1990 AMC 12 Problem 28
Attempt Problem 28 of the 1990 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1990 AMC 12 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
28.
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 divides that side into segments of lengths and Find
Answer: B
Small hint:
Assign one tangent length to each vertex, so adjacent pairs sum to the four side lengths
Big hint:
For supplementary opposite angles, the products of the tangent lengths at opposite vertices are equal
Solution:
Let the tangent lengths from the four consecutive vertices be Then Thus and If the inradius is a vertex with angle has tangent length Opposite angles are supplementary, so Therefore giving Hence the two segments of the -side are and whose difference is
Thus the correct answer is B.
Problem 28 in other years
- 1950 AMC 12
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