Problem:
In △ABC,tan(∠CAB)=22/7 and the altitude from A divides BC into segments of length 3 and 17. What is the area of △ABC?
Solution:
Let x be the length of the altitude from A. Then
tan−1(22/7)=tan−1(3/x)+tan−1(17/x)
Taking tangents of both sides, and using the formula for tan(α+β), we obtain.
722=1−(3/x)(17/x)(3/x)+(17/x)
which simplifies to
11x2−70x−561=(x−11)(11x+51)=0
Since x must be positive, we conclude that x=11, and that the area of △ABC is (21)(3+17)(11)=110.
The problems on this page are the property of the MAA's American Mathematics Competitions
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