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.
