The intended value of this write-up is to show examples of the actual problem solving thought process someone has followed, not just a polished solution.

**Problem 19**

Since the circle has area 156 π the radius squared is 156, which is 4 * 3* 13. That seems like a strange number, I'm curious to see where it will make calculations come out nicely, later in the problem.

We're told OA has length 4 sqrt(3). Squaring that only gives 48, so A is well within the circle. That means triangle ABC has vertex pointing toward O along the perpendicular bisector of side BC.

This lets me set up a picture of a right triangle with legs length $x$ (half the side of the equilateral triangle), $x$ sqrt(3) + 4 sqrt(3) (the altitude of the equilateral triangle plus OA) and hypotenuse r (2 sqrt(3*13)).

Applying the pythagorean theorem and simplifying along the way:

$$x^2 + 3(x+4)^2 = 156$$

$$4x^2 + 24x + 48 - 156 = 0$$

$$x^2 + 6x - 27 = 0$$

Visually factoring gets me $x$ is either -9 or 3. 3 is much more reasonable for half the length of the side of a triangle, so my answer is 6.

Note: If this weren't a contest problem, I might try to think about the -9 root a little more carefully.

**Equiangular hexagon**

I had already solved this problem from a recent tweet of Mikes, so I can't fully recreate my thought process. Here were some of the highlights:

- wonder why 70%
- draw a picture: fail to notice that triangle ACE is equilateral
- Split the hexagon into two trapezoids by line CF
- Calculate the area of those two trapezoids
- Calculate the length from C the intersection with AE and segment CF.
- Calculate the area of ACE based on the two subtriangles split by CF.
- Obtain the quadratic equation for r based on the formulae for the two areas and the given 70% parameter.

**School Competition**

Let's call Andrea's rank $m$ for median. Since she is the unique median, there must be $2m - 1$ total contestants. We also know:

- $m \leq 36$ because Andrea scored higher than Beth who was ranked 37th
- $64 \leq 2m -1 $ because Carla ranked 64th, so there were at least that many competitors
- $3 \mid (2m - 1)$ since every school sent three competitors

The first two inequalities tell us that $33 \leq m \leq 36$. Because $2m -1$ is a multiple of three, $m$ can't be a multiple of 3, so it has to be 34 or 35. Calculating mod 3, we can quickly check both and see that $m$ has to be 35.

That means the competition had 69 competitors from 23 schools.

After putting together these notes, I saw a commenter on Mike's blog use "solution by multiple choice." When I was doing timed tests/contests, I would make use of the options as part of my strategy. However, I don't do that now, since I'm much more interested in exploring the mathematics of the problems than getting the final answer.

**Using multiple choice**After putting together these notes, I saw a commenter on Mike's blog use "solution by multiple choice." When I was doing timed tests/contests, I would make use of the options as part of my strategy. However, I don't do that now, since I'm much more interested in exploring the mathematics of the problems than getting the final answer.

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