Note that $\measuredangle$ denotes a directed angle considered modulo $\pi$. Let $M$ denote the midpoint of $BC$, $D$ denote the foot of the $A$-altitude (of $\triangle ABC$) and $\omega$ denote the nine-point circle of $\triangle ABC$. Also, $Z$ is the midpoint of $PX$. Lemma 1: $A,X,O,M,T$ are concyclic. Call this circle $\Omega$ Proof: Simply note that $\measuredangle OXT=90^o=\measuredangle OAT$ and $\measuredangle OMT=90^o=\measuredangle OAT$. Hence. $X, M$ both lie on $\odot(AOT)$, as desired. $\square$ Lemma 2: $AM$ bisects $PO$. Proof: To prove this, we will show that $APMO$ is a parallelogram. Vectorially, this is equivalent to $\vec{AO}=\vec{PM}$. But this readily follows as $$\vec{PM}=\vec{M}-\vec{P}=\frac{1}{2}\left(\vec{B}+\vec{C}\right)-\frac{1}{2}\left( \vec{A}+(\vec{A}+\vec{B}+\vec{C}-2\vec{O}) \right)=\vec{O}-\vec{A}=\vec{AO}$$as desired. $\square$ Lemma 3: $P$ is the orthocenter of $\triangle ATM$. Proof: Note that $OA \perp AT$ and $OM \perp TM$ and $O$ lies on the $\odot(ATM)$. Hence, $AM$ is the $l=O\text{-Simson line}$ of $\triangle ATM$. Also, $OP$ is bisected by $l$ and $P$ lies on the $A\text{-Altitude}$ of this triangle. Hence $P$ must be the orthocenter of $\triangle ATM$, as desired $\square$ Lemma 4: $\triangle AXP$ is isosceles. In particular, $AZ \perp PZ$. Proof: Let $E=AH \cap \Omega$. Since $P$ is the orthocenter of $\triangle ATM$ by lemma 3, hence $E$ is the reflection of $P$ over $TM$. Hence, $\measuredangle AXT \overset{\Omega}{=} \measuredangle AET=\measuredangle PET=\measuredangle TPE=\measuredangle XPA$, as desired $\square$ Lemma 5: $Z$ lies on $AM$. Proof: By lemma 3, we get $TP \perp AM$. Also by lemma 4, we get $PZ \perp ZA$. Since $Z$ lies on $TP$, hence $PZ \perp ZA \Leftrightarrow TP \perp AZ$. Hence we must have that $Z$ lies on $AM$. $\square$ Hence, by lemma 3, we get that $TP \perp AM \Leftrightarrow \measuredangle PZM=90^o$. Since $PM$ is a diamater of the $\omega$, we get that $Z$ lies on $\omega$, as desired. $\blacksquare$

In the following diagram, $P$ is relabelled as $H_A$, $M$ is the midpoint of $BC$ and $D$ is the second intersection of $AM$ with the nine-point circle (call it $\omega$). Lemma. $H_A$ is the orthocenter of $\triangle AMT$, with $D$ the foot from $T$ onto $AM$. Clearly $AH_A \perp BC$, so it suffices to show $MH_A \perp AT$. Take a homothety centered at $H$ with scale factor $2$: $MH_A$ is sent to line $AO$, so $AT \perp AO \parallel MH_A$, proving the first part. Meanwhile, note that $H_A$ is the antipode of $M$ w.r.t. $\omega$, meaning that $D$ is the foot from $H_A$ onto $AM$. It follows from a well-known property of the orthocenter that the reflection $X'$ of $H_A$ over $D$ lies on $(AMOT)$, which is cyclic by right angles. Yet this implies $X'$ is the foot from $O$ onto $TH_A$, so $X = X'$ and we are done.

Let $M$ denote the midpoint of $BC$. Let $\Gamma$ be the circle with diameter $BC$. Note that $AP^2=OM^2=OC^2-MC^2=PM^2-MC^2$, thus $P$ lies on the radical axis of $A$ and $\Gamma$, note that $T$ also lies on the same radical axis, thus $TP \perp AM$, and hence the intersection of lines $TP$ and $AM$ , $V$, lies on the nine-point circle, but because $OX || AM$ we have that $MV$ is clearly the $P$-midline in $\triangle{POX}$.

From the point where one proves that $AM \perp TX$ . We have that $XO \parallel AM$ . Thus $AXOM$ is an isosceles trapezium (since $TAXOM$ is cyclic). However, $AX=OM=AH/2=AP$. Thus $AM \bigcap TX$ is the midpoint of $PX$ ($N$). So $N, P, D, M$ are cocyclic. However, one can easily prove that $P$ belongs to the nine point circle. Hence, the result.

