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 $HD:HA?$

$3:11$

$5:11$

$1:2$

$2:3$

$25:33$

Answer: B

Concepts:altitude coordinate geometry Heron’s Formula

Difficulty rating: 1850

Small Hint:

The $13$ - $14$ - $15$ triangle has area $84$ , so the altitude to side $14$ is $12$

Big Hint:

The altitude foot divides the side of length $14$ into segments $5$ and $9$ ; use coordinates to locate $H$

Solution:

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.

1964 AMC 12 Problem 35

Answer: B

Difficulty rating: 1850

Small Hint:

Big Hint:

Solution:

Problem 35 in Other Years