Let $P_0$ be the reflection of $P$ wrt $O$. let $O'$ be the reflection of $P_0$ wrt $N$ the nine-point center. We prove that $O'$ is in fact the circumcenter of $XYZ$ and $H$ is also on that circle $H$ is as always the orthocenter.and then the similarity will be trivial as we have (since H is the reflection of A' wrt M_a, the midpoint of BC) : $HX || BC$ and the same for the other sides. now it sufficies to prove that $O'$ is on the perpendicular bisector of $HX$ and we'll do the same thing for the other sides(HY,HZ) and we'll be done. For that let $D'$ be the reflection of $D$ wrt $M_a$.let $d'$ pass thru $D'$ and be perpendicular to $BC$.obviously, $P_0$ is on $d'$.we have to prove that $d'$ is in fact the reflection of $s_X$ wrt $N$ where $s_x$ is the perpendicular bisector of $HX$.but obviously $s_x || d'$.if $H_a$ is $(AH,BC)$ and $X'=(s_X,BC)$ then we have $H_aX'=1/2 HX=M_aD=M_aD'$and since $N$ is the midpoint of $OH$ therefor it is on the perpendicular bisector of $M_aH_a$ and so of $X'D'$.we r done.

The problem is true for any point $P$, not necessarily inside the $\triangle{ABC}$

See here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=321333&hilit=pedal

Dear Mathlinkers, one basic idea (in order to make a link) is to consider this problem like a particular development of the O-Hagge's circle. You can see this circle in a more general situation on http://perso.orange.fr/jl.ayme Sincerely Jean-Louis

Let $X'$, $Y'$, and $Z'$ be the midpoints of $XO$, $YO$, and $ZO$, respectively. Let $O'$ be the nine point center. $A''$, $B''$, and $C''$ are the midpoints opposite $A$, $B$, and $C$, respectively. It suffices to show that $\triangle X'Y'Z' \sim \triangle ABC$. We have $X'D \parallel AO$ and $X'D = \dfrac {AO} {2}$. Therefore, $A''O' || X'D$ and $A''O' = X'D$, so $O'A''X'D$ is a parallelogram. Similarly, we get $Z'EC''O'$ and $Y'FB''O'$ are parallelograms. Extend lines through $O$ parallel to $BC$, $CA$, and $BA$ to intersect $PD$, $PE$, and $PF$ at $D'$, $E'$, and $F'$, respectively. Note that $OD'E'F'$ is homothetic to $NX'Y'Z'$, therefore it suffices to show that $\triangle D'E'F \sim \triangle ABC$. Since $\angle OD'P = \angle OE'P = \angle OF'P = 90$, we get $O'D'E'F'$ cyclic, so $\angle E'F'D' = 180 - \angle E'PD' = \angle ACB$, and so on, hence $\triangle D'E'F' \sim \triangle ABC$, as desired.

We use complex numbers. Let $a,b,c$ lie on the unit circle so that $x = -a$, et cetera. Now it is well known that \[ d = \frac{1}{2} \left( p + \frac{(b-c)\bar p + \bar b c - b \bar c}{\bar b - \bar c} \right) \] Hence \begin{align*} x &= 2d - (-a) \\ &= p + \frac{(b-c)\bar p + \bar b c - b \bar c}{\bar b - \bar c} + a \\ &= p + \frac{(b-c) \bar p + \frac{c^2-b^2}{bc}}{\frac{c-b}{bc}} + a \\ &= p - bc \bar p + a+b+c \\ y &= p - ca \bar p + a+b+c \\ z &= p - ab \bar p + a+b+c \end{align*} Therefore, \[ \frac{x-y}{a-b} = \bar p c \implies \frac{\lvert XY \rvert}{\lvert AB \rvert} = \lvert \bar p c \rvert = \lvert \bar p \rvert. \] Hence we get the conclusion.

