1978 AHSME Problems/Problem 26
Problem
In and . Circle is the circle with smallest radius which passes through and is tangent to . Let and be the points of intersection, distinct from , of circle with sides and , respectively. The length of segment is
Solution
We know that triangle is similar to triangle . We draw a line to point on hypotenuse such that is and that is a rectangle. Since triangle is similar to triangle , let be and be . Now we have line segment = , and line segment = . Since , we use simple algebra and Pythagorean Theorem to get + = . Expanding and simplifying gives us + = .
Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by . Now, we can square both sides and simplify to get . Dividing both sides by , we get = . We then add to both sides to get . Since this is very messy, let . Squaring both sides, we get . Solving for , we have . Plugging in as , we have . Using the quadratic equation, we get . Therefore, .
Remember that our desired answer is the hypotenuse of the triangle . Since is the hypotenuse, our answer is
~Arcticturn
See Also
1978 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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