IMO2024 Shortlisted Problems

Let $X$ be the intersection of $I_B{J}_C$ and $J_B{I}_C$. We will prove that, provided that $AB<AC<BC$, the following two conditions are equivalent: (1) $AX$ bisects $\angle BAC$; (2) $I_B,I_C,J_B$, and $J_C$ are concyclic. Let circles $AIB$ and $AIC$ meet $BC$ again at $P$ and $Q$, respectively. Note that $AB=BQ$ and $AC=CP$ because the centres of circles $AIB$ and $AIC$ lie on $CI$ and $BI$, respectively. Thus, $B,P,Q$, and $C$ are collinear in this order as $BQ+PC=AB+AC>BC$ by the triangle inequality.

Claim 1. Points $P,J_B$, and $I_C$ are collinear, and points $Q,I_B$, and $J_C$ are collinear.

Proof. We have that $$\angle A{J}_{B}B=90^{\circ }+\frac{1}{2}\angle AEB=90^{\circ }+\frac{1}{2}\angle ACB=\angle AIB=\angle APB,$$ so $AB{J}_{B}P$ is cyclic. As $A$ is the centre of spiral similarity between $ABE$ and $ADC$, it is also the centre of spiral similarity between $AB{J}_B$ and $AD{I}_C$. Hence, $A$ is the Miquel point of self-intersecting quadrilateral $BD{I}_C{J}_B$, so $P$ lies on $J_B{I}_C$. Analogously, we have that $Q$ lies on $I_B{J}_C$. $\square$



Throughout the rest of the solution, we will use directed angles.

Proof of (1) $⟹$ (2). We assume that (1) holds. Claim 1 and the similarities $ABD{I}_B\sim AEC{J}_C$ and $ABE{J}_B\sim ADC{I}_C$ tell us that $$⋡I_{B}X{I}_C=\text{Varangle}{J}_{C}QC+\text{Varangle}BP{J}_B=\text{Varangle}{J}_{C}AC+\text{Varangle}BA{J}_B=\text{Varangle}{I}_{B}AD+\text{Varangle}DA{I}_C=\text{Varangle}{I}_{B}A{I}_C,$$ so $A{I}_{B}X{I}_C$ is cyclic. Also, as $X\in AI$, we have that $$⋡I_{B}AX=\text{Varangle}BAI-\text{Varangle}BA{I}_B=\text{Varangle}{I}_{B}A{I}_C-\text{Varangle}I{B}_{A}D=\text{Varangle}DA{I}_C.$$ Using these, we have that $$⋡I_B{I}_{C}P=\text{Varangle}{I}_{B}AX=\text{Varangle}DA{I}_C=\text{Varangle}BA{J}_B=\text{Varangle}BP{J}_B,$$ so $I_B{I}_C\parallel BC$. Hence, $$⋡I_B{I}_C{J}_B=\text{Varangle}BP{J}_B=\text{Varangle}BI{J}_B=\text{Varangle}{I}_{B}I{J}_B,$$ so $I{I}_B{J}_B{I}_C$ is cyclic. Analogously, we have that $I{I}_C{J}_C{I}_B$ is cyclic, so $I_B{J}_B{J}_C{I}_C$ is cyclic, thus proving (2).

Proof of (2) $⟹$ (1). We assume that (2) holds.

Claim 2. Circles $IBC,I{J}_B{I}_C$, and $I{I}_B{J}_C$ are tangent at $I$.

Proof. Using the cyclic quadrilateral $BI{J}_{B}P$, we have that $$⋡IBC=\text{Varangle}IBP=\text{Varangle}I{J}_{B}P=\text{Varangle}I{J}_B{I}_C.$$ As $C,I_C$, and $I$ are collinear, the tangents to circles $I_B{I}_C$ and $IBC$ at $I$ coincide, so circles $I{J}_B{I}_C$ and $IBC$ are tangent at $I$. Analogously, circles $I{I}_B{J}_C$ and $IBC$ are tangent at $I$, so all three circles are tangent at $I$.

