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Source: Iranian TST 2021, first exam day 2, problem 6

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Mr.C

#1 h May 17, 2021, 3:15 PM • 1 Y

Y by centslordm

Point $D$ is chosen on the Euler line of triangle $ABC$ and it is inside of the triangle. Points $E,F$ are were the line $BD,CD$ intersect with $AC,AB$ respectively. Point $X$ is on the line $AD$ such that $\angle EXF =180 - \angle A$ , also $A,X$ are on the same side of $EF$ . If $P$ is the second intersection of circumcircles of $CXF,BXE$ then prove the lines $XP,EF$ meet on the altitude of $A$

Proposed by Alireza Danaie

This post has been edited 3 times. Last edited by Mr.C, May 17, 2021, 5:22 PM

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#2 h May 19, 2021, 6:44 PM • 5 Y

Y by centslordm, KST2003, math_comb01, Deadline, luutrongphuc

First we prove that
$\frac{BD}{CD}\cdot \frac{\sin(\angle A-\angle ABD)}{\sin(\angle A-\angle ACD)}=\frac{\cos \angle B}{\cos \angle C}\quad (\star)$ Let $R$ and $Q$ be intersections of the perpendicular bisectors of $AC$ and $AB$ with $AB$ and $AC$ , respectively. It's well-known that $CR$ and $BQ$ meet on $OH$ so let $S$ be the concurrency point.
$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(1.2) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ /* draw figures */ draw((-3.92,1.58)--(-5.6,-3.42), red); draw((-5.6,-3.42)--(1.6,-3.54), red); draw((1.6,-3.54)--(-3.92,1.58), red); draw((-3.9741182710156453,-1.667096260938743)--(4.204510574550921,-2.4409752970756866), orange); draw((4.204510574550921,-2.4409752970756866)--(-5.6,-3.42), orange); draw((4.204510574550921,-2.4409752970756866)--(-6.836456291650331,-7.099929439435513), orange); draw((-5.6,-3.42)--(-6.836456291650331,-7.099929439435513), orange); draw((-5.6,-3.42)--(-3.1641386039243344,-1.7437382379614483), orange); draw((-3.1641386039243344,-1.7437382379614483)--(1.6,-3.54), orange); /* dots and labels */ dot((-3.92,1.58),linewidth(4pt) + dotstyle); label("$A$", (-4.02,1.8), NE * labelscalefactor); dot((-5.6,-3.42),linewidth(4pt) + dotstyle); label("$B$", (-6.1,-3.64), NE * labelscalefactor); dot((1.6,-3.54),linewidth(4pt) + dotstyle); label("$C$", (1.72,-3.84), NE * labelscalefactor); dot((-1.9729408644921775,-1.8564518695306287),linewidth(4pt) + dotstyle); label("$O$", (-2,-1.74), NE * labelscalefactor); dot((-3.9741182710156453,-1.667096260938743),linewidth(4pt) + dotstyle); label("$H$", (-4.16,-1.48), NE * labelscalefactor); dot((-3.1641386039243344,-1.7437382379614483),linewidth(4pt) + dotstyle); label("$D$", (-3.25,-1.62), NE * labelscalefactor); dot((0.7834143473147777,-2.7825872206977653),linewidth(4pt) + dotstyle); label("$Q$", (.7,-2.66), NE * labelscalefactor); dot((-6.836456291650331,-7.099929439435513),linewidth(4pt) + dotstyle); label("$R$", (-7.35,-7.28), NE * labelscalefactor); dot((4.204510574550921,-2.4409752970756866),linewidth(4pt) + dotstyle); label("$S$", (4.28,-2.32), NE * labelscalefactor); [/asy]$
Now it's obvious that $\angle A-\angle ACD=180^\circ-\angle DCS$ and $\angle A-\angle ABD=\angle SBC$ . So by law of sines
$\frac{\sin\angle DSB}{\sin\angle DSC}=\frac{BD}{CD}\cdot\frac{\sin \angle DBS}{\sin\angle DCS}=\frac{BD}{CD}\cdot \frac{\sin(\angle A-\angle ABD)}{\sin(\angle A-\angle ACD)}\quad (1)$ $\frac{\sin\angle OSB}{\sin\angle OSC}=\frac{BO}{CO}\cdot\frac{\sin\angle OBS}{\sin\angle OCS}=\frac{\cos\angle B}{\cos\angle C}\quad (2)$ From $(1)$ and $(2)$ , $(\star)$ follows.
$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(1.2) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ /* draw figures */ draw((-3.92,1.58)--(-5.6,-3.42), red); draw((-5.6,-3.42)--(1.6,-3.54), red); draw((1.6,-3.54)--(-3.92,1.58), red); draw((-3.9741182710156453,-1.667096260938743)--(-1.9729408644921775,-1.8564518695306287), orange); draw(circle((-3.3462354145413116,-2.349170102648458), 2.495221728770749), blue); draw(circle((-1.8124091878146156,-2.8571326757341287), 3.4800638280973577), blue); draw((-3.9741182710156453,-1.667096260938743)--(-3.92,1.58), orange); draw((-5.6,-3.42)--(-1.5409066091042478,-0.6266953190917119), orange); draw((1.6,-3.54)--(-4.826216091785365,-1.117071701742159), orange); draw((-5.575871790829434,-1.2289678417765875)--(0.56189817785975,-0.3128235787618109), orange); draw((-3.92,1.58)--(-2.7722169191116968,-3.4671297180148053), orange); draw((-2.7495590486944725,0.494373610383279)--(-3.5911611695136862,0.13400183064188284), orange); draw((-3.5911611695136862,0.13400183064188284)--(-1.5409066091042478,-0.6266953190917119), orange); draw((-3.5911611695136862,0.13400183064188284)--(-4.826216091785365,-1.117071701742159), orange); draw((-3.5911611695136862,0.13400183064188284)--(-4.496141504417192,-0.1347068583845015), orange); /* dots