It is a particular case for $AD, BE, CF$ are concurrent at a point $P$.

It follows from the same lemma that was used to prove problem 2 from 2nd China TST 2006.

Can someone bash this problem with complex numbers?

it s special case of hagge circles when $P$ is an infinity point

Electron_Madnesss wrote: Actually, this problem was there in Evan's textbook and my solution to it was (surprisingly!) the same as his, so I am posting it here. $\rightarrow$ We $(ABC)$ to be the unit circle with $a=1,b=-1 $ and $k=\dfrac{-1}{2}$. Also, let $s,t \in (ABC)$. We claim that $K$ is the center of $(STUH)$. Notice that $x = \dfrac{1}{2}(s+t-1+\dfrac{s}{t}) \implies 2\cdot(2\text{Re}x +1 ) = s+t+\dfrac{1}{s} +\dfrac{1}{t} + \dfrac{s}{t} + \dfrac{t}{s}$, which does not depend on $X$. Also, $|k+\dfrac{s+t}{2}|^2 = 3+ 2\cdot(2Rex+2) \implies \dfrac{s+t}{2} $ has a fixed distance with $k$ as desired$\blacksquare$ I do not understand. You're not allowed to put $a=1,b=-1$ since $ABC$ is not a right triangle. And what is $X$??

Electron_Madnesss wrote: Actually, this problem was there in Evan's textbook and my solution to it was (surprisingly!) the same as his, so I am posting it here. $\rightarrow$ We $(ABC)$ to be the unit circle with $a=1,b=-1 $ and $k=\dfrac{-1}{2}$. Also, let $s,t \in (ABC)$. We claim that $K$ is the center of $(STUH)$. Notice that $x = \dfrac{1}{2}(s+t-1+\dfrac{s}{t}) \implies 2\cdot(2\text{Re}x +1 ) = s+t+\dfrac{1}{s} +\dfrac{1}{t} + \dfrac{s}{t} + \dfrac{t}{s}$, which does not depend on $X$. Also, $|k+\dfrac{s+t}{2}|^2 = 3+ 2\cdot(2Rex+2) \implies \dfrac{s+t}{2} $ has a fixed distance with $k$ as desired$\blacksquare$ Is this a solution to USAMO 2015 P2 or am I wrong?

Oh Darn, I copied the wrong solution from my notebook. @above you are correct this the solution of USAMO 2015-2

$a,b,c,d$, with the circumcircle as the unit circle, will be our free variables. Note that due to parallel conditions we have that $e=\frac{ad}{b}$ and $f=\frac{ad}{c}$ Then, \[s=b+c-bc\overline{d}=b+c-\frac{bc}{d}\]\[t=a+c-ac\overline{e} = a+c - ac \cdot \frac{b}{ad}=a+c-\frac{bc}{d}\]\[u=a+b-ab\overline{f} = a+b - ab\cdot \frac{c}{ad}=a+b-\frac{bc}{d}\]\[h =a+b+c\]Now, we translate by $\frac{bc}{d}-a-b-c$ to get \[s'=-a,t'=-b,u'=-c,h'=\frac{bc}{d}\]These 4 points are clearly all on the unit circle, so we have shown that $S,T,U,H$ can be translated to a unit circle, and are therefore concyclic $\blacksquare$

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/vol5.html then P-hagge circle, p. 44-48. Sincerely Jean-Louis

We use complex number. $(ABC)$ is our unit circle. Consider $A=a,B=b,C=c$. So $h=a+b+c$. And suppose $D$ is any point on the unit circle. From the parallel condition we get a relation between points $a,b,c,d,e,f$ that \begin{eqnarray*} ad=be=cf \end{eqnarray*}The co-ordinate of $s=b+c-\frac{bc}{d}$. Similarly $t=a+c-\frac{ac}{e},u=a+b-\frac{ab}{f}$. For being $S,T,U,H$ concyclic, the quantity must be \begin{eqnarray*} \frac{s-u}{t-u}\div \frac{s-h}{t-h}&=&\frac{\frac{abd-bcf}{df}+c-a}{\frac{abe-acf}{ef}+c-b}\div \frac{c+d}{c+e},\hspace{2em}[abd=bcf, abe=acf]\\ &=&\frac{c-a}{c-b}\cdot \frac{c+e}{c+d}\in \mathbb{R} \end{eqnarray*}Substituting the conjugates of all points (1 minute computation) we get the same quantity. Therefore it must be a real number and we are done. $\square$

