Difference between revisions of "1988 AHSME Problems/Problem 23"
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\textbf{(D)}\ 27\qquad | \textbf{(D)}\ 27\qquad | ||
\textbf{(E)}\ 36 </math> | \textbf{(E)}\ 36 </math> | ||
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==Solution== | ==Solution== | ||
− | + | 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 14:20, 27 February 2018
Problem
The six edges of a tetrahedron measure
and
units. If the length of edge
is
, then the length of edge
is
Solution
By the triangle inequality in , we find that
and
must sum to greater than
, so they must be (in some order)
and
,
and
,
and
,
and
, or
and
. We try
and
, and now by the triangle inequality in
, we must use the remaining numbers
,
, and
to get a sum greater than
, so the only possibility is
and
. This works as we can put
,
,
,
,
, so that
and
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
, which is
.
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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