just use complex numbers. it's fairly easy.

this problem was long forgotten, until bugzpoder attained my attention to it. let me give a more detailed hint the condition about the sum of the angles show that the number N = (a-b)(c-d)(e-f) / (c-b)(e-d)(a-f) is real, where a,b,c,d,e,f are the affixes of their respective vertices denoted with capital letters. but this means nothing else than N=AB.CD.EF / BC.DE.FA ... I will leave you to make the computations (and judgements) from here on ..

Is there another proof for this problem please post it if you have ... Darij Grinberg do you have any fantastic proof for this problem or any ides RESPECT : Davron

err...i'm not darij, and my proof isn't fantastic, but i see a solution with sine law and...that's it.

Let $ABCDEF$ be the labelled in clockwise order. Let $A$ be represented by the complex number $a$ from an arbitary origin, $B$ be represented by $b$ and so forth. Also let $\angle A = \alpha$, $\angle B = \beta$ and so forth. Therefore: \begin{eqnarray*} (b - a) \times e^{i\beta} \times \frac{|CB|}{|AB|} &=& (b - c) \\ (d - c) \times e^{i\delta} \times \frac{|ED|}{|CD|} &=& (d - e) \\ (f - e) \times e^{i\zeta} \times \frac{|AF|}{|EF|} &=& (f - a) \\ \therefore (b - a)(d - c)(f - e) \times e^{i(\beta+\delta+\zeta)} \times \frac{BC.DE.FA}{AB.CD.EF} &=& (b - c)(d - e)(f - a) \\ \therefore (b - a)(d - c)(f - e) &=& (b - c)(d - e)(f - a) \\ \therefore -(ace + adf - acf - ade - bce - bdf + bcf + bde) &=& abe + acd - abd - ace - bef - cdf + bdf + cef \\ \therefore -(adf - acf - ade - bce + bcf + bde - abf + cde) &=& abe + acd - abd - bef - cdf + cef + abf - cde \\ \therefore -(f - e)(c - a)(b - d) &=& (c - b)(f - d)(e - a) \\ \therefore |f - e||c - a||b - d| &=& |c - b||f - d||e - a| \\ \therefore EF.AC.DB &=& BC.DF.AE \\ \end{eqnarray*}

i dont kidding many years ago , i have solve it by inversion... u know i didnt want to solve it by that. but the problem was in my inversive problems notes. these years i have exams is school and cant come to ML too. i will promise to send its proof later. i only posted these sentenses to say that there exists an inversive proof for it.

Hi my dears. I'm here, because I want to post my solutions for this nice, and may be old($1998$) problem. First I should write the invrsive proof as I promised recently. Ok………………………………………………………….. First consider this fact that the circum circles of triangles $ABC, CDE, EFA$ have a common point $P$. Because, if $P$ is the intersection of circum circles $CDE, EFA$ , then: $\angle EPC=180-\angle D$ $\angle APE=180-\angle F$ then: $\angle APC=360-(\angle D+\angle F)=\angle B$ and it means that $P$ is on circum circle of $ABC$, too. Now……………………………………………………. I will invert this shape with the center $P$ and an arbitrary radius. circum circles of $ABC, CDE, EFA$, Changes to three line, which each two of them intersects each other at inversion form of $A,E,C$ means $A',E',C'$. and if $K$ is the radius of inversion,: $A'B'=K.\frac{BC}{PB.PC}$. Thus: $\frac{A'B'}{B'C'}=\frac{AB}{BC}.\frac{PC}{PA}$.and with a similar way we can conclude that: $\frac{C'D'}{D'E'}=\frac{CD}{DE}.\frac{PE}{PC}$ $\frac{E'F'}{F'A'}=\frac{EF}{FA}.\frac{PA}{PE}$ and if we multiple these equalities we should have: $\frac{A'B'}{B'C'}.\frac{C'D'}{D'E'}.\frac{E'F'}{F'A'}=(\frac{AB}{BC}.\frac{CD}{DE}.\frac{EF}{FA}).(\frac{PC}{PA}.\frac{PE}{PC}.\frac{PA}{PE})= 1$ Then according to Menelaus theorem, for triangle $A'C'E'$, the points $F',D',B',$ should be collinear. And according to this theorem and for the triangle $A'B'F'$ and the line $E'D'C'$ we know: $\frac{A'C'}{B'C'}.\frac{B'D'}{D'F'}.\frac{E'F'}{A'E'}=1$ there for: $\frac{AC}{AE}.\frac{BD}{DF}.\frac{EF}{BC}=(\frac{AC}{BC}.\frac{PB}{PA}).(\frac{BD}{DF}.\frac{PF} {PB}).(\frac{EF}{AE}.\frac{PA}{PF})=\frac{A'C'}{B'C'}.\frac{B'D'}{D'F'}.\frac{E'F'}{A'E'}=1$ and we conclude that: $AC.BD.EF=AE.DF.BC$. ooooooooooooooooooooooooooofffff, finally it is proved by $INVERSION$. Really an enjoyable problem. Very nice, very nice, congratulations to the designer of this problem.

