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Source: Iranian 3rd round Geometry exam P3 - 2014

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#1 h Sep 28, 2014, 12:33 PM • 2 Y

Y by Adventure10, Mango247

Distinct points $B,B',C,C'$ lie on an arbitrary line $\ell$ . $A$ is a point not lying on $\ell$ . A line passing through $B$ and parallel to $AB'$ intersects with $AC$ in $E$ and a line passing through $C$ and parallel to $AC'$ intersects with $AB$ in $F$ . Let $X$ be the intersection point of the circumcircles of $\triangle{ABC}$ and $\triangle{AB'C'}$ ( $A \neq X$ ). Prove that $EF \parallel AX$ .

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#2 h Sep 28, 2014, 6:05 PM • 3 Y

Y by wiseman, Adventure10, Mango247

Let $X_{\infty},$ $Y_{\infty},$ $Z_{\infty},$ $A_{\infty},$ $B_{\infty},$ $C_{\infty}$ denote the infinite points of $BE,$ $EF,$ $FC,$ $BC,$ $CA,$ $AB.$ Parallel from $A$ to $EF$ cuts $BC$ at $P.$ By Desargues involution theorem, the opposite sidelines of the complete quadrangle $BEFC$ form an involution on the line at infinity $\Longrightarrow$ $A(X_{\infty}, Y_{\infty},C_{\infty} ) \ \overline{\wedge} \ A(Z_{\infty}, A_{\infty}, B_{\infty})$ $\Longrightarrow$ $(B',P,B) \ \overline{\wedge} \ (C',A_{\infty},C)$ $\Longrightarrow$ $P$ is center of this involution $\Longrightarrow$ $PB \cdot PC=PB' \cdot PC'$ $\Longrightarrow$ $P$ is on radical axis $AX$ of $\odot(ABC)$ and $\odot(AB'C')$ $\Longrightarrow$ $EF \parallel AXP,$ as desired.

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#3 h Oct 29, 2014, 7:00 PM • 2 Y

Y by wiseman, Adventure10

Let $D$ is intersection point of $|AX|$ and $|FC|$ . Then we have $\angle{DB'A}=\angle{FBE}$ and we have
$DB'/FB=AB'/EB$ . So we have $AX||EF$ .

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jred

#4 h Nov 29, 2016, 9:40 AM • 1 Y

Y by Adventure10

Mathematicalx wrote:

Let $D$ is intersection point of $|AX|$ and $|FC|$ . Then we have $\angle{DB'A}=\angle{FBE}$ and we have
$DB'/FB=AB'/EB$ . So we have $AX||EF$ .

Why $\angle{DB'A}=\angle{FBE}$ , would you give more details?
BTW, is there any non-projective solution?

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370 posts

#5 h Apr 24, 2017, 5:54 PM • 2 Y

Y by Adventure10, Mango247

jred wrote:

Mathematicalx wrote:

Let $D$ is intersection point of $|AX|$ and $|FC|$ . Then we have $\angle{DB'A}=\angle{FBE}$ and we have
$DB'/FB=AB'/EB$ . So we have $AX||EF$ .

Why $\angle{DB'A}=\angle{FBE}$ , would you give more details?
BTW, is there any non-projective solution?

Note that B'XCD is concyclic

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#6 h Aug 15, 2019, 6:55 AM • 1 Y

Y by Adventure10

wiseman wrote:

Distinct points $B,B',C,C'$ lie on an arbitrary line $\ell$ . $A$ is a point not lying on $\ell$ . A line passing through $B$ and parallel to $AB'$ intersects with $AC$ in $E$ and a line passing through $C$ and parallel to $AC'$ intersects with $AB$ in $F$ . Let $X$ be the intersection point of the circumcircles of $\triangle{ABC}$ and $\triangle{AB'C'}$ ( $A \neq X$ ). Prove that $EF \parallel AX$ .

Claim : $\frac{XC}{XC’} = \frac{BA}{B’A} \cdot \frac{B’C}{BC’}$

Proof : By Ceva’s theorem, $\frac{\sin XAB’}{\sin B’CX} = \frac{\sin AXB’}{\sin B’XC} \cdot \frac{\sin B’AC}{\sin B’CA}$ By noting that $\measuredangle{XAB’} = \measuredangle{XCC’}$ , $\measuredangle{B’XC} = 180 - \measuredangle{BAC’}$ , $\measuredangle{AXB’} = \measuredangle{B’CA}$ and by using sine rule, we get the desired result. $\square$

So, as $EB \parallel AB’$ and $FC \parallel AC’$ , $\frac{XC}{XC’} = \frac{BA \cdot BC}{BC’} \cdot \frac{B’A \cdot BC}{B’C} = \frac{BF}{BE}$ So, combing this with $\measuredangle{CXC’} = \measuredangle{FBE}$ , we get $\triangle CXC’ \sim \triangle FBE$ . So, $\measuredangle{BEF} = \measuredangle{B’AX}$ which implies $EF \parallel AX$ . $\blacksquare$

This post has been edited 4 times. Last edited by MathDelicacy12, Aug 20, 2019, 2:33 PM

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#7 h Apr 24, 2022, 2:45 PM • 2 Y

Y by Mango247, Mango247

Let $O_1$ , $O_2$ are the center of circles $\odot ABC$ , $\odot AB'C'$ and $K = AB \cap \odot AB'C'$ , $J = EA \cap \odot AB'C'$
Cause $AX \bot O_1O_2$ so we need to prove it $EF \bot O_1O_2$
with The Perpendicularity Lemma : $EF \bot O_1O_2 \leftrightarrow EO_1^2 - EO_2^2 = FO_1^2 - FO_2^2 \leftrightarrow {P_{O_1}^{E}}^2 - {P_{O_2}^{E}}^2 = {P_{O_1}^{F}}^2 - {P_{O_2}^{F}}^2$
if $F$ in the $\odot AB'C$ we have: $EA.EC-EA.EJ=FA.FB-FA.FK \leftrightarrow EA(EC-EJ) = FA(FB-FK) \leftrightarrow EA(CJ) = FA(BK)$
if $F$ out the $\odot AB'C$ we have: $EA.EC-EA.EJ=FA.FB+FA.FK \leftrightarrow EA(EC-EJ) = FA(FB+FK) \leftrightarrow EA(CJ) = FA(BK)$
So we just need to prove it $EA.CJ = FA.BK$
with Thales's theorem we have $CJ=\frac{B'C.CC'}{AC}$ and $FA=\frac{CC'.AB}{C'B}$
then $EA.CJ = \frac{EA}{AC}.B'C.CC'= \frac{BB'}{B'C}.B'C.CC' = BB'.CC'$
and then $FA.BK = \frac{CC'}{C'B}.AB.BK = \frac{CC'}{C'B}.BB'.C'B= BB'.CC'$
So $EA.CJ= FA.BK$ .
$Q.E.D$

This post has been edited 1 time. Last edited by TheMathematics, Apr 24, 2022, 2:46 PM

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#8 h Apr 24, 2022, 3:14 PM

Y by

Let $AX$ meet $CF$ at $S$ . $\angle CSX = \angle C'AX = \angle CB'X \implies CSB'X$ is cyclic. Now we have $\frac{CA}{CE} = \frac{CB'}{CB}$ so we need to prove $FB || SB'$ . $\angle CBF = \angle CBA = \angle CXA = \angle CXS = \angle CB'S \implies FB || SB'$ .
we're Done.

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