Let $M_A$ be the midpoint of $BC$, and $\triangle DEF$ be the orthic triangle of $\triangle ABC$. Let $M$ be the midpoint of $PX$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(15); defaultpen(dps); /* default pen style */ pen dotstyle = green; /* point style */ real xmin = -16.62, xmax = 9.58, ymin = -4.7, ymax = 7.34; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((0.6012190533078552,6.245840367777759)--(0.7153786855300965,5.987059421085614)--(0.974159632222241,6.101219053307855)--(0.86,6.36)--cycle, linewidth(1.6)); draw((2.1473823505951684,4.17200291842619)--(2.215332324429749,3.8974436332874793)--(2.48989160956846,3.96539360712206)--(2.421941635733879,4.239952892260771)--cycle, linewidth(1.6)); draw((2.8721950772975338,-1.3203223722714752)--(2.592517449569009,-1.2781272949739417)--(2.5503223722714754,-1.5578049227024662)--(2.83,-1.6)--cycle, linewidth(1.6)); /* draw figures */ draw((0.86,6.36)--(3.222841081768753,1.0038312727493026), linewidth(2) + wrwrwr); draw((-12.083581619768571,0.6500181167155676)--(-2.34,-0.82), linewidth(2) + wrwrwr); draw((-1.2065428110900374,1.7231945676167293)--(8,-2.38), linewidth(2) + wrwrwr); draw((-12.083581619768571,0.6500181167155676)--(0.86,6.36), linewidth(2) + wrwrwr); draw((0.86,6.36)--(-0.2703584479813297,-1.132247661619838), linewidth(2) + wrwrwr); draw((-12.083581619768571,0.6500181167155676)--(2.421941635733879,4.239952892260771), linewidth(2) + wrwrwr); draw((2.421941635733879,4.239952892260771)--(3.222841081768753,1.0038312727493026), linewidth(2) + wrwrwr); draw((3.0970212111124376,3.621685520290938)--(-2.34,-0.82), linewidth(2) + wrwrwr); draw((0.86,6.36)--(2.83,-1.6), linewidth(2) + wrwrwr); draw((2.83,-1.6)--(3.222841081768753,1.0038312727493026), linewidth(2) + wrwrwr); draw((0.86,6.36)--(-2.34,-0.82), linewidth(2) + wrwrwr); draw((-2.34,-0.82)--(8,-2.38), linewidth(2) + wrwrwr); draw((8,-2.38)--(0.86,6.36), linewidth(2) + wrwrwr); /* dots and labels */ dot((0.86,6.36),dotstyle); label("$A$", (0.56,6.78), NE * labelscalefactor); dot((-2.34,-0.82),dotstyle); label("$B$", (-2.7,-1.36), NE * labelscalefactor); dot((8,-2.38),dotstyle); label("$C$", (7.92,-2.9), NE * labelscalefactor); dot((3.222841081768753,1.0038312727493026),linewidth(4pt) + dotstyle); label("$O$", (3.46,1.12), NE * labelscalefactor); dot((-12.083581619768571,0.6500181167155676),linewidth(4pt) + dotstyle); label("$T$", (-12.22,0.86), NE * labelscalefactor); dot((-0.2703584479813297,-1.132247661619838),linewidth(4pt) + dotstyle); label("$D$", (-0.3,-1.82), NE * labelscalefactor); dot((3.0970212111124376,3.621685520290938),linewidth(4pt) + dotstyle); label("$E$", (3.3,3.7), NE * labelscalefactor); dot((-1.2065428110900374,1.7231945676167293),linewidth(4pt) + dotstyle); label("$F$", (-1.66,1.74), NE * labelscalefactor); dot((0.0743178364624929,1.152337454501396),linewidth(4pt) + dotstyle); label("$H$", (0.32,0.46), NE * labelscalefactor); dot((0.46715891823124644,3.756168727250698),linewidth(4pt) + dotstyle); label("$P$", (0.62,3.26), NE * labelscalefactor); dot((2.421941635733879,4.239952892260771),linewidth(4pt) + dotstyle); label("$X$", (1.82,4.48), NE * labelscalefactor); dot((1.4445502769825629,3.998060809755734),linewidth(4pt) + dotstyle); label("$M$", (0.76,4.1), NE * labelscalefactor); dot((2.83,-1.6),linewidth(4pt) + dotstyle); label("$M_A$", (2.78,-2.18), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that since $\angle OAT = 90^{\circ}$, $AT$ is tangent to $\odot (ABC)$. Also, $\measuredangle OXT = \measuredangle OAT = \measuredangle OM_AT = 90^{\circ} \Rightarrow O, X, A, T, M_A$ lie on a circle. $\Rightarrow$ We can redefine $X$ as the second intersection point of $TP$ and $\odot (ATM_A)$. Now, $EF$ is the radical axis of $\odot (AEHF)$ and $\odot (BFEC) \Rightarrow PM_A$ is the perpendicular bisector of $EF$. And, $\angle TAB = \angle ACB \Rightarrow \angle TAF = \angle AFE \Rightarrow TA \parallel EF$ But, $M_AP$ is perpendicular to $EF \Rightarrow M_AP \bot TA$ Also, $AP \bot TM_A \Rightarrow P$ is the orthocenter of $\triangle ATM_A \Rightarrow TP \bot AM_A$ Now, It is well known that the point where the altitude of a triangle meets its circumcircle again is the reflection of the orthocenter about the foot of that altitude on the opposite side. Thus, In $\triangle AM_AT$, the midpoint of $PX$ is the foot of the $T$-altitude on $AM_A$ $\Rightarrow \measuredangle PMM_A = 90^{\circ} = \measuredangle PDM_A \Rightarrow M$ lies on $\odot (PDM_A)$ $\Rightarrow M$ lies on the nine-point circle of $\triangle ABC$. $\blacksquare$

Let $Y$ be the midpoint of $PX$ and $M$ be the midpoint of $BC.$ It is well known that $AP=MO,$ and $AP\parallel MO,$ so $APMO$ is a parallelogram. Thus $AO\parallel PM\perp AT.$ Also, $AP\perp TM,$ so $P$ is the orthocenter of $\triangle ATM$ and thus $AM\perp TX.$ Note that pentagon $TAXOM$ is cyclic in that order with diameter $TO,$ and $OX\parallel AM,$ so $PM=AO=MX.$ Then $MA$ is the perpendicular bisector of $PX,$ so $Y$ lies on the circle with diameter $PM,$ as desired.

