1953 AHSME Problems/Problem 50
Contents
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 1
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 .
Solution 2
Label the tangent points on as respectively. Let , , and The problem is a matter of solving for . To this, we use the fact that if are the angles of a triangle, then We know that , , and so we have the equation Solving this equation yields , so the shortest side has length .
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See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 49 |
Followed by Last Question | |
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