Let $T'$ be the second intersection of $PT$ with the circumcircle of $ABC$. Since $AT'\perp TT'=PT$, we have to show that $XY\| AT'\ (\#)$. Since $T'A,T'P$ are the internal, and external bisectors of $\angle BT'C$ (respectively), it means that $(AB,AC;AT',AP)=-1\ (*)$, and, since $XY$ is cut by $AB,AC,AP$ in three points one of which is the midpoint of the segment formed by the other two, it means that $XY$ must be parallel to the harmonic conjugate of $AP$ wrt $AB,AC$, which is $AT'$, according to $(*)$, so we have proven $(\#)$.

There is also a different solution: Problem (slightly extended). Let ABC be an isosceles triangle such that AB = AC. Let P be a point on the line BC, and let X and Y be two points on the lines AB and AC such that PX || AC and PY || AB. Finally, let T be the midpoint of the arc BC on the circumcircle of triangle ABC. Prove that $PT\perp XY$. Solution. Since PX || AC, Thales yields BX : XA = BP : PC. Since PY || AB, Thales yields AY : YC = BP : PC. Thus, BX : XA = AY : YC. Let O be the circumcenter of triangle ABC. Then, since the triangle ABC is isosceles with AB = AC, the triangles AOB and AOC are congruent. But since the triangle AOC is isosceles (in fact, OA = OC, since the point O is the circumcenter of triangle ABC), the triangle AOC is congruent to the triangle COA (please do look at the order of the vertices), and thus, the triangles AOB and COA are congruent. Since the points X and Y lie on the respective sides BA and AC of these two triangles and divide they in the same ratio BX : XA = AY : YC, they are corresponding points of these two triangles; hence, since corresponding points of congruent triangles have equal distances, we have OX = OY. Thus, the point O lies on the perpendicular bisector of the segment XY. In other words, if R is the midpoint of the segment XY, then $RO\perp XY$. We have PX || AC and PY || AB; in other words, PX || AY and PY || AX. Thus, the quadrilateral AXPY is a parallelogram. Since the diagonals of a parallelogram bisect each other, therefore, the segments AP and XY must bisect each other. In other words, the midpoint R of the segment XY is also the midpoint of the segment AP. Since the triangle ABC is isosceles with AB = AC, symmetry observations show that its apex A, its circumcenter O and the midpoint T of the arc BC on the circumcircle of the triangle ABC all lie on the perpendicular bisector of the segment BC. Hence, the segment AT, being a chord of the circumcircle of triangle ABC and passing through the center O of this circumcircle, is a diameter of this circumcircle. Hence, the center O of the circumcircle is the midpoint of the segment AT. On the other hand, we know that the point R is the midpoint of the segment AP. Hence, the line RO is a centers line in triangle APT, and thus it is parallel to the side PT of this triangle. Thus, $RO\perp XY$ yields $PT\perp XY$. Problem solved. By the way, this problem is equivalent to Argentina 5 / geo problem 2 from the op02 project. Darij

$TX^2+PY^2=BX^2+BT^2+CY^2=PX^2+CT^2+CY^2=PX^2+TY^2$. Thus, $PT\perp XY$.

mecrazywong wrote: $TX^2+PY^2=BX^2+BT^2+CY^2=PX^2+CT^2+CY^2=PX^2+TY^2$ Sorry but how did you get that ?

erdos wrote: mecrazywong wrote: $TX^2+PY^2=BX^2+BT^2+CY^2=PX^2+CT^2+CY^2=PX^2+TY^2$ Sorry but how did you get that ? Just apply Pythagoras repeatedly.

Oh yes, now it's clear Let me say that it's a really nice solution

draw circles $C_1(A,AB),C_2(X,XB),C_3(Y,YC)$ and radical center is$T$ and from here it's easy. was is it Final Problem???!!

