Tag Archives: triangle

Calculating the Incircle Radius

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.

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. 

Screen Shot 2014-05-25 at 6.44.11 PM



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.

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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.Screen Shot 2014-05-28 at 8.31.00 PM

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:Screen Shot 2014-05-28 at 8.47.42 PM


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.

Pythagorean Theorem:

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.

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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.