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Problem 925

AMC 12 late, AIME early

Geometry Difficulty 4.7 Find the answer HMMT November

Now a ball is launched from a vertex of an equilateral triangle with side length 5. It strikes the opposite side after traveling a distance of $\sqrt{19}$ . Find the distance from the ball's point of first contact with a wall to the nearest vertex.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

Consider the diagram above, where $M$ is the midpoint of $BC$ . Then $AM$ is perpendicular to $BC$ since $ABC$ is equilateral, so by the Pythagorean theorem $AM = \frac{5 \sqrt{3}}{2}$ . Then, using the Pythagorean theorem again, we see that $MY = \frac{1}{2}$ , so that $BY = 2$ .

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.