It's indeed simple: In triangle $AQC$ $QB$ is the symmedian, so $\frac{BA}{BC}=\frac{QA^2}{QC^2}$ and it's fixed, so $\frac{QA}{QC}$ must also be fixed, and we're done.

This was exactly my solution. Some participants used Cartesian coordinates , others used the trigonometric form of Ceva's theorem. Darij

yes this one is trivial just use some trig. it turns out that the intersection is the square root of B wrt to AC. (btw this was in australian tst 2).

Solution Let $D=G \cap PQ$. Because $ADCQ$ is cyclic, we have the following. \[\frac{AD}{CQ}=\frac{BD}{BC} \quad \textbf{and} \quad \frac{AB}{BD}=\frac{AQ}{CD}\] Multiplying these yields the following. \[\frac{AB}{BC}=\frac{AQ}{QC} \cdot \frac{AD}{CD} \quad (1)\] Since $\triangle{ADP} \sim \triangle{QAP}$ and $\triangle{CDP} \sim \triangle{QCP}$, we also have the following. \[\frac{AD}{AQ}=\frac{DP}{AP}=\frac{DP}{CP}=\frac{CD}{CQ}\] Substitution into (1) yields that \[\frac{AQ}{CQ}=\sqrt{\frac{AB}{BC}}\] Let $X$ be the intersection of the bisector of $\angle{AQC}$ with $AC$. By the angle bisector theorem we have that \[\frac{AX}{XC}=\frac{AQ}{CQ}=\sqrt{\frac{AB}{BC}}\] Therefore the location of $X$ on segment $AC$, is dependent only on the location of $B$.

The coordinate bash is not too bad if you're careful:

We see that $BQ$ is a symmedian. Let the angle-bisector intersect $AC$ at $K$. Then by Steiner's theorem $\frac{AB}{BC}=\frac{AQ^2}{AC^2}$, implying that $\frac{AQ}{QC}$ is fixed. But by the angle-bisector theroem, $\frac{AK}{KC}=\frac{AQ}{QC}$ and we have that $K$ is fixed.

Let the second intersection of the angle bisector with the circle be $R$. Also , Let the intersection of $AC$ and the bisector be $Y$ Then from the similarity of $\Delta ARP$ and $\Delta QAP$ , we get , $\frac{AR}{RQ}=\frac{RP}{PA}$. Similarly ,we get $\frac{RP}{PC}=\frac{RC}{QC}$. So , $AQ/QC=AR/RC$ . Now , $\frac{AB}{BC}=\frac{AR\sin{\angle{ARQ}}}{RC\sin{\angle{CRQ}}}$ So , $\frac{AQ}{QC}.\frac{AR}{RC}=\frac{AB}{BC}$ hence, $\frac{AY}{YC}=\frac{AQ}{QC}= (\frac{AB}{BC})^{1/2}$ , which is fixed. hence done

Apply Ceva(sinus) to triangle ACP and after law of sinus to tirangles APM and CPM and we are done. M is the intersection of PQ and AC.

FantasyLover wrote: $\sqrt{\frac{AQ\cdot AR}{CQ\cdot CR}}=\sqrt{\frac{[AQR]}{[CQR]}}=\sqrt{\frac{AB}{BC}}$, which is indeed constant. We are done. $\blacksquare$ Why $\sqrt{\frac{[AQR]}{[CQR]}}$ gives us $\sqrt{\frac{AB}{BC}}$ ?

sjaelee wrote: Then by Steiner's theorem $\frac{AB}{BC}=\frac{AQ^2}{AC^2}$, implying that $\frac{AQ}{QC}$ is fixed. What does Steiner's theorem say ?

samirka259 wrote: sjaelee wrote: Then by Steiner's theorem $\frac{AB}{BC}=\frac{AQ^2}{AC^2}$, implying that $\frac{AQ}{QC}$ is fixed. What does Steiner's theorem say ? I didn't read proofs above, but if you are asking for Steiner's theorem, it may be the next theorem(I am too lazy to read the problem and solutions ): for $\triangle{ABC}$ and points $X$ and $Y$ that lies on side $AC$, $BX$ and $BY$ are symmetric wrt(I am not sure for this word, may be "about" suits more) angle bissector of $\angle{ABC}$ if and only if $\frac{AX*AY}{CX*CY}=\frac{AB^2}{CB^2}$. I am pretty sure, that I used this theorem when I was solving it on the training

I guess this was before symmedians were well known... Let the bisector of angle $AQC$ intersect $AC$ at $T$. By the angle-bisector theorem, $\frac{AT}{CT}=\frac{AQ}{QC}$. By the Symmedian Lemma, we have that $BQ$ is the $Q$-symmedian of $AQC$. Thus, Steiner's theorem and our work before yields that $\frac{AT}{CT}=\frac{AQ}{QC}=\sqrt{\frac{AB}{BC}}$, which is fixed.