Claim 3. Point $I$ lies on circle $I_B{J}_B{J}_C{I}_C$.

Proof. Suppose that $I$ does not lie on circle $I_B{J}_B{J}_C{I}_C$. Then the circles $I{I}_B{J}_C,I{J}_B{I}_C$, and $I_B{J}_B{J}_C{I}_C$ are distinct. We apply the radical axis theorem to these three circles. By Claim 2, the radical axis of circles $I{I}_B{J}_C$ and $I{J}_B{I}_C$ is the tangent to circle $IBC$ at $I$. As $I_B{J}_C$ and $J_B{I}_C$ intersect at $X$, $IX$ must be tangent to circle $IBC$.

However, by Claim 1 we have that $X$ is the intersection of $P{I}_C$ and $Q{I}_B$. As $D$ lies on the interior of segment $BC$, $I_B$ lies on the interior of segment $BI$ and $I_C$ lies on the interior of segment $CI$. Hence, $I_B,P,Q$, and $I_C$ all lie on the perimeter of triangle $IBC$ in this order, so $X$ must be in the interior of triangle $IBC$. This means that $IX$ cannot be tangent to circle $BIC$, so $I$ must lie on circle $I_B{J}_B{J}_C{I}_C$.

By Claims 2 and 3, circles $I{I}_B{I}_C$ and $IBC$ are tangent, so $I_B{I}_C\parallel BC$. Since $I_B{J}_B{J}_C{I}_C$ is cyclic, we have that $$⋡P{J}_B{J}_C=\text{Varangle}{I}_C{J}_B{J}_C=\text{Varangle}{I}_C{I}_B{J}_C=\text{Varangle}PQ{I}_B=\text{Varangle}PQ{J}_C,$$ so $P{J}_B{J}_{C}Q$ is cyclic. By the radical axis theorem on circles $AIP{J}_B,AIQ{J}_C$, and $P{J}_B{J}_{C}Q$, we have that $AI,I_B{J}_C$, and $J_B{I}_C$ concur at $X$, thus proving (1).

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Alternative solution.

Let $X$ be the intersection of $I_B{J}_C$ and $J_B{I}_C$. As in Solution 1, we will prove that conditions (1) and (2) are equivalent. To do so, we introduce the new condition: (3) $I_B{I}_C\parallel BC$ and show that (3) is equivalent to both (1) and (2), provided that $AB<AC<BC$. Note that $ABD\stackrel{+}{\sim }AEC$ and $ABE\stackrel{+}{\sim }ADC$, where $\stackrel{+}{\sim }$ denotes positive similarity. We will make use of the following fact.

Fact. For points $P,P_1,P_2,P_3$, and $P_4$, the positive similarities $$P{P}_1{P}_2\stackrel{‡}{\sim }P{P}_3{P}_4\text{ and }P{P}_1{P}_3\stackrel{‡}{\sim }P{P}_2{P}_4$$ are equivalent.

Proof of (1) $⟺$ (3). Let $A{I}_B$ and $A{I}_C$ meet $BC$ at $S$ and $T$, respectively. Let $A{J}_B$ meet $BE$ at $K$, $A{J}_C$ meet $CE$ at $L$, and $KT$ and $SL$ meet at $Y$.



Claim 1. Line $AY$ bisects $\angle BAC$.

Proof. Let $Y^{\prime }$ be the intersection of $KT$ and the bisector of $\angle BAC$. As $$\angle BAK=\frac{1}{2}\angle BAE=\frac{1}{2}\angle DAC=\angle TAC,$$ $A{Y}^{\prime }$ also bisects $\angle KAT$. Hence, $Y^{\prime }$ is the foot of the bisector of $\angle KAT$ in triangle $AKT$. Using the Fact, we have that $$\begin{array}{rl}ABE\sim ADC& ⟹ABEK\sim ADCT\\ & ⟹ABD\sim AKT\sim AEC\\ & ⟹ABDS\sim AKT{Y}^{\prime }\sim AECL\\ & ⟹ABEK\sim ASL{Y}^{\prime }\sim ADCT.\end{array}$$ As $K$ lies on $BE$, we have that $Y^{\prime }$ lies on $SL$, so $Y=Y^{\prime }$ and $AY$ bisects $\angle BAC$. $\square$