Let $H_A$ be the $A-$Humpty point and $P’-$ the midpoint of $AH_A$. We know that $H_A$ is on the $A-$ Apollonius circle of $\triangle{ABC}$. As $HH_A \parallel PP’$ and $TP’ \perp AM$, we have that $T-P-P’$. Now, $\angle{XAM}=\angle{XTM}=\angle{HAM} \rightarrow P’$ is the midpoint of $PX$

whatshisbucket wrote: Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be the midpoint of $\overline{AH}$ and let $T$ be on line $BC$ with $\angle TAO=90^{\circ}$. Let $X$ be the foot of the altitude from $O$ onto line $PT$. Prove that the midpoint of $\overline{PX}$ lies on the nine-point circle* of $\triangle ABC$. *The nine-point circle of $\triangle ABC$ is the unique circle passing through the following nine points: the midpoint of the sides, the feet of the altitudes, and the midpoints of $\overline{AH}$, $\overline{BH}$, and $\overline{CH}$. Proposed by Zack Chroman Solution. Let $N_9$ be the center of the nine point circle,$D$ be the foot of the $A-$altitude, $K$ be the midpoint of $PX,$ $M$ be the midpoint of $BC$,$L$ be the foot of the perpendicular from $O$ to $AD$, $P'$ be the reflection of $P$ on $D.$ It is well known that $N_9$ is the midpoint of $OH,$ and that $P,D$ and $M$ lie on the nine-point circle. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -11.72, xmax = 11.4, ymin = -7.6, ymax = 8.48; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((-2.26,4.08)--(-3.2,-2.12), linewidth(0.5) + wrwrwr); draw((-2.26,4.08)--(6.13,-2.34), linewidth(0.5) + wrwrwr); draw((xmin, -0.023579849946409406*xmin-2.19545551982851)--(xmax, -0.023579849946409406*xmax-2.19545551982851), linewidth(0.5) + wrwrwr); /* line */ draw(circle((1.5254779442162991,0.33481463426398006), 5.325059313272605), linewidth(0.5) + wrwrwr); draw((xmin, 42.409090909090956*xmin + 99.92454545454557)--(xmax, 42.409090909090956*xmax + 99.92454545454557), linewidth(0.5) + wrwrwr); /* line */ draw((1.5254779442162991,0.33481463426398006)--(-2.26,4.08), linewidth(0.5) + wrwrwr); draw((xmin, 1.0107585004600594*xmin + 6.364314211039734)--(xmax, 1.0107585004600594*xmax + 6.364314211039734), linewidth(0.5) + wrwrwr); /* line */ draw((xmin, 0.5903328050713154*xmin + 2.885039619651347)--(xmax, 0.5903328050713154*xmax + 2.885039619651347), linewidth(0.5) + wrwrwr); /* line */ draw((0.014828758983379346,2.893793522537732)--(1.5254779442162991,0.33481463426398006), linewidth(0.5) + wrwrwr); draw((1.5254779442162991,0.33481463426398006)--(1.465,-2.23), linewidth(0.5) + rvwvcq); draw((-8.27559930220558,-2.0003181300658923)--(-2.4927944304726974,-5.792600165046676), linewidth(0.5) + rvwvcq); draw((-2.4927944304726974,-5.792600165046676)--(1.465,-2.23), linewidth(0.5) + rvwvcq); draw((-2.346158243128199,0.4261072346086519)--(1.5254779442162991,0.33481463426398006), linewidth(0.5) + wrwrwr); draw((-2.3204779442162993,1.51518536573602)--(1.465,-2.23), linewidth(0.5) + wrwrwr); draw((-2.380955888432599,-1.0496292685279605)--(1.5254779442162991,0.33481463426398006), linewidth(0.5) + linetype("4 4") + wrwrwr); draw((-1.15282459261646,2.204489444136876)--(1.465,-2.23), linewidth(0.5) + linetype("4 4") + wrwrwr); draw((-8.27559930220558,-2.0003181300658923)--(1.5254779442162991,0.33481463426398006), linewidth(0.5) + rvwvcq); draw(circle((-0.4277389721081497,-0.35740731713199014), 2.6625296566363024), linewidth(0.8) + linetype("4 4") + sexdts); draw(circle((-3.3750606789946396,-0.8327517479009561), 5.037706824969461), linewidth(0.8) + linetype("4 4") + dtsfsf); /* dots and labels */ dot((-2.26,4.08),dotstyle); label("$A$", (-2.18,4.28), NE * labelscalefactor); dot((-3.2,-2.12),dotstyle); label("$B$", (-3.12,-1.92), NE * labelscalefactor); dot((6.13,-2.34),dotstyle); label("$C$", (6.22,-2.14), NE * labelscalefactor); dot((-2.4066361873444984,-2.138707399655328),linewidth(4pt) + dotstyle); label("$D$", (-2.32,-1.98), NE * labelscalefactor); dot((1.5254779442162991,0.33481463426398006),linewidth(4pt) + dotstyle); label("$O$", (1.6,0.5), NE * labelscalefactor); dot((-8.27559930220558,-2.0003181300658923),linewidth(4pt) + dotstyle); label("$T$", (-8.2,-1.84), NE * labelscalefactor); dot((-2.380955888432599,-1.0496292685279605),linewidth(4pt) + dotstyle); label("$H$", (-2.3,-0.88), NE * labelscalefactor); dot((-2.3204779442162993,1.51518536573602),linewidth(4pt) + dotstyle); label("$P$", (-2.24,1.68), NE * labelscalefactor); dot((0.014828758983379346,2.893793522537732),linewidth(4pt) + dotstyle); label("$X$", (0.1,3.06), NE * labelscalefactor); dot((-1.15282459261646,2.204489444136876),linewidth(4pt) + dotstyle); label("$K$", (-1.08,2.36), NE * labelscalefactor); dot((1.465,-2.23),linewidth(4pt) + dotstyle); label("$M$", (1.54,-2.08), NE * labelscalefactor); dot((-2.4927944304726974,-5.792600165046676),linewidth(4pt) + dotstyle); label("$P'$", (-2.42,-5.64), NE * labelscalefactor); dot((-2.346158243128199,0.4261072346086519),linewidth(4pt) + dotstyle); label("$L$", (-2.26,0.58), NE * labelscalefactor); dot((-0.42773897210815004,-0.3574073171319902),linewidth(4pt) + dotstyle); label("$N_9$", (-0.34,-0.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that, \begin{align*} \angle OMP' &=90^{\circ}+\angle DMP'\\ &=90^{\circ}+\angle PMD\\ &=90^{\circ}+\angle AOL\\ &=90^{\circ}+(90^{\circ}-\angle LAO)\\ &=180^{\circ}-\angle P'AO. \end{align*}Therefore, $AOMP'$ is cyclic, but as $\angle OXT=\angle TAO=90^{\circ}$, so $AXOT$ is cyclic as well also $\angle TAO=\angle OMT=90^{\circ},$ so $AOMT$ is cyclic too. Hence, the points $A,O,M,T,P',X$ are all concyclic. Thus, \begin{align*} \angle PXM &=\angle TXM\\ &=\angle TOM\\ &=180^{\circ}-\angle TP'M\\ &=180^{\circ}-\angle TPM\\ &=\angle XPM. \end{align*}Which means $MK\perp PX\implies PKMD\text{ is cyclic.}$ And we are done. $\blacksquare$

