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Geometry Difficulty 5.4 AIME, harder Find the answer

9. (16th Russian Mathematical Olympiad) Given a circle and a point MM inside it, consider all possible rectangles MKTPM K T P, with vertices K,PK, P on the circle. Find the locus of point TT.

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9. (16th Russian Mathematical Olympiad) Given a circle and a point MM inside it, consider all possible rectangles MKTPM K T P, with vertices K,PK, P on the circle. Find the locus of point TT.

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Solution

9. Let the equation of the circle be x2+y2=r2x^{2}+y^{2}=r^{2}, and the points be M(a,b),K(x1,y1),P(x2,y2),T(x,y)M(a, b), K\left(x_{1}, y_{1}\right), P\left(x_{2}, y_{2}\right), T(x, y), then a2+b2<r2,x+a2=a^{2}+b^{2}<r^{2}, \frac{x+a}{2}= x1+y22(1),y+b2=y1+y22(2),(xa)2+(yb)2=(x1x2)2+(y1y2)2\frac{x_{1}+y_{2}}{2}(1), \frac{y+b}{2}=\frac{y_{1}+y_{2}}{2}(2),(x-a)^{2}+(y-b)^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}, squaring (1) and (2) and using xi2+yi2=x_{i}^{2}+y_{i}^{2}= r2(i=1,2)r^{2}(i=1,2) we get (x+a)2+(y+b)2=2r2+2x1x2+2y1y2(x+a)^{2}+(y+b)^{2}=2 r^{2}+2 x_{1} x_{2}+2 y_{1} y_{2} (4), adding (3) and (4) yields x2+y2=2r2(a2+b2)x^{2}+y^{2}=2 r^{2}-\left(a^{2}+b^{2}\right), hence the required locus is a circle centered at (0,0)(0,0) with radius 2r2(a2+b2)\sqrt{2 r^{2}-\left(a^{2}+b^{2}\right)}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.