Maths Olympiad Prep

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Geometry Difficulty 4.7 AIME Find the answer

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 19\sqrt{19}. Find the distance from the ball's point of first contact with a wall to the nearest vertex.

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

Solution

Consider the diagram above, where MM is the midpoint of BCBC. Then AMAM is perpendicular to BCBC since ABCABC is equilateral, so by the Pythagorean theorem AM=532AM = \frac{5 \sqrt{3}}{2}. Then, using the Pythagorean theorem again, we see that MY=12MY = \frac{1}{2}, so that BY=2BY = 2.

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