Might have the same basic ideas as the inversion solution, but here's a purely synthetic way: Let P be the point such that EDP ~ EFA. Note that since $\angle B+\angle D+\angle F = 360, \angle CDP = \angle CBA$. Also, $\frac{PD}{CD}= \frac{AF}{CD}\cdot \frac{DE}{EF}= \frac{AB}{BC}$. Hence CDP ~ CBA. Now using spiral similarity, $BD \cdot \frac{AC}{BC}= AP = FD \cdot \frac{AE}{FE}$ $\frac{BD \cdot AC \cdot FE}{BC \cdot FD \cdot AE}= 1$

probability1.01 wrote: Might have the same basic ideas as the inversion solution, but here's a purely synthetic way: Let P be the point such that EDP ~ EFA. Note that since $ \angle B + \angle D + \angle F = 360, \angle CDP = \angle CBA$. Also, $ \frac {PD}{CD} = \frac {AF}{CD}\cdot \frac {DE}{EF} = \frac {AB}{BC}$. Hence CDP ~ CBA. Now using spiral similarity, $ BD \cdot \frac {AC}{BC} = AP = FD \cdot \frac {AE}{FE}$ $ \frac {BD \cdot AC \cdot FE}{BC \cdot FD \cdot AE} = 1$ can anyone explain how do you choose point P with more details? I'm not how does it imply thet angles CDP and CBA are equal...

felixx wrote: probability1.01 wrote: Might have the same basic ideas as the inversion solution, but here's a purely synthetic way: Let P be the point such that EDP ~ EFA. Note that since $ \angle B + \angle D + \angle F = 360, \angle CDP = \angle CBA$. Also, $ \frac {PD}{CD} = \frac {AF}{CD}\cdot \frac {DE}{EF} = \frac {AB}{BC}$. Hence CDP ~ CBA. Now using spiral similarity, $ BD \cdot \frac {AC}{BC} = AP = FD \cdot \frac {AE}{FE}$ $ \frac {BD \cdot AC \cdot FE}{BC \cdot FD \cdot AE} = 1$ can anyone explain how do you choose point P with more details? I'm not how does it imply thet angles CDP and CBA are equal... It works if the point $P$ is chosen to be outside of the convex polygon near vertex $D$. Since $\angle EDP+\angle PDC+\angle CDE = 360^\circ$, it follows that $\angle PDC= 360^\circ - \angle EDP - \angle CDE = 360^\circ - \angle EFA - \angle CDE = \angle ABC$.

al.M.V. wrote: err...i'm not darij, and my proof isn't fantastic, but i see a solution with sine law and...that's it. Well...its easy to write like that.Please post the proof al.M.V.To be fair,I don't think there's a proof using the same....and did you do all the calculations. Ashegh's proof gives us the glimpse of how powerful inversion can be.Well done incredible Ashegh!!