My solution was similar to those above: Note that $QP$ is a symmedian of $\triangle QAC$, so $\frac{AB}{CB} = \frac{QA^2}{QC^2}$, so $\frac{QA}{QC}$ is fixed, and thus the point of intersection does not depend on $G$.

$BQ$ is a symmedian of $\triangle AQB$. Then let the intersection of $PQ$ and $AC$ be $G$ and $M$ be the midpoint of $AC$. We get $\angle AQB = \angle MQC$ and $\angle AQM = \angle BQC$. Firstly $\frac{AB}{MC}=\frac{|AQB|}{|MQC|}=\frac{0.5AQ \cdot BQ \sin{AQB}}{0.5MQ \cdot CQ \sin {MQC}} = \frac{AQ \cdot BQ}{MQ \cdot CQ}$ In a similar way $\frac{AM}{BC} = \frac{AQ \cdot MQ}{BQ \cdot CQ}$ Note the two above quantities are constant because $A,M,B,C$ are fixed. Multiplying the two quantities, we get $\frac{AQ^2}{CQ^2}$ and therefore $\frac{AQ}{CQ}$ is constant. Since $\frac{AG}{GC}=\frac{AQ}{CQ}$ by angle bisector theorem on $QG$, we get $G$ is fixed.

Huh? ISL G2?!! darij grinberg wrote: Three distinct points $A$, $B$, and $C$ are fixed on a line in this order. Let $\Gamma$ be a circle passing through $A$ and $C$ whose center does not lie on the line $AC$. Denote by $P$ the intersection of the tangents to $\Gamma$ at $A$ and $C$. Suppose $\Gamma$ meets the segment $PB$ at $Q$. Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ does not depend on the choice of $\Gamma$. Let $T$ denote the intersection of the angle bisector of $\angle AQC$ with the fixed line $\overline{AC}$. $ $ From the angle bisector theorem in $\triangle AQC$, we have that $\frac{AT}{CT} = \frac{ AQ}{CQ}$. Also, since $\overline{QB}$ is the $Q-$ symmedian of $\triangle AQC$, by Steiner Ratio theorem we have that : $\frac{AB}{CB}= \left(\frac{AQ}{CQ}\right)^2$. $ $ Therefore we have that $$\frac{AT}{CT} = \underbrace{\sqrt{\frac{AB}{CB}}}_{\text{Fixed}}$$So $T$ is fixed irrespective of choice of $\Gamma$.

Let $I$ be the intersection of the angle bisector with line $AC$ and let $X$ be the intersection of segment $PB$ with the circle. Then we have $\frac{AI}{IC} = \frac{AQ}{QC}$ by the Angle Bisector Theorem, and $AQ = \frac{PA \cdot AX}{PX}$ and $QC = \frac{PQ \cdot CX}{PX}$ and thus $\frac{AQ}{QC} = \frac{PA \cdot AX \cdot PC}{PX \cdot PQ \cdot CX} = \frac{PQ}{PA} \cdot \frac{AX}{PQ} \cdot \frac{PC}{CX} = \frac{AX}{CX}$. But by the Ratio Lemma we have $\frac{AX}{CX} \cdot \frac{\sin \angle AXB}{\sin \angle CXB} = \frac{AX}{CX} \cdot \frac{AQ}{QC} = \frac{AB}{BC}$. Thus, $\frac{AX}{CX} = \frac{\frac{AB}{BC}}{\frac{AQ}{QC}} = \frac{AQ}{QC}$, so $\frac{AB}{BC} = \left(\frac{AQ}{QC}\right)^2$. But this means that $\frac{AQ}{QC}$ is determined, and thus so is $\frac{AI}{IC}$ as desired.

Because $\overline{QBP}$ is the $Q$-Symmedian of $ACQ$, we have $$\frac{QA}{QC} = \sqrt{ \frac{BA}{BC} }$$which finishes via the Angle Bisector Theorem. $\blacksquare$ Remark: Because the center of $\Gamma$ doesn't lie on $\overline{ABC}$, we know $P$ isn't at infinity.

Nice problem and a good exercise in symmedians. We observe that $QP$ is a $Q-\text{symmedian}$ of $\triangle QBC$. Thus $\left(\frac{QA}{QC}\right)^2=\frac{AB}{AC} \implies \frac{QA}{QC}=\sqrt{\frac{AB}{AC}}$ so by angle bisector theorem the intersection of the bisector of $\angle AQC$ with $AC$ is a fixed point $X$ with $\frac{AX}{XC}=\sqrt{\frac{AB}{AC}}$.

Let $X$ be the desired point. Note that $QB$ is the Q-symmedian in $\triangle ACQ$. This means that the ratio $$QA/QC=\sqrt{AB/BC}$$does not depend on $\Gamma$. By the Angle Bisector Theorem, this also means that $AX/XC$ does not depend on $\Gamma$, so the location of $X$ does not depend on $\Gamma$, QED.

why is QB a symmedian?

