Two circles are congruent if tangent

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Two circles are congruent if tangent

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Source: CSMO 2019 Grade 10 Problem 2

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#1 h Jul 30, 2019, 7:09 AM • 1 Y

Y by Adventure10

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.

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#2 h Jul 31, 2019, 12:06 PM • 1 Y

Y by Adventure10

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RC.

#3 h Jul 31, 2019, 12:44 PM • 1 Y

Y by Adventure10

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,

This post has been edited 1 time. Last edited by RC., Jul 31, 2019, 12:44 PM

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#4 h Aug 3, 2019, 3:46 PM • 2 Y

Y by Adventure10, Mango247

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"?

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#5 h Aug 3, 2019, 6:52 PM • 2 Y

Y by lemon94, Adventure10

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

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