Hi. I have a nice solution . Consider the circles $C_1,C_2,C_3$ be the circles with center $A$ and radius $AB=AC$ ,circle with center $X$ and radius $XP = XB$, circle with center $Y$ and radius $YC=YP$ respectively. $BT$ is the radical axis of $C_1,C_2$ . And $CT$ is the radical axis of $C_1,C_3$. We know the radical axis of $C_2,C_3$ passes through $P$. So $PT$ is radical axis of $C_2,C_3$ and so $PT$ is perpendicular to $XY$ .

kvmedya U R a little late!!

Let $O$ be the midpoint of $AT$. Note that $\triangle OAX\cong\triangle OCY$ hence $\angle OCA=\angle OXA$, and $O,A,Y,X$ are concyclic. This also implies that $OX=OY$. Let $A',O',T'$ be the reflections of $A,O,T$ across line $XY$. Hence $A',O',P,X,Y$ are concyclic. Since $\angle A'XY=\angle AXY=\angle PYX$, which means $XYPA'$ is a trapezoid. Since $OX=OY$ then $O'X=O'Y$ and $O'A'=O'P$, which implies that $O'$ is the circumcenter of $(A'PT')$, therefore $PT'\perp A'P$. From $A'P\parallel XY$ and $TT'\perp XY$, we get $PT\perp XY$.

Suppose $B=1,C=-1,\angle{BAC}=2\theta$ and so now we've $A=iCot(\theta),t=-itan(\theta),p=a$ where $a\in\mathbb R$ now we get $x=(a+1)(Sin\theta+iCos\theta)-1,y=(a-1)(Sin\theta+iCos\theta)+1$ and now it's very easy to show that $i\frac {p-t}{x-y}\in\mathbb R$ and that's what we're required to show.

here my solution: Let $PT\cap (ABC)=S,T$ and $M\in AB$ $CM\parallel PA$, $CM\cap AS=N$, $AS\cap BC=Q$. We have $SQ$ is internal bisector, $ST$ is external bisector of $ \angle BSC $. Then $ \{ B,Q,C,P \} $ are harmonic. By Menelaus's theorem for the triangle $BCM$ we get that $CN=MN$. Since $AXPY$ is parallelogram, we have $XY$ passes through the midpoint of $AP$. The triangles $AMC$ and $XAP$ are similar $ \Rightarrow $ so $AN\parallel XY$, $AS\parallel XY$. We have $AT$ is diameter of $(ABC)$ and $AS\perp PT$. Hence $XY\perp PT$.

Let $AP \cap XY=E$ and $PT$ $\cap$ $\odot (ABC)$ $=$ $D$. Let $O$ be center of $\odot (ABC)$, then $AY$ $\stackrel{O}{\mapsto}$ $BX$ $\implies$ $OX=OY$ or inshort, $XY||AD$ $\implies$ $PT \perp XY$ $\qquad \blacksquare$

$TX^2=AT^2+AX^2-2.AT.AX.\cos\angle XAT$ and $TY^2=AT^2+AY^2-2.AT.AY.\cos\angle ATY$. Subtracting both the the equations we get $$TX^2-TY^2=YP^2-YX^2-2AB(AY-AX)$$Making use of the fact that $\cos\angle XAT=-\cos\angle BAT=-\cos\angle TAC$ and $\cos\angle BAT=\frac{AB}{AT}$. Now again after some easy calculations we get that $2AB(AY-AX)=YP^2-PX^2$. Hence we get that $$TX^2-TY^2=PX^2-PY^2$$, hence by Converse of Baudhyana's Theorem we get that $PT\perp XY$.

I'm not confident in this stuff, so please tell me if it's wrong Animate $P$ on $BC$. Let $P' = PT \cap XY , P''=PT \cap (XBT)$. $\angle XP''T=90^\circ$. $P \mapsto P'$ and $P \mapsto P''$ are projective maps (they are just a projections of $P$ on line and on circle). So to prove that $P'$ and $P''$ coincide, we need to prove it for 3 cases. It's obvious in cases $P=[B,C,M]$, (where $M$ - midpoint of $BC$) . Hence, proved.

