# Mr. 23

23 is everywhere!
- Greg Funchess
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## Solution to the “Extreme” problem

This wasn’t meant to be an extremely hard problem, the word extreme refers to a group of my friends that I randomly saw in public the day I wrote the problem. The problem is fairly difficult though so I will explain the solution. Initially I thought this problem would be harder since I was thinking of the calculus approach to finding the area of an ellipse, but I realized there is a simple equation for the area of an ellipse which vastly simplifies this problem.

I will try to be very systematic in solving this problem so that it makes sense. The problem asks to find the probability within R’s orbit but not T’s orbit, and you can be placed anywhere within M’s orbit. The logical first step is find the various areas of the orbits so a probability can be found.

Starting from the center, X’s orbit has an area of π super-start units squared. Note the units won’t matter in the end. When T’s orbit is closest to X’s orbit it is 1 unit away, thus the shortest radius of T is 2 units. T is a 2 to 1 ellipse thing so it’s radii are 2 and 4 units. The formula for area of an ellipse is π(r1r2), plugging in the values gives 8π units squared. Next find R’s area. With radii 5 (one more than T’s longest, 4) and 10 (double 5), R’s area comes out to be 50π. And finally calculate M’s area, 11 squared, yielding 141π. Now the area in R’s orbit but not T’s orbit is R-T = 50π-8π = 42π. The probability is then the desired area over all possibilities, that is, 42π over 141π = 42 / 121 since the π’s cancel. So the answer is 42/121 ≈ 0.358

P.S. Thanks fuckyeahmath for following me and reblogging me.

EDIT: I wrote this very late and realized that I calculated M like an ellipse, not like a circle that the problem says, this solution is now correct.

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