Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Find the answer

4. Given a right triangle ABCABC with a right angle at AA. On the leg ACAC, a point DD is marked such that AD:DC=1:3AD: DC = 1: 3. Circles Γ1\Gamma_{1} and Γ2\Gamma_{2} are then constructed with centers at AA and CC respectively, passing through point DD. Γ2\Gamma_{2} intersects the hypotenuse at point EE. Circle Γ3\Gamma_{3} with center BB and radius BEBE intersects Γ1\Gamma_{1} inside the triangle at a point FF such that angle AFBAFB is a right angle. Find BCBC if AB=5AB = 5.

(P. D. Mulyenko)

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

Solution

Answer: 13.

Solution. Let AC=xAC = x. Then AD=x/4,DC=CE=3x/4,BE=BCCE=x2+253x/4AD = x / 4, DC = CE = 3x / 4, BE = BC - CE = \sqrt{x^2 + 25} - 3x / 4. According to the problem, AFB=90\angle AFB = 90^\circ, so AF2+FB2=AB2AF^2 + FB^2 = AB^2, which means AD2+BE2=25AD^2 + BE^2 = 25. Expressing everything in terms of xx and simplifying, we get 13x=12x2+2513x = 12 \sqrt{x^2 + 25}. Squaring both sides, we obtain x2=144x^2 = 144, so x=12x = 12 and

!
BC=13BC = 13.

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