parmenides51 wrote: Santa decorates his Christmas tree with a triangular decoration. Suppose the triangular decoration can be represented by an acute-angled scalene triangle $\vartriangle ABC$ with circumcircle $\omega$. $D$ is the foot of $B$-altitude of $\vartriangle ABC$. $E$ and $F$ are the points of projection of $D$ on $AB$ and $BC$ respectively. $A'$ is the point of reflection of $A$ over $BC$. $AA'$ meets $\omega$ again at $T$. $TD$ meets $\omega$ again at $X$. $Y$ is a point on $\omega$ such that $\angle TY A' = 90^o$. Prove that$ AY$ , $BX$ and $EF$ concur. one observation that i proved: $AY$ is the $A$-symmedian of $\triangle{ABC}$. i'll post the full solution later

this is a problem involving inversion and crazy length bash, so i think it's a little bit over for P2 (?) Denote $\angle{A},\angle{B},\angle{C}$ as the angles formed by corresponding vertices in $\triangle{ABC}$. Also, denote $a,b,c$ as the sides opposing angles $\angle{A},\angle{B},\angle{C}$, respectively.. Define $\varphi$ as the an inversion with center $A$ and radius $\sqrt{AB\times AC}$ followed by the reflection over the angle bisector of $\angle{A}$ (a.k.a. $\sqrt{bc}$ inversion). Note that $Y$ lies on a circle with diameter $TA'$, let it be $\Gamma$. Since $\Gamma$ doesn't pass through $A$, then it will be inverted to another circle which is not passing through $A$ also. It is well known that $\varphi:A'\leftrightarrow O$. Now let $T^*=AO\cap BC$. Since that $AH$ and $AO$ are isogonal lines w.r.t. $\angle{A}$ and $T\in AH$, then the image of $T$ under this inversion must lies on $AO$. Notice that $\angle{BAT}=\angle{T^*AC}$ and $\angle{BTA}=\angle{T^*AC}$ so $\triangle{ABT}\sim\triangle{AT^*C}$ by $AA$ similarity. So we have $\frac{AB}{AT}=\frac{AT^*}{AC}\Longleftrightarrow AT\cdot AT^*=AB\cdot AC$, meaning that $T^*$ is the image of $T$ under this inversion. And so $\Gamma$ is inverted to a circle with diameter $OT^*$, denote it as $\gamma$. Defining $M$ as the midpoint of $BC$, we see that $M$ lies on $\gamma$ since that $\angle{OMT^*}=90^\circ$. Well known that $\varphi: \omega\leftrightarrow BC$, so $\varphi: (\omega\cap\Gamma)\leftrightarrow(BC\cap\gamma)$. Since that $Y=\Gamma\cap\omega$, then $\varphi:Y\leftrightarrow M$. But it's also well known that $M$ will mapped to the point of intersection between $A$-symmedian of $\triangle{ABC}$ and $\omega$ by this inversion, so we must have $Y$ to be that point. As a consequence, $AY$ is the $A$-symmedian of $\triangle{ABC}$. Next, define $K=BX\cap EF$ and $J=AY\cap BC$. It's easy to see by triangles similarities that $BE\cdot BA=BD^2=BF\cap BC$, so $AEFC$ is cyclic by Power of Point's Theorem. So we have $\angle{AEK}=\angle{AEF}=180^\circ-\angle{ACF}=180^\circ-\angle{ACB}=180^\circ-\angle{AXB}=180^\circ-\angle{AXK}\Longleftrightarrow$ $AEKX$ is cyclic. Similarly, $CFKX$ is also cyclic as well. By Menelause's Theorem, it suffices to prove that $\frac{FJ}{JB}\cdot\frac{BA}{AE}\cdot\frac{EK}{KF}=1$ to show that $\overline{A,K,J,Y}$. From our previous founding, we have \[\frac{JB}{JC}=\frac{AB^2}{AC^2}\]\[\frac{JB}{a-JB}=\frac{c^2}{b^2}\]\[JB=\frac{ac^2}{b^2+c^2}\]and \[FB\cdot BC=BD^2\]\[FB=\frac{BD^2}{a}\]so \[\frac{FJ}{JB}=\frac{FB-JB}{JB}\]\[\frac{FJ}{JB}=\frac{BD^2(b^2+c^2)}{a^2c^2}-1\]Next, we also get \[BD^2=BE\cdot BA\]\[BE=\frac{BD^2}{c}\]\[c-EA=\frac{BD^2}{c}\]\[EA=\frac{c^2}{c^2-BD^2}\]From the three previous cyclic quadrilaterals, we have $\triangle{BEK}\sim\triangle{BAX}$, $\triangle{BKF}\sim\triangle{BCX}$ and $\triangle{BEF}\sim\triangle{ABC}$. Observing cyclic quadrilaterals $AXCT$ will also gives us $\triangle{ADX}\sim\triangle{CDT}$ and $\triangle{XDC}\sim\triangle{ADT}$. Hence we have the followings \[\frac{EK}{AX}=\frac{BE}{BX}\Longleftrightarrow EK=\frac{BE\cdot AX}{BX}\dots(1)\]\[\frac{FK}{XC}=\frac{BF}{BX}\Longleftrightarrow FK=\frac{BF\cdot XC}{BX}\dots(2)\]\[\frac{AX}{TC}=\frac{XD}{CD}\Longleftrightarrow AX=\frac{XD\cdot TC}{CD}\dots(3)\]\[\frac{XC}{AT}=\frac{XD}{AD}\Longleftrightarrow XC=\frac{XD\cdot AT}{AD}\dots(4)\]Dividing $(3)$ by $(4)$ yields \[\frac{AX}{XC}=\frac{TC}{AT}\cdot\frac{AD}{CD}\]Dividing $(1)$ by $(2)$ yields \[\frac{EK}{FK}=\frac{BE}{BF}\cdot\frac{AX}{XC}\]\[\frac{EK}{FK}=\frac{a}{c}\cdot\frac{TC}{AT}\cdot\frac{AD}{CD}\]Now here comes the trigonometry bashes. Firstly, it's obvious that $BD=\sin{\angle{A}}\cdot c=\sin{\angle{C}}\cdot a$. By Law of Sine in $\triangle{ABC}$, we also have $b=\frac{\sin{\angle{B}}\cdot a}{\sin{\angle{A}}}$. Thus, we can derive the followings \[\frac{FJ}{JB}=\frac{\sin^2{\angle{B}}\cdot a^2c^2+\sin^2{\angle{C}}\cdot a^2c^2}{a^2c^2}-1\]\[\frac{FJ}{JB}=\sin^2{\angle{B}}+\sin^2{\angle{C}}-1=1-\cos^2{\angle{B}}-\cos^2{\angle{C}}\]\[\frac{BA}{AE}=\frac{c^2}{c^2-\sin^2{\angle{A}}}=\frac{1}{\cos^2{\angle{A}}}\]After that, applying Law of Sine in $\triangle{ATC}$ yields $\frac{TC}{AT}=\frac{\sin{90^\circ-\angle{C}}}{\sin{90^\circ-(\angle{B}-\angle{C})}}=\frac{\cos{\angle{C}}}{\cos{\angle{B}-\angle{C}}}$. So \[\frac{EK}{KF}=\frac{a}{c}\cdot\frac{AD}{CD}\]\[\frac{EK}{KF}=\frac{a}{c}\cdot\frac{\cos{\angle{A}}\cdot c}{\cos{\angle{C}}\cdot a}\cdot\frac{\cos{\angle{C}}}{\cos{\angle{B}-\angle{C}}}\]\[\frac{EK}{KF}=\frac{\cos{\angle{A}}}{\cos{\angle{B}-\angle{C}}}\]Multiplying the three terms proceeds to \[\frac{FJ}{JB}\cdot\frac{BA}{AE}\cdot\frac{EK}{KF}=(1-\cos^2{\angle{B}}-\cos^2{\angle{C}})(\frac{1}{\cos^2{\angle{A}}})(\frac{\cos{\angle{A}}}{\cos{\angle{B}-\angle{C}}})\]\[\frac{FJ}{JB}\cdot\frac{BA}{AE}\cdot\frac{EK}{KF}=(1-\cos^2{\angle{B}}-\cos^2{\angle{C}})(\frac{1}{-\cos{\angle{B}+\angle{C}}})(\frac{1}{\cos{\angle{B}-\angle{C}}})\]\[\frac{FJ}{JB}\cdot\frac{BA}{AE}\cdot\frac{EK}{KF}=(\cos^2{\angle{B}}+\cos^2{\angle{C}}-1)(\frac{1}{\cos^2{\angle{B}}\cos^2{\angle{C}}-(1-\cos^2{\angle{B}})(1-\cos^2{\angle{C}})})\]\[\boxed{\frac{FJ}{JB}\cdot\frac{BA}{AE}\cdot\frac{EK}{KF}=\frac{\cos^2{\angle{B}}+\cos^2{\angle{C}}-1}{\cos^2{\angle{B}}+\cos^2{\angle{C}}-1}=1}\]as desired. Therefore, $AY,BX,EF$ are concurrent at $K$. $\blacksquare$

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Cool problem! Here's a more synthetical solution First, let $H$ be the orthocenter of $\triangle ABC$ and let $P$ be the projection of $A$ onto $BC$. Let $R$ be the projection of $C$ onto $AB$. I claim that the intersection point is the midpoint of $DR$. Claim: $AY$ is the $A$-symmedian in $\triangle ABC$ It is well-known that $HP = PT$. Now, looking at $\triangle A'BC$, we get that $T$ is its orthocenter. Thus, $Y$ is the $A'$-Humpty point wrt $\triangle A'BC$. To prove the claim, it's enough to prove that $ABYC$ is harmonic $\iff AB \cdot CY = AC \cdot BY$. Let $M$ be the midpoint of $BC$, thus $M$ is on line $A'Y$. From Sin Law, we have $\frac{BY}{CY} = \frac{sin \angle BCY}{sin \angle CBY} = \frac{sin \angle CA'M}{sin \angle BA'M} = \frac{AB}{AC}$, what we wanted to prove, so claim $1$ is true. Let $S$ be the midpoint of $RD$. Let's prove that the $3$ line must pass through $S$. Step $1$, $S$ is on the line $AY$: Since $S$ is the midpoint of $RD$ and triangles $ABC$ and $ADP$ are similar, we easily get that $S$ is on the $A$-symmedian, thus $AY$ passes through $S$. Step $2$, $S$ is on the line $BX$: It is enough to prove that $\angle ABS = \angle ABX \iff \angle ABS = ATX$. We'll prove that $BRS \sim THD$. First, it's easy to see that $\angle THD = 180 - \angle C = \angle BRS$. We want $\frac{BR}{HT} = \frac{RS}{HD} \iff \frac{BR}{2HP} = \frac{RD}{2HD} \iff \frac{BR}{HP} = \frac{RD}{HD}$ $\frac{BR}{HP} = \frac{BC}{HC}$ and $HD = HC sin(90 - \angle A)$, so I want to prove that $BC = \frac{RD}{sin(90 - \angle A)}$ But $\frac{RD}{BC} = \frac{AR}{AC} = sin(90 - \angle A)$, thus the $2$ triangles are similar, so this step is done. Step $3$, $S$ is on the line $EF$: Since $\angle RED = 90$ and $S$ is the midpoint of $RD$, we get that $SE = SD \implies \angle DES = \angle EDS = 90 - \angle ARD = 90 - \angle C$. But since $\angle BED = 90 = \angle DFB \implies EDFB$ is cyclic $\implies \angle DEF = \angle DBF = 90 - \angle C$, thus $\angle DES = \angle DEF \implies$ the points $E, S, F$ lie on the same line, so this step is done. Thus, we have proved that $AY$, $BX$ and $EF$ pass through $S$, so we are done.

