# Assignment 6: The Polya Problem

For my sixth assignment, I investigated a problem originally posed by Polya. Given an angle ACB, locate a point X on AC and Y on BC so that AX = XY = YB.

There are many ways to solve this problem, and in a week and a half I only came up with two. And even that took some help from my friends. The rest of this write up will explain the two solutions I came up with.

The idea in both is to construct a scale model of the answer I want and then blow it up to the needed size. To start with I choose a random point M on AC. AM will represent AX in the final solution. I also need to locate a point N, which will become Y in the final solution. Since we want MN = AM, I will make a circle centered at M with radius AM. We know N will be somewhere on that circle. I also need to find a point O, which will correspond to B in the final solution. O should lie on AB in order for my scale model to be similar to my final model. Hence I will draw in AB. From there the two methods diverge.

## Method 1: The Rhombus method

If we imagine already having NO drawn in, we can see that MN and NO would form two sides of a rhombus, whose fourth vertex would be the point between A and B where the circle hits. So next we construct the needed rhombus. You can do this by making a circle centered at the intersection between circle M and AB and with radius of length AM. The point of intersection of AB and the new circle that is closest to B will be O. Then construct a parallel line to BC that passes through O. This line will intersect circle M at N.

Now skip down to the Common End to finish up.

## Method 2: The Parallel method

In this method we note NO will be the same length as AM, parallel to BC and N will lie on circle M. So we will measure off a segment with length AM on BC, starting from B, and then make a parallel line to AB which passes through the other endpoint of our new segment.

The point of intersection between this new line and circle M is N. Complete the scale model by constructing a parallel line to BC through N. This line will intersect AB at O.

Common End:

To enlarge our miniature solution, we will dilate from point A. We know Y will be on BC, so just find the point where ray AY intersects BC. In order to find X, we can use a circle centered at A with radius YB. This circle will intersect AC at X. And we are done!