Geometry Problem

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

Click the figure below to see the complete problem 793.

## Monday, July 30, 2012

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

Labels:
altitude,
circumcircle,
circumradius,
incenter,
metric relations,
right triangle

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1)angle DBE=90/2=45 degrees

ReplyDelete2)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

Note that angle IDM = angle IEN = 45 degrees

ReplyDeleteR

= [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

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

ReplyDeleteTrue or false?

ReplyDelete(i)The incircle of ∆ABC, the circmcircle of ∆DEI and the circumcircle of ∆BDE are of equal radii

To Pravin: It is true. See problem 794

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