# 1996 AJHSME Problems/Problem 8

## Problem

Points $A$ and $B$ are 10 units apart. Points $B$ and $C$ are 4 units apart. Points $C$ and $D$ are 3 units apart. If $A$ and $D$ are as close as possible, then the number of units between them is $\text{(A)}\ 0 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 11 \qquad \text{(E)}\ 17$

## Solution

If $AB = 10$ and $BC=4$, then $(10 - 4) \le AC \le (10 + 4)$ by the triangle inequality. In the triangle inequality, the equality is only reached when the "triangle" $ABC$ is really a degenerate triangle, and $ABC$ are collinear.

Simplifying, this means the smallest value $AC$ can be is $6$.

Applying the triangle inequality on $ACD$ with $AC = 6$ and $CD = 3$, we know that $6 - 3 \le AD \le 6 + 3$ when $AC$ is minimized. If $AC$ were larger, then $AD$ could be larger, but we want the smallest $AD$ possible, and not the largest. Thus, $AD$ must be at least $3$, but cannot be smaller than $3$. Therefore, $\boxed{B}$ is the answer.

This answer comes when $ABCD$ are all on a line, with $A=0, B=10, C = 6,$ and $D=3$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 