# 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.

In this post we will calculate the incircle radius of a triangle given only the sides. For this we need heron’s formula which says, that when the sides of a triangle are a,b,c that then:
Let s=(a+b+c)/2.
Then the area is the square root of s*(s-a)*(s-b)*(d-c). This can easily be proven with the pythagorean theorem.

We connect the incircle center with the corners of the triangle:

The triangle is now devided into 3 triangles. The small triangles have the base sides of lengths a,b and c. The hight is always the incircle radius which we denote r. So the Area A of the big triangle is equal to a*r/2+b*r/2+c*r/2. This equals (a+b+c)*r/2. We have A=(a+b+c)*r/2. We devide by (a+b+c)/2 and get 2A/ (a+b+c)=r.
The formula for the incircle radius is two times the area devided by the circumference.

# Napoleon’s Theorem

First I want to thank an anonymous Person who Formulated this Post for me.

Napoleon’s Theorem:

Let ABC be a triangle. Let equilateral triangles be constructed outward at each side of the triangle ABC. Then the midpoints of these equilateral triangles are the vertices of another equilateral triangle.

Proof:

We use vector calculus. Let A, B, C denote the vectors pointing to the vertices.  By translating the tirangle so that the barycenter of it becomes the origin, we may assume that

(1)       A+B+C=0

Let R be a the linear map which turns each vector by 60 degrees to the right. Note that R^3 = -1and for every vector X we have the identity

(2)     R^2 X  = RX – X

The tip of the equilateral triangle above the side from A to B is

B+R(A-B)

The center of this equilateral triangle is then

(A+B+(B+R(A-B)))/3

Using (1), we may write this as

(3)    ((B-C)+ R(A-B))/3

By symmetry the centers of the other equilateral triangles are

(4)   ((C-A)+R(B-C))/3

(5)   ((A-B)+R(C-A))/3

We need to prove that these three points are permuted when we rotate by 120 degrees aound the origin, that is R^2. For simplicity we may multiply the three vectors by 3. Applying R^2 to 3 times expression (3) gives

R^2(B-C)+R^3(A-B)

Applying (2) and the fact R^3=-1 turns this into

R(B-C)-(B-C)-(A-B)=(C-A)+R(B-C)

This evidently is three times (4). By symmetry, the three points (3),(4),(5) are permuted when rotated by R^2, which proves that they form an equilateral triangle.

# The Farmer Problem

The problem goes like this:

There are 2 points on a plane, which are not the same. One represents a farmer and the other point represents a pig. Then there is a straight line on the plane, which does not go through either of the points. The line represents a river. The farmer wants to give the pig water. To do his, he has to first go to the river and then to the pig. How can you find the fastest way for him to do so.

The answer is astonishingly easy. You have to reflect the (pig) point at the river as in the picture (a=b) :

It is known, that the shortest path between two points on a plane is a straight line. Now we connect the farmer with the reflected point is a straight line. Then we call the point where the line cuts the river c. Now we just have to connect the farmer with c and the pig with c. This is the shortest path, since the left side of the river was just reflected to the right side and we had the shortest path before we reflected back..

# 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.

# A Math Problem

Let ABCE be a square. Let D be a point, so that CED is right angle and ED has the same length as CD. Use 2 straight lines to cut the pentagon into three pieces and arrange these, so that they form a square.