2013 AIME I Problems/Problem 13

Problem 13

Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ , $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$ . The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $q$ .

Solution

Using Heron's Formula we can get the area of the triangle $\Delta AB_0C_0 = 90$ . Since $\Delta AB_0C_0 \sim \Delta B_0C_1C_0$ then the scale factor for the dimensions of $\Delta B_0C_1C_0$ to $\Delta AB_0C_0$ is $\dfrac{17}{25}.$ Therefore, the area of $\Delta B_0C_1C_0$ is $(\dfrac{17}{25})^2(90)$ . Also, the dimensions of the other sides of the $\Delta B_0C_1C_0$ can be easily computed: $\overline{B_0C_1}= \dfrac{17}{25}(12)$ and $\overline{C_1C_0} = \dfrac{17^2}{25}$ . This allows us to compute one side of the triangle $\Delta AB_0C_0$ , $\overline{AC_1} = 25 - \dfrac{17^2}{25} = \dfrac{25^2 - 17^2}{25}$ . Therefore, the scale factor $\Delta AB_1C_1$ to $\Delta AB_0C_0$ is $\dfrac{25^2 - 17^2}{25^2}$ , which yields the length of $\overline{B_1C_1}$ as $\dfrac{25^2 - 17^2}{25^2}(17)$ . Therefore, the scale factor for $\Delta B_1C_2C_1$ to $\Delta B_0C_1C_0$ is $\dfrac{25^2 - 17^2}{25^2}$ . Some more algebraic manipulation will show that $\Delta B_nC_{n+1}C_n$ to $\Delta B_{n-1}C_nC_{n-1}$ is still $\dfrac{25^2 - 17^2}{25^2}$ . Also, since the triangles are disjoint, the area of the union is the sum of the areas. Therefore, the area is the geometric series $\dfrac{90 \cdot 17^2}{25^2} \sum_{n=0}^{\infty} (\dfrac{25^2-17^2}{25^2})^2$ At this point, it may be wise to "simplify" $25^2 - 17^2 = (25-17)(25+17) = (8)(42) = 336$ . So the geometric series converges to $\dfrac{90 \cdot 17^2}{25^2} \dfrac{1}{1 - \dfrac{336^2}{625^2}} = \dfrac{90 \cdot 17^2}{25^2} \dfrac{625^2}{625^2 - 336^2}$ . Using the diffference of squares, we get $\dfrac{90 \cdot 17^2}{25^2}\dfrac{625^2}{(625 - 336)(625 + 336)}$ , which simplifies to $\dfrac{90 \cdot 17^2}{25^2} \dfrac{625^2}{(289)(961)}$ . Cancellling all common factors, we get the reduced fraction $= \dfrac{90 \cdot 25^2}{31^2}$ . So $\frac{p}{q}=1-\frac{90 \cdot 25^2}{31^2}=\frac{90 \cdot 336}{961}$ , yielding the answer $\fbox{961}$ .

2013 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2013 AIME I Problems/Problem 13

Problem 13

Solution

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