Let $\Omega$ and $\Gamma$ two circumferences such that $\Omega$ is in interior of $\Gamma$. Let $P$ a point on $\Gamma$. Define points $A$ and $B$ distinct of $P$ on $\Gamma$ such that $PA$ and $PB$ are tangentes to $\Omega$. Prove that when $P$ varies on $\Gamma$, the line $AB$ is tangent to a fixed circunference.
Problem
Source: 2016 Olympic Revenge, Problem 4
Tags: geometry
toto1234567890
02.01.2017 06:26
Are you sure? I think it should be $ P $ varies ON $ \Gamma $?
LittleGlequius
19.01.2017 17:22
toto1234567890 wrote: Are you sure? I think it should be $ P $ varies ON $ \Gamma $? Sorry. It's edited.
chrono223
31.05.2021 16:05
Nice problem!
defLet) $I$, $O$: center of $\Omega$ and $\Gamma$
$r$, $R$: radius of $\Omega$ and $\Gamma$
$\Omega'$: a circle with center $I$ and radius $r'=\frac{R^2-OI^2}{2R}$
$C$, $D$: intersection points of $\Gamma$ and two tangent lines of $\Omega$ that passes $P$
$M$: midpoint of arc $AB$
$H$: foot of perpendicular from $I$ to $CD$
$Q$: intersection point of $HM$ and $AB$
$J$: point on $IO$ such that $QJ\bot AB$
$X$, $Y$: foot of perpendicular from $M$ to $AB$ and $CD$
by ponslet porism $CD$ is tangent to $\Omega'$
Claim) $HM$: $QM$=$r'^2$: $r^2$
pf$HM$: $QM$=$YM$: $XM$=$CM$ $sin\angle DCM$: $AM$ $sin \angle BAM$
=$sin\angle CPM ^2$: $sin \angle APM^2$
=$r'^2$: $r^2$
by Claim) $IO$: $JO$=$r'^2$: $r^2$, so $J$ is a constant point
as $IH$, $OM$, $IO$: $JO$ are all constant,
$a=JQ$ is also constant'
$\to$ a circle with center $J$ and radius $a$ is the circle we want. $\blacksquare$
mayiyahi mayiyaho mayiyaha mayiyahaha
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aaabc123mathematics
09.12.2022 17:45
chrono223 wrote: Nice problem!
defLet) $I$, $O$: center of $\Omega$ and $\Gamma$
$r$, $R$: radius of $\Omega$ and $\Gamma$
$\Omega'$: a circle with center $I$ and radius $r'=\frac{R^2-OI^2}{2R}$
$C$, $D$: intersection points of $\Gamma$ and two tangent lines of $\Omega$ that passes $P$
$M$: midpoint of arc $AB$
$H$: foot of perpendicular from $I$ to $CD$
$Q$: intersection point of $HM$ and $AB$
$J$: point on $IO$ such that $QJ\bot AB$
$X$, $Y$: foot of perpendicular from $M$ to $AB$ and $CD$
by ponslet porism $CD$ is tangent to $\Omega'$
Claim) $HM$: $QM$=$r'^2$: $r^2$
pf$HM$: $QM$=$YM$: $XM$=$CM$ $sin\angle DCM$: $AM$ $sin \angle BAM$
=$sin\angle CPM ^2$: $sin \angle APM^2$
=$r'^2$: $r^2$
by Claim) $IO$: $JO$=$r'^2$: $r^2$, so $J$ is a constant point
as $IH$, $OM$, $IO$: $JO$ are all constant,
$a=JQ$ is also constant'
$\to$ a circle with center $J$ and radius $a$ is the circle we want. $\blacksquare$
mayiyahi mayiyaho mayiyaha mayiyahaha nice solution