Problem:
Triangle AB0C0 has side lengths AB0=12,B0C0=17, and C0A=25. For each positive integer n, points Bn and Cn are located on ABn−1 and ACn−1, respectively, creating three similar triangles △ABnCn∼△Bn−1CnCn−1∼ △ABn−1Cn−1. The area of the union of all triangles Bn−1CnBn for n≥1 can be expressed as qp, where p and q are relatively prime positive integers. Find q.
Solution:
By Heron's Formula, the area of △AB0C0 is 90. Triangle B0C1C0 is similar to △AB0C0 with ratio of similarity r=AC0B0C0=2517, and △AB1C1 is similar to △AB0C0 with ratio of similarity AC0AC1=2525−17r=1−r2. Therefore the area of △B0C1B1 is 90(1−r2−(1−r2)2). For n>1, the ratio of similarity between △ABnCn and △ABn−1Cn−1 is also 1−r2, so the ratio of their areas is (1−r2)2. The ratio of the area of △Bn−1CnBn to that of △Bn−2Cn−1Bn−1 is also (1−r2)2. Hence the areas of the triangles Bn−1CnBn for n≥1 form a geometric sequence with initial term 90r2(1−r2) and ratio (1−r2)2. Because the triangles have disjoint interiors, the area of their union is the sum of the series, which is
1−(1−r2)290r2(1−r2)=90(1−2−r21)=90(1−961625)=96190⋅336
Thus q=961.
The problems on this page are the property of the MAA's American Mathematics Competitions