Sharygin 2025 CR P17

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Gengar_in_Galar

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Gengar_in_Galar

#1 h Mar 10, 2025, 12:11 PM

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Let $O$ , $I$ be the circumcenter and the incenter of an acute-angled scalene triangle $ABC$ ; $D$ , $E$ , $F$ be the touching points of its excircle with the side $BC$ and the extensions of $AC$ , $AB$ respectively. Prove that if the orthocenter of the triangle $DEF$ lies on the circumcircle of $ABC$ , then it is symmetric to the midpoint of the arc $BC$ with respect to $OI$ .
Proposed by: P.Puchkov,E.Utkin

This post has been edited 1 time. Last edited by Gengar_in_Galar, Mar 11, 2025, 9:56 AM

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#2 h Mar 10, 2025, 12:25 PM

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what a problem
sol

This post has been edited 1 time. Last edited by InterLoop, Mar 10, 2025, 12:26 PM

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#3 h Mar 10, 2025, 12:29 PM

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Solution -
Following the given notations we have the following extra notations $I_A$ the $A$ -excenter or the circumcenter of $DEF$ , $M$ is the midpoint of arc $BC$ not containing $A$ , $H_A$ is the orthocenter of the triangle $DEF$ , $R$ , $r$ , $r_a$ is the circumradius , inradius and exradius corresponding to A , respectively of triangle $ABC$ , $X$ be the intersection of $H_AI_A$ and circumcircle of triangle $ABC$ and $N_A$ the nine point center of triangle $DEF$

TO prove that $H_A$ is symmetric to $M$ with respect to $OI$ it suffices to prove that $IH_A=IM$ because then triangles $OH_AI$ and $OMI$ will be congruent by $SSS$ and $HM$ will be the perpendicular to $OI$ and both will be symmetric with respect to $OI$

$[asy] import olympiad ; size(800); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(08); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.019307430904394, xmax = 19.417005852771155, ymin = -13.268453059361399, ymax = 5.830285049224888; /* image dimensions */ pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ccqqqq = rgb(0.8,0,0); pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures // special point */ draw(circle((-5.679867559939484,1.2084027102619619), 4.168222045751201), linewidth(0.4) + linetype("2 2")); /* special point // special point */ draw((xmin, 2.120386263690971*xmin + 15.68744302822356)--(xmax, 2.120386263690971*xmax + 15.68744302822356), linewidth(0.8)); /* line // special point */ draw((xmin, -1.3714363791428523*xmin-1.4142257053584768)--(xmax, -1.3714363791428523*xmax-1.4142257053584768), linewidth(0.8)); /* line */ draw((xmin, 0.397253041143568*xmin + 1.3141560965262242)--(xmax, 0.397253041143568*xmax + 1.3141560965262242), linewidth(0.8)); /* line // special point */ draw((xmin, -10.530837339331121*xmin-46.27361778836229)--(xmax, -10.530837339331121*xmax-46.27361778836229), linewidth(0.8)); /* line // special point // special point */ draw(circle((-3.7389711889453943,-6.899120381132856), 6.2526580644719365), linewidth(0.4) + linetype("2 2")); /* special point // special point */ draw((-9.394260166587916,-4.232017186548713)--(-6.047382797768048,-1.0881851108464315), linewidth(1.2) + fuqqzz); draw((-6.047382797768048,-1.0881851108464315)--(1.3132335952056033,-3.215242032135999), linewidth(1.2) + fuqqzz); draw((1.3132335952056033,-3.215242032135999)--(-9.394260166587916,-4.232017186548713), linewidth(1.2) + fuqqzz); /* special point */ draw((-6.650466991259582,5.262796432734572)--(-3.7389711889453943,-6.899120381132856), linewidth(1.2) + ccqqqq); /* special point */ draw((-5.679867559939484,1.2084027102619619)--(-4.141006420175933,-2.6653527563636685), linewidth(1.2) + ccqqqq); draw((-4.543041651406472,1.5684148684055257)--(-6.650466991259582,5.262796432734572), linewidth(1.2) + qqwuqq); draw((-4.543041651406472,1.5684148684055257)--(-4.141006420175933,-2.6653527563636685), linewidth(1.2) + qqwuqq); /* special point */ /* dots and labels */ label("$O$", (-5.566208321703934,1.3653097887040404), NE * labelscalefactor); label("$A$", (-4.456416782961756,5.210866515973441), NE * labelscalefactor); label("$B$", (-8.250355299126877,-1.8350192997617696), NE * labelscalefactor); label("$C$", (-1.4367514333609452,0.8491276776611678), NE * labelscalefactor); label("$I$", (-4.430607677409612,1.7266372664340512), NE * labelscalefactor); label("$I_A$", (-3.630525405293158,-6.738749354669058), NE * labelscalefactor); label("$F$", (-10.082801793329079,-4.18364790500684), NE * labelscalefactor); label("$D$", (-7.0631364437282675,-1.1639825554060352), NE * labelscalefactor); label("$E$", (1.4280592829270033,-3.04804726071252), NE * labelscalefactor); label("$H_A$", (-7.630936765875428,5.030202777108435), NE * labelscalefactor); label("$M$", (-4.043471094127457,-2.506056044117504), NE * labelscalefactor); label("$X$", (-4.611271416274618,-2.686719782982509), NE * labelscalefactor); dot((-5.194719090102486,-0.8181619741991425),dotstyle); label("$N_A$", (-5.901726693881802,-0.5703731277067318), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

