## Thursday, November 14, 2013

### Geometry Problem 935: Triangle, Cevian, Equal Incircles, Inscribed Circles

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

Click on the figure below to enlarge it.

1. Angela Drei has an excellence solution for this equal incircles problem.
See link below for his solution.
http://www.cut-the-knot.org/triangle/EqualIncirclesTheorem.shtml

2. Since I prefer basic geometrical argument, here is my proposal...

http://bleaug.free.fr/gogeometry/935.png

First, forget about the 3 equal circles, it's a strawman ;-) Let's simplify the picture and keep A, B, C, D, O1 and Q2 (which I renamed Q1 for simplicity in this post). Let's add S as the center of ABC incircle. By construction O1 lies on AS and Q1 lies on CS.

Bisectors of ACxBD in D are orthogonal, therefore O1-D-Q1 is a right triangle. Let's build P1 such as O1-P1-Q1-D is a rectangle. It has the following properties:
- PS is orthogonal to AC (simple angle calculation)
- altitudes relative to AC obey: alt(P1) = alt(O1)+alt(Q1) because of rectangle projection property

Conclusion: alt(O1) and alt(Q1) play a symmetric role. Setting O1' on CS such as alt(O1')=alt(O1) would imply corresponding alt(Q1')=alt(Q1). Therefore in original problem wordings: alt(O1)=alt(O3) implies alt(Q1)=alt(Q2) irrespective of alt(O2).

bleaug