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Difference between revisions of "1988 AHSME Problems/Problem 23"

(Created page with "==Problem== The six edges of a tetrahedron <math>ABCD</math> measure <math>7, 13, 18, 27, 36</math> and <math>41</math> units. If the length of edge <math>AB</math> is <math>41<...")
 
(Added a solution with explanation)
 
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\textbf{(D)}\ 27\qquad
 
\textbf{(D)}\ 27\qquad
 
\textbf{(E)}\ 36  </math>   
 
\textbf{(E)}\ 36  </math>   
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\textbf{(E)}\ \text{more than } 7  $ 
 
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==Solution==
 
==Solution==
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By the triangle inequality in <math>\triangle ABC</math>, we find that <math>BC</math> and <math>CA</math> must sum to greater than <math>41</math>, so they must be (in some order) <math>7</math> and <math>36</math>, <math>13</math> and <math>36</math>, <math>18</math> and <math>27</math>, <math>18</math> and <math>36</math>, or <math>27</math> and <math>36</math>. We try <math>7</math> and <math>36</math>, and now by the triangle inequality in <math>\triangle ABD</math>, we must use the remaining numbers <math>13</math>, <math>18</math>, and <math>27</math> to get a sum greater than <math>41</math>, so the only possibility is <math>18</math> and <math>27</math>. This works as we can put <math>BC = 36</math>, <math>AC = 7</math>, <math>AD = 18</math>, <math>BD = 27</math>, <math>CD = 13</math>, so that <math>\triangle ADC</math> and <math>\triangle BDC</math> 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 <math>CD = 13</math>, which is <math>\boxed{\text{B}}</math>.
  
  

Latest revision as of 13:20, 27 February 2018

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 $\boxed{\text{B}}$.


See also

1988 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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