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

Labels:
common tangent,
incenter,
incircle,
inradius,
parallel,
right triangle,
tangency point

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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