Tow circles $C,D$ are exterior tangent to each other at point $P$. Point $A$ is in the circle $C$. We draw $2$ tangents $AM,AN$ from $A$ to the circle $D$ ($M,N$ are the tangency points.). The second meet points of $AM,AN$ with $C$ are $E,F$, respectively. Prove that $\frac{PE}{PF}=\frac{ME}{NF}$.
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Tags: geometry, geometry proposed
jgnr
23.09.2010 14:41
Invert about $A$, the rest is easy.
forevercontinue
23.09.2010 15:18
Dear, Johan Gunardi ,sorry for my restraint of solving geometry problem. How could you use inversion in this one?
jgnr
23.09.2010 16:32
$AE'M'$, $AF'N'$, $E'P'F'$ become lines tangents to $(M'P'N')$. Hence $E'M'=E'P'$ and $F'N'=F'P'$, which gives $\frac{P'E'}{P'F'}=\frac{M'E'}{N'F'}$, so $\frac{\frac{r^2\cdot PE}{AP\cdot AE}}{\frac{r^2\cdot PF}{AP\cdot AF}}=\frac{\frac{r^2\cdot ME}{AM\cdot AE}}{\frac{r^2\cdot NF}{AN\cdot AF}}$, therefore $\frac{PE}{PF}=\frac{ME}{NF}$ (because $AM=AN$).
huynguyen
20.11.2016 09:11
My solution without inversion: Equivalently, we need to show that $\frac{PE}{ME}=\frac{PF}{NF}$. Apply Sine Theorem for triangles $EPM, FPN$ repsectively: $\frac{PE}{ME}=\frac{sinEMP}{sinEPM}$ $\frac{PF}{NF}=\frac{sinPNF}{sinFPN}$ It suffices to show that $\frac{sinEMP}{sinEPM}=\frac{sinPNF}{sinFPN}$ Now construct the common tangent line of $C,D$ passing through $P$, call it $Px, Py$ (as the diagram). Then note that: $\widehat{EPM}=\widehat{EPx}+\widehat{xPM}=\widehat{EAP}+\widehat{EMP}=180-\widehat{APM}$ $\Rightarrow sinEPM=sinAPM$ Similarly $sinFPN=sinAPN$ The Sine theorem for triangles $AMP,APN$ shows that: $\frac{sinEMP}{sinAPM}=\frac{AP}{AM}=\frac{AP}{AN}=\frac{sinPNF}{sinAPN}$ This completes the proof.
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reveryu
08.05.2017 10:45
sororak wrote: Point $A$ is in the circle $C$ I spent my time for 2 hours solving this problem(and fail) because I think that A is an INTERIOR point.
CT17
05.12.2021 18:46
Is this a fakesolve?? My solution doesn't seem to require $A$ to be on the circumference of $C$. The problem statement is equivalent to showing $\frac{\text{pow}_D(E)}{\text{pow}_D(F)} = \frac{PE^2}{PF^2}$ for 2 arbitrary points $E$ and $F$ on $C$. Let $EP$ and $FP$ intersect $D$ again at $E'$ and $F'$. By negative homothety at $P$, $\triangle PEF\sim \triangle PE'F'$. Then $\frac{PE^2}{PF^2} = \frac{PE}{PF}\cdot \frac{PE + PE'}{PF + PF'} = \frac{EP\cdot EE'}{FP\cdot FF'} = \frac{\text{pow}_D(E)}{\text{pow}_D(F)}$ as desired.
Mahdi_Mashayekhi
09.01.2022 14:32
we will prove PE/ME = PF/MF. PE/ME = sin ∠PME/ sin ∠MPE and PF/MF = sin ∠PNF/ sin ∠NPF. ∠MPE = ∠PAM + ∠PMA = 180 - ∠APM ---> sin ∠MPE = sin ∠APM ∠NPF = ∠PAN + ∠PNA = 180 - ∠APN ---> sin ∠NPF = sin ∠APN PE/ME = sin ∠PME/ sin ∠MPE = sin ∠PME/ sin ∠APM = AP/AM = AP/AN = sin ∠PNF/ sin ∠APN = sin ∠PNF/ sin ∠NPF = PF/MF. we're Done.