IMOC 2020 G6 2 tangents of different circle concurrent with a line

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IMOC 2020 G6 2 tangents of different circle concurrent with a line

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Source: https://artofproblemsolving.com/community/c6h2254883p17398793

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#1 h Sep 1, 2020, 8:33 PM

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Let $ABC$ be a triangle, and $M_a, M_b, M_c$ be the midpoints of $BC, CA, AB$ , respectively. Extend $M_bM_c$ so that it intersects $\odot (ABC)$ at $P$ . Let $AP$ and $BC$ intersect at $Q$ . Prove that the tangent at $A$ to $\odot(ABC)$ and the tangent at $P$ to $\odot (P QM_a)$ intersect on line $BC$ .

(Li4)

This post has been edited 2 times. Last edited by parmenides51, Dec 15, 2022, 7:26 PM

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#2 h Sep 1, 2020, 8:49 PM • 2 Y

Y by Kagebaka, Gaussian_cyber

$\textbf{LEMMA:-}$ $ABC$ be a triangle and let $\Delta M_AM_BM_C$ be the Medial Triangle of $\Delta ABC$ . Let the tangent to $\odot(ABC)$ at $C$ intersect $\overline{M_BM_C}$ at $P$ and let $\overline{M_AM_B}$ hit $\odot(ABC)$ at $\{X\}$ . Then $\{\overline{XA},\overline{XP}\}$ are Isogonal Conjugates WRT $\Delta CXM_C$ .

Let $\overline{XP}\cap\overline{BC}=\{Q\}$ and $\overline{AC}\cap\overline{XM_C}=\{R\}$ . Then $\frac{RX}{RM_C}=\frac{M_BX}{M_AM_B}=\frac{PX}{PQ}$ . Now, $\frac{PX}{PQ}=\frac{CX}{CQ}\cdot\frac{\sin\angle PCX}{\sin\angle QCP}=\frac{CX}{CQ}\cdot\frac{\sin\angle XAR}{\sin\angle M_CAR}=\frac{RX}{RM_C}=\frac{XA}{AM_C}\cdot\frac{\sin\angle XAR}{\sin\angle M_CAR}\implies\frac{CX}{CQ}=\frac{AX}{AM_C}$ also we know that $\measuredangle X_AMC=\measuredangle XCQ\implies\Delta XAM_C\cup R\stackrel{+}{\sim}\Delta XCQ\cup P$ . So, $\{\overline{XA},\overline{XP}\}$ are Isogonal Conjugates WRT $\Delta CXM_C$ .
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Coming back to the problem. Let the tangent to $\odot(ABC)$ at $A$ intersect $\odot(ABC)$ at $V$ and let $\overline{M_AM_B}\cap\overline{AV}=\{U\}$ . Now by Dual of Desargues Involution Theorem on the Complete Quadrilateral $\mathbf{Q}\equiv\{\overline{M_AM_B},\overline{AM_B},\overline{AV},\overline{M_AV}\}$ we get that there is an Involution $\Phi$ which swaps $\{(\overline{PA},\overline{PM_A}),(\overline{PM_B},\overline{PV}),(\overline{PU},\overline{PC})$ . Now by $\textbf{LEMMA}$ we already know that $\{\overline{PU},\overline{PC}\}$ are Isogonal Conjugates WRT $\Delta APM_A$ . So, from $\Phi$ $(\overline{PM_B},\overline{PB})$ are Isogonal Conjugates too WRT $\Delta APM_A$ . So, $\measuredangle VPM_A=\measuredangle APM_C=\measuredangle PQM_A\implies\overline{PV}$ is tangent to $\odot(PM_AQ)$ . $\blacksquare$

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#3 h Sep 2, 2020, 2:18 AM • 2 Y

Y by Beginner2004, soryn

Nice problem!