See my solution to this problem in the video on my Youtube channel here: https://www.youtube.com/watch?v=MesO2HcLrh4

One more solution. Let $O$ - circumcenter of $\triangle ABC$, $S$ - midpoint of $XY$. Let perpendicular to $BC$ through $P$ cut $AB, AC$ at $B',C'$. Wee easily can see that $X,Y$ are midpoints of $BB', CC'$ and $\triangle AB'C' \sim \triangle TBC$. So by gliding principle $\triangle OXY$ are also similar to them and so $OX=OY$. Clearly $S$ - midpoint of $AP$. So $PT \parallel OS \perp XY$.

zuss77 wrote: I'm not confident in this stuff, so please tell me if it's wrong Animate $P$ on $BC$. Let $P' = PT \cap XY , P''=PT \cap (XBT)$. $\angle XP''T=90^\circ$. $P \mapsto P'$ and $P \mapsto P''$ are projective maps (they are just a projections of $P$ on line and on circle). So to prove that $P'$ and $P''$ coincide, we need to prove it for 3 cases. It's obvious in cases $P=[B,C,M]$, (where $M$ - midpoint of $BC$) . Hence, proved. I think that your solution is wrong, because your points $P',P''$ are on $XY$, which is a curve/line that is not fixed, so you cannot apply the Moving Points Lemma. But here is a solution using Moving Points that I hope that is correct: Animate $P$ on $BC$. Let $Q_1=PT \cap (ABC)$, $Q_1 \neq T$ and $Q_2$ is the intersection between the line throught $T$ perpendicular to $XY$ with $(ABC)$. Then, note that the composed map $$ Q_2 \overset{A}{\mapsto} AQ_2 \cap BC \overset{B}{\mapsto} P \overset{T}{\mapsto} Q_1 $$is projective. $(*)$ Thus, by the Moving Points Lemma, we only have to check that $Q_1=Q_2$ for three choices of $P$. Choosing $P=B,C,M$, where $M$ is the midpoint of $BC$, we have that $Q_1=Q_2$ $\implies$ from $(*)$, $Q_1=Q_2$ for all $P$. Then, $PT \perp XY$, as desired. $\blacksquare$

Am I missing something or does the following just work?

