Tangent circles

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Tangent circles

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Source: Own-inspired from RMO 2016

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#1 h May 9, 2016, 5:56 PM • 5 Y

Y by hoangA1K44PBC, kun1417, uraharakisuke_hsgs, Adventure10, Mango247

Given triangle $ABC$ inscribed in $(O)$ . Two tangent of $(O)$ at $B,C$ meet at $P$ . Let $A'$ be the antipode of $A$ . $BA',CA'$ intersect $AM$ at $L,K$ , respectively. The lines through $L$ and perpendicular to $BA',$ through $K$ and perpendicular to $CA'$ and line $OP$ intersect and form triangle $XYZ$ . Prove that $(XYZ)$ is tangent to $(AMP).$

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#2 h May 9, 2016, 10:35 PM • 7 Y

Y by livetolove212, aopser123, langkhach11112, Letteer, enhanced, Adventure10, Mango247

Let the A-median $AM,$ A-symmedian of $\triangle ABC$ cuts $\odot (O)$ at $U,$ $V,$ respectively and let $E$ $\equiv$ $A'B$ $\cap$ $OM.$ From $\measuredangle CEM$ $=$ $\measuredangle MEB$ $=$ $90^{\circ}$ $-$ $\measuredangle EBC$ $=$ $\measuredangle CBA$ $=$ $\measuredangle CUM$ $\Longrightarrow$ $C,$ $E,$ $M,$ $U$ are concyclic, so we get $\measuredangle LUE$ $=$ $\measuredangle MCE$ $=$ $\measuredangle EBC$ $=$ $\measuredangle LXE$ $\Longrightarrow$ $E,$ $L,$ $U,$ $V,$ $X$ are concyclic, hence $X$ lies on $CU$ and $BV.$ Similarly, we can prove $Y$ $\in$ $BU,$ $CV.$

Let $T$ be the Miquel point of the complete quadrilateral with the sides $BC,$ $OM,$ $BX,$ $CY.$ Since $A,$ $P,$ $V$ are collinear, so we get $\measuredangle PAT$ $=$ $\measuredangle VBT$ $=$ $\measuredangle PMT$ $\Longrightarrow$ $T$ $\in$ $\odot (AMP),$ hence notice $\measuredangle MTX$ $+$ $\measuredangle MTY$ $=$ $\measuredangle CBV$ $+$ $\measuredangle BCV$ $=$ $\measuredangle CAV$ $+$ $\measuredangle MAC$ $=$ $\measuredangle MAV$ $=$ $\measuredangle MTP$ $\Longrightarrow$ $\measuredangle MTX$ $=$ $\measuredangle YTP$ we get $\odot (VXY)$ is tangent to $\odot (AMP)$ at $T.$ Finally, it's easy to see $V$ lies on $\odot (XYZ),$ so we conclude that $\odot (XYZ)$ is tangent to $\odot (AMP)$ at $T.$

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uraharakisuke_hsgs

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uraharakisuke_hsgs

#3 h May 10, 2016, 11:09 AM • 4 Y

Y by livetolove212, langkhach11112, HolyMath, Adventure10

My solution :
Let the $A$ - symmedian line of $\triangle ABC$ cuts $(O)$ at $P$ , The line perpendicular with $AP$ cuts $BC$ at $G$
$P'$ reflects with $P$ about $O$ , $Q$ is the intersection of 2 tangents at $B$ and $C$ of $(O)$
$GP$ cuts $(O)$ at $S$ . $AS$ cut $PP'$ at $R$ . By Pascal => $R$ lies on $GG'$ with $G'$ is the intersection of 2 tangents at $A$ and $P$ => $R$ lies on $BC$
We have : $(GRBC) = -1 => P(GRBC) = -1 => P(SP'BC) = -1 => (SP'BC) = -1$ => $P,S,Q$ are collinear so $A,M,S,Q,G$ are coincyclic
$\angle MCY = \angle MKY = \angle MAC = \angle PAB$ => $C,Y,P$ are collinear
=> $\angle YCS = \angle PAS = \angle QMS$ => $Y,M,C,K,S$ are concyclic
Similary => $S,B,M,L,X$ are concyclic
$\angle LKZ = \angle LA'Z = \angle BCP$ => $A' , Z , P$ are collinear
$\angle ZA'S = \angle SAP = \angle PCS = \angle YKS$ => $L,Z,K,S,A'$ are concyclic => $A,Z,S$ are collinear
$\angle YZX = \angle LZK = 180 - \angle LA'K = 180 - \angle YPX$ so $YPXZ$ is cyclic
$\angle ZSP = 90 - \angle P'SZ = 90 - \angle ACP' = 90 - \angle PCK = \angle CYK$ => $P,Y,Z,X,S$ are concyclic
In $\triangle SAG$ , we have : $XP \parallel AG$ => $(SZP)$ tangent $(SAG)$
=> $(XYZ)$ tangent $(ASQ)$
P/S : In my solution , the intersection of 2 tangents at $B$ and $C$ is $Q$ not $P$ , sorry for that mistake

