Today I will introduce a Japanese problem. In Japan problems like this are called a sangaku. Monks used to write sangaku’s on wood plates, which then would be sold to samurai for them to solve. In this post I will solve the problem and in the next one, I will construct the figure which will be presented.

The Problem:

Let abcd be a square. We construct a semicircle over ad with the radius of half the side length of the square. Let K be a circle which touches the semicircle and bc and cd. As the picture below shows:

Solution:

We construct a parallel line through the middle point of ad parallel to ab. Also we construct a parallel line parallel to ad through the middle point of K. Let the interception point be denoted as y. Last we construct the line from the middle point of ad and the center of K.

This forms a rightangeled triangle y and the center points of the two circles.

We denote the radius of K as r and the radius of the semicircle as a/2.

The sides of the triangle which we constructed are:

r+a/2 , a/2-r and a-r.

After the pythagorean theorem we have the following equlity.

(r+a/2)^2=(a/2-r)^2+(a-r)^2.

Solving the equality offers

r=(2-sqr (3))a.