Difference between revisions of "1993 AHSME Problems/Problem 29"
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<math>b^2+c^2=j^2,</math> | <math>b^2+c^2=j^2,</math> | ||
<math>c^2+a^2=k^2</math> | <math>c^2+a^2=k^2</math> | ||
− | where <math>\{ i,j,k \}</math> are the given diagonal lengths. WLOG, let <math>i \leq j \leq k.</math> It suffices to check if <math>i+j \geq k.</math> We see that for <math>\boxed{B},</math> 4^2+5^2 = 16 + 25 = 41 < 7^2 = 49, so this case is impossible. | + | where <math>\{ i,j,k \}</math> are the given diagonal lengths. WLOG, let <math>i \leq j \leq k.</math> It suffices to check if <math>i^2+j^2 \geq k.</math> We see that for <math>\boxed{B},</math>4^2+5^2 = 16 + 25 = 41 < 7^2 = 49$, so this case is impossible. |
== See also == | == See also == |
Revision as of 23:12, 18 July 2020
Contents
[hide]Problem
Which of the following could NOT be the lengths of the external diagonals of a right regular prism [a "box"]? (An is a diagonal of one of the rectangular faces of the box.)
Solution
Let and be the side lengths of the rectangular prism. By Pythagoras, the lengths of the external diagonals are and If we square each of these to obtain and we observe that since each of and are positive, then the sum of any two of the squared diagonal lengths must be larger than the square of the third diagonal length. For example, because
Thus, we test each answer choice to see if the sum of the squares of the two smaller numbers is larger than the square of the largest number. Looking at choice (B), we see that so the answer is
--goldentail141
Solution 2
Let and be the side lengths of the rectangular prism. We then have: where are the given diagonal lengths. WLOG, let It suffices to check if We see that for 4^2+5^2 = 16 + 25 = 41 < 7^2 = 49$, so this case is impossible.
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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All AHSME Problems and Solutions |
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