Monday, July 30, 2012

Problem 793: Right Triangle, Altitude, Incircles, Incenters, Circumcircle, Circumradius, Perpendicular, Distance, Metric Relations

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

Click the figure below to see the complete problem 793.

1. 1)angle DBE=90/2=45 degrees
2)points A,D,I,N are in the same line and this line is perpendicular to BE. This way the angle BDE is 45 degrees.
Similarly, points C,E,I,M are in the same line and this liner is perpendicular to BD. This way the angle BEI=45 degrees
3)bxc=2RxHa(b,c are triangle sides and Ha is altitude)
IDxIE=2RxIF
IDxIE=2IMxIN
IMxIN=RxIF, => R=(IMxIN)/IF.
Erina

2. Note that angle IDM = angle IEN = 45 degrees

R
= [ID*IE*DE] / [4*S(IDE)]
= [sqrt(2)*IM*sqrt(2)*IN*DE] / [4*S(IDE)]
= [IM*IN*DE] / [2*S(IDE)]
= IM*IN/IF

3. Let IK is perpendicular to AC. From similarity of IKM and INF, we have that IK/IM=IN/IF or R/IM=IN/IF.

4. True or false?
(i)The incircle of ∆ABC, the circmcircle of ∆DEI and the circumcircle of ∆BDE are of equal radii

1. To Pravin: It is true. See problem 794