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -126.5949778516833, xmax = 106.33722818117934, ymin = -56.83197312872014, ymax = 88.71259482383863; /* image dimensions */ pen ffxfqq = rgb(1,0.4980392156862745,0); /* draw figures */ draw((0,34.61959274272818)--(23.186285888611817,0), linewidth(0.8) + blue); draw((0,34.61959274272818)--(-23.186285888611817,0), linewidth(0.8) + blue); draw((23.186285888611817,0)--(-23.186285888611817,0), linewidth(0.8) + blue); draw((-57.88865541688788,0)--(-23.186285888611817,0), linewidth(0.8) + blue); draw((-57.88865541688788,0)--(-40.537470652749846,-25.90716568509682), linewidth(0.8) + blue); draw((-40.537470652749846,-25.90716568509682)--(-23.186285888611817,0), linewidth(0.8) + blue); draw((-17.35118476413803,60.526758427825)--(0,34.61959274272818), linewidth(0.8) + blue); draw((-17.35118476413803,60.526758427825)--(-57.88865541688788,0), linewidth(0.8) + blue); draw(circle((0,9.545351288147986), 25.074241454580196), linewidth(0.8) + red); draw((-57.88865541688788,0)--(0,-15.528890166432213), linewidth(0.8) + linetype("4 4") + ffxfqq); draw((-17.35118476413803,60.526758427825)--(-40.537470652749846,-25.90716568509682), linewidth(0.8) + linetype("4 4") + ffxfqq); /* dots and labels */ dot((0,34.61959274272818),dotstyle); label("$A$", (0.680397470848183,36.18866601250727), NE * labelscalefactor); dot((23.186285888611817,0),dotstyle); label("$B$", (23.82137480221754,1.4772000154535045), NE * labelscalefactor); dot((-23.186285888611817,0),linewidth(4pt) + dotstyle); label("$C$", (-22.61282313243808,1.1727134716196996), NE * labelscalefactor); dot((-57.88865541688788,0),dotstyle); label("$P$", (-57.324289129492115,1.4772000154535045), NE * labelscalefactor); dot((-17.35118476413803,60.526758427825),linewidth(4pt) + dotstyle); label("$X$", (-16.675335527678836,61.76553569454689), NE * labelscalefactor); dot((-40.537470652749846,-25.90716568509682),linewidth(4pt) + dotstyle); label("$Y$", (-39.9685561309651,-24.708642754253724), NE * labelscalefactor); dot((0,-15.528890166432213),linewidth(4pt) + dotstyle); label("$T$", (0.680397470848183,-14.356100263904354), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice for bashing, so why not bash??? $\color{black}\rule{25cm}{1pt}$ By construction we must have that $PXAY$ is a parallelogram. Throw this configuration onto the complex plane where the circumcircle of $ABC$ will be the unit circle and we set $a=1$ and $c=\frac{1}{b}$. Obviously $p$ must be a point which satisfies $\overline{p}=b+\frac{1}{b}-p$, since this implies that $P \in BC$ and also it's obvious that $t=-1$. Since we have that $Y$ is on $AC$, we must have that: $$\frac{y-a}{\overline{y-a}}=\frac{c-a}{\overline{c-a}}=-ca$$now when simplified and when we plug in what we have we get that $\overline{y}=b+1-yb$. But since we have that $PY \parallel AB$ we have that: $$\frac{p-y}{\overline{p-y}}=\frac{a-b}{\overline{a-b}}=-ab$$when simplified and when we plug in what we got we get that $y=\frac{p+1}{b+1}$. Since $PXAY$ is a parallelogram we have that $p+a=x+y$, which directly implies that $x=\frac{(p+1)b}{b+1}$. But now notice that we must have that: $$\frac{x-y}{\overline{x-y}}=-\frac{p-t}{\overline{p-t}}$$thus we have that $PT \perp XY$. . .

Coord bash goes brr Let $B = (-1, 0), A = (0, a), C = (1, 0), P = (p, 0)$. By power of a point, $T = (0, -\frac{1}{a})$. Since $\frac{BP}{PC} = \frac{p+1}{2}$, we have $X = (\frac{p+1}{2}(-1), \frac{p+1}{2}\cdot a)$. Similarly, $Y = (\frac{1-p}{2}\cdot 1, \frac{1-p}{2}\cdot a)$. Finally, the slope of $XY$ is \[\frac{a(\frac{1-p}{2} - \frac{p+1}{2})}{\frac{1-p}{2} + \frac{p+1}{2}} = \frac{a(-p)}{1} = -ap\]But the slope of $PT$ is $\frac{1}{ap}$, so $PT\perp XY$.

Let $O$ be the midpoint of $AT$. Note that $\triangle OAX\cong\triangle OCY$ hence $OX=OY$. Let $M$ be the midpoint of $XY$. Then $OM\perp XY$. As $AXPY$ is a parallelogram, $M$ is also the midpoint of $AP$. Hence $OM$ is the midsegment of the triangle $APT$ and $OM \parallel PT$ Then we get $PT\perp XY$.

Wait this problem is amazing. Let $O$ be the circumcenter of $ABC$ and $M$ be the midpoint of $XY$. Note that since $BX \cdot XA = AY \cdot YC$, we have that $OX = OY$. But this indicates $OM \perp XY$, and now taking a $\times 2$ homothety proves the result.