Let $a$, $b$ and $c$ lie on the unit circle, corresponding to points $A$, $B$, and $C$. Also, we have that $A'$, $B'$, and $C'$ are equal to $-a$, $-b$ and $-c$ respectively. We can calculate that: $$d=\frac{1}{2}(b+c+p-bc\overline{p})$$$$e=\frac{1}{2}(a+c+p-ac\overline{p})$$$$f=\frac{1}{2}(a+b+p-ab\overline{p})$$ We have that $\frac{x-a}{2}=d \rightarrow x=2d+a=a+b+c+d-bc\overline{p}$. Exploiting symmetry, we find that: $$x=a+b+c+p-bc\overline{p}$$ $$y=a+b+c+p-ac\overline{p}$$ $$z=a+b+c+p-ab\overline{p}$$ We have that $$\frac{x-y}{x-z}=\frac{ac\overline{p}-bc\overline{p}}{ab\overline{p}-bc\overline{p}}=\frac{ac-bc}{ab-bc}=\frac{\frac{1}{b}-\frac{1}{a}}{\frac{1}{c}-\frac{1}{a}}=\overline{\bigg(\frac{b-a}{c-a}\bigg) }=\overline{\bigg( \frac{a-b}{a-c}\bigg) }$$ implying that $\triangle{XYZ} \sim \triangle{ABC}$ as desired. $\square$

hi Delray, could you please explain why the last "=" holds? there seem to be an extra c/b term. thanks.

how to write proofs using computer

niwobin wrote: hi Delray, could you please explain why the last "=" holds? there seem to be an extra c/b term. thanks. It doesn't hold, in fact, because the triangles are actually oppositely oriented. (That's why I had to use an absolute value in my solution.)

thanks to all. After some thought, to deal with the reverse-oriented, we can do a conjugate in the last step in Delray's proof. If you are reading the book EGMO and thinking applying theorem 6.16 as I did, all you need to do is add a conjugate.

Thanks guys! I will edit my proof accordingly.

enhanced wrote: how to write proofs using computer What do you mean by this? If you are wondering how to write math stuff, like you see here, you should learn latex.

Delray wrote: Let $a$, $b$ and $c$ lie on the unit circle, corresponding to points $A$, $B$, and $C$. Also, we have that $A'$, $B'$, and $C'$ are equal to $-a$, $-b$ and $-c$ respectively. We can calculate that: $$d=\frac{1}{2}(b+c+p-bc\overline{p})$$$$e=\frac{1}{2}(a+c+p-ac\overline{p})$$$$f=\frac{1}{2}(a+b+p-ab\overline{p})$$ We have that $\frac{x-a}{2}=d \rightarrow x=2d+a=a+b+c+d-bc\overline{p}$. Exploiting symmetry, we find that: $$x=a+b+c+p-bc\overline{p}$$ $$y=a+b+c+p-ac\overline{p}$$ $$z=a+b+c+p-ab\overline{p}$$ We have that $$\frac{x-y}{x-z}=\frac{ac\overline{p}-bc\overline{p}}{ab\overline{p}-bc\overline{p}}=\frac{ac-bc}{ab-bc}=\frac{\frac{1}{b}-\frac{1}{a}}{\frac{1}{c}-\frac{1}{a}}=\overline{\bigg(\frac{b-a}{c-a}\bigg) }=\overline{\bigg( \frac{a-b}{a-c}\bigg) }$$ implying that $\triangle{XYZ} \sim \triangle{ABC}$ as desired. $\square$ opps this is cool solution...

