Maths Olympiad Prep

Library / /702 of 740

, 2023

Geometry Difficulty 5.7 AIME, harder Prove it United States

Problem:

Let ω1\omega_{1} and ω2\omega_{2} be two non-intersecting circles. Suppose the following three conditions hold:
- The length of a common internal tangent of ω1\omega_{1} and ω2\omega_{2} is equal to 1919.
- The length of a common external tangent of ω1\omega_{1} and ω2\omega_{2} is equal to 3737.
- If two points XX and YY are selected on ω1\omega_{1} and ω2\omega_{2}, respectively, uniformly at random, then the expected value of XY2XY^{2} is 20232023.

Compute the distance between the centers of ω1\omega_{1} and ω2\omega_{2}.

Solutions — 2

Solution 1

Solution:

The key claim is that E[XY2]=d2+r12+r22\mathbb{E}\left[XY^{2}\right] = d^{2} + r_{1}^{2} + r_{2}^{2}.

To prove this claim, choose an arbitrary point BB on ω2\omega_{2}. Let r1,r2r_{1}, r_{2} be the radii of ω1,ω2\omega_{1}, \omega_{2} respectively, and O1,O2O_{1}, O_{2} be the centers of ω1,ω2\omega_{1}, \omega_{2} respectively. Thus, by the law of cosines, O1B=d2+r222r2dcos(θ)\overline{O_{1}B} = \sqrt{d^{2} + r_{2}^{2} - 2 r_{2} d \cos(\theta)}, where θ=O1O2B\theta = \angle O_{1}O_{2}B. Since the average value of cos(θ)\cos(\theta) is 00, the average value of (O1B)2\left(\overline{O_{1}B}\right)^{2} is d2+r22d^{2} + r_{2}^{2}.

Now suppose AA is an arbitrary point on ω1\omega_{1}. By the law of cosines, AB2=(O1B)2+r122r1dcos(θ)\overline{AB}^{2} = \left(\overline{O_{1}B}\right)^{2} + r_{1}^{2} - 2 r_{1} d \cos(\theta), where θ=AO1B\theta = \angle AO_{1}B. Thus, the expected value of AB2\overline{AB}^{2} is the expected value of (O1B)2+r12\left(\overline{O_{1}B}\right)^{2} + r_{1}^{2}, which becomes d2+r12+r22d^{2} + r_{1}^{2} + r_{2}^{2}. This proves the key claim.

Thus, we have d2+r12+r22=2023d^{2} + r_{1}^{2} + r_{2}^{2} = 2023. The lengths of the internal and the external tangents give us d2(r1+r2)2=361d^{2} - (r_{1} + r_{2})^{2} = 361, and d2(r1r2)2=1369d^{2} - (r_{1} - r_{2})^{2} = 1369. Thus,
d2(r12+r22)=(d2(r1+r2)2)+(d2(r1r2)2)2=361+13692=865 d^{2} - (r_{1}^{2} + r_{2}^{2}) = \frac{\left(d^{2} - (r_{1} + r_{2})^{2}\right) + \left(d^{2} - (r_{1} - r_{2})^{2}\right)}{2} = \frac{361 + 1369}{2} = 865
Thus, d2=865+20232=1444d=38d^{2} = \frac{865 + 2023}{2} = 1444 \Longrightarrow d = 38.

Solution 2

Solution:

We present another way of showing that E[XY2]=d2+r12+r22\mathbb{E}\left[XY^{2}\right] = d^{2} + r_{1}^{2} + r_{2}^{2} using complex numbers. The finish is the same as Solution 1.

Let the center of ω1\omega_{1} and ω2\omega_{2} be 00 and kk, respectively. Select Z1Z_{1} and Z2Z_{2} uniformly random on unit circle. Then,
E[XY2]=Ek+r1Z1+r2Z22=E(k+r1Z1+r2Z2)(kˉ+r1Z1+r2Z2) \begin{aligned} \mathbb{E}\left[XY^{2}\right] &= \mathbb{E}\left|k + r_{1} Z_{1} + r_{2} Z_{2}\right|^{2} \\ &= \mathbb{E}\left(k + r_{1} Z_{1} + r_{2} Z_{2}\right)\left(\bar{k} + r_{1} \overline{Z_{1}} + r_{2} \overline{Z_{2}}\right) \end{aligned}
Then, observe that
E[Z1]=E[Z2]=E[Z1Z2]=E[Z2Z1]=0 \mathbb{E}\left[Z_{1}\right] = \mathbb{E}\left[Z_{2}\right] = \mathbb{E}\left[Z_{1} \overline{Z_{2}}\right] = \mathbb{E}\left[Z_{2} \overline{Z_{1}}\right] = 0
so when expanding, six terms vanish, leaving only
E[XY2]=E[kkˉ+r12Z1Z1+r22Z2Z2]=d2+r12+r22 \mathbb{E}\left[XY^{2}\right] = \mathbb{E}\left[k \bar{k} + r_{1}^{2} Z_{1} \overline{Z_{1}} + r_{2}^{2} Z_{2} \overline{Z_{2}}\right] = d^{2} + r_{1}^{2} + r_{2}^{2}
The rest of the computation is as in Solution 1, so d=38d = 38.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.