tag:blogger.com,1999:blog-6933544261975483399.post1008469663678736500..comments2022-11-29T01:21:52.908-08:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1184: Right Triangle, Square, Inscribed Circle, Tangent, Concurrent LinesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-14904517169176773192016-01-30T23:30:16.307-08:002016-01-30T23:30:16.307-08:00OQ, TH meet at P. We must prove that AC also passe...OQ, TH meet at P. We must prove that AC also passes thro P<br /><br />If OT = R and QH = r, then OA = sqrt2 R and QC = sqrt2 r<br /><br />So OA/QC = OT/QH which in turn = PO/PQ <br /><br />Now consider the right triangles POA and PQC in which PO/PQ = OA/QC and < O < Q = 90<br /><br />It follows that the 2 right triangles are similar and so < PAO =PCQ and for this to happnd PCA must be collinear <br /><br />Hence AC, OQ and TH meet in a point P<br /><br />Sumith Peiris<br />Moratuwa<br />Sri Lanka Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-24524450104665768742016-01-30T12:17:50.937-08:002016-01-30T12:17:50.937-08:00Ang CFB=45=P1+FCP, Ang BFG=45=P2+FMP(M on GF)
Ang ...Ang CFB=45=P1+FCP, Ang BFG=45=P2+FMP(M on GF)<br />Ang FMP=CNP (N on DB) => in tr PNC = 180c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-34416934627955439792016-01-30T12:00:20.465-08:002016-01-30T12:00:20.465-08:00Let r1 and r2 are radii of circles Q and O
Since T...Let r1 and r2 are radii of circles Q and O<br />Since TH is the common tangent of 2 circles <br />so we have PH/PT= PQ/PO= r1/r2<br />Let AC meet OQ at P’<br />We have AO ⊥OQ and CQ⊥OQ<br />So P’C/P’A= QC/OA=r1/r2=P’Q/P’O<br />So P coincide to P and TH, EF and AC concur at P<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com