## Monday, August 5, 2019

### Geometry Problem 1443: Triangle Area, Incenter, Circumcenter, 90 Degree, Perpendicular, One-third, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below. 1. Let BI extended meet AC at D and circle O at E

< EAI = EIA = (A+B)/2 so AE = EI = IB since OI is perpendicular to BE

Further, since < EAD = < ABE = B/2, Tr.s ADE & ABE are similar and so

But AD = bc/(a+c), hence c/2 = bc/(a+c) from which, a+c= 2b

Now S1 = rb/2 = (S/s)*(b/2) = Sb/(a+b+c) = Sb/(3b) = S/3

Sumith Peiris
Moratuwa
Sri Lanka

2. https://photos.app.goo.gl/oBvDsrGZ2wwrWe9V7

let BI meet AC at N and arc AC at D
Let CI meet arc AB at M
We have I is the midpoint of BD
and Triangle DIC is isoceles
Triangle DNC simillar to DCB ( case AA)
so DC^2=DN.DB=2DN.DI=2DN.DC
so DN/DC=DN/DI=1/2 => N is the midpoint of ID
and NI/NB=1/3
Area(AIC)/Area(ABC)= NI/NB= 1/3

3. Draw BH=h, IM=r, altitudes of tr S and S1: Need to prove r/h=1/3
Extend BI to N, T, extend OI to P, join O to T.Easy OBPT rhombus
From OPHS r=((h-R)+(R-r))/2 (N midpoint of SM from P155) =>r/h=1/3
https://photos.app.goo.gl/V3QACpjdp1e7uyy3A