I didn't attempt ELMO, but tried this geometry problem. [asy][asy] import geometry; import olympiad; unitsize(3cm); pair A = dir(135), B = dir(210), C = -1/B; triangle tri = triangle((point) A, (point) B, (point) C); line t = tangents(circumcircle(tri), A)[0]; pair T = intersectionpoint(t, tri.BC); pair O = circumcenter(tri); pair H = orthocentercenter(tri); pair P = (A+H)/2; pair X = foot(O, P, T); pair M = (B+C)/2; pair A_ = -A; pair N = (O + H)/2; pair Y = (P+X)/2; draw(A--T--B); draw(tri); draw(A--M); draw(circle(M, abs(M-P)), 0.8blue); draw(T--X); draw(A--H--A_, dotted); draw(M--P^^A--A_, linetype("0 2")); draw(circumcircle(tri)); draw(circumcircle((point) O, (point) T, (point) A), 0.9red); draw(circle(N, abs(N - P)), 0.8green); dot(A^^B^^C); dot(T^^P^^X); dot(O^^H); dot(M); dot(Y); dot(N); label("A", A, A); label("B", B, B*dir(10)); label("C", C, C); label("T", T, dir(180)); label("P", P, 2dir(-10)); label("H", H, dir(45)); label("O", O, dir(180)); label("X", X, dir(60)); label("M", M, dir(-60)); label("N$_9$", N, dir(-90)); label("Y", Y, dir(-30)); [/asy][/asy] Let $M$ be the midpoint of side $BC$. Claim 1: $O, A, T, M, X$ are concyclic. \[\measuredangle OAT = \measuredangle OXT = \measuredangle OMT = 90^{\circ}\] Claim 2: $P$ is the orthocneter of $\triangle ATM$. Note that $AP\perp MT$. Now consider a homothety at $H$ with scale $2$, it's well-known that $P \to A$ and $M \to \overleftrightarrow{AO}\cap\odot(ABC)$, thus, $PM\perp AT$, therefore, $P$ is the orthocenter of $\triangle ATM$. Claim 3: $MP = MX$ It's well known that the reflection of orthocenter on a side lies on the circumcircle. Since, $TP\perp AM$ and $X = TP\cap\odot(ATM)$, $X$ is the reflection of $P$ in $AM$, therefore, $MP = MX$. Main Proof: Let the nine-point center be $N_9$ and the midpoint of $PX$ be $Y$. Let the nine-point circle be $\omega$. $\odot(A, AB)$ refers to the circle with center $A$ and radius $AB$. Note that $M$ is the antipode of $P$ in $\omega$. Consider the homothety at $P$ with scale $\tfrac12$. This sends $X$ to $Y$ and, the circle $\odot(M, MP)\to \odot(N_9, N_9P)\equiv \omega$. Since, $X\in\odot(M, MP)$, $Y\in\omega$. Therefore, $Y$ lies on the nine-point circle. 1111th post

Let $Q$ denote the foot of the altitude from $H$ to $AT$, $M$ denote the midpoint of $BC$, and $N$ denote the midpoint of $PX$. Lemma 1: $MQ = MA$. Proof: Note that $AOMP$ is in fact a parallelogram so $MP \parallel OA \parallel HQ$ and $P$ is the midpoint of $AH$; hence $MP \perp AQ$ and $MP$ bisects segment $AQ$ so the conclusion follows readily. Lemma 2: $P$ is the orthocenter of $\triangle ATM$. Proof: As shown in Lemma 1, $MP \perp AT$ and $AP \perp TM$, so the conclusion is obvious. Lemma 3: $M$ is the Miquel point of quadrilateral $AXPQ$. Proof: It suffices to show that $M$ lies on the circumcircles of both $\triangle AXT$ and $\triangle QPT$. The former is obvious, as $\angle AOT = \angle XOT = \angle OMT = 90^{\circ}$. To show the latter, note that $\measuredangle MQT = \measuredangle MQA = \measuredangle MAT = \measuredangle MHT$ where the second and third equalities follow from the previous lemmas. To the main problem. By Lemma 3, $M$ is the center of spiral similarity sending $AQ$ to $XP$; hence $\triangle MAQ \sim \triangle MXP \implies MX = MP$. Thus $\angle MNP = 90^{\circ}$, but $MP$ is a diameter of the nine-point circle of $\triangle ABC$ so it follows that $N$ must lie on it.