[asy][asy] defaultpen(fontsize(10pt)); unitsize(4.5cm); pair A = dir(125); pair B = dir(210); pair C = dir(330); pair A1 = -A; pair B1 = -B; pair C1 = -C; pair P = (-0.3,0.2); pair D = foot(P,B,C); pair E = foot(P,A,C); pair F = foot(P,A,B); pair X = 2*D-A1; pair Y = 2*E-B1; pair Z = 2*F-C1; pair L = -(A+B+C); pair P1 = 2P-L; pair H = A+B+C; pair M = 0.5*(B+C); draw(A--B--C--cycle,linewidth(1.5)); draw(P--D); draw(P--E); draw(P--F); draw(A1--X,red); draw(B1--Y,red); draw(C1--Z,red); draw(X--Y--Z--X,blue); draw(L--P1,dashed); draw(L--A1); draw(H--X); draw(X--P1); draw(circumcircle(A,B,C)); draw(circumcircle(X,Y,Z)); draw(H--A1,dashed); dot(A);dot(B);dot(C); dot(D);dot(E);dot(F);dot(P); dot(X);dot(Y);dot(Z); dot(A1);dot(B1);dot(C1); dot(P1);dot(H);dot(L);dot(M); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$A'$",A1,dir(A1)); label("$B'$",B1,dir(B1)); label("$C'$",C1,dir(C1)); label("$P$",P,SE); label("$L$",L,NE); label("$D$",D,S); label("$E$",E,NE); label("$F$",F,dir(200)); label("$H$",H,dir(15)); label("$P'$",P1,NW); label("$X$",X,SW); label("$Y$",Y,NW); label("$Z$",Z,NE); label("$M_a$",M,dir(76)); [/asy][/asy] Let $H, L$ be the orthocenter, de Longchamps point of $\triangle ABC$. Let $M_a, M_b, M_c$ be midpoints of $BC, CA, AB$. Notice that $DM_a$ is $A'$-midline of $\triangle AXH$ thus $XH\parallel BC$. Now notice that $A'L\perp BC\implies XP'\perp BC$ where $P'$ is reflection of $L$ across $P$. Hence $\angle P'XH=90^{\circ}$. Similarly, we get $H,X,Y,Z,P'$ concyclic. Finally, note that $$\measuredangle YXZ = \measuredangle(YH,HZ) = \measuredangle(AC,AB) = -\measuredangle BAC$$so we are done.

@above amazing solution! There is just one minor typo: instead of $\angle XP'H=90^{\circ}$ it should be $\angle P'XH=90^{\circ}$

Slightly different complex numbers. As before, let $(ABC)$ be unit circle, and let $A=a,B=b,C=c$, so that $A'=-a,B'=-b,C'=-c$. Let $P=p$. Then $d=\frac12(b+c+p-bc\overline{p})$. Since $\frac{x+a'}{2} = d$, then $x=2d-a'$, which is \begin{align*} x&=a+b+c+p-bc\overline{p} \\y&=a+b+c+p-ac\overline{p} \\z&=a+b+c+p-ab\overline{p}. \end{align*}Translate these points by $-(a+b+c+p)$ and apply the spiral similarity given by multiplication by $-\tfrac{1}{abc\overline{p}}$. Then \[ x'=1/a=\overline{a}, y=\overline{b}, z=\overline{c}. \]Now, $\triangle X'Y'Z'$ is similar to $\triangle ABC$, since in the complex plane, we have just reflected over the $x$-axis. Undoing our transformations preserves the similarity, so $\triangle XYZ \sim \triangle ABC$.

n00b.....

By using Ptolemy sinus theorem and Carnot theorem , we can easily find that $XYZH$ is concyclic , the else is easy .

I claim that $XYZ$ is in fact oppositely similar to $ABC$. Use complex numbers. Let $a, b, c$ lie on the unit circle as usual and $a'=-a$ etc. We also have $$d=\frac{1}{2}(b+c+p-bc\overline{p}),$$and similar formulas hold for $e$ and $f$, so $x=a+b+c+p-bc\overline{p}$ and similar formulas hold for $y$ and $z$. It suffices to show that $$0=\left|\begin{matrix} a&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\overline{p}-\frac{p}{bc} &1\\ b&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\overline{p}-\frac{p}{ca} &1\\ c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\overline{p}-\frac{p}{ab} &1\end{matrix}\right|=\left|\begin{matrix}a&-\frac{p}{bc}&1\\b&-\frac{p}{ca}&1\\c&-\frac{p}{ab}&1\end{matrix}\right|=\frac{-p}{abc}\left|\begin{matrix}a&a&1\\b&b&1\\c&c&1\end{matrix}\right|,$$which is certainly true. $\blacksquare$

