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
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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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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.