See complete Problem 157

Distance from the Circumcenter to the Excenter. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Tuesday, August 5, 2008

### Elearn Geometry Problem 157

Labels:
circumcenter,
circumradius,
distance,
excenter,
exradius,
triangle

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join A to E, A to O

ReplyDeletedraw OG perpendicular to AE

OG² = R² - (AD/2)² ( from tr AOG )

OG² = d² - (AD/2 + DE)² ( from tr OGE )

R² - (AD/2)² = d² - (AD/2)² - AD∙DE - DE²

R² = d² - DE∙( AD + DE )

d² = R² + DE∙AE

from P156 => DE∙AE = 2Rr

d² = R² + 2Rr

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http://ahmetelmas.wordpress.com/2010/05/15/geo-geo/

ReplyDeleteSolution to problem 157.

ReplyDeleteLine AE meets the circumcircle at D. Segment EO meets the circumcircle at F, and the extension of EO meets the circumcircle at G.

Taking the circle power of point E with relation to the circumcircle we have EF.EG = ED.EA. But EF.EG = (d-R)(d+R) = d^2 – R^2. By problem 156 we know that ED.EA = 2.R.r1. Hence d^2 – R^2 = 2.R.r1, and d^2 = R^2 + 2.R.r1.