1990 AHSME Problems/Problem 28
Problem
A quadrilateral that has consecutive sides of lengths and is inscribed in a circle and also has a circle inscribed in it. The point of tangency of the inscribed circle to the side of length 130 divides that side into segments of length and . Find .
Solution
Let , , , and be the vertices of this quadrilateral such that , , , and . Let be the center of the incircle. Draw in the radii from the center of the incircle to the points of tangency. Let these points of tangency , , , and be on , , , and , respectively. Using the right angles and the fact that the is cyclic, we see that quadrilaterals and are similar.
Let have length . Chasing lengths, we find that . Using Brahmagupta's Formula we find that has area and from that we find, using that fact that , where is the inradius and is the semiperimeter, .
From the similarity we have Or, after cross multiplying and writing in terms of the variables, Plugging in the value of and solving the quadratic gives , and from there we compute the desired difference to get .
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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