If $D', E', F'$ are the projections from $P$ to the perpendicular bisectors of $BC, AC, AB$ respectively then we claim that $\triangle{XYZ}$ is the image of $\triangle{D'E'F'}$ after a 2x dilation. ($\triangle{D'E'F'} \sim \triangle{ABC}$ since $D'E'F'OP$ is cyclic if $O$ is the circumcenter of $\triangle{ABC}$). Note that $HX = 2 \cdot M_AD$ if $M_A$ is the midpoint of $BC$ and $H$ is the orthocenter, so $HX = 2 \cdot PD'$ and $HX \parallel PD'$, thus $HX, HY, HZ$ correspond to $PD', PE', PF'$ and we are done.

WLOG, let $a$, $b$, and $c$ lie on the unit circle. We have that $d=\frac{1}{2}(p+b+c-bc\bar{p})$, $e=\frac{1}{2}(p+c+a-ca\bar{p})$, and $f=\frac{1}{2}(p+a+b-ab\bar{p})$. Now solving for $x$, $y$, and $z$, we find that \[x=a+b+c+p-bc\bar{p}\]\[y=a+b+c+p-ca\bar{p}\]\[x=a+b+c+p-ab\bar{p}\]Negating all three of these and adding $a+b+c$ and then dividing by $abc$(or rotating the triangle with around the origin at the angle of arg$(abc)$), we get a triangle with $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$. Since $a$, $b$, $c$ lie on the unit circle, this circle is just $\triangle{}ABC$ reflected across the $x$-axis, therefore $\triangle{}XYZ\sim\triangle{}ABC$, and we are done.

We have that $$d=\frac{1}{2}(b+c+p-bc\overline{p}),$$so $$x=2d-(-a)=a+b+c+p-bc\overline{p}.$$Similarly, $$y=a+b+c+p-ac\overline{p},z=a+b+c+p-ab\overline{p}.$$Since similarly does not care about shifting or scaling by -1, define $$x'=bc\overline{p},y'=ac\overline{p},z'=ab\overline{p}.$$ Claim: $\angle X=\angle A$. This is equivalent to $$\frac{y'-x'}{z'-x'}\div\frac{c-a}{b-a}\in R.$$Substituting $x'=bc\overline{p},y'=ac\overline{p},z'=ab\overline{p}$ makes this $$\frac{c(a-b)^2}{b(a-c)^2}\in R.$$If we take the conjugate, this becomes $$\frac{\frac{1}{c}(\frac{1}{a}-\frac{1}{b})^2}{\frac{1}{b}(\frac{1}{a}-\frac{1}{c})^2}=\frac{c(a-b)^2}{b(a-c)^2}$$after multiplying top and bottom by $a^2b^2c^2$, so we have shown the claim. Thus, similarly we also have $\angle Y=\angle B$, so we are done.

Yay!! Solved a China TST (with egmo hints) Let $T,Q,R$ denote $A',B',C'$ resp. Take $(ABC)$ as the unit circle. As $AOT$ is collinear, we get $\frac{a}{t}=\frac{a^*}{t^*}=\frac{t}{a}$ as $a\neq t$ we must have $a=-t$. Similarly it follows that $b=-q$ and $c=-r$. Now as $B,C,D$ are collinear $$\frac{b-d}{b-c}=\frac{b^*-d^*}{b^*-c^*} \implies \frac{b-d}{b-c}=-\frac{c(1-bd^*)}{b-c} \implies d^*bc=c+b-d.$$Now the condition $PD\perp BC$ gives us $$\frac{p-d}{b-c}+\frac{p^*-d^*}{b^*-c^*}=0 \implies \frac{p-d}{b-c}=\frac{bc(p^*-d^*)}{b-c} \implies p-d=bcp^*-bcd^*\implies p-d=bcp^*-c-b-d\implies 2d=p+c+b-bcp^*.$$$D$ being the midpoint of $T$ and $X$ we get $d=\frac{x-a}{2}$ so that $x=a+2d=a+b+c+p-bcp^*$. Analogously we find $y=a+b+c+p-acp^*$ and $z=a+b+c+p-abp^*$. Note that $$\frac{XY}{AB}=\frac{|x-y|}{|a-b|}=\left |\frac{cp^*(a-b)}{a-b}\right |=|cp^*|=|p^*|.$$It is now clear that $\frac{XY}{AB}=\frac{XZ}{AC}=\frac{YZ}{BC}=|p^*|$. So the triangles $XYZ$ and $ABC$ are similar, as desired.

