## Thursday, December 8, 2016

### Solving contest problems (challenge from Mike Lawler)

These are my notes working through problems posted by Mike Lawler on his blog. You'll have to go there to see the problem statements.

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.

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.