Maths Olympiad Prep

Library / /9 of 17

, 2021

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let OO and AA be two points in the plane with OA=30OA = 30, and let Γ\Gamma be a circle with center OO and radius rr. Suppose that there exist two points BB and CC on Γ\Gamma with ABC=90\angle ABC = 90^{\circ} and AB=BCAB = BC. Compute the minimum possible value of r\lfloor r \rfloor.

Solution

Solution:

Let f1f_1 denote a 4545^{\circ} counterclockwise rotation about point AA followed by a dilation centered at AA with scale factor 1/21/\sqrt{2}. Similarly, let f2f_2 denote a 4545^{\circ} clockwise rotation about point AA followed by a dilation centered at AA with scale factor 1/21/\sqrt{2}. For any point BB in the plane, there exists a point CC on Γ\Gamma such that ABC=90\angle ABC = 90^{\circ} and AB=BCAB = BC if and only if BB lies on f1(Γ)f_1(\Gamma) or f2(Γ)f_2(\Gamma). Thus, such points BB and CC on Γ\Gamma exist if and only if Γ\Gamma intersects f1(Γ)f_1(\Gamma) or f2(Γ)f_2(\Gamma). So, the minimum possible value of rr occurs when Γ\Gamma is tangent to f1(Γ)f_1(\Gamma) and f2(Γ)f_2(\Gamma). This happens when r/2+r=30/2r/\sqrt{2} + r = 30/\sqrt{2}, i.e., when r=302+1=30230r = \frac{30}{\sqrt{2} + 1} = 30\sqrt{2} - 30. Therefore, the minimum possible value of r\lfloor r \rfloor is 30230=12\lfloor 30\sqrt{2} - 30 \rfloor = 12.

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