jmc

Question

In triangle ABC, AB=13, BC=15 and CA=17. Point D is on AB, E is on BC, and F is on CA. Let AD=p AB, BE=q BC, and CF=r CA, where p, q, and r are positive and satisfy p+q+r=2/3 and p^2+q^2+r^2=2/5. The ratio of the area of triangle DEF to the area of triangle ABC can be written in the form m/n, where m and n are relatively prime positive integers. Find m+n.

Accepted Answer

FIGURE We let [] denote area; then the desired value is mn = [DEF][ABC] = [ABC] - [ADF] - [BDE] - [CEF][ABC] Using the formula for the area of a triangle 12ab C, we find that [ADF][ABC] = 12 p AB (1-r) AC CAB 12 AB AC CAB = p(1-r) and similarly that [BDE][ABC] = q(1-p) and [CEF][ABC] = r(1-q). Thus, we wish to findalign*[DEF][ABC] &= 1 - [ADF][ABC] - [BDE][ABC] - [CEF][ABC] &= 1 - p(1-r) - q(1-p) - r(1-q) &= (pq + qr + rp) - (p + q + r) + 1 align*We know that p + q + r = 23, and also that (p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) pq + qr + rp = ( 23)^2 - 252 = 145. Substituting, the answer is 145 - 23 + 1 = 1645, and m+n = 61.