We show that $X$ lies on $AY$ if and only if $I_B{I}_C\parallel BC$, which implies the equivalence of (1) and (3) by Claim 1. Let $AY$ meet $I_B{J}_C$ and $J_B{I}_C$ at $X_1$ and $X_2$, respectively. As $ABD$ and $AEC$ are similar, we have that $\frac{A{I}_B}{AS}=\frac{A{J}_C}{AL}$, so $I_B{J}_C\parallel SL$. Analogously, we have that $J_B{I}_C\parallel KT$. Hence, $X_1$ and $X_2$ coincide with $X$ if and only if $$\frac{A{I}_B}{AS}=\frac{A{X}_1}{AY}=\frac{A{X}_2}{AY}=\frac{A{I}_C}{AT},$$ which is equivalent to $I_B{I}_C\parallel BC$. $\square$

Proof of (2) $⟺$ (3). Let $A{J}_B$ and $A{J}_C$ meet circle $ABC$ at $M$ and $N$, respectively, and let $I_B^{\prime }$ and $I_C^{\prime }$ be the $A$-excentres of $ABD$ and $ADC$, respectively.



Claim 2. Lines $I_B{I}_C,I_B^{\prime }{I}_C^{\prime }$, and $BC$ are concurrent or pairwise parallel.

Proof. We work in the projective plane. Let $I_B{I}_C$ and $I_B^{\prime }{I}_C^{\prime }$ meet $BC$ at $Z$ and $Z^{\prime }$, respectively. Note that $Z$ is the intersection of the external common tangents of the incircles of $ABD$ and $ADC$ and $AD$ is a common internal tangent of the incircles of $ABD$ and $ADC$, so $\left(AD,AZ;A{I}_B,A{I}_C\right)=-1$. Applying the same argument to the $A$-excircles of $ABD$ and $ADC$ gives $\left(AD,A{Z}^{\prime };A{I}_B^{\prime },A{I}_C^{\prime }\right)=-1$, which means that $Z=Z^{\prime }$. Thus, $I_B{I}_C,I_B^{\prime }{I}_C^{\prime }$, and $BC$ concur, possibly at infinity. $\square$



Claim 3. Lines $J_B{I}_C$ and $CM$ are parallel, and lines $I_B{J}_C$ and $BN$ are parallel.

Proof. Using the Fact, we have that $$ABE\stackrel{+}{\sim }ADC⟹ABE{J}_B\stackrel{+}{\sim }ADC{I}_C⟹A{J}_B{I}_C\stackrel{+}{\sim }ABD.$$ Thus, $\angle (BD,J_B{I}_C)=\angle BA{J}_B=\angle BCM$, so $J_B{I}_C\parallel CM$. Similarly, we have that $I_B{J}_C\parallel BN$. $\square$

Claim 4. The centre of spiral similarity between $J_B{J}_C$ and $I_B^{\prime }{I}_C^{\prime }$ is $A$.

Proof. As $I_B$ and $I_B^{\prime }$ are respectively the incentre and $A$-excentre of triangle $ABD$, we have that $AB{I}_B^{\prime }\stackrel{+}{\sim }A{I}_{B}D$. Using the similarity $ABD\stackrel{+}{\sim }AEC$, this means that $AB{I}_B^{\prime }\stackrel{+}{\sim }A{J}_{C}C$, so $AB\cdot AC=A{I}_B^{\prime }\cdot A{J}_C$ and $\angle BA{I}_B^{\prime }=\angle J_{C}AC$. Similarly, we have that $AB\cdot AC=A{J}_B\cdot A{I}_C^{\prime }$ and $\angle BA{J}_B=\angle I_C^{\prime }AC$. Together, these imply that $A{I}_B^{\prime }\cdot A{J}_C=A{J}_B\cdot A{I}_C^{\prime }$ and $\angle J_{B}A{J}_C=\angle I_B^{\prime }A{I}_C^{\prime }$, so $A{J}_B{J}_C\stackrel{+}{\sim }A{I}_B^{\prime }{I}_C^{\prime }$. $\square$

