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
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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