Geo Final but hard to solve with Conics...

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Geo Final but hard to solve with Conics...

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Source: 2025 Korea Winter Program Practice Test P4

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#1 h Jan 18, 2025, 7:13 AM • 3 Y

Y by Rounak_iitr, sashamusta, ehuseyinyigit

Let $\omega$ be the circumcircle of triangle $ABC$ with center $O$ , and the $A$ inmixtilinear circle is tangent to $AB, AC, \omega$ at $D,E,T$ respectively. $P$ is the intersection of $TO$ and $DE$ and $X$ is the intersection of $AP$ and $\omega$ . Prove that the isogonal conjugate of $P$ lies on the line passing through the midpoint of $BC$ and $X$ .

This post has been edited 1 time. Last edited by Seungjun_Lee, Jan 18, 2025, 12:44 PM

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#2 h Jan 18, 2025, 7:23 AM

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What is $O$ ?

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#3 h Jan 18, 2025, 8:59 AM

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@above
$O$ is the center of circle $\omega$ . I’ll tell him to fix the statement

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#4 h Jan 18, 2025, 1:07 PM • 4 Y

Y by Nari_Tom, Stuffybear, ehuseyinyigit, L13832

Let $Y$ be the point on $\omega$ that $AX$ and $AY$ are isogonal, and $S, L$ be the intersection of $XM, YM$ and $\omega$ , respectively. We can see that $SL \parallel BC$ . All polars taken in the solution is a polar wrt the $A$ mixtilinear incircle.

Using Well Known Fact, it suffices to prove that $PM$ and $AQ$ are parallel. Now, as $-1 = (B,C;M, \infty_{BC}) = (YB,YC;YM,Y\infty_{BC}) = (B,C;L,X)$ , we know that $LX$ passes through the intersection of tangents from $B,C$ to $\omega$ . This implies that $O,M,X,L$ are concyclic. Since $\angle YAX = \angle YLX = \angle MLX$ , it suffices to prove that $P,M,X,L$ are concyclic. We will prove that $P,O,M,X$ are concyclic by proving that $\angle XPO = \angle XMO$ .

Since $P$ lies on the $A$ polar, $A$ lies on the polar of $P$ . Since $-1 = (B,C;L,X) = (AD, AE; AL, AP)$ , we can easily see that $AL$ is $P$ polar. This implies that $AL \perp PO$ . Hence, $\angle XPO = \angle (XP, OP) = \angle (XA, LA) + \angle (LA, OP)$ . From $\angle (LA, OP) = \angle (CM, MO)$ and $\angle (XA, LA) = -\angle LAX = -\angle LSX = -\angle CMX = \angle XMC$ , we obtain that $\angle XPO = \angle (XA, LA) + \angle (LA, OP) = \angle XMC + \angle CMO = \angle XMO$ , as desired.

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#5 h Jan 19, 2025, 11:31 AM • 2 Y

Y by ehuseyinyigit, L13832

i swear the motivation is so hard to find so i rate 40 mohs but @above is so orz :love:

:love:

:blush:

or maybe i didn't do enough taiwan-style config geo

https://i.imgur.com/CrDGUBa.png

Let $N$ be the midpoint of major arc $BC$ , and $F$ midpoint of minor arc $BC$ . Let $O'$ be the centre of the mixtilinear incircle. Let $I$ be the incentre of $\triangle ABC$ , and $M$ the midpoint of $BC$ .

We reconstruct $Q$ as the intersection of the reflection of $AP$ over $AI$ with $MX$ . It suffices to show that $QC$ and $CP$ are isogonal in $\angle BCA$ to prove $P$ and $Q$ are isogonal conjugates.

The following results are well-known or trivial to see:

$DE\perp AI$ , and $D,I,E$ collinear.
$O,O', T$ collinear.
$T,I,N$ collinear.
$DBTI$ and $ECTI$ cyclic.
$N,O,M,F$ collinear.
$A,I,O',F$ collinear.

Our solution depends on the following claim:
Claim: $XMOP$ is cyclic.
Proof:
We will show that $\triangle TIX\sim\triangle FMX$ . Clearly, $\angle XFN=\angle XTN$ by Bowtie, hence we just need
$\frac{IT}{MF}=\frac{TX}{FX}$ To conclude by $SAS$ similarity.

Let us proceed by trigonometric bash. First evaluate the right hand side,
$\frac{TX}{FX}=\frac{\sin\angle XFT}{\sin\angle FTX}=\frac{\sin\angle TAX}{\sin\angle FAX}$ However, by Ratio Lemma in triangle $\triangle ATO'$ , we have
$\frac{\sin\angle TAX}{\sin\angle FAX}=\frac{TP}{PO'}\times \frac{AO'}{TA}$ Now, by Ratio Lemma in triangle $\triangle TIO'$ now, we have
$\frac{TP}{PO'}=\frac{IT}{IO'}\times\frac{\sin\angle DIT}{\sin 90^\circ}$ Hence we have
$\frac{TX}{FX}=\frac{AO'}{TA}\times \frac{ IT\sin\angle DIT}{IO'}$ Now, let's move to the left hand side of our initial equation. First of all, we prove an important similarity: $\triangle IMF\sim\triangle AIT$ .