We proceed using directed angles. By Claim 3, we have that $I_B{J}_B{J}_C{I}_C$ is cyclic if and only if $$\begin{array}{rl}⋠I_B{I}_C{J}_B=\text{Varangle}{I}_B{J}_C{J}_B& ⟺\text{Varangle}{I}_B{I}_C{J}_B+\nrightarrow MCB=\text{Varangle}{I}_B{J}_C{J}_B+\text{Varangle}MNB\\ & ⟺\text{Varangle}(I_B{I}_C,BC)=\text{Varangle}(MN,J_B{J}_C).\end{array}$$ By Claim 4, we have that $$\begin{array}{rl}⋡(J_B{J}_C,I_B^{\prime }{I}_C^{\prime })& =\text{Varangle}{J}_{B}A{I}_B^{\prime }\\ & =\text{Varangle}BA{I}_B+\text{Varangle}MAB\\ & =\text{Varangle}EA{J}_C+\text{Varangle}MAB\\ & =\text{Varangle}NAC+\text{Varangle}MAB\\ & =\text{Varangle}(MN,BC),\end{array}$$ which is equivalent to $⋡(BC,I_B^{\prime }{I}_C^{\prime })=\text{Varangle}(MN,J_B{J}_C)$. Thus, $I_B{J}_B{J}_C{I}_C$ is cyclic if and only if $$\begin{array}{c}⋡(I_B{I}_C,BC)=\text{Varangle}(BC,I_B^{\prime }{I}_C^{\prime }).\end{array}\text{\hspace{1em}(*)}$$ Suppose that $I_B{I}_C$ is parallel to $BC$. By Claim 2, $I_B^{\prime }{I}_C^{\prime }$ is also parallel to $BC$, so we have that $\text{Varangle}(I_B{I}_C,BC)=\text{Varangle}(BC,I_B^{\prime }{I}_C^{\prime })=0^{\circ }$. Thus, () is satisfied, so $I_B{J}_B{J}_C{I}_C$ is cyclic.



Suppose now that $I_B{I}_C$ is not parallel to $BC$ while $I_B{J}_B{J}_C{I}_C$ is cyclic. By Claim 2, $I_B{I}_C$, $I_B^{\prime }{I}_C^{\prime }$, and $BC$ concur at a point $Z$. As $I_B$ and $I_C$ lie on segments $BI$ and $CI$, $Z$ must lie outside segment $BC$. Since $A$ is the intersection of the common external tangents of the incircle and $A$-excircle of $ABD$ and $ZD$ is a common internal tangent of the incircle and $A$-excircle of $ABD$, we have that $\left(ZA,ZD;Z{I}_B,Z{I}_B^{\prime }\right)=-1$. By (), $ZD$ bisects $\angle I_{B}Z{I}_B^{\prime }$, so $\angle AZD=90^{\circ }$: that is, $Z$ is the foot from $A$ to $BC$. But this implies that $\angle ABC$ or $\angle BCA$ is obtuse, contradicting the fact that $AB<AC<BC$. $\square$

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Alternative solution.

Let ${\omega }_B$ and ${\omega }_C$ denote circles $AIB$ and $AIC$, respectively. Introduce $P,Q$ and $X$ as in Solution 1 and recall from Claim 1 in Solution 1 that $P,J_B$ and $I_C$ are collinear with $J_B$ lying on ${\omega }_B$. From this, we can define $J_B$ and $I_C$ in terms of $X$ by $I_C=XP\cap CI$ and $J_B\ne P$ as the second intersection of line $XP$ with ${\omega }_B$. Similarly, we can define $I_B=XQ\cap BI$ and $J_C\ne Q$ as the second intersection of line $XQ$ with ${\omega }_C$. Note that this now detaches the definitions of points $I_B,I_C,J_B$, and $J_C$ from points $D$ and $E$.

