MHB Problem of the Week # 221 - Jun 21, 2016

[Ackbach]

Here is this week's POTW:

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Let $C_1$ and $C_2$ be circles whose centers are 10 units apart, and whose radii are 1 and 3. Find, with proof, the locus of all points $M$ for which there exists points $X$ on $C_1$ and $Y$ on $C_2$ such that $M$ is the midpoint of the line segment $XY$.

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[Ackbach]

Re: Problem Of The Week # 221 - Jun 21, 2016

This was Problem A-2 in the 1996 William Lowell Putnam Mathematical Competition.

Congratulations to kiwi for his correct solution, which follows:

Spoiler

Let's decribe \(C_2\) as \(x_2=-5+3\cos\theta;y_2=3\sin\theta\) and \(C_1\) as \(x_1=5+\cos\phi;y_2=\sin\phi\). Where theta and phi range from zero to 2pi.

Now any point on the locus is located at the mean of X and Y. It is given by:

\(2x=3\cos \theta + \cos \phi\)
and
\(2y=3\sin \theta + \sin \phi\)

Therefore:
\(3\cos \theta = 2x - \cos \phi\) and \(3\sin \theta = 2y - \sin \phi\)

Squaring and adding these:
\(9=(2x-\cos \phi)^2+(2y-\sin \phi)^2=4x^2-x4\cos\phi+y^2-y4\sin\phi+1\)

So
\(2=x^2-x\cos\phi+y^2-y\sin\phi\)

Now changing to polar coordinates with angle variable gamma and requiring r to be non-negative:
\(2=r^2-r\cos\gamma\cos\phi-r\sin\gamma\sin\phi=r^2-r\cos(\gamma+\phi)\)

So
\(r^2-ar-2 =0 \) where \(a=\cos(\gamma+\phi)\)

Note: \(a^2\) can take on any value between 0 and 1 independent of gamma. Likewise a is between -1 and 1.

So: \(r=\frac{a \pm \sqrt{a^2+8}}{2}\)

The absolute maximum and minimum of this occur when a=-1, a=1 or \(\frac{d r}{d a}=0\).

\(\frac{d r}{d a}=\frac 12 \pm \frac{a}{2\sqrt{a^2+8}}=0\) if

\(1 = \pm \frac{a}{\sqrt{a^2+8}}=0\) if \(a^2=a^2+8\) which has no solution.

So r is maximum or minimum when a = -1 or 1. These correspond to r= 1 and r= 2 respectively.

So for any gamma the radius can be between +1 and +2. The locus can pass through any point on the closed punctured disc with outer radius 2 and inner radius 1.