1988 AHSME Problems/Problem 23

Problem

The six edges of a tetrahedron $ABCD$ measure $7, 13, 18, 27, 36$ and $41$ units. If the length of edge $AB$ is $41$ , then the length of edge $CD$ is

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

Solution

By the triangle inequality in $\triangle ABC$ , we find that $BC$ and $CA$ must sum to greater than $41$ , so they must be (in some order) $7$ and $36$ , $13$ and $36$ , $18$ and $27$ , $18$ and $36$ , or $27$ and $36$ . We try $7$ and $36$ , and now by the triangle inequality in $\triangle ABD$ , we must use the remaining numbers $13$ , $18$ , and $27$ to get a sum greater than $41$ , so the only possibility is $18$ and $27$ . This works as we can put $BC = 36$ , $AC = 7$ , $AD = 18$ , $BD = 27$ , $CD = 13$ , so that $\triangle ADC$ and $\triangle BDC$ also satisfy the triangle inequality. Hence we have found a solution that works, and it can be verified that the other possibilities don't work, though as this is a multiple-choice competition, you probably wouldn't do that in order to save time. In any case, the answer is $CD = 13$ , which is $\boxed{\text{B}}$ .

1988 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1988 AHSME Problems/Problem 23

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