Difference between revisions of "1953 AHSME Problems/Problem 50"
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== Solution == | == Solution == | ||
| − | Let the triangle | + | Let the triangle have side lengths <math>14, 6+x,</math> and <math>8+x</math>. The area of this triangle can be computed two ways. We have <math>A = rs</math>, and <math>A = \sqrt{s(s-a)(s-b)(s-c)}</math>, where <math>s = 14+x</math> is the semiperimeter. Therefore, <math>4(14+x)=\sqrt{(14+x)(x)(8)(6)}</math>. Solving gives <math>x = 7</math> as the only valid solution. This triangle has sides <math>13,14</math> and <math>15</math>, so the shortest side is <math>\boxed{\textbf{(B) \ } 13 \mathrm{\ units}}</math>. |
== See Also == | == See Also == | ||
{{AHSME 50p box|year=1953|num-b=49|after=Last Question}} | {{AHSME 50p box|year=1953|num-b=49|after=Last Question}} | ||
Revision as of 11:31, 6 August 2012
Problem
One of the sides of a triangle is divided into segments of 
and 
units by the point of tangency of the inscribed circle. If the radius of the circle is 
, then the length of the shortest side is
Solution
Let the triangle have side lengths 
and 
. The area of this triangle can be computed two ways. We have 
, and 
, where 
is the semiperimeter. Therefore, 
. Solving gives 
as the only valid solution. This triangle has sides 
and 
, so the shortest side is 
.
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 49 |
Followed by Last Question | |
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| All AHSME Problems and Solutions | ||