Alas, I am wayyy too late to this thread. Let $D$, $E$, $F$ be the feet of the altitudes from $A$, $B$, $C$ respectively, $M$ be the midpoint of $\overline{BC}$, $Q=\overline{AM}\cap\overline{TP}$, and $\omega$ be the circle with diameter $\overline{BC}$. To begin, we know that $PE=PF=PA$ and $\overline{PE}$, $\overline{PF}$ are tangent to $\omega$. This tells us that $P$ has equal power wrt $A$ and $\omega$, which used in conjunction with the fact that $$\text{Pow}_A T=TA^2=TB\cdot TC=\text{Pow}_\omega T$$implies that $\overline{TP}$ is, in fact, the radical axis of $A$ and $\omega$; therefore, $\overline{OX}\parallel\overline{AM}\perp\overline{TP}$. Notice that this places $Q$ on the nine-point circle because $\overline{PM}$ is one of the diameters. [asy][asy] size(12cm); defaultpen(fontsize(9pt)); pair A=(-40,80), B=(-80,-40), C=(80,-40), D=(-40,-40), E=(0,40), F=(-64,8), H=(-40,0), P=(-40,40), T=(-280,-40), M=(0,-40), O=(0,0), X=(-16,48), N=(-28,44); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(H); dot(P); dot(T); dot(M); dot(O); dot(X); dot(N); draw(circumcircle(A,B,C), linewidth(0.5)); draw(A--B--C--cycle, linewidth(0.5)); draw(A--D, linewidth(0.5)); draw(B--E, linewidth(0.5)); draw(C--F, linewidth(0.5)); draw(circumcircle(B,E,C), linewidth(0.7)+dashed); draw(A--T--B, linewidth(0.5)); draw(T--X--O, linewidth(0.5)); draw(circumcircle(T,A,O), linewidth(0.8)+dotted); draw(A--M, linewidth(0.4)+dashed); label("$A$", A, (0,1)); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, (0,-1)); label("$M$", M, (0.5,-1)); label("$O$", O, (1,0)); label("$T$", T, SW); label("$E$", E, (0.5,1)); label("$F$", F, (-1,0)); label("$P$", P, (-0.5,0.5)); [/asy][/asy] Now, we have that $T$, $A$, $X$, $O$, and $M$ are concylic by perpendicularities; furthermore, $MOXA$ must be an isosceles trapezoid based on the parallelism we established above. It follows that $AX=OM=AP$, so $APX$ is an isosceles triangle; therefore, $Q$ must actually be the midpoint of $\overline{PX}$. That's it.

TheDarkPrince wrote:

This was exactly what I did. But alas! I forgot to submit my solutions.

whatshisbucket wrote: Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be the midpoint of $\overline{AH}$ and let $T$ be on line $BC$ with $\angle TAO=90^{\circ}$. Let $X$ be the foot of the altitude from $O$ onto line $PT$. Prove that the midpoint of $\overline{PX}$ lies on the nine-point circle* of $\triangle ABC$. *The nine-point circle of $\triangle ABC$ is the unique circle passing through the following nine points: the midpoint of the sides, the feet of the altitudes, and the midpoints of $\overline{AH}$, $\overline{BH}$, and $\overline{CH}$. Proposed by Zack Chroman Let $K$ be the $A$-HM point of $\triangle ABC$. It is well-known that $K$ lies on the $A$-appollonius circle of $\triangle ABC$ hence $TA=TK$. Moreover, $\angle AKH=90^{\circ}$ hence $PA=PK$. Consequently, $\overline{TP}$ is the perpendicular bisector of $\overline{AK}$. Let $Y$ be the midpoint of $\overline{PO}$ and $Z$ be the midpoint of $\overline{PX}$. Then $\angle PZY=90^{\circ}$. Now $\overline{ZY} \perp \overline{TP}$ hence $\overline{ZY} \parallel \overline{AM}$ where $M$ is the midpoint of $\overline{BC}$. Note that $APMO$ is a parallelogram hence $Y$ lies on $\overline{AM}$; thus $\angle PZM=90^{\circ}$ proving the claim. $\blacksquare$

$M$ is the midpoint of $BC$, $Z$ be $TX$ intersected with the Euler Circle, $D$ is $AO$ intersected with $BC$. $\angle ADT=\angle PAT=90-B+C$, and using that $DOM\sim DTA$, we get $TOD\sim TPA$, so $\angle ATX=\angle OTM$, so $AXOM$ is a cyclic trapezium, so $AO=MX=MP$. Since $\angle PZM=90$, we get $PZ=ZX$.

Not adding much, but anyways : By the homothety at $H$ sending the nine-point circle of $ABC$ to its circumcircle, we have $PM \parallel AO \perp AT$, so $P$ is the orthocenter of $AMT$. So $AM \perp PT$ and thus, since $PAOM$ is a parallelogram, $AM$ is the prependicular bisector of $PX$, so done.

