Problem:
Triangle AB0​C0​ has side lengths AB0​=12,B0​C0​=17, and C0​A=25. For each positive integer n, points Bn​ and Cn​ are located on ABn−1​​ and ACn−1​​, respectively, creating three similar triangles △ABn​Cn​∼△Bn−1​Cn​Cn−1​∼ △ABn−1​Cn−1​. The area of the union of all triangles Bn−1​Cn​Bn​ 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 △AB0​C0​ is 90. Triangle B0​C1​C0​ is similar to △AB0​C0​ with ratio of similarity r=AC0​B0​C0​​=2517​, and △AB1​C1​ is similar to △AB0​C0​ with ratio of similarity AC0​AC1​​=2525−17r​=1−r2. Therefore the area of △B0​C1​B1​ is 90(1−r2−(1−r2)2). For n>1, the ratio of similarity between △ABn​Cn​ and △ABn−1​Cn−1​ is also 1−r2, so the ratio of their areas is (1−r2)2. The ratio of the area of △Bn−1​Cn​Bn​ to that of △Bn−2​Cn−1​Bn−1​ is also (1−r2)2. Hence the areas of the triangles Bn−1​Cn​Bn​ 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