Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Triangle ABCABC is an equilateral triangle with side length 11. Let X0,X1,X_{0}, X_{1}, \ldots be an infinite sequence of points such that the following conditions hold:

- X0X_{0} is the center of ABCABC.
- For all i0i \geq 0, X2i+1X_{2i+1} lies on segment ABAB and X2i+2X_{2i+2} lies on segment ACAC.
- For all i0i \geq 0, XiXi+1Xi+2=90\measuredangle X_{i} X_{i+1} X_{i+2} = 90^{\circ}.
- For all i1i \geq 1, Xi+2X_{i+2} lies in triangle AXiXi+1A X_{i} X_{i+1}.

Find the maximum possible value of i=0XiXi+1\sum_{i=0}^{\infty} |X_{i} X_{i+1}|, where PQ|PQ| is the length of line segment PQPQ.

Solution

Solution:

63\boxed{\sqrt{\dfrac{\sqrt{6}}{3}}}

Let YY be the foot of the perpendicular from AA to X0X1X_{0} X_{1}: note that the sum we wish to maximize is simply X0Y+YAX_{0}Y + YA. However, it is not difficult to check (for example, by AM-GM) that AY+YX02AX0=63AY + YX_{0} \geq \sqrt{2} \cdot AX_{0} = \dfrac{\sqrt{6}}{3}. This may be achieved by making YX0A=45\angle Y X_{0} A = 45^{\circ}, so that AX1X0=105\angle A X_{1} X_{0} = 105^{\circ}.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.