Cyclic points [variations on a Fuhrmann generalization]

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shobber

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shobber

#1 h Jun 18, 2006, 5:16 AM • 3 Y

Y by mathematicsy, Adventure10, Mango247

$H$ 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.

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treegoner

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treegoner

#2 h Jun 18, 2006, 6:41 PM • 4 Y

Y by Adventure10, Mango247, khina, LNHM

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

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iura

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iura

#3 h Jun 20, 2006, 10:46 AM • 1 Y

Y by Adventure10

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

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User335559

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User335559

#4 h Jun 10, 2018, 12:18 PM • 2 Y

Y by Adventure10, Mango247

Can someone bash this problem with complex numbers?

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PROF65

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PROF65

#6 h Jun 10, 2018, 3:00 PM • 2 Y

Y by amar_04, Adventure10

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

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User335559

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User335559

#7 h Jun 10, 2018, 7:58 PM • 2 Y

Y by Adventure10, Mango247

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 $X$ ??

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Wictro

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Wictro

#8 h Jun 10, 2018, 8:24 PM • 1 Y

Y by Adventure10

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?

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e_plus_pi

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e_plus_pi

#9 h Jun 11, 2018, 5:18 AM • 1 Y

Y by Adventure10

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

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AlastorMoody

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AlastorMoody

#10 h Jan 3, 2020, 7:09 AM • 2 Y

Y by Adventure10, Mango247

Solution

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AwesomeYRY

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AwesomeYRY

#11 h Apr 9, 2021, 5:12 AM

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$a,b,c,d$ , 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
$\blacksquare$

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jayme

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jayme

#12 h Apr 9, 2021, 6:57 AM

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Dear Mathlinkers,

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

Sincerely
Jean-Louis

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Tafi_ak

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Tafi_ak

#13 h Dec 2, 2021, 1:11 PM

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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. $\square$

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HoRI_DA_GRe8

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HoRI_DA_GRe8

#14 h Apr 24, 2022, 11:32 AM • 1 Y

Y by D_S

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 $\blacksquare$

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LLL2019

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LLL2019

#15 h Oct 23, 2022, 5:41 AM

Y by

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" :maybe:

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eibc

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eibc

#16 h Mar 11, 2023, 1:32 AM

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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.

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peppapig_

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peppapig_

#17 h Mar 16, 2023, 5:28 PM • 2 Y

Y by mulberrykid, john0512

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.

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john0512

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john0512

#18 h Aug 11, 2023, 8:07 AM

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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.

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Ritwin

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Ritwin

#19 h Feb 28, 2024, 4:31 AM • 1 Y

Y by LLL2019

There's a quicker finish to the complex bash than checking the angle condition 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 $1$ . $\blacksquare$

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hexapr353

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hexapr353

#20 h Feb 28, 2024, 7:12 AM

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

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OronSH

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OronSH

#21 h May 9, 2024, 8:04 PM • 1 Y

Y by megarnie

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.

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RedFireTruck

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RedFireTruck

#22 h Jun 14, 2024, 4:21 PM

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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 $\overline{(\frac{(c-a)(b+abc)}{(c-b)(a+abc)})}=\frac{(\frac1c-\frac1a)(\frac1b+\frac1{abc})}{(\frac1c-\frac1b)(\frac1a+\frac1{abc})}=\frac{(ab-bc)(ac+1)}{(ab-ac)(bc+1)}=\frac{(c-a)(b+abc)}{(c-b)(a+abc)}.$

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Ianis

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Ianis

#23 h Jun 14, 2024, 8:08 PM

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Synthetic

Complex

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Nuterrow

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Nuterrow

#24 h Jan 3, 2025, 8:57 PM

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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, $\overline{\frac{(be-ae)(cf+ab)}{(cf-af)(be+ac)}} = \frac{c}{b}\times\frac{(b-a)(cf+ab)}{(c-a)(be+ac)}$

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lpieleanu

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lpieleanu

#25 h Mar 10, 2025, 11:21 PM

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Solution

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ali123456

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ali123456

#26 h Apr 19, 2025, 4:24 PM

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My solution

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Ilikeminecraft

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Ilikeminecraft

#27 h Apr 25, 2025, 7:07 AM

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Let us redefine $S, T, U$ as $D', E', F'.$ Let $A, B, C, D$ be $a, b, c, d.$ We have that the foot from $D$ to $\overline{BC}$ is $\frac12(d + b + c - \overline dbc).$ Thus, $D'$ is $b + c - \overline dbc.$ Now, using the fact that $\overline{AD} \parallel \overline{BE} \parallel \overline{CF}$ and the reflections, we have that $BF'E'C, AF'D'C, AE'D'B$ are all parallelograms. Thus, we can compute $E' = A + D' - B = a + c - \overline dbc, F' = A + D' - C = a + b - \overline dbc.$ To prove cyclic, we can just prove that $\angle D'F'E' = \angle D'HE'.$ We can do this by proving that $\frac{\frac{D' - F'}{E' - F'}}{\frac{D' - H}{E' - H}}\in\mathbb R.$ Here is the computation:
$\begin{align*} \frac{\frac{D' - F'}{E' - F'}}{\frac{D' - H}{E' - H}} & = \frac{c - a}{c - b}\cdot\frac{-\overline dbc - b}{-\overline dbc - a} \\ & = \frac{c - a}{c - b} \cdot \frac{bc + bd}{bc + ad} \\ \overline{\left(\frac{c - a}{c - b} \cdot \frac{bc + bd}{bc + ad}\right)} & = \frac{a - c}{b - c} \frac ba \cdot \frac{a(d + c)}{bc + ad} \\ & = \frac{c - a}{c - b} \cdot \frac{bc + bd}{bc + ad} \end{align*}$ and hence we are done.

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