Proving this we will start with our first lemma ,

Lemma 1 : In such a configuration $3R=2r_a$

Proof : For the proof of this fact we will use the well known fact that there is a homothety between the nine point center of the $A$ - excentral triangle and the circumcircle of triangle $ABC$ centered at $I_A$ with the ratio $r_a/2R$ that sends $N_A$ to $O$ using this and combinig the fact that $H_A$ lies on the euler line of $DEF$ we have $H_A-O-I_A$

Let $H_A N_A = N_A I_A = x$ by homothety we have that $\frac{I_A O}{I_A N_A} = \frac{2R}{r_a}$

$\frac{2x - R}{x} = \frac{2R}{r_a}$

which gives $x = \frac{Rr_a}{2(r_a - R)}$

we also know by euler theorem or otherwise that $Pow( I_A, (ABC)) = 2R \cdot r_a = I_A X \cdot I_A H_A$

also $I_A H_A = 2x$ as $N_A$ is the midpoint of $I_A H_A$

so we have $2R \cdot r_a = 2x \cdot I_A X$
using the value of $x$ obtained above we have
$I_A X = 2(r_a - R)$

but $I_A X + XH_A = 2x$

$2(r_a - R) + 2R = 2x$ which gives $x = r_a$
using this value in the equation $x = \frac{Rr_a}{2(r_a - R)}$
we have $2r_a = 3R$

follows from this that $I_AX=R$ and $I_AH_A=3R$

Now in the next lemma , with the repeated use of cosine law we will prove that $IH_A=IM$

Lemma 2 : $IH_A=IM$

first by trident lemma we have 2 $IM$ = $II_A$ = 2 $y$ and let $H_AI=z$ for the moment

$\cos(\angle OI_A I)$ using triangle $OI_AM$ is $\frac{OI_A^2 + I_AM^2 - OM^2}{2 \cdot OI_A \cdot I_AM}$

which is equal to $\frac{4R^2 + y^2 - R^2}{2 \cdot 2R \cdot y}$

$\cos(\angle OI_A I)$ using triangle $H_AI_AI$ is $\frac{I_AH_A^2 + I_AI^2 - H_AI^2}{2 \cdot I_AH_A \cdot I_AI}$

which is equal to $\frac{9R^2 + 4y^2 - z^2}{2 \cdot 3R \cdot 2y}$

equating both we get $y = z$ and we are done $\boxed{}$

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58 posts

#4 h Mar 11, 2025, 2:07 PM

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When I first solved this question"synthetic" I got came up with a point $"Q"=OI \cap C(M,MB) \cap DR$ such that
$DR \perp EF$ if $H$ is the orthocenter lying on circumcircle than this $\implies$ to prove concurrency of point " $Q$ "

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#5 h Mar 11, 2025, 8:10 PM • 2 Y

Y by HotSinglesInYourArea, Calamarul

There exists a (kinda) nice complex bash to this problem:

Take the A-excircle to be the unit circle. So we have the following:

$|d|=|e|=|f|=1$ , $i_a=0$ , $a=\frac{2ef}{e+f}$ , $b=\frac{2df}{d+f}$ , $c=\frac{2de}{d+e}$ , $o=\frac{2def(d+e+f)}{(d+e)(e+f)(f+d)}$ .