We will use complex bashing. Let $\Gamma$(circumcenter of $\triangle ABC$) be the unit circle, and $a$, $b$, and $c$ be points $A$, $B$ and $C$ and so on. Then \[d = \frac{1}{2}(b + c + p - bc\overline{p})\]\[e = \frac{1}{2} (a + c + p - ac\overline{p})\]\[f = \frac{1}{2} (a + b + p - ab\overline{p})\] Then \[x = \frac{1}{2} (a + b + c + p - bc\overline{p})\]\[y = \frac{1}{2} (a + b + c + p - ac\overline{p})\]\[z = \frac{1}{2} (a + b + c + p - ab\overline{p})\] We can subtract $a + b + c + p$ from all of these equations, since $\triangle XYZ \sim \triangle ABC$ does not depend on that. Similarly, we divide by $-abc\overline{p}$, to get $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$. Since $a$, $b$, and $c$ are on the unit circle, the new triangle is just the complex conjugate of $\triangle ABC$, so we are done.

hi complecks It suffices to show that $\frac{\frac{b-a}{c-a}}{\frac{z-x}{y-x}}$ is a real number (and cyclic variants). Note that the bottom fraction is flipped because $\triangle ABC$ and $\triangle XYZ$ are in opposite orientation. We have \begin{align*} d &= \frac 12 (p+b+c-bc\overline{p}) \\ e &= \frac 12 (p+a+c-ac\overline{p}) \\ f &= \frac 12 (p+a+b-ab\overline{p}) \\ x &= 2d + a \\ y &= 2e + b \\ z &= 2f + c \\ \end{align*} Thus, we have \begin{align*} \frac{\frac{b-a}{c-a}}{\frac{z-x}{y-x}} &= \frac{(b-a)(y-x)}{(c-a)(z-x)} \\ &= \frac{(b-a)((p+a+c)-ac\overline{p}-a+b-(p+b+c)+bc\overline{p})}{(c-a)((p+a+b)-ab\overline{p}-a+c-(p+b+c)+bc\overline{p})} \\ &= \frac{(b-a)(bc\overline{p}-ac\overline{p})}{(c-a)(bc\overline{p}-ab\overline{p})} \\ &= \frac{(b-a)^2c}{(c-a)^2b} \\ \end{align*} Thus, we want to show $\frac{(b-a)^2c}{(c-a)^2b} \in \mathbb{R}$. However, this is equivalent to saying $2\measuredangle CAB + \measuredangle BOC = 0$, which is obvious. We can finish the other two angles in a similar fashion.

Use complex bash Get that $d=\frac{b+c+p-bc\overline{p}}{2}$ $e=\frac{a+c+p-ac\overline{p}}{2}$ $f=\frac{a+b+p-ab\overline{p}}{2}$ $x=2d+a=a+b+c+p-bc\overline{p}$ $y=2e+b=a+b+c+p-ac\overline{p}$ $z=2f+c=a+b+c+p-ab\overline{p}$ $\frac{z-y}{y-x}=\frac{\overline{p}(ac-ab)}{\overline{p}(bc-ac)}=\frac{ac-ab}{bc-ac}=\frac{\frac{1}{b}-\frac{1}{c}}{\frac{1}{a}-\frac{1}{b}}$, so $XYZ$ is similar to $ABC$.

