

HD:HA=5:11.

Therefore, the correct answer is B.

Problem 35 in other years