Abhaysingh2003 wrote: @above can you please teach me what is a prallelogram? a quadrilateral that has each side is parallel to this opposite one

I really like this problem! The key observation is that since $QX \perp AP$ and $PX \perp AQ,$ it follows that $QXPH$ is a parallelogram, since $HP \perp AP$ and $HQ \perp AQ.$ Thus we can redefine point $X$ as the point satisfying $QXPH$ is a parallelogram. Now let $\omega$ be the nine point circle of $(ABC).$ Let $HX \cap PQ = Y.$ I claim that $Y \in \omega,$ which finishes since $Y$ is the midpoint of $XH.$ Since $P$ and $Q$ are both points on $(AH),$ the projection from the center of $AH$ onto $PQ$ is $Y,$ which always lies on $\omega$ since the center of $(AH)$ lies on $\omega$ and $PQ$ passes through the diameter of $\omega.$ $\blacksquare$

[asy][asy] size(10cm); pair A,B,C,F,O,T,M,P,Z,X; A=dir(130); B=dir(220); C=dir(320); M=(B+C)/2; O=(0,0); F=A+dir(220); T=extension(A,F,B,C); P=A+M; Z=extension(P,T,A,M); X=2Z-P; draw(A--B--C--cycle);draw(unitcircle);draw(A--T--B);draw(A--M);draw(T--X--O);draw(circumcircle((A+B)/2,M,P),dotted); dot("$A$",A,dir(A));dot("$B$",B,dir(B));dot("$C$",C,dir(C));dot("$O$",O,W);dot("$M$",M,dir(M));dot("$T$",T,dir(T));dot("$P$",P,N);dot("$Z$",Z,dir(Z));dot("$X$",X,dir(X)); [/asy][/asy] Letting $M$ be the midpoint of $BC$, it's well known $APMO$ is a parallelogram. Note $$AT^2-AP^2=(OT^2-AO^2)-OM^2=MT^2-MP^2$$hence $PT \perp AM$ by perpendicularity lemma. Thus $P$ is the orthocenter of $\triangle ATM$; let the foot from $T$ to $AM$ be $Z$, which lies on the nine point circle since $\angle PZM= 90^\circ$. Finally since $MZ \perp PX \perp OX$ and $APMO$ is a parallelogram we have $$PZ=\text{distance}(O,\overline{AM})=XZ$$so $Z$ is the midpoint of $PX$, the end.

The title is wrong - this problem was worth a total of 7 points on contest.

brutal problem. Let $M$ be the midpoint of segment $BC$ so $X,A,M$ lie on the circle with diameter $OT$. Let $D$ denote the foot from $A$ to line $BC$. Remark that $AP=OM$ because $MP\parallel AO$ by homothety at $H$ and $OM\parallel AP$. WLOG assume $AB<AC$. We claim $PT\perp AM$. It is equivalent to show the length from $A$ to the projection of $P$ onto $AM$ is the same as the length from $A$ to the projection of $T$ onto $AM$: that is, to show \[AP\cos\angle PAM=AT\cos\angle TAM\iff OM\cos\angle DAM=OM\cdot\frac{\sin\angle AMT}{\sin\angle OAM}\sin\angle OAM\iff\]\[\cos\angle DAM=\sin\angle AMD,\]which is trivial. Let $PT\cap AM=Y$. It is clear that $Y$ lies on the nine-point circle, so the problem amounts to showing that $AX=AP=OM$. But since $OX\perp PT\perp AM$, then $AXOM$ is an isosceles trapezoid because it is inscribed in a circle and the result follows.

Beautiful problem.. And very rich configuration. Let $D,D_B,D_C$ be the foot of perpendicular from $A,B,C$ respectively, $E=AH \cap TX$ and let $M,M_B,M_C$ be midpoints of $BC, AB, AC$ respectively Now as $OM||AP, OM=AP$ so $AOPM$ is a parallelogram so $PM||AO$ So $\angle MAO=\angle AMP$ And now construct $\triangle CFD \cong\triangle CPD$ Now As $AP||MP, MP=NF$ so $F\in \odot ACF$ so $\angle FTM=\angle PTD=\angle MAF$ So $AM\perp TX$ and $\angle ATX=\angle MAO=\angle AMP$ Hence $E$ is also midpoint of $PX$ and as $\angle AEP=\angle ADM=90^\circ$ so $PEDM$ is also cyclic and Hence $E\in$ nine point circle of $\triangle ABC$ Also $E$ is the intersection of $\odot(CD_BK_1), \odot(BD_CK_2), \odot(DD_BD_C)$ where $K_1=CM_C\cap \odot(ABC), K_2=BM_B\cap\odot(ABC)$

Let $D = AH\cap BC$ and $M$ the midpoint of $BC$. Note that $AP\perp TM$ and $MP\parallel OA\perp AT$, so $P$ is the orthocenter of $\triangle ATM$. In particular, since $A$, $T$, and $M$ all lie on the circle with diameter $TO$, $X$ is the reflection of $P$ over $AM$. Now treating points as their projections onto line $TP$, we have $$N_9 = \frac{1}{2}H + \frac{1}{2}O = \frac{1}{2}(2P-A) + \frac{1}{2}X = \frac{1}{2}(2P-\frac{1}{2}P-\frac{1}{2}X)+\frac{1}{2}X = \frac{3}{4}P+\frac{1}{4}X$$ so if $K$ is the midpoint of $PX$, the foot from $N_9$ to $PX$ is the midpoint of $PK$, as desired.

