Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Find the answer Italy

Let $ABC$ be a right triangle with the right angle at $C$ , with sides $BC=3$ and $AB=12$ . Let $M$ be the midpoint of $AB$ , and $D$ the intersection between $AC$ and the circle circumscribed about $BCM$ . Finally, let $P$ be the point of intersection between $BC$ and $MD$ . What is the length of the segment $PA$ ?

Pick one

Solution

The answer is (C). Since $BCDM$ is cyclic, we have $\angle BMD = \angle BCD = 90^{\circ}$ . The triangles $ABC$ and $BMP$ are therefore similar, since they are right triangles sharing the angle at $B$ . We thus have $BM : BC = BP : BA$ , that is $6 : 3 = BP : 12$ , from which $BP = 24$ and $CP = 21$ . Finally, triangle $ACP$ is also a right triangle, and applying the Pythagorean theorem we get
$AP = \sqrt{CP^{2} + CA^{2}} = \sqrt{441 + (AB^{2} - BC^{2})} = \sqrt{576} = 24.$

Second solution: As in the first solution, we have $DM \perp AB$ , so $PM$ turns out to be the perpendicular bisector of segment $AB$ . In particular $ABP$ is isosceles, so $AP = BP = 24$ , as was computed in the first solution.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.