Problem:
Triangle ABC has side lengths AB=12,BC=25, and CA=17. Rectangle PQRS has vertex P on AB, vertex Q on AC, and vertices R and S on BC. In terms of the side length PQ=w, the area of PQRS can be expressed as the quadratic polynomial
Area(PQRS)=Ξ±wβΞ²β
w2
Then the coefficient Ξ²=nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
By Heron's formula, the area of β³ABC is 90. Then the altitude from A has length h=252β
90β. The altitude from A in β³APQ has length BCPQβh=25wβh. It follows that PS=hβ25wβh, so
Area(PQRS)=PQβ
PS=w(hβ25wβh)=hwβ25hβw2=hwβ2522β
90βw2
and Ξ²=2522β
90β=12536β. The requested sum is 36+125=161β.
OR
Let f(w) denote the area of the rectangle of side w. Because f(0)=f(25)=0,
f(w)=Ξ±wβΞ²w2=Ξ²w(25βw)
It is easy to check that if w=225β, then Area(PQRS)=21ββ
90=45. Therefore
45=f(225β)=Ξ²β
225β(25β225β)=4252βΞ².
Hence
Ξ²=2524β
45β=625180β=12536β.
OR
The Law of Cosines can be used to calculate cos(β ABC)=54β and cos(β ACB)= 8577β. Then tan(β ABC)=43β and tan(β ACB)=7736β. Let h=PS. Then 25=BC=tan(β ABC)hβ+w+tan(β ACB)hβ, from which h=12536(25βw)β. Then the area of the rectangle is wh=536βwβ12536βw2.
The problems on this page are the property of the MAA's American Mathematics Competitions