Let $a,b,c,d,e,f$ be the coordinates of $A,B,C,D,E,F$ in the complex plane.By the problem we have \[ \frac{a-b}{b-c} \cdot \frac{c-d}{d-e} \cdot \frac{e-f}{f-a}=-1. \] We also have the identity $(a-b)(c-d)(e-f)+(b-c)(d-e)(f-a)=(b-c)(a-e)(f-d)+(c-a)(e-f)(d-b)$ which can be verified by expanding the terms of LHS ans RHS,so we get \[ \frac{b-c}{c-a} \cdot \frac{a-e}{e-f} \cdot \frac{f-d}{d-b}=-1. \] Taking modulus on both sides does the job... I think this problem is made for using affixes.In my opinion the problem creators could not think of an inversive proof.

sayantancha..., why the first relation happens? you know that the product of the modules is $1$. How you deduce that the product of the explicitates is $-1$ ? And where you involve the angle condition? Please tell me.

i think i see a proof with jacobi and trig ceva

can anyone explain to me?

sayantanchakraborty wrote: Let $a,b,c,d,e,f$ be the coordinates of $A,B,C,D,E,F$ in the complex plane.By the problem we have \[ \frac{a-b}{b-c} \cdot \frac{c-d}{d-e} \cdot \frac{e-f}{f-a}=-1. \] Why do we have this thing and how is the angle condition used here?

A beautiful synthetic solution can be found if one sees that the hexagon "almost" tesselates the plane. I can post it if someone requests it.

butter67 wrote: A beautiful synthetic solution can be found if one sees that the hexagon "almost" tesselates the plane. I can post it if someone requests it. Would you please post it?

Valentin Vornicu wrote: this problem was long forgotten, until bugzpoder attained my attention to it. let me give a more detailed hint the condition about the sum of the angles show that the number N = (a-b)(c-d)(e-f) / (c-b)(e-d)(a-f) is real, where a,b,c,d,e,f are the affixes of their respective vertices denoted with capital letters. but this means nothing else than N=AB.CD.EF / BC.DE.FA ... I will leave you to make the computations (and judgements) from here on .. Could you explain your answer a bit more?How did you get the condition equelent to that?

Taha1381 wrote: Valentin Vornicu wrote: this problem was long forgotten, until bugzpoder attained my attention to it. let me give a more detailed hint the condition about the sum of the angles show that the number N = (a-b)(c-d)(e-f) / (c-b)(e-d)(a-f) is real, where a,b,c,d,e,f are the affixes of their respective vertices denoted with capital letters. but this means nothing else than N=AB.CD.EF / BC.DE.FA ... I will leave you to make the computations (and judgements) from here on .. Could you explain your answer a bit more?How did you get the condition equelent to that? It's obvious, since for every real number, the argument is congruent to 0 modulo pi

This task appeared on the final of $L$ Polish Math Olympiad.

can anyone post a synthetic proof?

WLOG $ABCDEF$ clockwise. The problem's conditions translates to $\frac{a-b}{b-c}\cdot \frac{c-d}{d-e}\cdot \frac{e-f}{a-f}=1$. Let $z=\frac{b-c}{c-a}\cdot \frac{a-e}{b-d}\cdot \frac{d}{e}$; we desire to prove $|z|=1$. Now, WLOG $f=0$, so the problem's condition is equivalent to $e=\frac{ad(b-c)}{ab+bd-ad-bc}$, and $z=\frac{b-c}{c-a}\cdot \frac{a-e}{b-d}\cdot \frac{d}{e}$. But substituting our expression for $e$ into this, we get $z=-1$, so $|z|=1$, as desired.