Let $M_b M_c$ intersect $\odot (ABC)$ at $R$ , and $AR$ intersect $BC$ at $S$ .
Also, define $K$ as the intersection of the tangent to $\odot (RSM_a)$ at $R$ and the tangent to $\odot (ABC)$ at $A$ .
Then, $\angle KAP=180^\circ -\angle ARP=180^\circ -\angle ASQ = \angle KRM_a$ .
$\frac{KR}{KA}=\frac{\sin\angle RAK}{\sin\angle KRA}=\frac{\sin\angle APR}{\sin\angle M_a RP} =\frac{\triangle APR}{\triangle RM_a P}\times\frac{PM_a }{AP}=\frac{RM_a }{AP}$

So, $\triangle KRM_a \sim\triangle KAP \rightarrow \triangle KPM_a \sim \triangle KAR$
Then $\angle KPM_a=\angle KAR=\angle APR=\angle PQM_a$ , which shows that $KP$ is a tangent to $\odot (PQM_a)$
Also, $\angle KM_a P=\angle KRA =\angle RM_a C=\angle BM_aP$ ,and hence, $K$ lies on $BC$

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#5 h Oct 1, 2020, 8:59 AM • 2 Y

Y by SerdarBozdag, tiendung2006

$\textbf{Notation:}$
Let $K$ be the meeting point of $BC$ and the tangent at $A$ to $(ABC)$ .
$A'\in (ABC)$ such that $AA'\parallel BC$ .
$N$ the midpoint of $AA'$

$\textbf{Solution:}$
1- with just angle working you get $\triangle AQK\sim A'AP$ , then $\angle KPQ = \angle PNA$
2- line $PM_cM_a$ bisects $NM_a$ then $\angle PNA = \angle PM_aB$
3- Combining the previous $\angle KPQ = \angle PM_aB$ , then $KP$ is tangent to $(PQM_a)$ .

This post has been edited 1 time. Last edited by hectorraul, Oct 1, 2020, 8:59 AM

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#6 h Aug 15, 2022, 9:04 AM

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Let $M_bM_c$ meets $(ABC)$ again at $X$ .
Let $AP\cap XM_a=R$ , and $AX\cap PM_a=S$
Since $PX$ passes through the midpoint of $AM_a$ , by Ceva on $\triangle APM_a$ and point $X$ we have $AM_a\parallel RS$ .
By Menelaus on $\triangle RPM_a$ and $\overline{AXS}$ , we have $\frac{RA}{AP}\cdot\frac{PS}{SM_a}\cdot\frac{M_aX}{XR}=1$
$\frac{XM_a}{SM_a}=\frac{PM_a}{SM_a}=\frac{PA}{RA}$ , so $PS=XR$ .

Let perpendicular bisector of $PR$ and $XS$ meet at $K$ .
$PK=RK$ and $XK=SK$ , so $\triangle KSP\equiv\triangle KXR$ , and therefore $\triangle KSX\sim\triangle KPR$
$\angle KSX=\angle KPR$ , $\angle KXS=\angle KRP$ so we have that $APSK, ARKX$ is cyclic.
It shows that $K$ is a Miquel point of quadrilateral $APM_aX$ .

$\angle PM_aX=\angle PKR$ , so $\triangle PM_aX\sim\triangle PKR\sim\triangle SKX$
$\angle XM_aC=\angle M_aXP=\angle KXS=\angle XSK=\angle XM_aK$ , so $K$ is on line $BC$ .
$\angle KAX=\angle KPS=\angle APX=\angle AQK$ , so $KA, KP$ is the tangent of $(APX), (PQM_a)$ , respectively.
Therefore, $K$ is the concurrency point.

Remark : This is the similar idea to 2019 RMM #2 :-D

:-D

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#8 h Mar 23, 2024, 3:15 PM • 1 Y

Y by Vahe_Arsenyan

We let $T$ be the intersection of the tangent through $A$ with $BC$ and wish to show a tangency. Let $K$ be the intersection of the $A$ -symmedian with $(ABC)$ . Now it is well-known that $KM_A \cap (ABC) = A'$ is such that $AA'\parallel BC$ . This implies (after an easy angle chase) that $QPM_AK$ is cyclic. Now we perform $\sqrt{bc}$ inversion. Notice how this sends $T\to A'$ , $M_A \to K$ , and we'll call $Q'$ and $P'$ to the images of $Q$ and $P$ respectively. Clearly $P\neq Q'=M_BM_C \cap (ABC)$ and $P'=AQ'\cap BC$ . So $(QPM_AK) \to (KM_AQ'P')$ and $PT \to (AA'P)$ . We will show they both are tangent at $P'$ , which will end the problem. Let $E$ be the antipode of $P'$ in $(KM_AQ'P')$ . Clearly $\angle EM_AP'=90^\circ$ , so $E$ lies on the perpendicular bisector of $BC$ , which coincides with the perpendicular bisector of $AA'$ . So we just need to show that $EA=EP'$ to end the problem. But this is triavial as $AQ'=Q'P'$ and $\angle AQ'E=\angle EQ'P'=90^\circ$ implies that triangles $AQ'E$ and $PQ'E$ are congruent, so we are done.