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Nice geo for Christmas Day ! Here is a solution without any length/trig bashes. Claim: $AY$ is the symmedian of $\triangle ABC$. Proof. Let $A''$ be the antipode of $T$ in $\omega$, note that $\angle TYA'' = 90^\circ$ so $Y$ is the intersection of $A'A''$ with $\omega$. Now it is clear that $A'A'' \cap BC$ is $M$, the midpoint of $BC$. Thus, $A''Y$ is the median of $\triangle A''BC$. Hence $AY$ is clearly the symmedian of $\triangle ABC$. $\Box$ Let $R$ be the foot of the $C$-altitude, and let $S$ be the midpoint of $RD$. It is clear that $S \in AY$, since $AS$ is the median of $\triangle ARD$ which is oppositely similar to $\triangle ABC$ with symmedian $AY$. Furthermore, a quick angle chase yields that $S \in EF$. Thus, it suffices to show the following claim. Claim: $BSX$ are collinear. Proof. It suffices to show that $\angle ABS = \angle ATX$. We will show that $\triangle BRS \sim \triangle THD$. Let $K$ be the $A$-queue point; it is clear that $K$ is the center of spiral similarity sending $RB$ to $HT$. Now, to show that $K$ is the center of spiral similarity sending $RS$ to $HD$, it suffices to prove that circle $(KRS)$ is tangent to line $CR$. By the spiral similarity at $K$ taking $RD$ to $BC$, we know that $\angle SKR = \angle MKB$. But by Brokard's Theorem on quadrilateral $BCDR$, we know that $MB^2 = MH\cdot MK$, thus $\angle MKB = \angle MBH = \angle CRD$. Since $\angle SKR = \angle CRD$, it is clear that $CR$ is tangent to $KRS$. In conclusion, $\triangle BRS$ and $\triangle THD$ are spirally similar wrt $K$. The claim follows. $\Box$ Hence, $AY \cap BX \cap EF = S$. We are done.

Curious_Droid wrote: Nice geo for Christmas Day ! Here is a solution without any length/trig bashes. Claim: $AY$ is the symmedian of $\triangle ABC$. Proof. Let $A''$ be the antipode of $T$ in $\omega$, note that $\angle TYA'' = 90^\circ$ so $Y$ is the intersection of $A'A''$ with $\omega$. Now it is clear that $A'A'' \cap BC$ is $M$, the midpoint of $BC$. Thus, $A''Y$ is the median of $\triangle A''BC$. Hence $AY$ is clearly the symmedian of $\triangle ABC$. $\Box$ Let $R$ be the foot of the $C$-altitude, and let $S$ be the midpoint of $RD$. It is clear that $S \in AY$, since $AS$ is the median of $\triangle ARD$ which is oppositely similar to $\triangle ABC$ with symmedian $AY$. Furthermore, a quick angle chase yields that $S \in EF$. Thus, it suffices to show the following claim. Claim: $BSX$ are collinear. Proof. It suffices to show that $\angle ABS = \angle ATX$. We will show that $\triangle BRS \sim \triangle THD$. Let $K$ be the $A$-queue point; it is clear that $K$ is the center of spiral similarity sending $RB$ to $HT$. Now, to show that $K$ is the center of spiral similarity sending $RS$ to $HD$, it suffices to prove that circle $(KRS)$ is tangent to line $CR$. By the spiral similarity at $K$ taking $RD$ to $BC$, we know that $\angle SKR = \angle MKB$. But by Brokard's Theorem on quadrilateral $BCDR$, we know that $MB^2 = MH\cdot MK$, thus $\angle MKB = \angle MBH = \angle CRD$. Since $\angle SKR = \angle CRD$, it is clear that $CR$ is tangent to $KRS$. In conclusion, $\triangle BRS$ and $\triangle THD$ are spirally similar wrt $K$. The claim follows. $\Box$ Hence, $AY \cap BX \cap EF = S$. We are done. orz, you even provide a shorter solution than the official one