We use complex numbers. We know that $$\frac{|b-a|}{|b-c|}\cdot\frac{|d-c|}{|d-e|}\cdot\frac{|f-e|}{|f-a|}=1$$and $$\arg\left(\frac{b-a}{b-c}\right)+\arg\left(\frac{d-c}{d-e}\right)+\arg\left(\frac{f-e}{f-a}\right)=2\pi,$$so $$\frac{(b-a)(d-c)(f-e)}{(b-c)(d-e)(f-a)}=e^{i2\pi}=1.$$Thus, \begin{align*}0&=(b-a)(d-c)(f-e)-(b-c)(d-e)(f-a)\\&=(f-e)(c-a)(d-b)-(f-e)(a-d)(b-c)\\&-(b-c)(a-e)(f-d)-(b-c)(e-f)(a-d)\\&=(c-a)(f-e)(d-b)-(b-c)(a-e)(f-d).\end{align*}$\square$

A more motivated synthetic solution . Solved with kevinmathz. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -7.999564740377847, xmax = 12.530973744417288, ymin = -2.2789482545010245, ymax = 7.102367247578972; /* image dimensions */ /* draw figures *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point *//* special point */ draw((0,2.309401076758504)--(0,0), linewidth(1)); draw((0,0)--(2,-1.1547005383792515), linewidth(1)); draw((2,-1.1547005383792515)--(4,0), linewidth(1)); draw((4,0)--(4,2.3094010767585025), linewidth(1)); draw((2,3.4641016151377553)--(6,1.15470053837925), linewidth(1)); draw((6,1.15470053837925)--(6,-1.1547005383792528), linewidth(1)); draw((4,2.3094010767585025)--(2,-1.1547005383792515), linewidth(1)); draw((2,-1.1547005383792515)--(0,2.309401076758504), linewidth(1)); draw((2,3.4641016151377553)--(-2,1.1547005383792524), linewidth(1)); draw((-2,1.1547005383792524)--(-2,-1.154700538379252), linewidth(1)); /* special point */ draw((2,5.77350269189626)--(2,3.4641016151377553), linewidth(1)); draw((2,5.77350269189626)--(6,-1.1547005383792528), linewidth(1)); draw((2,5.77350269189626)--(-2,-1.154700538379252), linewidth(1)); /* dots and labels */ label("$E$", (0.05582015122579975,0.11628123539174046), NE * labelscalefactor); label("$C$", (4.062126619439295,0.11628123539174046), NE * labelscalefactor); label("$A$", (2.051844726136438,3.580809604700918), NE * labelscalefactor); label("$F$", (0.05582015122579975,2.425966814931192), NE * labelscalefactor); label("$B$", (4.062126619439295,2.425966814931192), NE * labelscalefactor); label("$D$", (2.051844726136438,-1.2385615543779856), S * labelscalefactor); label("$E'$", (-1.9402044236848384,1.2711240251614664), NE * labelscalefactor); label("$C'$", (6.058151194349933,1.2711240251614664), NE * labelscalefactor); label("$D'_2$", (-1.8689178317237443,-1.2528188727702044), S * labelscalefactor); label("$D'_1$", (6.058151194349933,-1.2385615543779856), S * labelscalefactor); label("$P$", (2.051844726136438,5.89049518424037), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Extend $AB$ past $B$ to $C'$ such that $BC = BC'$. Denote point $D'_1$ such that $\triangle BC'D'_1$ is directly similar to $\triangle BCD$. Define points $E'$ and $D'_2$ similarly. Note that $\angle D'_1C'A + \angle C'AE' + \angle AE'D'_2 = 360^{\circ}\implies C'D'_1 \parallel E'D'_2$. Let $l$ be the line through $A$ parallel to $C'D'_1$. Let $P_1 = l\cap BD'_1$ and $P_2 = l\cap FD'_2$. Then, since $\triangle BC'D'_1\sim \triangle BAP_1$ and $\triangle FE'D'_2\sim FAP_2$, $$1 = \frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA} = \frac{AB}{BC'} \cdot \frac{CD}{DE} \cdot \frac{E'F}{FA} = \frac{AP_1}{C'D'_1} \cdot \frac{CD}{DE} \cdot \frac{E'D'_2}{AP_2}=\frac{AP_1}{AP_2},$$and $P_1=P_2=P$. Then, since $\triangle FED$ and $\triangle FAP$ are spirally similar about $F$, $\triangle FAE\sim \triangle FPD$. Similarly, $\triangle BAC\sim \triangle BPD$. Thus $$ \frac{BC}{CA} \cdot \frac{AE}{EF} \cdot \frac{FD}{DB} = \frac{BD}{DP} \cdot \frac{PD}{DF} \cdot \frac{FD}{DB}=1.$$$\blacksquare$

