Proposed Problem
See also:Complete Problem 27, Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Friday, April 10, 2009
Problem 27: Right triangle, Incenter, Parallel, Incircles, Inradius
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Hints: Two tangent segments to a circle are congruent. Square...
ReplyDeleteSee also Problem 668
http://img98.imageshack.us/img98/3497/problem27.png
ReplyDeleteConnect CO and AO
Note that CO and CO2 are angle bisector of angle (BCA) so
C, O, O2 are collinear
Triangle OGC congruence with triangle OKC ( right triangles with common hypotenuse and CO is angle bisector)
So OC is the angle bisector of angle (GOK) >>Incircle O2 tangent to OG also tangent to OK and O2N=r2
Similarly AO is angle bisector of angle HOK) and incircle O1 tangent to both OH and OK and O1P=r1
DE=DK+KE=O1P+O2N=r1+r2
Peter Tran
Using Antonio's hint above....
ReplyDeleteIf X is the point of tangency of circle O with AC
AK = c - r and KC = a - r
Similarly AD = c-r-r1 and CE = a-r-r2
So AC - AD - EC = r1+r2 + (b-c-a + 2r) from using the above expressions for AD and CE.
The expression within the bracket = 0 and the result follows
Sumith Peiris
Moratuwa
Sri Lanka
Define O1 and O2 the center of incirles with radius r1 and r2
ReplyDeletePoints A, O1 and O are aligned on the angle bisector AD=AD’ and CE=CE’
D’H=r1, E’G=r2 and HB=HG=r
AB=AD’+r1+r and CB=CE’+r2+r
Since r is the incircle of ΔABC right in B, then 2r=AB+BC-AC
Therefore 2r=(AD+r1+r)+(CE+r2+r) – AC
2r=r1+r2+2r+AD+CE-AC
But AD+DE+EC=AC
Therefore 2r=r1+r2+2r-DE
=> DE=r1+r2