Sketch :Note that $S\in \odot(\triangle BHC)$ and similarly other cases hold. Prove that $DS,ET,FU$ are concurrent and they lie on $\odot(\triangle ABC)$ If these 3 lines concurr at $G$ prove that $AGTU$ and similarly other symnetrical quadrilaterals are cyclic. Now use this cyclicities to prove the final concyclicity. All of these can be proved by angels $\blacksquare$

shobber wrote: $H$ is the orthocentre of $\triangle{ABC}$. $D$, $E$, $F$ are on the circumcircle of $\triangle{ABC}$ such that $AD \parallel BE \parallel CF$. $S$, $T$, $U$ are the semetrical points of $D$, $E$, $F$ with respect to $BC$, $CA$, $AB$. Show that $S, T, U, H$ lie on the same circle. Cited in Titu's book as "MOP 2006"

The parallel condition gives us $ad = be = cf = z$ for some complex number $z$ of magnitude $1$, so $d = \tfrac{z}{a}$, $e = \tfrac{z}{b}$, $f = \tfrac{z}{c}$. Then, using the complex foot formula, we can find that $$\frac{\tfrac{z}{a} + s}{2} = \frac{1}{2}\left(b + c + \frac{z}{a} - \frac{abc}{z}\right).$$This means $s = b + c - \tfrac{abc}{z}$, and we similarly find that $t = a + c - \tfrac{abc}{z}$, $u = a + b - \tfrac{abc}{z}$. Now, to show that $STUH$ is concyclic, it suffices to prove that $\tfrac{h - s}{u - s} \div \tfrac{h - t}{u - t}$ is real, or it equals its conjugate. First, we evaluate the quantity as it is; noting that $h = a + b + c$, we get $$\begin{aligned} \frac{h - s}{u - s} \div \frac{h - t}{u - t} &= \frac{a + \tfrac{abc}{z}}{a - c} \div \frac{b + \tfrac{abc}{z}}{b - c} \\ &= \frac{a(z + bc)(b - c)}{b(z + ac)(a - c)}.\end{aligned}$$The conjugate of this is $$\begin{aligned} \frac{\tfrac{1}{a}(\tfrac{1}{z} + \tfrac{1}{bc})(\tfrac{1}{b} - \tfrac{1}{z})}{\tfrac{1}{b}(\tfrac{1}{c} + \tfrac{1}{ac})(\tfrac{1}{a} - \tfrac{1}{c})} &= \frac{a(z + bc)(b - c)}{b(z + ac)(a - c)} \\ &= \frac{h - s}{u - s} \div \frac{h - t}{u - t}, \end{aligned}$$so we are done.

WLOG, let $(ABC)$ be the unit circle and rotate $ABC$ so that $AD$ is parallel to the $y$-axis. Therefore, we have that $d=\frac{1}{a}$, $e=\frac{1}{b}$, and $f=\frac{1}{c}$. Using reflection formulas, we find that $s=b+c-abc$, $t=c+a-abc$, and $u=a+b-abc$. From here, we find that since $\frac{s-t}{h-t}*\frac{h-u}{s-u}$ is equal to its conjugate, meaning that it's real, $S$, $T$, $H$, and $U$ are concyclic, and we are done.

Let the real line be the line through $O$ perpendicular to all of $AD,BE,CF$. Thus, we have $d=\frac{1}{a}$ and so on. We have that $$D'=b+c-\frac{bc}{d}=b+c-abc,$$and similarly $$E'=c+a-abc,F'=a+b-abc.$$We wish to show that these are concyclic with $a+b+c$. Of course, shift by $-a-b-c+abc$, so we wish to show $$-a,-b,-c,abc$$are concyclic. Thus, it suffices to show that $$\frac{c(b-a)(1+ab)}{b(c-a)(1+ac)}$$is real. This is just because $$\frac{\frac{1}{c}(\frac{1}{b}-\frac{1}{a})(1+\frac{1}{ab})}{\frac{1}{b}(\frac{1}{c}-\frac{1}{a})(1+\frac{1}{ac})}$$$$=\frac{b(ac-bc)(abc+c)}{c(ab-bc)(abc+b)}=\frac{c(a-b)(ab+1)}{b(a-c)(ac+1)},$$as desired.

