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Due:

10/6/2018

Problem:

1996 Putnam | A2

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 \(XY\).

My Work:

The green dots are midpoints.

Without loss of generality, I am going to assume that centres of both the circles lie on \(X\)-axis.

Any point on \(C_1\) would be \((cos\theta, sin\theta)\) and any point on \(C_2\) would be \((10 + 3cos\phi, 3sin\phi)\), assuming \(\theta, \phi\) are angles.

\(\therefore\) midpoint \(\mathcal{M} = \big( 5 + \frac{cos\theta + 3cos\phi}{2} , \frac{sin\theta + 3sin\phi}{2} \big)\)

Rewriting as locus of \(\mathcal{M} = (x,y)\), we get, \(2(x-5) = cos\theta + 3cos\phi\) and \(2y = sin\theta + 3sin\phi\)

Squaring and adding the above equations, we get, \((x-5)^2 + y^2 = \frac{10 + 6cos(\theta - \phi)}{4}\)

We see that it’s a circle equation which is centered at \((5,0)\) and radiusn which depends on phase difference between the points taken on \(C_1, C_2\) respectively.

We also note that since \(cos\) is bounded function, hence, \(\frac{10 + 6cos(\theta - \phi)}{4} \in [1,2]\)

Therefore, we conclude that, locus of \(\mathcal{M}\) is the area between two concentric circles of radii \(1, 2\) centered at \((5,0)\) which I learned is called annulus.

Verdict:

Honeslty, I was able to do the Math in no time, but was stuck at eliminating the relative phase dependency in the locus equation.

This seemed like an easy problem but I nevertheless took a week for this.

But the solution given uses terminology I am not at all familiar with.