Use complex numbers with $(ABC)=\mathbb{S}^1$. Let $N$ be the midpoint of $OH$ and let $D\in (ABC)$ be the point such that $ABDC$ is harmonic. Then$$\frac{b-a}{c-a}\frac{c-d}{b-d}=-1,$$so$$d=\frac{2bc-ca-ab}{b+c-2a}.$$Also, the tangents to $(ABC)$ through $A$ and $D$ intersect at $T$, so$$t=\frac{2ad}{a+d}=\frac{2a\frac{2bc-ca-ab}{b+c-2a}}{a+\frac{2bc-ca-ab}{b+c-2a}}=a\frac{2bc-ca-ab}{bc-a^2}.$$Clearly$$p=\frac{a+h}{2}=\frac{a+a+b+c}{2}=a+\frac{b+c}{2},$$so\begin{align*}x & =\frac{\overline pt-p\overline t}{2\left (\overline p-\overline t\right )} \\ & =\frac{\left (\frac{1}{a}+\frac{b+c}{2bc}\right )a\frac{2bc-ca-ab}{bc-a^2}-\left (a+\frac{b+c}{2}\right )\frac{1}{a}\frac{\frac{2}{bc}-\frac{1}{ca}-\frac{1}{ab}}{\frac{1}{bc}-\frac{1}{a^2}}}{2\left (\frac{1}{a}+\frac{b+c}{2bc}-\frac{1}{a}\frac{\frac{2}{bc}-\frac{1}{ca}-\frac{1}{ab}}{\frac{1}{bc}-\frac{1}{a^2}}\right )} \\ & =\frac{\frac{(b-c)^2(a^2+bc)}{2bc(a^2-bc)}}{\frac{(a^2+bc)(ca+ab-2bc)}{abc(a^2-bc)}} \\ & =\frac{a(b-c)^2}{2(ca+ab-2bc)}. \end{align*}Now observe that\begin{align*}\frac{p+x}{2}-n & =\frac{a+\frac{b+c}{2}+\frac{a(b-c)^2}{2(ca+ab-2bc)}}{2}-\frac{a+b+c}{2} \\ & =\frac{bc(2a-b-c)}{2(2bc-ca-ab)} \\ & =\frac{1}{2a}\frac{2a-b-c}{\overline{2a-b-c}}, \end{align*}so\begin{align*}\left |\frac{p+x}{2}-n\right | & =\left |\frac{1}{2a}\frac{2a-b-c}{\overline{2a-b-c}}\right | \\ & =\frac{1}{2|a|}\frac{|2a-b-c|}{\left |\overline{2a-b-c}\right |} \\ & =\frac{1}{2}, \end{align*}implying that the midpoint of $PX$ lies on the nine-point circle of $ABC$.

Let $M$ the midpoint of $BC$ and let $PT$ meet the NPC at $Q$, then clearly $PM \perp EF \parallel AT$ and $AP \perp TM$ so $P$ is orthocenter of $\triangle ATM$, in fact $PM$ is diameter in The NPC so $\angle PQM=90$ and $T,P,Q$ are colinear so $A,Q,M$ colinear, now note $ATMOX$ is cyclic so by final angle chase we get $\angle XMA=\angle ATX=\angle PMA$ and $AM \perp PX$ so $PQ=QX$ as desired thus we are done .

Solved with OronSH and GrantStar: Let $P'$ be the reflection of $P$ about line $BC$; let $M$ be the midpoint of $\overline{BC}$. Claim: $ATP' MOX$ is cyclic with diameter $\overline{TO}$; in particular, $TP'MA$ is cyclic. Proof: We have \[\angle P'MT = \angle PMT = 90^{\circ} - \angle PMO = 90^{\circ} - \angle PAO = \angle TAP'.\]What is a configuration issue $\blacksquare$ From there it follows that the radius of $(TPM)$ is equal to the radius of $(TP'M)$, which in turn is equivalent to $(TMX)$. Segments $\overline{MP}$ and $\overline{MX}$ inscribe an angle of $\angle T$ in congruent circles $(TPM)$ and $(TMX)$ respectively, hence $MP = MX$. It follows that if $N$ is the midpoint of $\overline{PX}$, $\angle PNM = 90^{\circ} = \angle PDM$, so $N$ lies on the nine-point circle.

Let $D$ be the foot of the $A$-altitude, $K \neq A$ be the intersection of the $A$-altitude with $(ABC)$, and $M$ be the midpoint of $\overline{BC}$. We begin with the following key claim. Claim: $\overline{MP} \perp \overline{AT}$. Proof: It suffices to show that $\measuredangle TAD=\measuredangle PMD$, since this implies $A,M,D,\overline{MP} \cap \overline{AT}$ cyclic. Indeed, we have $$\measuredangle TAD=\measuredangle TAK=\measuredangle AH'K=\measuredangle PMD$$where $H'$ is the reflection of $H$ over $M$ (which is well-known to lie on $(ABC)$). $\blacksquare$ This claim, along with the obvious $\overline{AP} \perp \overline{TM}$, implies that $P$ is the orthocenter of $\triangle AMT$, so $\overline{TP} \perp \overline{AM}$ as well. On the other hand, clearly $TAMXO$ is cyclic, so $X$ is the reflection of the orthocenter $P$ over $\overline{AM}$. It follows that the midpoint $N$ of $\overline{PX}$ is the foot of the altitude from $T$ (as well as $P$) to $\overline{AM}$, hence $\angle PNM=90^\circ=\angle PDM$, hence $PDMN$ is cyclic. Since $PDM$ is the nine-point circle of $\triangle ABC$ this finishes the problem. $\blacksquare$

