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Source: 2025 Israel TST Test 1 P2

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395 posts

#1 h Oct 28, 2024, 8:05 PM • 2 Y

Y by ehuseyinyigit, Rounak_iitr

Triangle $\triangle ABC$ is inscribed in circle $\Omega$ . Let $I$ denote its incenter and $I_A$ its $A$ -excenter. Let $N$ denote the midpoint of arc $BAC$ . Line $NI_A$ meets $\Omega$ a second time at $T$ . The perpendicular to $AI$ at $I$ meets sides $AC$ and $AB$ at $E$ and $F$ respectively. The circumcircle of $\triangle BFT$ meets $BI_A$ a second time at $P$ , and the circumcircle of $\triangle CET$ meets $CI_A$ a second time at $Q$ . Prove that $PQ$ passes through the antipodal to $A$ on $\Omega$ .

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#2 h Oct 28, 2024, 8:29 PM • 1 Y

Y by ehuseyinyigit

Trivially from I-E Lemma it's enough to prove that $PQ$ is the perpendicular bisector of $II_A$ , let $M$ midpoint of minor arc $BC$ and let $NI_A \cap EF=U$ , first note that:
$\angle UTB=90-\frac{\angle A}{2}=\angle AFE=\angle AEF=\angle UTC \implies TBFU, TCEU \; \text{cyclic}$ Now we also have $\angle UIM=90=\angle UTM$ so $IUTM$ is cyclic, so by PoP:
$I_AB \cdot I_AP=I_AU \cdot I_AT=I_AC \cdot I_AQ=I_AI \cdot I_AM$ Which imply that $BPMI, CQMI$ are both cyclic and also since $M$ is center of $(BICI_A)$ we notice that this means $\angle IMP=90=\angle IMQ$ therefore not only $P,M,Q$ are colinear, in fact they lie on the perpendicular bisector of $II_A$ , thus we are done :cool:

:cool:

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#3 h Oct 28, 2024, 10:19 PM

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Is P3 posted? Or is it from ISL?

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#4 h Oct 29, 2024, 10:07 AM • 1 Y

Y by ehuseyinyigit

Let $S = NI_a \cap EF$ . From easy angle chase, we have cyclic pentagons $(ESTQC)$ and $(FSTPB)$ . So we are done if we prove image of $A'$ in $\sqrt{I_a T \cdot I_a S}$ inversion lies on $(BIC)$ . Let $I_aA' \cap (BIC)$ meet again in point $X$ and $M$ the midpoint of $II_a$ . From Reim's, $SIMT$ cyclic, and also $MIA'X$ cyclic by angle chase. So the $I_aT \cdot I_aS = I_aA' \cdot I_a X$ , done.

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cursed_tangent1434

586 posts

cursed_tangent1434

#5 h Oct 31, 2024, 4:30 PM

Y by

I thought I was smart when I noted that $T$ was the $A-$ mixtillinear extouch point, but it actually turns out that knowing that doesn't help you much with the solution. Pretty straightforward for Israel standards. We denote by $X$ the intersection of lines $I_AN$ and $EF$ . We start off with some easy observations.

Claim : Point $X$ lies on $(BFT)$ and $(CET)$ .
Proof : Note that,
$\measuredangle BTX = \measuredangle BTN = \measuredangle BCN = \measuredangle AFE = \measuredangle AFX$ Thus, $BFXT$ must be cyclic. Similarly, $CEXT$ is also cyclic, and we have our claim.

Now note that since lines $\overline{BP}$ , $\overline{CQ}$ and $\overline{XT}$ are concurrent at $I_A$ , by the converse of the Radical Center Theorem, it follows that $BCQP$ is cyclic. Next, we can observe the following.

Claim : Lines $\overline{EF}$ and $\overline{PQ}$ are parallel.
Proof : Note that,
$\begin{align*} \measuredangle (BI_A,EF) &= \measuredangle EFA + \measuredangle I_ABC \\ &= \frac{\pi}{2} + \measuredangle IBC + \frac{\pi}{2} + \measuredangle CAI\\ &= \frac{\pi}{2} + \measuredangle ICB \\ &= \measuredangle I_ACB\\ &= \measuredangle QPI_A \end{align*}$ from which the claim follows. Next, we have the following rather convenient cyclic quadrilaterals.

Claim : Quadrilaterals $BPMI$ and $CQMI$ are cyclic.
Proof : Since $\measuredangle NAI = \frac{\pi}{2}$ it is clear that $AN \parallel EF$ . Thus, $\triangle I_AIX \sim \triangle I_AAN$ . Now,
$\begin{align*} \frac{I_AI}{I_AA} &= \frac{I_AX}{I_AN}\\ \frac{I_AI \cdot I_AM}{I_AA \cdot I_AM}&= \frac{I_AX \cdot IA_T}{I_AN \cdot IA_T}\\ I_AI \cdot I_AM &= I_AX \cdot I_AT\\ I_AI \cdot I_AM &= I_AB \cdot I_AP \end{align*}$ which implies that $BPMI$ is cyclic. Similarly it also follows that $CQMI$ is cyclic, proving the claim.