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#4 h May 10, 2016, 9:36 PM • 2 Y

Y by livetolove212, Adventure10

Let $S\equiv A'Z\cap (ABC),Q\equiv (XYZ)\cap (MYKC$ i.e the miquel point of $MLYZ$ then $Q\in (LZKA')$ that 's means $Q$ is the Miquel point of $MLCA'$ so $Q$ is on $(BA'C)$ and on $(BLM)$ thus BLMXQ cyclic.
$\widehat{SAB}=\widehat{SA'B}=\widehat{ZA'L}=\widehat{LKZ}\overset{||}{=\widehat{LAC}}$ thus AS is the $A$ -symmedian .
$\widehat{QSZ}=\widehat{QSA'}=\widehat{QBA'}\widehat{QBL}=\widehat{QXL}=\widehat{QXZ}$ thus SXYZ cyclic.
$\widehat{A'QZ}=\frac{\pi}{2}$ so $A,Q,Z$ are colinear .
$\widehat{ZQS}=\widehat{AQS}=\widehat{AA'S}=\widehat{OA'S}=\widehat{OSA'}=\widehat{OSZ}$ which means $OS$ is tangent to $(XYZ)$ then $OQ$ is tangent to $(XYZ)$ .
$\widehat{QAS}=\widehat{QA'S}=\widehat{QA'Z}=\widehat{QKZ}=\widehat{QKY}=\widehat{QMY}$ so $AMQP$ are cyclic but we know $OM\cdot OP=R^2$ hence $OQ$ is tangent to $(AMP)$

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#5 h May 11, 2016, 5:03 AM • 2 Y

Y by Adventure10, Mango247

Actually $XYZ$ and $BCA'$ are paralogic triangles. Then $(XYZ)$ is orthogonal to $(O)$ . Let $E$ and $F$ be the midpoints of arc $BC$ and $BAC$ we get $OE^2=OM\cdot OP$ then $(AMP)$ is also orthogonal to $(O)$ . So we only need to prove that $(XYZ), (AMP)$ and $(O)$ are concurrent.
We have some properties of paralogic triangles as following:
$(XYZ)$ intersects $(O)$ at two points, one is the intersection of $BX,CY,A'Z$ and another is Miquel point of complete quadrilateral $BCA'MLK$ . For proof you can see my article (in Vietnamese).
Let $T$ and $R$ be the intersections of $(XYZ)$ and $(O)$ ( $T$ is Miquel point of $BCA'MLK$ , $R$ is the intersection of $BX,CY,A'Z$ ).
We have $\angle RAB=\angle RCB=\angle MKY=\angle MAC$ then $R\in AP$ . Therefore $\angle PMT=\angle RCT=\angle PAT$ or $T\in (AMP).$ We are done.

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#6 h May 11, 2016, 5:30 AM • 2 Y

Y by Adventure10, Mango247

From these properties of paralogic triangles, we have generalization:
Given triangle $ABC$ inscribed in $(O)$ . Let $P$ be an arbitrary point on $BC$ , $A'$ be the antipode of $A$ . $BA',CA'$ intersect $AP$ at $K,L$ , respectively. $BY$ meets $CZ$ at $R$ . $AR$ meets line through $P$ and perpendicular to $BC$ at $Q$ . Prove that $(APQ)$ is tangent to $(XYZ)$ .

