My 10th problem

Geometry Level 5

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 $q$ .

The answer is 961.

1 solution

Johanz Piedad
Sep 16, 2015

By Heron's formula, the area of $AB_0C_0$ is $\sqrt{27(27-25)(27-17)(27-12)}=90$ . Now the area of the union divided by the area of $AB_0C_0$ is simply $\frac{[B_0C_1C_0]}{[B_0B_1C_1BC_0]}=\frac{[B_0C_1C_0]}{[B_0C_1C_0]+[B_0B_1C_1]}$ . Then it suffices to find $\frac{B_1C_1}{B_0C_0}$ , which is equal to $[B_0B_1C_1]/[B_0C_1C_0]$ . By POP, $C_0C_1\cdot 25=17^2$ , so $C_0C_1=\frac{17^2}{25}\Rightarrow AC_1=25-\frac{17^2}{25}=\frac{336}{25}$ . Then $\frac{B_1C_1}{B_0C_0}=\frac{AC_0}{AC_1}=\frac{336}{625}$ . Thus the area is $90\cdot \frac{625}{625+336}=90\cdot \frac{625}{961}$ Which can't be simplified, so the answer is $\boxed{961}$ .

My 10th problem

The answer is 961.

1 solution

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