Is this the easiest isl? Let the intersection be E. It's a well known property that QP is a symmedian of AQC, so then AB/BC=AQ^2/AC^2, and by angle bisector theorem the length AE/EC=AQ/AC is constant. @above because symmedians can be constructed by drawing the equal tangents' intersection, QB==BP.

$\color{red} \boxed{\textbf{SOLUTION}}$ Let the angle bisector of $\angle AQC$ intersect $AC$ at $K$ $QB$ is the symmedian of $\triangle AQC$ So, $$\frac{AB}{BC}={\frac{AQ^2}{QC^2}} \implies \frac{AQ}{QC}=\sqrt{\frac{AB}{BC}} \implies \frac{AK}{KC}= \underbrace{\sqrt{\frac{AB}{BC}}}_{\text{Fixed}} \blacksquare$$

why is g1 and g2 both not hard symmedian problems Clearly $QB$ is the $Q$-symmedian in $\triangle AQC$, so if we let $D$ be the foot of the $Q$-angle bisector in $\triangle AQC$, then \[ \frac{AD}{DC} = \frac{AQ}{QC} = \sqrt{\frac{AB}{BC}} \]which is fixed, so we're done. $\square$

We consider the converse: let $K,L$ lie on $\overline{AC}$ such that $(A,C;K,L)=-1$, and consider the circle $\Gamma$ with diameter $\overline{KL}$—some Apollonius circle. Let $Q$ be a variable point on $\Gamma$ and $P$ be the intersection of the tangents to $(ACQ)$ at $A$ and $C$. Then prove that $\overline{AC} \cap \overline{QP}:=X$ is fixed. Clearly $X$ is the foot of the $Q$-symmedian. If $M$ is the midpoint of $\overline{AC}$, then $\overline{QK}$ and $\overline{QL}$ are the $Q$-angle bisectors in $\triangle QXM$, hence $(X,M;K,L)=-1$. Since $K,L,M$ are fixed independent of $Q$, it follows that $X$ is as well. Returning to the main problem, as we vary $K$ and $L$ around $\overline{AC}$ such that $(A,C;K,L)$ holds, $X$ can achieve any point; in particular we can have $X=B$ (and $K$ lying inside segment $\overline{AC}$). Then as $Q$ varies for this fixed choice of $K,L$, $\Gamma$ can become any circle through $A$ and $C$. Hence in the original problem, the intersection of the $\angle AQC$-bisector and $\overline{AC}$ is fixed at $K$. $\blacksquare$

??? Let $D$ be the point where the angle bisector of $\angle AQC$ hits $\overline{AC}$. By construction, $QB$ is the $Q$-symmedian in $\triangle AQC$, so by the angle bisector theorem we have $$\frac{AD}{DC} = \frac{AQ}{QC} = \sqrt{\frac{AB}{AC}},$$and $D$ is fixed, as needed.

Let $PB$ meets the circle $\Gamma$ at $D$ and $Q$, respectively. Then note that $ADCQ$ is harmonic (from the construction of the tangents). Let the angle bisector of $\angle AQC$ meet $AC$ at a point $X$. Then note that \[ \left( \frac{AX}{XC} \right)^2 = \left( \frac{AQ}{QC} \right)^2 = \frac{AB}{BC} \]where we use Angle Bisector theorem and the definition of a harmonic quadrilateral. Thus, $X$ only depends on $A$, $B$, and $C$, as desired. $\blacksquare$

Let $R$ be the intersection point : By isogonal ratios : $\frac{AB}{AC} = \frac{AQ^2}{CQ^2}$ Using the angle bissector theorem : $\frac{AQ^2}{CQ^2} = \frac{AR^2}{CR^2}$ Since $B$ is fixed, then $R$ is fixed, we are done.

Note that due to the Angle Bisector Theorem, the problem is equivalent to showing that $\frac{AQ}{QC}$ is fixed. But now notice that $BQ$ is the symmedian of $\Delta ABC$, so $\frac{AB}{BC} = \left( \frac{AQ}{QC} \right)^2 \implies \frac{AQ}{QC} = \sqrt{\left(\frac{AB}{BC}\right)}$, which is fixed. $\square$

Im liek 95% sure this is wrong Observe that all points are determined by $A$, $B$, $C$, $P$. With $O$ the center of $\Gamma$ being defined as the harmonic conjugate of $P$ wrt to $AC$ on $(APC)$. And the arc midpoint of $AC$ on $(AQC)$ can be defined as the second intersection of $PO$ with $\Gamma$, and the intersection with the angle bisector can be defined as $QM \cap PB$. So we can affine transform the problem with cross ratios preserved. Notice we can affine transform so $P$ is any point on the perpendicular bisector of $AC$ and the cross ratio $(A,C; B, QM \cap AC)$ is fixed. Which implies the fixed-ness.

Ok now the symmedian B's foot is fixed (tangent going through A and C joined with B is basically the foot of symmedian , cite egmo or smth idk), the midpoint of AC is fixed so the foot of angle bisector is fixed QED