Problem 1910

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
$$\angle AJ_{B}B = 90^{\circ} + \frac{1}{2} \angle AEB = 90^{\circ} + \frac{1}{2} \angle ACB = \angle AIB = \angle APB,$$
so $ABJ_{B}P$ is cyclic. As $A$ is the centre of spiral similarity between $ABE$ and $ADC$, it is also the centre of spiral similarity between $ABJ_{B}$ and $ADI_{C}$. Hence, $A$ is the Miquel point of self-intersecting quadrilateral $BDI_{C}J_{B}$, 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) $\implies$ (2).** We assume that (1) holds.
Claim 1 and the similarities $ABDI_{B} \sim AECJ_{C}$ and $ABEJ_{B} \sim ADCI_{C}$ tell us that
$$\nsucceq I_{B}XI_{C} = \Varangle J_{C}QC + \Varangle BPJ_{B} = \Varangle J_{C}AC + \Varangle BAJ_{B} = \Varangle I_{B}AD + \Varangle DAI_{C} = \Varangle I_{B}AI_{C},$$
so $AI_{B}XI_{C}$ is cyclic. Also, as $X \in AI$, we have that
$$\nsucceq I_{B}AX = \Varangle BAI - \Varangle BAI_{B} = \Varangle I_{B}AI_{C} - \Varangle IB_{A}D = \Varangle DAI_{C}.$$
Using these, we have that
$$\nsucceq I_{B}I_{C}P = \Varangle I_{B}AX = \Varangle DAI_{C} = \Varangle BAJ_{B} = \Varangle BPJ_{B},$$
so $I_{B}I_{C} \parallel BC$. Hence,
$$\nsucceq I_{B}I_{C}J_{B} = \Varangle BPJ_{B} = \Varangle BIJ_{B} = \Varangle I_{B}IJ_{B},$$
so $II_{B}J_{B}I_{C}$ is cyclic. Analogously, we have that $II_{C}J_{C}I_{B}$ is cyclic, so $I_{B}J_{B}J_{C}I_{C}$ is cyclic, thus proving (2).

**Proof of (2) $\implies$ (1). We assume that (2) holds.

Claim 2.** Circles $IBC, IJ_{B}I_{C}$, and $II_{B}J_{C}$ are tangent at $I$.

Proof. Using the cyclic quadrilateral $BIJ_{B}P$, we have that
$$\nsucceq IBC = \Varangle IBP = \Varangle IJ_{B}P = \Varangle IJ_{B}I_{C}.$$
As $C, I_{C}$, and $I$ are collinear, the tangents to circles $I_{B}I_{C}$ and $IBC$ at $I$ coincide, so circles $IJ_{B}I_{C}$ and $IBC$ are tangent at $I$. Analogously, circles $II_{B}J_{C}$ 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 $II_{B}J_{C}, IJ_{B}I_{C}$, 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 $II_{B}J_{C}$ and $IJ_{B}I_{C}$ 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 $PI_{C}$ and $QI_{B}$. 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 $II_{B}I_{C}$ 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
$$\nsucceq PJ_{B}J_{C} = \Varangle I_{C}J_{B}J_{C} = \Varangle I_{C}I_{B}J_{C} = \Varangle PQI_{B} = \Varangle PQJ_{C},$$
so $PJ_{B}J_{C}Q$ is cyclic. By the radical axis theorem on circles $AIPJ_{B}, AIQJ_{C}$, and $PJ_{B}J_{C}Q$, we have that $AI, I_{B}J_{C}$, and $J_{B}I_{C}$ concur at $X$, thus proving (1).

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
$$PP_{1}P_{2} \stackrel{\ddagger}{\sim} PP_{3}P_{4} \quad \text{ and } \quad PP_{1}P_{3} \stackrel{\ddagger}{\sim} PP_{2}P_{4}$$
are equivalent.

**Proof of (1) $\iff$ (3).** Let $AI_{B}$ and $AI_{C}$ meet $BC$ at $S$ and $T$, respectively. Let $AJ_{B}$ meet $BE$ at $K$, $AJ_{C}$ meet $CE$ at $L$, and $KT$ and $SL$ meet at $Y$.

