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

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

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## Saturday, January 30, 2016

### Geometry Problem 1184: Right Triangle, Square, Inscribed Circle, Tangent, Concurrent Lines

Labels:
circle,
concurrent,
inscribed,
right triangle,
square,
tangent

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Let r1 and r2 are radii of circles Q and O

ReplyDeleteSince TH is the common tangent of 2 circles

so we have PH/PT= PQ/PO= r1/r2

Let AC meet OQ at P’

We have AO ⊥OQ and CQ⊥OQ

So P’C/P’A= QC/OA=r1/r2=P’Q/P’O

So P coincide to P and TH, EF and AC concur at P

Ang CFB=45=P1+FCP, Ang BFG=45=P2+FMP(M on GF)

ReplyDeleteAng FMP=CNP (N on DB) => in tr PNC = 180

OQ, TH meet at P. We must prove that AC also passes thro P

ReplyDeleteIf OT = R and QH = r, then OA = sqrt2 R and QC = sqrt2 r

So OA/QC = OT/QH which in turn = PO/PQ

Now consider the right triangles POA and PQC in which PO/PQ = OA/QC and < O < Q = 90

It follows that the 2 right triangles are similar and so < PAO =PCQ and for this to happnd PCA must be collinear

Hence AC, OQ and TH meet in a point P

Sumith Peiris

Moratuwa

Sri Lanka