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#7 h Nov 20, 2019, 3:42 PM • 1 Y

Y by Adventure10

livetolove212 wrote:

Given triangle $ABC$ inscribed in $(O)$ . Two tangent of $(O)$ at $B,C$ meet at $P$ . Let $A'$ be the antipode of $A$ . $BA',CA'$ intersect $AM$ at $L,K$ , respectively. The lines through $L$ and perpendicular to $BA',$ through $K$ and perpendicular to $CA'$ and line $OP$ intersect and form triangle $XYZ$ . Prove that $(XYZ)$ is tangent to $(AMP).$

Let $Q$ $\equiv$ $(AMP)$ $\cap$ $(O)$
We have: $\angle{BQM} = \angle{BQA} + \angle{AQM} = \angle{ACB} + \angle{APM} = \angle{ACB} + 90^o - \angle{AMB} - \angle{PAM} = 90^o - \angle{MAC} - \angle{PAM}$ $= 90^o - \angle{PAC} = 90^o - \angle{BAM} = \angle{BLM}$
But: $\angle{XMB} = \angle{XLB} = 90^o$ then: $X$ , $Q$ , $L$ , $M$ , $B$ lie on a circle
Similarly: $K$ , $Q$ , $Y$ , $M$ , $C$ lie on a circle
Hence: $Q$ is Miquel point of completed quadrilateral $YMLZ.XK$ or $Q$ $\in$ $(XYZ)$
Let $C'$ be reflection of $C$ through $O$
We have: $\angle{OQY} = \angle{QC'O} = \angle{KA'Q} = \angle{KZQ} = \angle{QXY}$
So: $OQ$ is common tangent of $(AMP)$ and $(XYZ)$

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#8 h Nov 21, 2019, 2:47 AM • 1 Y

Y by Adventure10

livetolove212 wrote:

From these properties of paralogic triangles, we have generalization:
Given triangle $ABC$ inscribed in $(O)$ . Let $P$ be an arbitrary point on $BC$ , $A'$ be the antipode of $A$ . $BA',CA'$ intersect $AP$ at $K,L$ , respectively. $BY$ meets $CZ$ at $R$ . $AR$ meets line through $P$ and perpendicular to $BC$ at $Q$ . Prove that $(APQ)$ is tangent to $(XYZ)$ .

Let $U$ $\equiv$ $(APQ)$ $\cap$ $(O)$
We have: $\angle{BRC} = 180^o - \angle{RBC} - \angle{RCB} = \angle{BYP} + \angle{CZP} = \angle{BKP} + \angle{CLP} = 180^o - \angle{BAP} - \angle{CAP}$ $= 180^o - \angle{BAC}$
Then: $R$ $\in$ $(O)$
But: $\angle{CAR} = \angle{CBR} = 90^o - \angle{BYP} = 90^o - \angle{BKA} = \angle{BAP}$ so: $AP$ , $AQ$ are isogonal conjugate in $\triangle ABC$
We also have: $\angle{RAO} = 90^o - \angle{ACR} = 90^o - \angle{APB} = \angle{APQ}$
Then: $AO$ tangents $(APQ)$ at $A$
But: $\angle{BUP} = \angle{BUA} + \angle{AUP} = \angle{BCA} + \angle{APQ} + \angle{PAQ} = \angle{BCA} + 90^o - \angle{APB} + \angle{PAQ}$ $= 90^o - \angle{QAC} = 90^o - \angle{BAP} = \angle{BKP}$ so: $U$ , $Y$ , $K$ , $P$ , $B$ lie on a circle
Similarly: $U$ , $P$ , $C$ , $L$ , $Z$ lie on a circle
Hence: $U$ is Miquel point of completed quadrilateral $KLZY.PX$ or $U$ $\in$ $(XYZ)$
We also have: $\angle{YUO} = \angle{BUY} - \angle{BUO} = 90^o - \angle{BUO} = \angle{BAU} = \angle{BCU} = \angle{PZY} = \angle{UXY}$
Then: $OU$ is common tangent $(APQ)$ and $(XYZ)$

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