Let $H$ be the orthocenter of the $\Delta DEF$ triangle. So $h=d+e+f$ .

Due to Incenter-Excenter Lemma we know that $L$ - the midpoint of arc $\widehat{BC}$ on the $(ABC)$ circle is the center of the $(IBI_AC)$ circle.

Compute $L$ using the circumcenter formula for $\Delta I_ABC$ and we get $\ell=\frac{2def}{(d+e)(d+f)}$ .

Since $L$ is the midpoint of $[II_A]$ we have $i=2\ell=\frac{4def}{(d+e)(d+f)}$ .

$H\in(ABC)\implies \frac{a-h}{b-h}\cdot\frac{b-c}{a-c}\in\mathbb{R}\implies...\implies \frac{d^2(e^2+f^2+de+df)(f-e)}{e^2(d^2+f^2+de+ef)(f-d)}\in\mathbb{R}\implies$
$\implies \frac{d^2(e^2+f^2+de+df)(f-e)}{e^2(d^2+f^2+de+ef)(f-d)}=\frac{(de^2+df^2+ef^2+e^2f)(e-f)}{(d^2e+ef^2+d^2f+df^2)(d-f)}\implies$
$\implies d^2(e^2+f^2+de+df)(d^2e+ef^2+d^2f+df^2)=e^2(d^2+f^2+de+ef)(de^2+df^2+ef^2+e^2f)\implies...\implies$
$\implies (d-e)(d+e)(e+f)(f+d)(d^2e+de^2+e^2f+ef^2+d^2f+df^2-def)=0$ .

Obviously $(d-e)(d+e)(e+f)(f+d)\neq0\implies d^2e+de^2+e^2f+ef^2+d^2f+df^2-def=0\quad(1)$ .

Now, since $H, L\in(ABC)$ , $L$ will be the reflection of $H$ through $OI$ if and only if the midpoint $M$ of $[HL]$ belongs to line $OI$ , which is equivallent to $\frac{o-i}{\overline{o}-\overline{i}}=\frac{m-i}{\overline{m}-\overline{i}}$ .

We have $m=\frac{d+e+f}{2}+\frac{def}{(d+e)(d+f)}$ .

$o-i=\frac{2def}{(d+e)(d+f)}\left(\frac{d+e+f}{e+f}-2\right)=\frac{2def(d-e-f)}{(d+e)(e+f)(f+d)}$ , so $\frac{o-i}{\overline{o}-\overline{i}}=\frac{def(d-e-f)}{ef-de-df}$ .

$m-i=\frac{3def}{(d+e)(d+f)}-\frac{d+e+f}{2}=\frac{3def-d^3-2d^2e-2d^2f-de^2-df^2-e^2f-ef^2}{2(d+e)(d+f)}$ , so
$\frac{m-i}{\overline{m}-\overline{i}}=\frac{def(d^3+2d^2e+2d^2f+de^2+df^2+e^2f+ef^2-3def)}{d^2e^2+e^2f^2+f^2d^2+2de^2f+2def^2+d^3e+d^3f-3d^2ef}$

So $M, I, O$ are collinear if and only if $\frac{d-e-f}{ef-de-df}=\frac{d^3+2d^2e+2d^2f+de^2+df^2+e^2f+ef^2-3def}{d^2e^2+e^2f^2+f^2d^2+2de^2f+2def^2+d^3e+d^3f-3d^2ef}\iff ...\iff$
$\iff 2(d^2-ef)(d^2e+de^2+e^2f+ef^2+d^2f+df^2-def)=0$ , which is true by $(1)$ , so we're done.

This post has been edited 1 time. Last edited by Double07, Mar 11, 2025, 8:10 PM

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HVP

#6 h Mar 17, 2025, 1:54 PM • 1 Y

Y by Double07

2 solutions of mine, one is synthetic while the other is complex bashing

Attachments:

Problem 17 Sharygin Olympiad 24.25 complex bash.pdf (97kb)

Problem 17 Sharygin Olympiad 24.25 synthetic.pdf (94kb)

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