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)=αw−β⋅w2
Then the coefficient β=nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
By Heron's formula, the area of △ABC is 90. Then the altitude from A has length h=252⋅90. The altitude from A in △APQ has length BCPQh=25wh. It follows that PS=h−25wh, so
Area(PQRS)=PQ⋅PS=w(h−25wh)=hw−25hw2=hw−2522⋅90w2
and β=2522⋅90=12536. The requested sum is 36+125=161.
OR
Let f(w) denote the area of the rectangle of side w. Because f(0)=f(25)=0,
f(w)=αw−βw2=βw(25−w)
It is easy to check that if w=225, then Area(PQRS)=21⋅90=45. Therefore
45=f(225)=β⋅225(25−225)=4252β.
Hence
β=2524⋅45=625180=12536.
OR
The Law of Cosines can be used to calculate cos(∠ABC)=54 and cos(∠ACB)= 8577. Then tan(∠ABC)=43 and tan(∠ACB)=7736. Let h=PS. Then 25=BC=tan(∠ABC)h+w+tan(∠ACB)h, from which h=12536(25−w). Then the area of the rectangle is wh=536w−12536w2.
The problems on this page are the property of the MAA's American Mathematics Competitions