and labels */ dot((-3.92,1.58),linewidth(4pt) + dotstyle); label("$A$", (-4.02,1.8), NE * labelscalefactor); dot((-5.6,-3.42),linewidth(4pt) + dotstyle); label("$B$", (-5.94,-3.7), NE * labelscalefactor); dot((1.6,-3.54),linewidth(4pt) + dotstyle); label("$C$", (1.72,-3.84), NE * labelscalefactor); dot((-1.9729408644921775,-1.8564518695306287),linewidth(4pt) + dotstyle); label("$O$", (-2,-1.74), NE * labelscalefactor); dot((-3.9741182710156453,-1.667096260938743),linewidth(4pt) + dotstyle); label("$H$", (-4.2,-2.04), NE * labelscalefactor); dot((-3.1641386039243344,-1.7437382379614481),linewidth(4pt) + dotstyle); label("$D$", (-3.4,-2.1), NE * labelscalefactor); dot((-1.5409066091042478,-0.6266953190917119),linewidth(4pt) + dotstyle); label("$E$", (-1.55,-0.5), NE * labelscalefactor); dot((-4.826216091785365,-1.117071701742159),linewidth(4pt) + dotstyle); label("$F$", (-5.06,-1.04), NE * labelscalefactor); dot((-3.5911611695136862,0.13400183064188284),linewidth(4pt) + dotstyle); label("$X$", (-3.6,0.26), NE * labelscalefactor); dot((-5.024910010548472,-4.195296608728117),linewidth(4pt) + dotstyle); label("$P$", (-5.3,-4.6), NE * labelscalefactor); dot((-2.7722169191116968,-3.4671297180148053),linewidth(4pt) + dotstyle); label("$G$", (-2.92,-3.88), NE * labelscalefactor); dot((-2.7495590486944725,0.494373610383279),linewidth(4pt) + dotstyle); label("$K$", (-2.75,0.62), NE * labelscalefactor); dot((-4.496141504417192,-0.1347068583845015),linewidth(4pt) + dotstyle); label("$L$", (-4.78,-0.06), NE * labelscalefactor); dot((-5.575871790829434,-1.2289678417765875),linewidth(4pt) + dotstyle); label("$U$", (-5.94,-1.22), NE * labelscalefactor); dot((0.56189817785975,-0.3128235787618109),linewidth(4pt) + dotstyle); label("$V$", (0.64,-0.2), NE * labelscalefactor); dot((-3.962803263142782,-0.9881957885669385),linewidth(4pt) + dotstyle); label("$J$", (-3.93,-0.93), NE * labelscalefactor); [/asy]$
Suppose that the circumcircle of triangle $BXE$ intersects the lines $AB$ and $EF$ again at $L$ and $U$ , and the circumcircle of triangle $CXF$ intersects the lines $AC$ and $EF$ again at $K$ and $V$ . Let $AH$ and $AD$ intersects $EF$ and $BC$ at $J$ and $G$ , respectively. We just need to show that $\frac{JE}{JV}=\frac{JF}{JU}$ that is equivalent to $\frac{JE}{EV}=\frac{JF}{FU}$ . Notice that $\angle KXE=\angle KXF-\angle EXF=180^\circ-\angle ACD-(180^\circ-\angle A)=\angle A-\angle ACD.$ Now by law of sines
$\frac{EK}{EX}=\frac{\sin\angle KXE}{\sin\angle XKE}=\frac{\sin(\angle A-\angle ACD)}{\sin\angle XFC}.$ Similarly we have $\frac{FL}{FX}=\frac{\sin(\angle A-\angle ABD)}{\sin\angle XEB}$ . Therefore
$\frac{EK}{FL}=\frac{EX}{FX}\cdot\frac{\sin\angle XEB}{\sin\angle XFC}\cdot\frac{\sin(\angle A-\angle ACD)}{\sin(\angle A-\angle ABD)}= \frac{\sin\angle EDA}{\sin\angle FDA}\cdot\frac{\sin(\angle A-\angle ACD)}{\sin(\angle A-\angle ABD)}=\frac{BG}{CG}\cdot\frac{CD}{BD}\cdot\frac{\sin(\angle A-\angle ACD)}{\sin(\angle A-\angle ABD)}\overset{(\star)}{=}\frac{BG}{CG}\cdot\frac{\cos\angle C}{\cos\angle B},$ and from the other hand by Ceva's theorem
$\frac{EK}{FL}=\frac{\frac{EV\cdot EF}{CE}}{\frac{FU\cdot EF}{BF}}=\frac{EV}{FU}\cdot\frac{BG}{CG}\cdot\frac{AF}{AE}$ $\Longrightarrow \frac{EV}{FU}=\frac{AE}{AF}\cdot\frac{\cos \angle C}{\cos\angle B}=\frac{JE}{JF},$ as desired.

This post has been edited 1 time. Last edited by Dadgarnia, May 24, 2021, 2:31 PM

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#3 h May 21, 2021, 8:12 AM • 1 Y

Y by centslordm

We don't even need $H$ , define $Q$ , $R$ and $S$ as hover for tip did. and a point $P$ is such that $B,C,BP \cap AC,CP \cap AB$ are cyclic and redefine $D$ to be a point on the line $SP$ , then the verdict is still true. the proof is still the same as above. (I don't how much it generalizes the problem, I hope it helps)

This post has been edited 3 times. Last edited by iman007, Oct 10, 2021, 10:59 AM

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#4 h May 11, 2023, 3:39 AM

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Does this have a geometry proof?

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