Two circles $\Gamma_1$ and $\Gamma_2$ intersect at $A,B$. Points $C,D$ lie on $\Gamma_1$, points $E,F$ lie on $\Gamma_2$ such that $A,B$ lies on segments $CE,DF$ respectively and segments $CE,DF$ do not intersect. Let $CF$ meet $\Gamma_1,\Gamma_2$ again at $K,L$ respectively, and $DE$ meet $\Gamma_1,\Gamma_2$ at $M,N$ respectively. If the circumcircles of $\triangle ALM$ and $\triangle BKN$ are tangent, prove that the radii of these two circles are equal.
Problem
Source: CSMO 2019 Grade 10 Problem 2
Tags: geometry, circumcircle
hyx
31.07.2019 15:06
Who has a solution?
RC.
31.07.2019 15:44
Let \(DE \cap CF = X.\) By trivial angle chase \(\quad BNPK, ALPM\) are concyclic. So the two circles \(\odot ALM, \odot BNK\) are tangent at \(P\). So, \(\Delta PNK \sim \Delta PML \Rightarrow \frac{PM}{PN}=\frac{PL}{PK} - (1)\) Also, by Reim's \(EF\) is parallel to \(CD \Rightarrow \frac{PC}{PF}= \frac{PD}{PE} - (2)\). Now, using Po P, \(PK \times PC = PM \times PD; PL \times FP = NP \times EP.\) Dividing the two we have, \(\frac{PK \times PC}{PL \times FP} = \frac{PM \times PD}{PN \times PE} - (3)\) Dividing (3) by (2) you get \(\frac{PM}{PN}=\frac{PK}{PL}\) but at the same time from (1) you get converse so \(PM = PN\) and \(PL = PK\) so the radii are indeed equal. done,
lemon94
03.08.2019 18:46
RC. wrote: Let \(DE \cap CF = X.\) By trivial angle chase \(\quad BNPK, ALPM\) are concyclic. So the two circles \(\odot ALM, \odot BNK\) are tangent at \(P\). So, \(\Delta PNK \sim \Delta PML \Rightarrow \frac{PM}{PN}=\frac{PL}{PK} - (1)\) Also, by Reim's \(EF\) is parallel to \(CD \Rightarrow \frac{PC}{PF}= \frac{PD}{PE} - (2)\). Now, using Po P, \(PK \times PC = PM \times PD; PL \times FP = NP \times EP.\) Dividing the two we have, \(\frac{PK \times PC}{PL \times FP} = \frac{PM \times PD}{PN \times PE} - (3)\) Dividing (3) by (2) you get \(\frac{PM}{PN}=\frac{PK}{PL}\) but at the same time from (1) you get converse so \(PM = PN\) and \(PL = PK\) so the radii are indeed equal. done, What is the "Reim's"?
BabaAuRhum
03.08.2019 21:52
DA2004 wrote: RC. wrote: Let \(DE \cap CF = X.\) By trivial angle chase \(\quad BNPK, ALPM\) are concyclic. So the two circles \(\odot ALM, \odot BNK\) are tangent at \(P\). So, \(\Delta PNK \sim \Delta PML \Rightarrow \frac{PM}{PN}=\frac{PL}{PK} - (1)\) Also, by Reim's \(EF\) is parallel to \(CD \Rightarrow \frac{PC}{PF}= \frac{PD}{PE} - (2)\). Now, using Po P, \(PK \times PC = PM \times PD; PL \times FP = NP \times EP.\) Dividing the two we have, \(\frac{PK \times PC}{PL \times FP} = \frac{PM \times PD}{PN \times PE} - (3)\) Dividing (3) by (2) you get \(\frac{PM}{PN}=\frac{PK}{PL}\) but at the same time from (1) you get converse so \(PM = PN\) and \(PL = PK\) so the radii are indeed equal. done, What is the "Reim's"? Search "Reim's theorem" you will find