## Friday, August 3, 2012

### Problem 794: Right Triangle, Altitude, Three Incenters, Circumcircles, Incircle, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 794.

1. AC passes through the center of c3 and touches c1. So the radius of c3 and c1 must be the same.
By previous problem we know that I is the orthocenter of the triangle BDE, hence the circumcircle of IDE and BDE must be in same size. This implies c1 and c2 are congruent .

2. I'm not gonna post a picture for this proof but here is my plan:
First prove that I,D,A and I,E,C are collinear.
then use the law of sines to prove that the radii of C2 and C3 are the same.
As for the radius of C1, you can use the formula for the inradius of a right triangle to prove the radii of C1 and C2 are the same.
And it works! By the way, this proof requires some similarities between triangles.

3. Let r, a, b, are the inradiis of the triangles ABC, ABH and BHC respectively.
Hence we get:
(1) r^2 = a^2 + b^2 (Problem 25) ;
BH = r + a + b (Problem 23).
P and Q are the points at which incircles of Tr. ABH and BHC touche AB and BC respectively.
Then:
BP = BH - a = r + b ; BQ = BH - b = r + a.
Now we take BA & BC as the x-axis and y-axis.
B(0 ; 0) ; D(r + b ; a) ; E(b ; r + a).
The equation of the circumcircle of tr. BDE is:
X^2(a+b+r)+Y^2(a+b+r)+X(a^2-2ab-b^2-2br-r^2)-Y(a^2+2ab+2ar-b^2+r^2)=0.
By (1) we have:
a^2-r^2= -b^2 and r^2 - b^2 = a^2.
By the substitution in the equation we have:
X^2(a+b+r)+y^2(a+b+r)-2bX(a+b+r)-2aY(a+b+r)=0
or X^2 + Y^2 - 2bX - 2aY = 0.
The center of the circumcircle C2 is G(b ; a) and the radius of the circle C2 is R =(a^2+B^2) = r.
Therefore all the circles C1, C2 and C3 has radius r.
Infact also the radius of C1 and C3 is r (problem 792 - 793).

4. In the precedent solution of the problem 794 there is a careless mistake:
.... The center of the circumcircle C2 is G(b ; a) and the radius of the circle C2 is R =sqrt(a^2 + b^2) = r.