circumcircle tangent to given circle

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circumcircle tangent to given circle

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Source: INAMO Shortlist 2015 G8

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#1 h May 14, 2019, 4:24 PM • 2 Y

Y by brain_idea, Adventure10

$ABC$ is an acute triangle with $AB> AC$ . $\Gamma_B$ is a circle that passes through $A,B$ and is tangent to $AC$ on $A$ . Define similar for $\Gamma_C$ . Let $D$ be the intersection $\Gamma_B$ and $\Gamma_C$ and $M$ be the midpoint of $BC$ . $AM$ cuts $\Gamma_C$ at $E$ . Let $O$ be the center of the circumscibed circle of the triangle $ABC$ . Prove that the circumscibed circle of the triangle $ODE$ is tangent to $\Gamma_B$ .

This post has been edited 2 times. Last edited by parmenides51, Aug 25, 2021, 9:12 PM
Reason: typo see #1, AM cuts \Gamma_C at E, instead of AM cuts C at E

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#3 h May 15, 2019, 10:45 PM • 2 Y

Y by Adventure10, Mango247

Should say " $AM$ cuts $\Gamma_C$ at $E$ " rather than " $AM$ cuts $C$ at $E$ ."

Firstly, note that it's well-known that $D$ is the midpoint of the $A-$ symmedian chord of $\triangle ABC.$ Therefore, we have that $\angle EAC = \angle MAC = \angle DAB = \angle ACD.$ Therefore, we have that $DE || AC.$ Therefore, as $ADEC$ is an isosceles trapezoid and $OA = OC$ , we know that $OD = OE.$ Let $O_b, O_c$ be the centers of $\Gamma_C, \Gamma_B$ , respectively (yes we switched them up). The following lemma is now helpful.

Lemma. $O_bODO_c$ is an isosceles trapezoid.

Proof. Observe that $OO_c \perp AB, AO_b \perp AB, AO_c \perp AC, OO_b \perp AC$ imply that $AO_bOO_c$ is a parallelogram. Therefore, we have that $OO_c = AO_b = DO_c$ and $OO_b = AO_c = DO_c.$ Together with $DO = OD$ , we have that $\triangle O_cOD \cong \triangle O_bDO.$ This, together with $O_bO_c \perp AD, OD \perp AD$ , gives the desired conclusion.

$\blacksquare$

Let's now define $P = O_cD \cap O_bO.$ Since $P \in O_bO$ , we have that $PD = PE.$ On the other hand, the lemma implies by symmetry that $PO = PD.$ Therefore, $P$ is actually the circumcenter of $\triangle ODE.$ Hence, since $P \in DO_c$ , it's readily seen that $(\triangle ODE)$ is tangent to $\Gamma_B,$ as desired.

$\square$

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#4 h Jun 4, 2019, 5:32 AM • 1 Y

Y by Adventure10

Pathological wrote:

Should say " $AM$ cuts $\Gamma_C$ at $E$ " rather than " $AM$ cuts $C$ at $E$ ."

Firstly, note that it's well-known that $D$ is the midpoint of the $A-$ symmedian chord of $\triangle ABC.$ Therefore, we have that $\angle EAC = \angle MAC = \angle DAB = \angle ACD.$ Therefore, we have that $DE || AC.$ Therefore, as $ADEC$ is an isosceles trapezoid and $OA = OC$ , we know that $OD = OE.$ Let $O_b, O_c$ be the centers of $\Gamma_C, \Gamma_B$ , respectively (yes we switched them up). The following lemma is now helpful.

Lemma. $O_bODO_c$ is an isosceles trapezoid.

Proof. Observe that $OO_c \perp AB, AO_b \perp AB, AO_c \perp AC, OO_b \perp AC$ imply that $AO_bOO_c$ is a parallelogram. Therefore, we have that $OO_c = AO_b = DO_c$ and $OO_b = AO_c = DO_c.$ Together with $DO = OD$ , we have that $\triangle O_cOD \cong \triangle O_bDO.$ This, together with $O_bO_c \perp AD, OD \perp AD$ , gives the desired conclusion.