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

Proof. Let $Y'$ be the intersection of $KT$ and the bisector of $\angle BAC$. As
$$\angle BAK = \frac{1}{2} \angle BAE = \frac{1}{2} \angle DAC = \angle TAC,$$
$AY'$ also bisects $\angle KAT$. Hence, $Y'$ is the foot of the bisector of $\angle KAT$ in triangle $AKT$. Using the Fact, we have that
$$\begin{aligned} ABE \sim ADC &\Longrightarrow ABEK \sim ADCT \\ &\Longrightarrow ABD \sim AKT \sim AEC \\ &\Longrightarrow ABDS \sim AKTY' \sim AECL \\ &\Longrightarrow ABEK \sim ASLY' \sim ADCT. \end{aligned}$$
As $K$ lies on $BE$, we have that $Y'$ lies on $SL$, so $Y = Y'$ 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 $\frac{AI_{B}}{AS} = \frac{AJ_{C}}{AL}$, 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
$$\frac{AI_{B}}{AS} = \frac{AX_{1}}{AY} = \frac{AX_{2}}{AY} = \frac{AI_{C}}{AT},$$
which is equivalent to $I_{B}I_{C} \parallel BC$. $\square$

**Proof of (2) $\iff$ (3).** Let $AJ_{B}$ and $AJ_{C}$ meet circle $ABC$ at $M$ and $N$, respectively, and let $I_{B}'$ and $I_{C}'$ be the $A$-excentres of $ABD$ and $ADC$, respectively.

Claim 2. Lines $I_{B}I_{C}, I_{B}'I_{C}'$, and $BC$ are concurrent or pairwise parallel.

Proof. We work in the projective plane. Let $I_{B}I_{C}$ and $I_{B}'I_{C}'$ meet $BC$ at $Z$ and $Z'$, 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 $\left(AD, AZ ; AI_{B}, AI_{C}\right) = -1$. Applying the same argument to the $A$-excircles of $ABD$ and $ADC$ gives $\left(AD, AZ' ; AI_{B}', AI_{C}'\right) = -1$, which means that $Z = Z'$. Thus, $I_{B}I_{C}, I_{B}'I_{C}'$, 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
$$ABE \stackrel{+}{\sim} ADC \Longrightarrow ABEJ_{B} \stackrel{+}{\sim} ADCI_{C} \Longrightarrow AJ_{B}I_{C} \stackrel{+}{\sim} ABD.$$
Thus, $\angle(BD, J_{B}I_{C}) = \angle BAJ_{B} = \angle BCM$, 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 $I_{B}'I_{C}'$ is $A$.

Proof. As $I_{B}$ and $I_{B}'$ are respectively the incentre and $A$-excentre of triangle $ABD$, we have that $ABI_{B}' \stackrel{+}{\sim} AI_{B}D$. Using the similarity $ABD \stackrel{+}{\sim} AEC$, this means that $ABI_{B}' \stackrel{+}{\sim} AJ_{C}C$, so $AB \cdot AC = AI_{B}' \cdot AJ_{C}$ and $\angle BAI_{B}' = \angle J_{C}AC$. Similarly, we have that $AB \cdot AC = AJ_{B} \cdot AI_{C}'$ and $\angle BAJ_{B} = \angle I_{C}'AC$. Together, these imply that $AI_{B}' \cdot AJ_{C} = AJ_{B} \cdot AI_{C}'$ and $\angle J_{B}AJ_{C} = \angle I_{B}'AI_{C}'$, so $AJ_{B}J_{C} \stackrel{+}{\sim} AI_{B}'I_{C}'$. $\square$

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
$$\begin{aligned} \npreceq I_{B}I_{C}J_{B} = \Varangle I_{B}J_{C}J_{B} &\Longleftrightarrow \Varangle I_{B}I_{C}J_{B} + \nrightarrow MCB = \Varangle I_{B}J_{C}J_{B} + \Varangle MNB \\ &\Longleftrightarrow \Varangle(I_{B}I_{C}, BC) = \Varangle(MN, J_{B}J_{C}). \end{aligned}$$
By Claim 4, we have that
$$\begin{aligned} \nsucceq(J_{B}J_{C}, I_{B}'I_{C}') &= \Varangle J_{B}AI_{B}' \\ &= \Varangle BAI_{B} + \Varangle MAB \\ &= \Varangle EAJ_{C} + \Varangle MAB \\ &= \Varangle NAC + \Varangle MAB \\ &= \Varangle(MN, BC), \end{aligned}$$
which is equivalent to $\nsucceq(BC, I_{B}'I_{C}') = \Varangle(MN, J_{B}J_{C})$. Thus, $I_{B}J_{B}J_{C}I_{C}$ is cyclic if and only if
$$\begin{equation*} \nsucceq(I_{B}I_{C}, BC) = \Varangle(BC, I_{B}'I_{C}'). \tag{*} \end{equation*}$$
Suppose that $I_{B}I_{C}$ is parallel to $BC$. By Claim 2, $I_{B}'I_{C}'$ is also parallel to $BC$, so we have that $\Varangle(I_{B}I_{C}, BC) = \Varangle(BC, I_{B}'I_{C}') = 0^{\circ}$. 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, $I_{B}I_{C}$, $I_{B}'I_{C}'$, 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 $\left(ZA, ZD ; ZI_{B}, ZI_{B}'\right) = -1$. By (*), $ZD$ bisects $\angle I_{B}ZI_{B}'$, 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 $AB < AC < BC$. $\square$

