Problem:
In triangle ABC, point D is on BC with CD=2 and DB=5, point E is on AC with CE=1 and EA=3,AB=8, and AD and BE intersect at P. Points Q and R lie on AB so that PQ is parallel to CA and PR is parallel to CB. It is given that the ratio of the area of triangle PQR to the area of triangle ABC is m/n, where m and n are relatively prime positive integers. Find m+n.
Solution:
Draw the line through E that is parallel to AD, and let K be its intersection with BC. Because CD=2 and KC:KD=EC:EA=1:3, it follows that KD=3/2. Therefore,
AEQP=BEBP=BKBD=5+(3/2)5=1310
Thus
ACQP=43⋅1310=2615
Since triangles PQR and CAB are similar, the ratio of their areas is (15/26)2=225/676. Thus m+n=901.
OR
Use mass points. Assign a mass of 15 to C. Since AE=3⋅EC, the mass at C must be 3 times the mass at A, so the mass at A is 5, and the mass at E is 15+5=20. Similarly, the mass at B is (2/5)⋅15=6, so the mass at D is 15+6=21, and the mass at P is 6+20=26. Draw CP and let it intersect AB at F. The mass at F is 26−15=11, so PF/CF=15/26, and the ratio of the areas is (15/26)2=225/676.
Remark: Research mass points to find out more about this powerful method of solving problems involving geometric ratios.