$\blacksquare$

Let's now define $P = O_cD \cap O_bO.$ Since $P \in O_bO$ , we have that $PD = PE.$ On the other hand, the lemma implies by symmetry that $PO = PD.$ Therefore, $P$ is actually the circumcenter of $\triangle ODE.$ Hence, since $P \in DO_c$ , it's readily seen that $(\triangle ODE)$ is tangent to $\Gamma_B,$ as desired.

$\square$

sorry, can you please elaborate the well-known lemma? seems pretty foreign to me.. thankyou

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#6 h Jun 5, 2019, 8:08 AM • 5 Y

Y by TCITC2, Adventure10, Mango247, Mango247, Mango247

Let $D'=DO \cap \Gamma_B$ and $K=AD\cap CE.$ Note that $AD' \perp AC.$ Clearly, $D$ lies on the $A$ -symmedian of $\triangle{ABC} \implies$ $\angle{CAD}=\angle{BAM}=\angle{ACE} \implies \angle{ACK}=\angle{CAK} \implies OK \perp AC$ Hence, because $AD'\mid \mid OK,$ by homothety it is enough to show that $ODKE$ is cyclic
$\angle{OME}=90-\angle{AMC}=90-\angle{B}-\angle{MAB}=\angle{OCA}-\angle{ACE}=\angle{OCE}\implies \angle{OEK}=90$ Because $\angle{ADC}=\angle{ADB}=180-\angle{A},$ we have that $AD$ is the internal bisector of $\angle{CDB}$ and $CDOB$ is cyclic, and since $O$ is on the bisector of $BC$ this means that $DO$ is the external bisector of $\angle{CDB},$ implying that $\angle{KDO}=\angle{ADO}=90 \ \ \ \blacksquare$

This post has been edited 1 time. Last edited by rmtf1111, Jun 5, 2019, 8:13 AM

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#7 h Feb 15, 2021, 1:32 PM • 1 Y

Y by Muaaz.SY

well-known that $D$ is the $A$ -dumpty point so $ODBC$ is cyclic and $AD$ is $A$ -symmedian in $\triangle ABC$
let $F=BD \cap AM$ and $CE \cap (ABC) =C'$
cliam(1): $OEMC$ is cyclic
proof:
$\angle C'BA=\angle C'CA=\angle EAB$ so $C'B \parallel AM$ so $\angle OEC=90$
$\blacksquare$
claim(2): $DFOE$ is cyclic
proof:
$\angle FDO=\angle OCB=\angle OEF$
$\blacksquare$
$\angle BAD=\angle BCO=\angle FEO$ this will imply that the two circles are tangent

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#8 h Feb 18, 2021, 12:28 AM • 1 Y

Y by Eagle143

Let $K , L$ be the midpoint of $AC , AB , F$ the projection of $A$ on $BC , J$ the intersection of $ML$ with the $A$ -symmedian
Consider an inversion around $A$ with power of $\sqrt{AB.AC}$ followed by a reflection over the $\hat{A}$ bisector then $B$ and $C$ swap and $AM , \Gamma_B$ and $\Gamma_C$ map to the $A$ -symmedian , a line through $C$ parallel to $AB$ , and a line through $B$ parallel to $AC$ respectively
so $D$ maps to the reflection of $A$ over $M , O$ maps to the reflection of $A$ over $BC$ (well-known lemma) , $E$ maps to the intersection of the $A$ -symmedian with the line through $B$ parallel to $AC$
take a dilation at $A$ with ratio $0.5$ and consider the images of these under the last three transformations:
$O \mapsto F$
$D \mapsto M$
$E \mapsto J$
$\Gamma_B \mapsto MK$
$\angle{LAM}=\angle{CAJ}=\angle{AJM}$
So $LF^2=LA^2=LM.LJ$
Hence $\angle{KMF}=\hat{B}=\angle{LFM}=\angle{MJF}$
implying that $MK$ tangents $(MJF)$ as desired

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