Question

Right triangle $ABC$ has side lengths $BC = 6,$ $AC = 8,$ and $AB = 10.$ A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A.$ A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B.$ What is $OP?$

$\dfrac{23}{8}$

$\dfrac{29}{10}$

$\dfrac{35}{12}$

$\dfrac{73}{25}$

$3$

Solution:

Place $C = (0, 0),$ $B = (6, 0),$ and $A = (0, 8),$ so the right angle is at $C.$

Circle $O$ is tangent to line $BC$ (the $x$ -axis) at $B,$ so $O = (6, k).$ Setting $OA = OB$ gives $36 + (k - 8)^2 = k^2,$ so $k = \tfrac{25}{4}$ and $O = \left(6, \tfrac{25}{4}\right).$

Circle $P$ is tangent to line $AC$ (the $y$ -axis) at $A,$ so $P = (h, 8).$ Setting $PB = PA$ gives $(h - 6)^2 + 64 = h^2,$ so $h = \tfrac{25}{3}$ and $P = \left(\tfrac{25}{3}, 8\right).$

Then $OP = \sqrt{\left(\tfrac73\right)^2 + \left(\tfrac74\right)^2} = 7\sqrt{\tfrac{25}{144}} = \dfrac{35}{12}.$

Thus, the correct answer is C.

2021 AMC 12B Fall Problem 22

Solution:

Problem 22 in Other Years