Question

Circles $\mathcal{C}_1,$ $\mathcal{C}_2,$ and $\mathcal{C}_3$ have their centers at $(0, 0),$ $(12, 0),$ and $(24, 0),$ and have radii $1,$ $2,$ and $4,$ respectively. Line $t_1$ is a common internal tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ and has a positive slope, and line $t_2$ is a common internal tangent to $\mathcal{C}_2$ and $\mathcal{C}_3$ and has a negative slope. Given that lines $t_1$ and $t_2$ intersect at $(x, y),$ and that $x = p - q\sqrt{r},$ where $p,$ $q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p + q + r.$

Answer

Solution:

A common internal tangent meets the segment between the centers at the point dividing it in the ratio of the radii. For $\mathcal{C}_1$ and $\mathcal{C}_2$ that point is $(4, 0),$ at distance $4$ from $(0,0).$ If $t_1$ makes angle $\theta$ with the $x$ -axis, then $\sin\theta = \frac{1}{4},$ so $\tan\theta = \frac{1}{\sqrt{15}}$ and $t_1$ is $y = \frac{1}{\sqrt{15}}(x - 4).$ For $\mathcal{C}_2$ and $\mathcal{C}_3$ the point is $(16, 0),$ at distance $4$ from $(12, 0);$ here $\sin\theta = \frac{2}{4} = \frac{1}{2},$ so the slope is $-\frac{1}{\sqrt{3}}$ and $t_2$ is $y = -\frac{1}{\sqrt{3}}(x - 16).$

Setting the two expressions equal and multiplying by $\sqrt{15}$ gives $x - 4 = -\sqrt{5}\,(x - 16),$ so $x(1 + \sqrt{5}) = 4 + 16\sqrt{5}$ and $x = \frac{4 + 16\sqrt{5}}{1 + \sqrt{5}} = \frac{(4 + 16\sqrt{5})(\sqrt{5} - 1)}{4} = \frac{76 - 12\sqrt{5}}{4} = 19 - 3\sqrt{5}.$

Thus $p + q + r = 19 + 3 + 5 = 27.$

2006 AIME II Problem 9

Solution:

Problem 9 in Other Years