Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1993 AHSME Problems/Problem 29"

m (→‎Solution)
(→‎Solution 2)
 
(4 intermediate revisions by 3 users not shown)
Line 16: Line 16:
  
 
--goldentail141
 
--goldentail141
 +
 +
==Solution 2==
 +
 +
Let <math>a,</math> <math>b,</math> and <math>c</math> be the side lengths of the rectangular prism. We then have:
 +
<math>a^2+b^2=i^2,</math>
 +
<math>b^2+c^2=j^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^2+j^2 \geq k.</math> We see that for <math>\boxed{B}</math>, <math>4^2+5^2 = 16 + 25 = 41 < 7^2 = 49</math>, so this case is impossible.
  
 
== See also ==
 
== See also ==

Latest revision as of 14:07, 6 July 2022

Problem

Which of the following could NOT be the lengths of the external diagonals of a right regular prism [a "box"]? (An $\textit{external diagonal}$ is a diagonal of one of the rectangular faces of the box.)

$\text{(A) }\{4,5,6\} \quad \text{(B) } \{4,5,7\} \quad \text{(C) } \{4,6,7\} \quad \text{(D) } \{5,6,7\} \quad \text{(E) } \{5,7,8\}$

Solution

Let $a,$ $b,$ and $c$ be the side lengths of the rectangular prism. By Pythagoras, the lengths of the external diagonals are $\sqrt{a^2 + b^2},$ $\sqrt{b^2 + c^2},$ and $\sqrt{a^2 + c^2}.$ If we square each of these to obtain $a^2 + b^2,$ $b^2 + c^2,$ and $a^2 + c^2,$ we observe that since each of $a,$ $b,$ and $c$ 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, $(a^2 + b^2) + (b^2 + c^2) = (a^2 + c^2) + 2b^2 > a^2 + c^2$ because $2b^2 > 0.$

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 $4^2 + 5^2 = 41 < 7^2 = 49,$ so the answer is $\boxed{\text{(B) \{4, 5, 7\}}}.$

--goldentail141

Solution 2

Let $a,$ $b,$ and $c$ be the side lengths of the rectangular prism. We then have: $a^2+b^2=i^2,$ $b^2+c^2=j^2,$ $c^2+a^2=k^2$ where $\{ i,j,k \}$ are the given diagonal lengths. WLOG, let $i \leq j \leq k.$ It suffices to check if $i^2+j^2 \geq k.$ We see that for $\boxed{B}$, $4^2+5^2 = 16 + 25 = 41 < 7^2 = 49$, so this case is impossible.

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_29&oldid=175674"