Beautiful! The idea is to vary $P$ along a line through the circumcenter $O$ at constant velocity. The claim is that points $X,Y,Z$ move at constant velocity along lines through the orthocenter $H.$ This is easy to see, since $D$ moves at constant velocity and thus $X$ will as well, and similarly for the other points, and that when $P=O$ we have $X=Y=Z=H.$ Now we may assume $P$ lies on the circumcircle, because all the triangles $XYZ$ will be similar as $P$ moves along the line. We now prove a lemma: given a triangle $ABC$ let $B',C'$ be the $B,C$ antipodes and let $P$ be a point on the circumcircle. Let $E,F$ be the feet from $P$ to $AC,AB$ respectively. Then the projections of $C'F$ and $B'E$ onto $EF$ have the same directed length. To show this, first let $X,Y,Z$ be the feet from $O,B,C$ onto $EF.$ We want to show $XF-ZX=XE-YX$ with directed lengths, or $YF=ZE.$ To do this we show $YE$ and $ZF$ share a midpoint. We consider the Newton-Gauss line of complete quadrilateral $BCEF.$ By a well-known lemma (for example, 10.9 in EGMO) the Newton-Gauss line is perpendicular to the Simson line $EF.$ Thus it is parallel to $BY$ and $CZ.$ Now the Newton-Gauss line is the $E$-midline of $\triangle BEY$ so the midpoint of $YE$ lies on it. Similarly the midpoint of $ZF$ lies on it, but they both also lie on $EF$ so they are the same. This proves our lemma. To finish the problem, we consider the Simson line $DEF$ of $P,$ and let $A^*B^*C^*$ be the reflection of $A'B'C'$ over this Simson line. By homothety, the directed length $B^*Y$ is twice the directed length of the projection of $B'E$ onto $EF,$ which is twice the directed length of the projection of $C'F$ on to $EF,$ which is the same as the directed length $C^*Z.$ By symmetry these are the same as the directed length $A^*X,$ and these are all parallel to the Simson line, so $XYZ$ is a translation of $A^*B^*C^*,$ which is a reflection of $A'B'C'$ which is a rotation of $ABC.$ This finishes.

We invoke the complex plane. Let $o = (0,0)$ be the center of the unit circle, which we also allow to be $ABC$'s circumcircle. Let the lowercase versions of the vertices denote their respective complex numbers (e.g. $A = a$). $a'$, $b'$, $c'$ are (respectively) reflections of $a$, $b$, $c$, across $o$, so we have $a' = a$, $b' = b$, and $c' = c$. We also know that \begin{align*} d &= \dfrac{c+b+p-bc\overline{p}}{2} \\ e &= \dfrac{a+c+p -ac\overline{p}}{2} \\ f &= \dfrac{a+b+c-ab\overline{p}}{2} \end{align*}by the well known formula for the foot of an altitude onto a segment connected by two vertices on the unit circle. It follows that \begin{align*} x &= a+b+c+p - bc\overline{p} \\ y &= a+b+c+p - ac\overline{p} \\ z &= a+b+c+p - ab\overline{p}. \end{align*}Thus, we have \begin{align*} \dfrac{z-x}{z-y} &= \dfrac{bc\overline{p} - ab\overline{p}}{ac\overline{p} - ab\overline{p}} \\ &= \dfrac{c-a}{c-b} \end{align*}which directly implies the result. $\blacksquare$

Let $\triangle ABC$ lie on the unit circle. Then, $$d=\frac{\frac{p-b}{c-b}+\overline{(\frac{p-b}{c-b})}}{2}(c-b)+b=\frac{p+b+\frac{(\overline{p}-\frac1b)(c-b)}{\frac1c-\frac1b}}{2}=\frac{p+b+(\frac1b-\overline{p})bc}{2}=\frac{p+b+c-\overline{p}bc}{2}.$$ Therefore, $x=2d-(-a)=a+b+c+p-\overline{p}bc$. Similarly, $y=a+b+c+p-\overline{p}ac$ and $z=a+b+c+p-\overline{p}ab$. Note that $$\frac{x-z}{y-z}=\frac{ab-bc}{ab-ac}=\frac{\overline{a}-\overline{c}}{\overline{b}-\overline{c}}$$so $\triangle XYZ\sim \triangle ABC$, as desired.

um what

im too lazy to type up my bash even if the "bash" is really only two lines long