This post has been edited 1 time. Last edited by Bigtaitus, Mar 23, 2024, 3:17 PM

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#10 h Oct 11, 2024, 9:39 AM

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Let $A-$ symedian intersect $(ABC)$ at $D$ and $T$ be the intersection of the tangent to $(ABC)$ at $A$ and $BC$ .
Claim: $P,Q,M_A,D$ are concyclic.
Proof: If $S\in (ABC),AS\parallel BC,$ then $(A,D;B,C)=-1=(SA,SM_a;SB,SC)=(A,SM_a\cap (ABC);B,C)$ thus, $S,M_a,D$ are collinear.
$\measuredangle DPQ=\measuredangle DSA=\measuredangle DM_AQ$ Which completes the proof. $\square$
Let $O$ be the circumcenter of $(ABC)$ . Now invert around $(ABC)$ .
New Problem Statement: $ABC$ is a triangle with circumcenter $O$ and midpoints of $BC,CA,AB$ are $M_a,M_b,M_c$ respectively. The ray $M_bM_c$ intersects $(ABC)$ at $P$ and $(OAP)$ meets $(OBC)$ at $Q$ . $D$ is the intersection of $A-$ symedian with $(ABC)$ and $T$ is the midpoint of $AD$ . Prove that $(OPT)$ and $(PQD)$ are tangent to each other.
We observe that $T$ is $A-$ dumpty point. Take $\sqrt{bc}$ inversion and reflect over the angle bisector of $\measuredangle A$ .
New Problem Statement: $ABC$ is a triangle with midpoints $M_a,M_b,M_c$ . $K$ is the altitude from $A$ to $BC$ and the ray $M_cM_b$ intersects $(ABC)$ at $P$ . If $D$ is the midpoint of $M_bM_c$ and $KP$ meets $(M_aM_bM_cK)$ at $Q,$ then show that $(PM_aK)$ and $(PQD)$ are tangent.
Let $L$ be the reflection of $P$ with respect to $D$ and $H$ be the orthocenter of $\triangle ABC$ . Let $X$ be the midpoint of $XP$ . $H$ is the center of homothety sends the ninepoint circle to $(ABC)$ hence $N$ lies on $(M_aM_bM_cK)$ .Let $N$ be the midpoint of $AH$ . $AP\parallel NX$ and $PA=PK$ hence
$\measuredangle XNK=\measuredangle PAK=\measuredangle AKQ=\measuredangle NKQ$ And $X,Q,K,N$ are concyclic thus, $XQ\parallel NK$ . This implies $Q$ is the midpoint of $KP$ . Note that $PLKM_a$ is an isosceles trapezoid because the perpendicular to $M_bM_c$ at $D$ is the perpendicular bisector of both $KM_a$ and $LP$ . If $l$ is the tangent line to $(PM_aK)$ at $P$ , then
$\measuredangle PDQ=\measuredangle PLK=\measuredangle PM_aC=\measuredangle PKM_a+\measuredangle M_aPK=\measuredangle M_aPK+\measuredangle (l,PM_a)=\measuredangle (l,PQ)$ Thus, $l$ is tangent to $(PDQ)$ as desired. $\blacksquare$

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#11 h Feb 28, 2025, 10:05 AM • 1 Y

Y by trigadd123

Let $P'$ be the second intersection of $EF$ with $(ABC)$ and let $T$ be the intersection of the tangent in $A$ with $BC$ . Also, let $X$ be a random point on the extension of $AQ$ beyond $Q$ . By the cotangent lemma, $\cot (\angle TPQ)=\frac{\cot (\angle TQX) + \cot(\angle TAQ)}{2}=\frac{\cot(\angle APP') + \cot(\angle AP'P)}{2}=\frac{PP'}{h_a}.$ On the other hand, $\cot(\angle PDB)=\frac{\cot(\angle PBT)+\cot(\angle PCB)}{2}=\frac{\cot (\angle BPP') + \cot(\angle BP'P)}{2}=\frac{PP'}{h_a}.$ So anyways, $\cot(\angle TPQ)=\cot(\angle PDB)$ , implying that $\angle TPQ = \angle PDB$ , i.e. $TP$ is tangent to $(QPD)$ .

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