Let $N$ be the midpoint of $PX$, and notice how the problem is equivalent to showing that $MN\perp PT$, where $M$ is the midpoint of $BC$. Now, let $O'$ be the reflection of $O$ with respect to $M$. See how by Tales the problem is equivalent to showing that $O'P\perp PT$. We now recall ourselves that we know lots of parallelograms concerning this points, in particular $AHO'O$, $APMO$ and $PHMO$. So we get that $$O'P\perp PT \iff AM\perp PT,$$but this is true as we can easily show that $P$ si the orthocenter of $\triangle TAM$ by noticing that $MP\parallel AO \perp AT$, which ends the problem.

whatshisbucket wrote: Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be the midpoint of $\overline{AH}$ and let $T$ be on line $BC$ with $\angle TAO=90^{\circ}$. Let $X$ be the foot of the altitude from $O$ onto line $PT$. Prove that the midpoint of $\overline{PX}$ lies on the nine-point circle* of $\triangle ABC$. *The nine-point circle of $\triangle ABC$ is the unique circle passing through the following nine points: the midpoint of the sides, the feet of the altitudes, and the midpoints of $\overline{AH}$, $\overline{BH}$, and $\overline{CH}$. Proposed by Zack Chroman Let $J$ be the $A-HUMPTY$ point and $A'$ be the intercetion $AT,HJ$ then: It is well know that $T$ is the center of the Apullonius circle so we have $TA=TJ$ but $<AJA'=90$ so $TA=TA'$. Now $T$ is the midpoint of $AA'$,$P$ is the midpoint of $AH$ so we get that $HJ//TP$ so if we let $L$ be the intersection of $AM$ with $PT$ be the paralliles lines we get that $L$ belongs to $DEF$ . It is enought now to prove that $L$ is the midpoint of $PX$. $A,X,O,M,T$ lie on a circle with diameter $TO$ so we have that $<AXP=<AXT=<AMT=<LMD=<APL=<APX$ done

Let $\tau$ be the circle with diameter $\overline{OT}$. Then, $A\in\tau$. Since $OM\perp MT$, we have $M\in\tau$. Since $\angle OX\perp XT$, we have $X\in\tau$. Let $D$ be the foot of altitude of $A$ in $\triangle ABC$. Claim: $P$ is the orthocenter of $\triangle TAM$. Proof. Note that $AP\perp MT$. From Euler's Theorem, we have $AP=\frac{AH}2=OM$ and $AP\parallel OM$, which means $APMO$ is a parallelogram. Hence, $PM\parallel AO\perp AT$. So, $MP,AP$ are altitudes in $\triangle TAM$. Hence, $P$ is the orthocenter. $\blacksquare$ Since $P$ is orthocenter, $TP\perp AM$. Let $J$ be the foot of altitude of $T$ in $\triangle TAM$. Note that the reflection of $P$ in $J$ lies on the circumcircle of $TAM$, which is $\tau$. So, the reflection of $P$ in $J$ is the second intersection point of line $TP$ with $\tau$ but that is $X$ by definition. So, $J$ is the mid-point of $\overline{PX}$. Finally, note that $PJ\perp JM$ and $PD\perp DM$, we have $\angle PJM=\angle PDM$, which means $J,P,M,D$ are concyclic. Note that circumcircle of $PMD$ is the nine-point circle by definition. Hence, $J$ lies on the nine-point circle. $\blacksquare$ Remark. This is a very beautiful configuration. Here are a few facts from this configuration: If $Q_A$ is the $A-$Dumpty Point, then $J$ is the reflection of $Q_A$ in the $A-$midline. If $H_A$ is the $A-$Humpty Point, then $J$ is the mid-point of $\overline{AH_A}$. Let $A,K$ be the intersections of $\tau$ and circumcircle of $ABC$. Then, $X,H_A,K$ are collinear and $XH_A\perp BC$. Let $Q$ be the mid-point of $\overline{AT}$. Then, $\overline{QD},\overline{QP}$ are tangents from $Q$ to the nine-point circle.

Denote by $D$ the foot from $A$ to $\overline{BC}$, and similarly define $E$ and $F$. Let $N$ be the midpoint of $\overline{BC}$ and define $M = \overline{AN} \cap (DEF)$. Claim: $M$ lies on $\overline{PT}$. Proof. Complex bash with unit circle $(ABC)$. Note $p = a + \frac{b+c}{2}$ and $n = \frac{b+c}{2}$. We may compute from $T = \overline{AA} \cap \overline{BC}$, \begin{align*} t &= \frac{a^2(b+c) - 2abc}{a^2 - bc}\\ \end{align*}Now Note that $\overline{NP} \perp \overline{AT}$ as, \begin{align*} \overline{\left(\frac{t - a}{p - n} \right)} &= \overline{\left(\frac{ \frac{a^2(b + c) - 2abc}{a^2 - bc} - a}{p - n}\right)}\\ &= \overline{\left(\frac{\frac{a^2(b + c) - abc - a^3}{a^2 - bc}}{a} \right)}\\ &= -\overline{\left(\frac{(a - b)(a - c)}{a^2 - bc} \right)}\\ &= \frac{(a - b)(a - c)}{a^2 - bc} \end{align*}which proves the claim. However then $P$ is the orthocenter of $\triangle NAT$ and thus upon noting that $\angle PMN = \angle PDN = 90$, we conclude the three points are collinear. $\square$ Now note as $\angle OXA = \angle OAT = \angle ONT = 90$ we have $X \in (AONT)$. Claim: $\triangle APX$ is isosceles. Proof. Note $$\angle AXT = \angle ANT = \angle MND = \angle APM = \angle APX$$which proves the claim. $\square$ From here the finish is easy. Note from our previous claims that $\overline{AM} \perp \overline{PT}$ and hence $M$ is the midpoint of $\overline{PX}$ as $\triangle PAX$ is isosceles, finishing.