Geometry Problem. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Saturday, January 21, 2017

### Geometry Problem 1309 Triangle, Circle, Inradius, Excircle, Tangent, Exradius, Measurement

Labels:
circle,
exradius,
geometry problem,
inradius,
measurement,
triangle

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P, P' tg points ( on AC, BC). Join O2 to C

ReplyDeleteFrom AD=AE, PC=P'C, we get PC=11

Right tr O1PP'~tr O2CE => 6/11=7/x => x=77/6

https://goo.gl/photos/gUJk9ftBfbk8pmpQ9

ReplyDeleteDefine points I, F, G, H and M as per sketch

We have CG=CE=BF= 7 and BG=CF=CH=11

We also have IC⊥CM

Triangle IHC simillar to CEM ( case AA)

So IH/CH= CE/ME => 6/11=7/x

So x= 77/6

If U,V,W are the tangency points of BC, CA, AB respectively of the in circle,

ReplyDeleteThen WD = VE = BC = 18.

So easily VC = 11

Hence from similar right triangles O1VC and O2CE

6/11 = 7/x and so x = 77/6

Sumith Peiris

Moratuwa

Sri Lanka