(A. Myakishev, 8--9) In the plane, given two concentric circles with the center $ A$. Let $ B$ be an arbitrary point on some of these circles, and $ C$ on the other one. For every triangle $ ABC$, consider two equal circles mutually tangent at the point $ K$, such that one of these circles is tangent to the line $ AB$ at point $ B$ and the other one is tangent to the line $ AC$ at point $ C$. Determine the locus of points $ K$.
Problem
Source: Sharygin contest 2008. The correspondence round. Problem 6
Tags: geometry proposed, geometry
Vo Duc Dien
03.08.2011 06:25
Locus is a circle with center A and radius of sqrt[(R^2 + r^2)/2].
tarzanjunior
20.12.2016 18:27
How did you determine the locus? Proof?
tarzanjunior
01.01.2017 19:55
Anyone???
vanstraelen
02.01.2017 20:11
Let $A(0,0)$; choose the x-axis through the given point $B$, then $B(r,0)$. Point $C(a,b)$ on the circle with radius $R$, so $a^{2}+b^{2}=R^{2} \quad\quad\quad (1)$. The first circle, touching $AB$ in $B$, has midpoint $M_{1}(b,n)$ and radius $n$. The second circle, touching $AC$ in $C$, has midpoint $M_{2}$ and also radius $n$. Equation of the line $CM_{2}\ :\ y-b=-\frac{a}{b}(x-a)$; equation of the circle, midpoint $C$ and radius $n\ :\ (x-a)^{2}+(y-b)^{2}=n^{2}$. Solving the system of this two equations, gives us $M_{2}(a+\frac{bn}{R},b-\frac{an}{R})$. It is possible to calculate $n$ from $M_{}M_{2}=2n$ or $(a+\frac{bn}{R}-r)^{2}+(b-\frac{an}{R}-n)^{2}=4n^{2} \quad\quad\quad (2)$. Point $K(x,y)$ is the midpoint of $M_{}M_{2}$: $a+\frac{bn}{R}+r=2x$ and $b-\frac{an}{R}+n=2y \quad\quad\quad (3)$. $a+\frac{bn}{R}=2x-r$ and $b-\frac{an}{R}=2y-n$. From (2): $(2x-r-r)^{2}+(2y-n-n)^{2}=4n^{2}$ and $n = \frac{(x-r)^{2}+y^{2}}{2y} \quad\quad\quad (4)$. Solving $a$ and $b$ from (3): $a = \frac{R(2Rx-2ny+n^{2}-rR)}{n^{2}+R^{2}}$ and $b=\frac{R(2nx+2Ry-nr-nR)}{n^{2}+R^{2}}$. Using (1), we have $4x^{2}-4rx+4y^{2}-4ny+r^{2}-R^{2}=0$. Using (4), we have $2x^{2}+2y^{2}-r^{2}-R^{2}=0$.