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 $J_{B} \neq P$ as the second intersection of line $XP$ with $\omega_{B}$. Similarly, we can define $I_{B} = XQ \cap BI$ and $J_{C} \neq Q$ 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 $\triangle ABC$ 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 $I_{B}, J_{B} \neq B$ and $I_{C}, J_{C} \neq C$. 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 $U \neq J_{B}$ be the second intersection of $I_{B}J_{B}$ with $\omega_{B}$. We have by the law of sines that
$$\frac{\sin \measuredangle IJ_{B}U}{\sin \measuredangle UJ_{B}B} = \frac{\sin \measuredangle IJ_{B}I_{B}}{\sin \measuredangle I_{B}J_{B}B} = \frac{\sin \measuredangle J_{B}II_{B}}{\sin \measuredangle J_{B}BI_{B}} \cdot \frac{II_{B}}{I_{B}B} = \frac{\sin \measuredangle J_{B}IB}{\sin \measuredangle J_{B}BI} \cdot \frac{II_{B}}{I_{B}B} = \frac{\sin \measuredangle XPQ}{\sin \measuredangle XPI} \cdot \frac{II_{B}}{I_{B}B}.$$
We also have
$$\frac{II_{B}}{I_{B}B} = \frac{\sin \measuredangle IQI_{B}}{\sin \measuredangle I_{B}QB} \cdot \frac{|IQ|}{|BQ|} = \frac{\sin \measuredangle IQX}{\sin \measuredangle XQP} \cdot \frac{|IQ|}{|BQ|}.$$
Combining these and applying Ceva's Theorem in $\triangle PIQ$ with point $X$, we get
$$\frac{\sin \measuredangle IJ_{B}U}{\sin \measuredangle UJ_{B}B} = \frac{\sin \measuredangle XPQ}{\sin \measuredangle XPI} \cdot \frac{\sin \measuredangle IQX}{\sin \measuredangle XQP} \cdot \frac{|IQ|}{|BQ|} = \frac{\sin \measuredangle XIQ}{\sin \measuredangle XIP} \cdot \frac{|IQ|}{|BQ|} = \frac{\sin \measuredangle (\ell, IQ)}{\sin \measuredangle (\ell, IP)} \cdot \frac{|IQ|}{|BQ|},$$
which is independent of the choice of $X$ on $\ell$. As $\nsucceq IJ_{B}U + \measuredangle UJ_{B}B = \measuredangle IJ_{B}B = \measuredangle IAB$ is fixed, this is enough to show point $U$ is fixed on $\omega_{B}$.

Similarly, if we define $V \neq J_{C}$ to be the second intersection of $I_{C}J_{C}$ with $\omega_{C}$, we get that $V$ is fixed on $\omega_{C}$.

Let $G \neq X$ and $H \neq X$ 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
$$\begin{aligned} &\Varangle I_{C}J_{B}I_{B} = \Varangle PJ_{B}U = \Varangle PGU = \Varangle PGQ \\ &\Varangle I_{C}J_{C}I_{B} = \Varangle VJ_{C}Q = \Varangle VHQ = \Varangle PHQ. \end{aligned}$$
Thus $\nsucceq I_{C}J_{B}I_{B} = \Varangle I_{C}J_{C}I_{B} \Longleftrightarrow \Varangle PGQ = \measuredangle PHQ$ 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
$$PQGH \text{ concyclic } \Longleftrightarrow \Varangle QHG = \Varangle QPG \Longleftrightarrow \Varangle QHG = \Varangle BIG \Longleftrightarrow QH \parallel BI.$$
We also have that $\measuredangle CQH = \measuredangle CIH$, so
$$\ell \text{ tangent to circle } BIC \Longleftrightarrow \Varangle CIH = \Varangle CBI \Longleftrightarrow \Varangle CQH = \Varangle CBI \Longleftrightarrow QH \parallel BI.$$
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
$$\begin{aligned} \nsucceq XPI &= \Varangle J_{B}PI = \Varangle J_{B}AI = \Varangle BAI - \Varangle BAJ_{B} = \Varangle J_{B}AJ_{C} - \Varangle J_{B}AE \\ &= \Varangle EAJ_{C} = \Varangle J_{C}AC = \Varangle J_{C}QC = \Varangle XQP, \end{aligned}$$
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 $PI_{C}$ with $\Gamma$ and $IP$ is tangent to $\Gamma$, $X$ lies in the interior of $\overparen{PQ}$ 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 $\ell \neq t$. Thus, the only remaining possibility in Claim 4 is $\ell = AI$. We then have
$$I_{B}I_{C}J_{B}J_{C} \text{ concyclic } \stackrel{\text{ Claim 3}}{\Longleftrightarrow} PQGH \text{ concyclic } \stackrel{\text{ Claim 4}}{\Longleftrightarrow} X \text{ lies on } AI,$$
finishing the problem.