## Tuesday, April 7, 2020

### Geometry Problem 1469: Triangle, Circumradius, Inradius, Midpoints, Arcs, Sum of Distances

Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

1. D is midpoint of I and center of excircle opposite to C.
Hence d=(Rc+r)/2, similarly e=(Rb+r)/2 and f=(Ra+r)/2.
We have d+e+f=(Ra+Rb+Rc+3r)/2,
Using result of Prob.1468, Ra+Rb+Rc=4R+r,
We get d+e+f=2(R+r)

2. https://photos.app.goo.gl/HHPWGhSK45Tc7ZVXA

Let A’, B’, C’ are excenters and Ra, Rb,Rc are radius of excenters of triangle ABC
Note that F,E and D are midpoints of IA’, IB’ and IC’
We have 2d= r+ Rc (midsegment of trapezoid ITMC’)
And 2e=r+Rb…(midsegment of trapezoid ITLB’)
And 2f=r+Ra (midsegment of trapezoid)
Add above expression side by side we have 2(d+e+f)= 3r+Ra+Rb+Rc …(1)
Per the result of problem 1468 we have Ra+Rb+Rc= 4R+r…..(2)
Replace (2) in (1) we have Ra+Rb+Rc= 2(R+r)