Let $E=\overline{AP}\cap\overline{XY}$, let $D$ be the center of $(ABCT)$. Observe that $D$ is the midpoint of $\overline{AT}$, thus $\overline{DE}\parallel\overline{PT}$. Also, we have $BD=AD$ and $BX=PX=AY$ and $\measuredangle XBD=\measuredangle ABD=\measuredangle DAB=\measuredangle CAD=\measuredangle YAD$, which all in all gives $\triangle XBD\cong\triangle YAD\implies YD=DX\implies \overline{XY}\perp\overline{DE}\parallel\overline{PT}$, as desired.

rcorreaa wrote: Animate $P$ on $BC$. Let $Q_1=PT \cap (ABC)$, $Q_1 \neq T$ and $Q_2$ is the intersection between the line throught $T$ perpendicular to $XY$ with $(ABC)$. Then, note that the composed map $$ Q_2 \overset{A}{\mapsto} AQ_2 \cap BC \overset{B}{\mapsto} P \overset{T}{\mapsto} Q_1 $$is projective. $(*)$ Thus, by the Moving Points Lemma, we only have to check that $Q_1=Q_2$ for three choices of $P$. Choosing $P=B,C,M$, where $M$ is the midpoint of $BC$, we have that $Q_1=Q_2$ $\implies$ from $(*)$, $Q_1=Q_2$ for all $P$. Then, $PT \perp XY$, as desired. $\blacksquare$ I have a doubt in this solution. Why does $Q_2$ move projectively on $\odot(ABC)$ ? Anyways below is a solution using Untethered Moving Points. We prove the assertion more generally for any point $P$ on line $BC$. Animate $P$ linearly on $BC$. Then $X,Y$ also move linearly on lines $AB,AC$ (respectively). Let $\infty$ be the point at infinity along line $XY$ and $\infty_{\perp}$ be the point at infinity in the direction perpendicular to $XY$. Then, Degree of line $XY$ is $\le \deg X + \deg Y = 1 + 1 = 2$. Thus $\deg \infty \le 2 \implies \deg \infty_{\perp} \le 2$. We want to prove points $P,T, \infty_{\perp}$ are collinear. Now $\deg P = 1, \deg T = 0, \deg \infty_{\perp} \le 2$, hence it suffices to solve the problem for $4$ choices of $P$. The case $P=B,C$ and midpoint of segment $BC$ are fairly direct. We consider $P$ as the point at infinity along line $BC$ as the fourth case. Note line $XY$ tends to be perpendicular to $BC$ (say because $AX = CY$, so in the limiting case $AX = AY$). This completes the proof. $\blacksquare$

Does this work? Let $D=\overline{PT} \cap \overline{XY}.$ Fix $\triangle ABC$ and animate $P$ on $\overline{BC}.$ Then, $T$ is fixed and $X,Y$ move as a function of $P.$ By rotation, $P \mapsto X$ and $P \mapsto Y$ are projective. In particular, we have $\deg(X)=1$ and $\deg(Y)=1.$ By Zack's Lemma, $\deg(\overline{XY})=2.$ By Zack's Lemma again, we also have $\deg(\overline{PT})=1.$ Zack's Lemma for a third time gives $\deg(D)=3.$ It suffices to check four cases of $P.$ This is not hard to do: take $P=B, P=C,$ $P$ the midpoint of $BC,$ and $P=P_\infty,$ where $P_\infty$ is the point at infinity along $\overline{BC}.$

Toss the diagram onto the cartesian plane. Let $A = (0,a), B = (-c,0), C = (c,0), P = (p, 0)$. WLOG $p > c > 0$. The following can be easily seen: 1. $T = \left(0, -\frac{c^2}{a}\right)$ (by seeing that $T$ lies on the y-axis by symmetry and that $AB \perp BT$ 2. $X = \left(\frac{p-c}{2}, \frac{a(p+c)}{2c}\right)$ (by noticing that the x-coordinate of $X$ is just the average of those of $B, P$; then using slope formulae to compute its y-coordinate) 3. $Y = \left(\frac{p+c}{2}, \frac{a(c-p)}{2c}\right)$ (similar methods as for $X$) Calculating the slopes of $XY, TP$ and multiplying, it can be easily seen that we get $-1$, so $XY \perp PT$. $\square$