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.
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.
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.
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.
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:
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.
Solving the equality offers
We will do a proof by contradiction:
Lets assume, that the square root of 2 is rational. Then you can write the square root of 2 as a fraction a over b. Lets assume that a and b are hole numbers and that you can not reduce a over b. Then you can do this:
It follows, that a is even.
Now we prove that b is even. For a we can write 2k.
It follows that b is even. That means, that a/2 and b/2 are hole numbers. It follows, that you can reduce a over b with 2. This is a contradiction, since we assumed that a over b can not be reduced.
Def. a and b are in the golden ratio if a/b=b/a-b. The symbol is used for the golden ratio.
Lets calculate the golden ratio. To do this let b=1. We have a/1=1/a-1. Therefor a^2-a-1=0. If we solve this quadratic equation we get . But since we want the ratio to be positive . Now we want to know what the difference between the golden ratio and one over the golden ratio is:
So x is equal to 1.
This is a very famous formula.
Proof of the irrationality of the golden ratio by contradiction:
Lets assume that the golden ratio is rational. Then it can be written as x/y when x and y are integers. We assume that the x and y are the smallest numbers where x/y is the golden ratio. If x and y are integers then y and x-y are integers to. Per definition y/(x-y) is also the golden ratio. This is a contradiction.
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.
Let the side EA have the length 1. The area of the figure is 1 and 1/4, since EDC is 1/4 of AECB. The square that we want to construct has the area 1 and 1/4 and the side length (5/4)^1/2
We can construct this length by connecting A to the middle of CB. Now we can construct a line that connects the middle of BC with D. Now we can guese a solution.
First we show that the lines that are together have the same length. We know that the 2 and 3 fit together since the sides that are touching each other were created by halving the line BC. We know that 2 and 1 fit together because the sides that are touching each other from 2 and 1 are both sides from the square ABCE. We know that 3 and 1 fit together since DC is equal to DE. The angle where 3 2 and 1 come together is 360 degrees, since the angle at 3 is 135 degrees, the angle at 1 is 135 degrees and the angle at 2 is 90 degrees since it was a corner of the square. The angle where only 1 is is 90 degrees. To show this, lets draw this picture:
We will show that ABC is congruent to CDE. BA has the same length as CD per definition. BC has the same length as DE per definition. ABC and CDE are both 90 degrees. It follows that ABC is congruent to CDE. Since the triangle has a 90 degree angle, the other angles added together are also 90 degrees. Since BCD is a straight line, ACE has 90 degrees. It also follows that the angle at 2 and 3 is 90 degrees. Since the angle at 2 and 1 were 90 degrees before we arranged the square, they are 90 degrees afterwards too. It follows that our arrangement is a square. This completes the proof.
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.
Let a,b,c be the sides of an right-angled triangle. Let a,b be the sides at the right angle. Then a^2+b^2=c^2.
There are 2 ways to calculate the area of the figure. Either (a+b)^2 or 2ab+c^2. This gives us .
(a+b)^2=2ab+c^2 a^2+2ab+b^2=c^2+2ab a^2+b^2=c^2
This completes the Proof.