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 $D=3$ .

1996 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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

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1996 AJHSME Problems/Problem 8

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