Difference between revisions of "1993 AHSME Problems/Problem 13"
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== Solution == | == Solution == | ||
| − | <math>\ | + | Assume one of the segments bisected by the inscribed square has length <math>x</math>. Thus, the alternate segment has length <math>7-x</math>. Applying Pythagorean's Theorem, <math>x^2+(x-7)^2=5^2</math>. Simplifying, <math>(x-3)(x-4)=0</math>, so <math>x=3</math> or <math>x=4</math> (it does not matter, as rotations produce the same figure). The longest line that can be made forms a right triangle with legs |
| + | of <math>4</math> and <math>7</math>. <math>\sqrt{4^2+7^2}=\sqrt{65} \rightarrow \boxed{D}</math> | ||
== See also == | == See also == | ||
Latest revision as of 20:42, 7 October 2018
Problem
A square of perimeter 20 is inscribed in a square of perimeter 28. What is the greatest distance between a vertex of the inner square and a vertex of the outer square?
Solution
Assume one of the segments bisected by the inscribed square has length 
. Thus, the alternate segment has length 
. Applying Pythagorean's Theorem, 
. Simplifying, 
, so 
or 
(it does not matter, as rotations produce the same figure). The longest line that can be made forms a right triangle with legs
of 
and 
. 
See also
| 1993 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
