Question

Eight circles of radius $34$ can be placed tangent to $\overline{BC}$ of $\triangle ABC$ so that the circles are sequentially tangent to each other, with the first circle being tangent to $\overline{AB}$ and the last circle being tangent to $\overline{AC},$ as shown. Similarly, $2024$ circles of radius $1$ can be placed tangent to $\overline{BC}$ in the same manner. The inradius of $\triangle ABC$ can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Answer: 197

Concepts:tangent circles incircle, incenter, and inradius trigonometry

Difficulty rating: 2560

Solution:

For a chain of $n$ circles of radius $\rho$ tangent to $\overline{BC},$ the centers lie at height $\rho$ with consecutive centers $2\rho$ apart. The first circle is tangent to $\overline{AB}$ and $\overline{BC},$ so its center lies on the bisector from $B,$ at horizontal distance $\rho\cot\frac{B}{2}$ from $B;$ similarly the last center is $\rho\cot\frac{C}{2}$ from $C.$ Hence with $k = \cot\frac{B}{2} + \cot\frac{C}{2},$ $BC = \rho k + 2\rho(n - 1).$

The two chains give $34k + 34 \cdot 14 = BC$ $= k + 2 \cdot 2023,$ so $33k = 3570$ and $k = \frac{1190}{11},$ whence $BC = k + 4046 = \frac{45696}{11}.$

The incircle is a chain of one circle of radius $r:$ $BC = rk.$ Therefore $r = \frac{BC}{k} = \frac{45696}{1190} = \frac{192}{5},$ and $m + n = 192 + 5 = 197.$

2024 AIME I Problem 8

Answer: 197

Difficulty rating: 2560

Solution:

Problem 8 in Other Years