Let $\ell$ be a line passing through $I$. We now allow $X$ to vary along $\ell$ while fixing $△ABC$ and points $I,P$, and $Q$. We use the definitions from above to construct $I_B,I_C,J_B$, and $J_C$. We will classify all cases where these four points are concyclic. Throughout the rest of the solution we use directed angles and directed lengths.

For nondegeneracy reasons, we exclude cases where $X=I$ and $X$ lies on line $BC$, which means that $I_B,J_B\ne B$ and $I_C,J_C\ne C$. We also exclude the cases where $\ell$ is tangent to either ${\omega }_B$ or ${\omega }_C$. Similar results hold in these cases and they can be treated as limit cases.



Claim 1. Line $I_B{J}_B$ passes through a fixed point on ${\omega }_B$, and line $I_C{J}_C$ passes through a fixed point on ${\omega }_C$ as $X$ varies on $\ell$.

Proof. Let $U\ne J_B$ be the second intersection of $I_B{J}_B$ with ${\omega }_B$. We have by the law of sines that $$\frac{\sin \measuredangle I{J}_{B}U}{\sin \measuredangle U{J}_{B}B}=\frac{\sin \measuredangle I{J}_B{I}_B}{\sin \measuredangle I_B{J}_{B}B}=\frac{\sin \measuredangle J_{B}I{I}_B}{\sin \measuredangle J_{B}B{I}_B}\cdot \frac{I{I}_B}{I_{B}B}=\frac{\sin \measuredangle J_{B}IB}{\sin \measuredangle J_{B}BI}\cdot \frac{I{I}_B}{I_{B}B}=\frac{\sin \measuredangle XPQ}{\sin \measuredangle XPI}\cdot \frac{I{I}_B}{I_{B}B}.$$ We also have $$\frac{I{I}_B}{I_{B}B}=\frac{\sin \measuredangle IQ{I}_B}{\sin \measuredangle I_{B}QB}\cdot \frac{|IQ|}{|BQ|}=\frac{\sin \measuredangle IQX}{\sin \measuredangle XQP}\cdot \frac{|IQ|}{|BQ|}.$$ Combining these and applying Ceva's Theorem in $△PIQ$ with point $X$, we get $$\frac{\sin \measuredangle I{J}_{B}U}{\sin \measuredangle U{J}_{B}B}=\frac{\sin \measuredangle XPQ}{\sin \measuredangle XPI}\cdot \frac{\sin \measuredangle IQX}{\sin \measuredangle XQP}\cdot \frac{|IQ|}{|BQ|}=\frac{\sin \measuredangle XIQ}{\sin \measuredangle XIP}\cdot \frac{|IQ|}{|BQ|}=\frac{\sin \measuredangle (\ell ,IQ)}{\sin \measuredangle (\ell ,IP)}\cdot \frac{|IQ|}{|BQ|},$$ which is independent of the choice of $X$ on $\ell$. As $⋡I{J}_{B}U+\measuredangle U{J}_{B}B=\measuredangle I{J}_{B}B=\measuredangle IAB$ is fixed, this is enough to show point $U$ is fixed on ${\omega }_B$.

Similarly, if we define $V\ne J_C$ to be the second intersection of $I_C{J}_C$ with ${\omega }_C$, we get that $V$ is fixed on ${\omega }_C$.

Let $G\ne X$ and $H\ne X$ be the second intersections of $\ell$ with ${\omega }_B$ and ${\omega }_C$, respectively which exist as we are assuming $\ell$ is not tangent to either ${\omega }_B$ or ${\omega }_C$.

Claim 2. Points $U,G$ and $Q$ are collinear and points $V,H$ and $P$ are collinear.

Proof. Taking $X=G$, we have $J_B=G$ and $I_B=XQ\cap BI$. Both of these points lie on line $QG$ which, by Claim 1, shows that $U,G,Q$ are collinear. Similarly, $V,H,P$ are collinear.

Claim 3. Points $I_B,I_C,J_B,J_C$ are concyclic if and only if points $P,Q,G,H$ are concyclic. In particular, this depends only on $\ell$, not on the choice of $X$ on $\ell$.

