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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.