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Geometry Difficulty 4.8 AIME Find the answer

Let ABCA B C be a triangle with AB=5,BC=4A B=5, B C=4, and CA=3C A=3. Initially, there is an ant at each vertex. The ants start walking at a rate of 1 unit per second, in the direction ABCAA \rightarrow B \rightarrow C \rightarrow A (so the ant starting at AA moves along ray AB\overrightarrow{A B}, etc.). For a positive real number tt less than 3, let A(t)A(t) be the area of the triangle whose vertices are the positions of the ants after tt seconds have elapsed. For what positive real number tt less than 3 is A(t)A(t) minimized?

A number or a short expression. Spacing and $ signs are ignored.

Solution

We instead maximize the area of the remaining triangles. This area (using 12xysinθ\frac{1}{2} x y \sin \theta ) is 12(t)(5t)35+12(t)(3t)45+12(t)(4t)1=110(12t2+47t)\frac{1}{2}(t)(5-t) \frac{3}{5}+\frac{1}{2}(t)(3-t) \frac{4}{5}+\frac{1}{2}(t)(4-t) 1=\frac{1}{10}\left(-12 t^{2}+47 t\right), which has a maximum at t=4724(0,3)t=\frac{47}{24} \in(0,3).

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