The problem dies immediately if you know this (note: this is a Chinese Handout) https://wenku.baidu.com/view/e0239e01eff9aef8941e061e

i stumbled upon the solution with literally no intention to solve it - the problem was there, and now its gone. maybe i shouldnt do oly problems with no intention of actually solving and just playing around with them anymore now that i think about it Throw $ABCDEF$ on the complex plane. The given two conditions can be succinctly rewritten as $$\frac{c-b}{a-b}\cdot\frac{e-d}{c-d}\cdot\frac{a-f}{e-f}=1.\qquad(\spadesuit)$$It is sufficient to prove that $$\color{blue}\frac{b-c}{a-c}\cdot\frac{a-e}{f-e}\cdot\frac{f-d}{b-d}=1.$$Repeatedly using $(\spadesuit)$, \begin{align*}\frac{b-c}{a-c}\cdot\frac{a-e}{f-e}\cdot\frac{f-d}{b-d} &= \frac{(a-b)(c-d)(a-e)(f-d)}{(a-c)(e-d)(a-f)(b-d)} \\ &= \frac{(d-e)(b-c)(c-d)(a-e)(f-d)}{(a-c)(e-d)(f-c)(d-c)(b-d)} \\ &= \frac{(d-e)(c-d)}{(e-d)(d-c)} = 1,\end{align*}done. $\square$

This is very straightforward with complex numbers. The given conditions translate to $\frac{(a-b)(c-d)(e-f)}{(b-c)(d-e)(f-a)} = 1$ precisely. But this implies $$(b-c)(a-e)(d-f) + (c-d)(e-f)(b-d) = (a-b)(c-d)(e-f)-(b-c)(d-e)(f-a) = 0,$$so $$\frac{(b-c)(a-e)(d-f)}{(c-d)(e-f)(b-d)} = -1,$$which finishes.

We use complex numbers. Note that \begin{align*} 0^\circ &= \angle ABC+\angle CDE+\angle EFA \\ &=\text{arg}\left(\frac{a-b}{c-b}\right)+\text{arg}\left(\frac{c-d}{e-d}\right)+\text{arg}\left(\frac{e-f}{a-f}\right) \\ &=\text{arg}\left(\frac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)}\right) \end{align*}So $\tfrac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)}$ is a positive real number. However, we have \begin{align*} 1 &= \left(\frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA}\right)^2 \\ &= \frac{|a-b|^2|c-d|^2|e-f|^2}{|c-b|^2|e-d|^2|a-f|^2} \\ &= \frac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)} \cdot \overline{\frac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)}} \end{align*}Since $\tfrac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)}$ is a real number, its conjugate is equal to itself, so it is equal to $1$. We can expand this to \[ace-acf-ade+adf-bce+bcf+bde-bdf=cea-cef-cda+cdf-bea+bef+bda-bdf\]From both sides, subtract $ace-bdf$ and add $cde-abf$. Move $bde$ and $acf$ from the left to the right, and move $cdf$ and $bea$ to the left. We get \[bce-cde-bcf+cdf-abe+ade+abf-adf=-abf+abd+bef-bde+acf-acd-cef+cde\]which implies $(c-a)(e-f)(b-d)=(b-c)(a-e)(d-f)$ and we are done.