Thus, $\measuredangle PMI = \measuredangle PBI = \frac{\pi}{2}$ and $\measuredangle IMQ = \measuredangle ICQ = \frac{\pi}{2}$ so points $P$ , $M$ and $Q$ are collinear. Now, let $A'$ be the $A-$ antipodal point in $\Omega$ . Since $\measuredangle AMA' = \frac{\pi}{2}$ , it follows that $A'$ also lies on $\overline{PQ}$ , and we are done.

This post has been edited 1 time. Last edited by cursed_tangent1434, Oct 31, 2024, 4:30 PM
Reason: latex errors oops

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#6 h Nov 1, 2024, 5:15 PM • 1 Y

Y by BR1F1SZ

BR1F1SZ wrote:

Is P3 posted? Or is it from ISL?

P3 was RMMSL.

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#7 h Feb 18, 2025, 10:12 AM

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Davdav1232 wrote:

BR1F1SZ wrote:

Is P3 posted? Or is it from ISL?

P3 was RMMSL.

Now that RMMSL is posted, which problem was P3 in this paper?

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#8 h Feb 18, 2025, 2:56 PM

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Nice one!

https://imagizer.imageshack.com/img922/416/5XBW0f.png

Let $J$ be the reflection of $I_a$ about $T$ , $I_aB$ intersects $EF$ at $Z$ . Denote $M$ the midpoint of minor arc $BC$ , $K$ be the antipodal point of $A$ in $(ABC)$ . Define $P'$ as the midpoint of segment $I_aZ$ .
Claim 1. Triangles $BTJ$ and $BI_aC$ are similar.
It's easy to see that $P'T \parallel BJ$ . Notice how $T$ is the $I_a$ -dumpty point of $\triangle BI_aC$ , so we must have triangles $BTI_a$ and $CTI_a$ are similar. Therefore:
$\cfrac{BT}{TJ} = \cfrac{BT}{TI_a} = \cfrac{BI_a}{CI_a}$ Then, by an angle chase: $\angle BTJ = \angle BTN = \angle BMN = \angle BI_aC$ . That being said, we now have $\triangle BTJ \sim \triangle BI_aC$ .
Claim 2. Triangles $ZBF$ and $CBI_a$ are similar.
But then, $\angle ZFB = \angle AFE = \angle NBC = \angle BMN = \angle CI_aB$ and $\angle ZBF = 180^{\circ} - \angle ABI_a = \angle CBI_a$ .
So $\triangle ZBF \sim \triangle CBI_a$ .
Claim 3. $P' \equiv P$ .
Now, this is relatively simple as $\triangle ZBF \sim \triangle CBI_a \sim \triangle JBT$ . So $\triangle ZBJ \sim \triangle FBC$ , thus:
$\angle BFT = \angle BZJ = 180^{\circ} - \angle BP'T$ So $B, F, T, P'$ are concyclic, thus $P'$ coincides with $P$ .
Claim 4. $PQ$ passes through $K$ .
Now, we should have that $PI = PI_a$ . Similarly, we can show that $QI = QI_a$ . So $PQ$ is the perpidencular bisector of segment $II_a$ , thus $PQ$ passes through $M$ and $PQ$ is perpidencular to $AM$ . Now, $\angle AMK = 90^{\circ}$ so we should have that $P, Q, M, K$ are collinear, as desired.

This post has been edited 10 times. Last edited by Retemoeg, Feb 19, 2025, 12:36 AM

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optimusprime154

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optimusprime154

#9 h Apr 5, 2025, 7:04 AM

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let $S$ be the interssection point of $NI_A$ and $FE$ its obvious from Reim's theorem that $BFST$ is cyclic, same goes for $TSEC$ . since $IT$ is the radical axis of these two circles. we get $BPQC$ cyclic. let $L$ be the intersection of $PQ$ and $II_A$ since $\angle I_APQ = \angle BCQ = \angle BIL = 90 - \angle BI_AI$ we get that $PQ \perp IL$ . now i prove that $L$ belongs to our original circle. we know $\angle LIC = \angle PBC = 180 - \angle PQC$ so $LICQ$ is cyclic. from PoP, we get $ILTS$ cyclic then by reim's theorem we get $ANLT$ cyclic. now if we let $R$ be the second intersection of $PQ$ with the circle, we get $\angle RLA = 90$ so $R$ is the $A$ Antipode.

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