I'm so confused, why did everyone using complex finish by checking angle conditions for concyclicity? Set $ABC$ on the unit circle and rotate so that $AD$, $BE$, and $CF$ are vertical lines. It follows that $(d, e, f) = (\overline{a}, \overline{b}, \overline{c})$ and $(x, y, z) = (b+c-abc, c+a-abc, a+b-abc)$. Recalling $h = a+b+c$, quadrilateral $HXYZ$ has circumcenter $\omega = a+b+c-abc$ because \[ x-\omega = -a, \quad y-\omega = -b, \quad z-\omega = -c, \quad h-\omega = abc, \]and all four of these differences have magnitude $1$. $\blacksquare$

Sorry for bumping but I wonder if there exists a synthetic solution.

First consider the following $\measuredangle SBC=\measuredangle CBD=\measuredangle CAD=\measuredangle ADF=\measuredangle ABF$ so $BS,BF$ are isogonal, similarly $CS,CE$ are isogonal and thus the isogonal conjugate of $S$ is the intersection of $BF$ and $CE,$ which lies on the shared perpendicular bisector of $AD,BF,CE.$ Similarly for $T,U$ and thus we get that $S,T,U$ are on the isogonal conjugate of this perpendicular bisector which is a rectangular circumhyperbola. Now the center of this hyperbola lies on the nine point circle so the antipode of $A$ on this hyperbola lies on $(BHC)$ by homothety, but $S$ lies on the hyperbola and this circle (by angle chasing) so it is the antipode of $A$ (since $B,H,C$ are already on the hyperbola and two conics intersect at $4$ points). Then $AS,BT,CU$ concur at the center of the hyperbola, which lies on the nine point circle, and $ABC,STU$ are reflections over this point. But the reflection of the point where the hyperbola meets $(ABC)$ again over the center of the hyperbola will be $H$ which finishes.

Let $\triangle ABC$ lie on the unit circle. WLOG, let $D$, $E$, and $F$ be $\overline{a}$, $\overline{b}$, and $\overline{c}$, respectively. $S$ must lie at $$s=\overline{(\frac{\overline{a}-b}{c-b})}(c-b)+b=\frac{(a-\frac1b)(c-b)}{\frac1c-\frac1b}+b=(\frac1b-a)bc+b=b+c-abc$$. Similarly, $t=a+c-abc$ and $u=a+b-abc$. Also note that $h=a+b+c$. It suffices to prove that $\arg(\frac{s-u}{t-u})=\arg(\frac{s-h}{t-h})$ or $\arg(\frac{c-a}{c-b})=\arg(\frac{a+abc}{b+abc})$ which is equivalent to proving $$\frac{(c-a)(b+abc)}{(c-b)(a+abc)}\in \mathbb{R}.$$ This is true because $$\overline{(\frac{(c-a)(b+abc)}{(c-b)(a+abc)})}=\frac{(\frac1c-\frac1a)(\frac1b+\frac1{abc})}{(\frac1c-\frac1b)(\frac1a+\frac1{abc})}=\frac{(ab-bc)(ac+1)}{(ab-ac)(bc+1)}=\frac{(c-a)(b+abc)}{(c-b)(a+abc)}.$$

The parallel condition tells us that $ad=be=cf$. We can compute $s=b+c-\frac{bc}{d}$, we can similarly compute $t$ and $u$ as well and we know that $h=\frac{a+b+c}{2}$. For $STUH$ to be cyclic, we want $\frac{(t-s)(u-h)}{(u-s)(t-h)}$ to be real. So, $$\frac{(t-s)(u-h)}{(u-s)(t-h)} = \frac{(bed-aed)(cf+ab)}{(cfd-afd)(be+ac)}=\frac{(be-ae)(cf+ab)}{(cf-af)(be+ac)}$$Now, $$\overline{\frac{(be-ae)(cf+ab)}{(cf-af)(be+ac)}} = \frac{c}{b}\times\frac{(b-a)(cf+ab)}{(c-a)(be+ac)}$$