Indeed, by Shooting Lemma and Fact 5 we have $FB^2=FI^2=FM\times FN$ hence $(NIM)$ is tangent to $FI$ . Thus, $\measuredangle FIM=\measuredangle INF=\measuredangle TAF$ , and by Bowtie we have $\angle MFI=\angle ITA$ . This is sufficient due to $AA$ similarity.

Now, this implies that $\frac{IT}{MF}=\frac{AT}{IF}$ . Thus, it simply suffices to show that
$\frac{AT}{IF}=\frac{AO'}{TA}\times\frac{IT\sin\angle DIT}{IO'}$ Now, by Ratio Lemma in $\triangle ATI$ , we have
$\frac{AO'}{IO'}=\frac{AT}{TI}\times\frac{\sin\angle ATO'}{\sin\angle ITO'}$ Hence it simply suffices to show that
$\frac{AT}{IF}=\frac{\sin \angle DIT\sin\angle ATO'}{\sin\angle ITO'}$ Now, $BDIT$ is cyclic as mentioned before, hence
$\sin{\angle DIT}=\sin\angle TBA=\sin\angle ANT=\frac{AI}{NI}$ Because $\triangle ANI$ is right. However, we also have $\triangle AIT\sim\triangle NIF$ because of Bowties and $AA$ similarity. Hence, $\frac{AI}{NI}=\frac{AT}{NF}$ . Hence it suffices to prove
$\frac{NF}{IF}=\frac{\sin\angle ATO'}{\sin\angle ITO'}$ However, from $TO=ON$ by circumcentre definition, $\angle ITO'=\angle ONI$ . Further, we have
$\measuredangle ATO'=\frac{180^\circ-\measuredangle TOA}{2}=90^\circ-\measuredangle TCA=90^\circ-\measuredangle TNA=\measuredangle AIN$ due to angle at centre theorem.

This implies that
$\frac{\sin\angle ATO'}{\sin\angle ITO'}=\frac{\sin \angle AIN}{\sin\angle FNI}$ And the right hand side is $\frac{NF}{IF}$ due to sine law in $\triangle INF$ .

This concludes our trigonometric bash.

Hence, $\triangle TIX\sim\triangle FMX$ .

Note that Reim's Theorem on $(ABC)$ and parallel lines $AN$ and $DE$ (Both are perpendicular to $AI$ ) implies that $IPXT$ is cyclic.

Using our similarity,
$\measuredangle FMX=\measuredangle TIX=\measuredangle TPX=\measuredangle OPX$ Which finally gives $OPMX$ cyclic.

This concludes our claim. $\square$

Lemma: $PM\parallel AQ$
Proof:
We will angle chase:
$\measuredangle MPX=\measuredangle FOX=2\measuredangle FNX=2\measuredangle FAX=\measuredangle QAX$ Where we use $XMOP$ cyclic, angle at centre theorem, and the fact that $AI$ bisects $\angle QAP$ . This suffices by corresponding angles in parallel lines. $\square$ .

Now, in order to show that $CP$ and $CQ$ are isogonal, we will employ DDIT at $C$ on quadrilateral $AQMP$ . Note that $CA$ and $CM$ are clearly isogonal. If we show that $AP\cap QM=X$ and $PM\cap AQ=P_\infty$ are isogonal, this would characterize our involution, then imply $CP$ and $CQ$ also being isogonal.

Indeed, it suffices to show that $\measuredangle(AC,CX)=\measuredangle(NM,BC)$ to prove isogonality. By Pitot Theorem on $\triangle NOT$ , $NIMX$ is cyclic. In particular, it is tangent to $AI$ from our earlier result.

Now we claim that $I$ is the incentre of $\triangle AQX$ as well. Indeed, $AI$ bisects $\angle QAP$ . Further,
$\measuredangle AQX=\measuredangle PMX=\measuredangle POX=180^\circ-2\measuredangle INX=180^\circ-2\measuredangle FIX=2\measuredangle AIX-180^\circ$ Which suffices by well-known incentre angle formula, where we used angle at center theorem and $XMOP$ cyclic.

Now, let us angle chase:
$\begin{align*}\measuredangle ACX&=\measuredangle ANX\\ &=\measuredangle INX+\measuredangle ANI\\ &=\measuredangle FIX+\measuredangle AXT\\ &=\measuredangle FIX+\measuredangle IXF\\ &=180^\circ-\measuredangle XFA\\ &=180^\circ-\frac{1}{2}\measuredangle XOA\\ &=90^\circ+\measuredangle PXO\\ &=90^\circ+\measuredangle PMO\\ &=\measuredangle PMB \end{align*}$ As desired, where we used $(NIMX)$ tangent to $AI$ , $MOPX$ cyclic, and multiple Bowties, as well as because from $XI$ bisecting $\measuredangle AXM$ ,
$\measuredangle AXI=\measuredangle IXM=\measuredangle TXF,$ implying $\measuredangle AXT=\measuredangle IXF$ . (Similar switch gives $\triangle IMX\sim\triangle TFX$ , and our angle equality.)

This concludes the problem, as $PC$ and $QC$ are now isogonal, and hence $Q,P$ are isogonal conjugates, but $Q$ lies on $MX$ , so we are done.

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#6 h Apr 2, 2025, 8:30 PM

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Nice problem man! Cool solutions above :orz:

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