Track / Stage 5 / 86 of 400 #686 of 1964

Problem 686

AIME late

Geometry Difficulty 5.1 Find the answer

On a straight line, points $A, B$ , and $C$ are given. It is known that $A B=5$ , and segment $A C$ is one and a half times longer than $B C$ . Find the segments $A C$ and $B C$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Consider all possible cases of the arrangement of points $A, B$ , and $C$ .

## Solution

Let point $B$ lie on the segment $A C$ . Then $3 / 2 B C = A C = A B + B C = 5 + B C$ , from which $B C = 10, A C = 15$ .

Point $A$ cannot lie between points $B$ and $C$ , as in this case $A C < B C$ .

Let point $C$ lie between $A$ and $B$ . Then $\frac{5}{2} B C = A C + B C = A B = 5$ , from which $B C = 2, A C = 3$ .

## Answer

$B C = 10, A C = 15$ or $B C = 2, A C = 3$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.