Geometry Problem

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

Click the figure below to see the complete problem 668.

## Sunday, September 11, 2011

### Problem 668: Right triangle, Incircles, Parallel, Common Tangent, Mind Map

Labels:
common tangent,
incenter,
incircle,
parallel,
right triangle

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http://img819.imageshack.us/img819/5083/problem668.png

ReplyDeleteConnect CO and AO

Note that CO and CO2 are angle bisector of angle (BCA) so

C, O, O2 are collinear

Triangle OEC congruence with triangle OHC ( right triangles with common hypotenuse and CO is angle bisector)

So OC is the angle bisector of angle (HOE) >>Incircle O2 tangent to OE also tangent to OH.

Similarly AO is angle bisector of angle HOF) and incircle O1 tangent to both OF and OH

Peter Tran

Let circle O₁ touch AB at J, AC at K & FG at L.

ReplyDeleteComplete the square O₁JHM

O₁M = KH = AH - AK = AF - AJ = KF = radius of O₁ (note FJ O₁K is a square)

follows OH touches circle O₁.

Similarly we can show OH touches circle O₂

Refer my solution to Problem 668:

ReplyDelete"FJO₁L is a square" (not FJO₁K as previously typed)

another point of vue

ReplyDeletein ABC r=s-b

triangles AFG and ABC are homothetic,center A ratio (s-a)/c

the incircle of AFG is the image of the incircle of ABC;r1=(s-a)r/c=(s-b)(s-a)/c

line OH becomes the parallel line O1H1

distance H1H=(s-a)-(s-a)²/c=(s-a)(s-b)c=r1

incircle O1 tangent to OH

The same way CDE and ABC are homothetic with center C and ratio (s-c)/a