# Constructing the Japanese problem figure

In this post we will construct the figure of the last post. I will leave the proof open. But i will post it if i get asked.

Problem:
Let a square abcd be given. A semicircle k in the square with the diameter ab is given. Construct a circle in the square, which touches k , cb and dc.

Solution:

First we construct the line through a and c.  We extend the side cb to a point e so that cb=be. Then we construct the tangent line to the semicircle k. Let the point on which the tangent line touches k be denoted as x. We construct a perpendicular line to the tangent line through x. The point where this line crosses the line ac ,we denote as n.
Then n is the center of the circle and nx is the radius of the circle we were searching for.

# A Japanese problem

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.

# Feuerbach’s circle proof

Feuerbach’s circle:

Let ABC be a triangle. Let D,E,F be the midpoints of AC,BC,BA respectively. Let G,H,I be the feet of each altitude in the triangle. Let J be the point where the altitudes meet. Denote the points M,L,K as the midpoints of JA,JC,JB. Then all points D,E,F,G,H,I,K,L,M lie on a circle.

Proof:

Due to the intercept theorem we can say, that ED is parallel to AB and therefore GF, EF is parallel to AC (DH),and DF is parallel to BC (EI). Let’s prove, that EDGF is isosceles. Since CGA is 90 degrees and D is half of AC, by the Thales theorem (which says, that D is the midpoint of the circle around ACG) DG is equal to AD. By the intercept theorem FE is also half of AC. Because the diagonals have the same length DGFE is an isosceles trapezoid and its vertices lie on a circle. It is possible to use the same argument on DHEF and DFEI. Since D,E,F are in each trapezoid, D,E,F,G,H,I are on a circle. Now we use this information on the triangle BCJ. We proved, that E,I,L,K are on one circle, since E,K,L are the midpoints of the sides of the triangle BCJ and I is the foot of the altitude from J. We can use the same arguments for triangles BJA and CAJ. This completes the proof.

# Thales Theorem

Thales theorem: Let A,B,C be points on a circle. Let AB be the Diameter of the circle. Then ACB is a right angle.  Proof: Let x be angle DAC. Since DA and DC are the radius of the circle they have the same length. It follows, that DAC has the same angle as DCA. Let DBC be y. We can use the same argument to show that DCB has the angle y. Since the sum of all angles in a triangle is 180 degrees, BDC is 180-2y and ADC is 180-2x. If we add these Angles we get 180 degrees since A,D,B lie on one line. We add all angles in both triangles. 180-2x+180-2y+2x+2y=360. 180+2x+2y=360. 2x+2y=180. Now we divide by 2 on both sides: x+y=90. This completes the proof since x+y is the size of the angle ACB.