Problem:
In β³ABC,tan(β CAB)=22/7 and the altitude from A divides BC into segments of length 3 and 17. What is the area of β³ABC?
Solution:
Let x be the length of the altitude from A. Then
tanβ1(22/7)=tanβ1(3/x)+tanβ1(17/x)
Taking tangents of both sides, and using the formula for tan(Ξ±+Ξ²), we obtain.
722β=1β(3/x)(17/x)(3/x)+(17/x)β
which simplifies to
11x2β70xβ561=(xβ11)(11x+51)=0
Since x must be positive, we conclude that x=11, and that the area of β³ABC is (21β)(3+17)(11)=110β.
The problems on this page are the property of the MAA's American Mathematics Competitions
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