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.

## Tuesday, April 7, 2020

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

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D is midpoint of I and center of excircle opposite to C.

ReplyDeleteHence 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)

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

ReplyDeleteLet 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)