Proof. We have that $$\begin{array}{rl}& \text{Varangle}{I}_C{J}_B{I}_B=\text{Varangle}P{J}_{B}U=\text{Varangle}PGU=\text{Varangle}PGQ\\ & \text{Varangle}{I}_C{J}_C{I}_B=\text{Varangle}V{J}_{C}Q=\text{Varangle}VHQ=\text{Varangle}PHQ.\end{array}$$ Thus $⋡I_C{J}_B{I}_B=\text{Varangle}{I}_C{J}_C{I}_B⟺\text{Varangle}PGQ=\measuredangle PHQ$ which proves the claim.

Claim 4. $P,Q,G,H$ are concyclic if and only if $\ell \in \{IA,IP,IQ,t\}$ where $t$ is the tangent to circle $BIC$ at $I$.

Proof. When $\ell =IA$, we have $G=H=A$ so the cyclic condition from Claim 3 holds. Similarly, when $\ell =IP$ or $\ell =IQ$, $G=P$ or $H=Q$, respectively, so again the cyclic condition holds.

Now, consider the case where $\ell \notin \{IP,IQ\}$. In this case it is straightforward to see that the four points $P,Q,G$, and $H$ are distinct. We then have that $\measuredangle QPG=\measuredangle BPG=\measuredangle BIG$, so $$PQGH\text{ concyclic }⟺\text{Varangle}QHG=\text{Varangle}QPG⟺\text{Varangle}QHG=\text{Varangle}BIG⟺QH\parallel BI.$$ We also have that $\measuredangle CQH=\measuredangle CIH$, so $$\ell \text{ tangent to circle }BIC⟺\text{Varangle}CIH=\text{Varangle}CBI⟺\text{Varangle}CQH=\text{Varangle}CBI⟺QH\parallel BI.$$ Hence, in this case $P,Q,G,H$ are concyclic if and only if $\ell$ is tangent to circle $BIC$, as claimed.

We now revert to using points $D$ and $E$ to define points $I_B,I_C,J_B,J_C$, and $X$, returning to the original set-up.

Claim 5. Let $\Gamma$ be the circle passing through $P$ and $Q$ that is tangent to $IP$ and $IQ$, which exists as $IP=IQ=IA$. Then $X$ lies on $\Gamma$. Furthermore, $X$ lies on the same side of $BC$ as $A$ and does not lie on line $BC$.

Proof. We have that $$\begin{array}{rl}⋡XPI& =\text{Varangle}{J}_{B}PI=\text{Varangle}{J}_{B}AI=\text{Varangle}BAI-\text{Varangle}BA{J}_B=\text{Varangle}{J}_{B}A{J}_C-\text{Varangle}{J}_{B}AE\\ & =\text{Varangle}EA{J}_C=\text{Varangle}{J}_{C}AC=\text{Varangle}{J}_{C}QC=\text{Varangle}XQP,\end{array}$$ so circle $XPQ$ is tangent to $IP$. Similarly, circle $XPQ$ is tangent to $IQ$, so $X$ lies on $\Gamma$.

As $D$ lies in the interior of segment $BC$, $I_C$ lies in the interior of segment $CI$. Since $X$ is the second intersection of $P{I}_C$ with $\Gamma$ and $IP$ is tangent to $\Gamma$, $X$ lies in the interior of $\text{overparen}PQ$ on $\Gamma$ on the same side of $BC$ as $A$. This implies the second part of the claim. $\square$

By Claim 5, we cannot have $\ell \in \{IP,IQ\}$ in the original problem. Furthermore, as shown in Claim 2 of Solution 1, we have that $X$ lies inside triangle $IBC$, which means that $\ell \ne t$. Thus, the only remaining possibility in Claim 4 is $\ell =AI$. We then have $$I_B{I}_C{J}_B{J}_C\text{ concyclic }\stackrel{\text{ Claim 3}}{⟺}PQGH\text{ concyclic }\stackrel{\text{ Claim 4}}{⟺}X\text{ lies on }AI,$$ finishing the problem.