The sum not depends to x

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The sum not depends to x

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Source: All russian olympiad 2016,Day2,grade 9,P7

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#1 h May 1, 2016, 10:47 AM • 1 Y

Y by Adventure10

In triangle $ABC$ , $AB<AC$ and $\omega$ is incirle.The $A$ -excircle is tangent to $BC$ at $A^\prime$ .Point $X$ lies on $AA^\prime$ such that segment $A^\prime X$ doesn't intersect with $\omega$ .The tangents from $X$ to $\omega$ intersect with $BC$ at $Y,Z$ .Prove that the sum $XY+XZ$ not depends to point $X$ .(Mitrofanov)

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#2 h May 1, 2016, 4:11 PM • 3 Y

Y by aops29, guptaamitu1, Adventure10

Let $\omega \equiv (I)$ touch $BC$ at $D$ and let $D'$ be the antipode of $D$ on $(I).$ It's well-known that $D' \in AA'.$ WLOG assume that $Z$ is between $D$ and $Y$ and let the Z-excircle $(J)$ of $\triangle XYZ$ touch $BC$ at $A''.$ Since $X$ is the insimilicenter of $(I) \sim (J)$ and $ID' \parallel JA'',$ then $X,D',A''$ are collinear $\Longrightarrow$ $A'' \equiv A'.$ Therefore $XY+XZ=DA'=AC-AB=\text{const}.$

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#3 h May 1, 2016, 4:28 PM • 3 Y

Y by aops29, Adventure10, Mango247

Define $\Omega$ the A-excircle of $\triangle ABC$ , $\Gamma$ the Y-excircle of $\triangle XYZ$ and let $D\equiv BC\cap \omega$ and $D$ , $Y$ , $Z$ lie on $BC$ in this order.
$\Longrightarrow$ The insimilicenter of $\omega$ and $\Gamma$ is $X$ , the insimilicenter of $\Omega$ and $\Gamma$ lie in $BC$ , the exsimilicenter of $\Omega$ and $\omega$ is $A$ $\Longrightarrow$ by Monge-D'alemberg theorem we get the insimilicenter of $\Omega$ and $\Gamma$ is $AX\cap BC=A'$ hence the $\Gamma$ is tangent to $BC$ in $A'$ $\Longrightarrow$ $ZA'=YD$ $\Longrightarrow$ since $ZA'=YD$ we get $XZ+XY=A'D$ which it is fixed.

This post has been edited 1 time. Last edited by FabrizioFelen, May 1, 2016, 4:35 PM

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#4 h May 2, 2016, 2:11 AM • 4 Y

Y by baopbc, buratinogigle, Adventure10, Mango247

I think the problem is based on the following lemma.
Lemma. Triangle $ABC$ with $\omega$ is incircle. The A-excircle is tangent to $BC$ at $A'$ . $X$ is in line $AA'$ . Tangents from $X$ to $\omega$ intersect $BC$ at $Y,Z$ then $BY=CZ$ .

I have a proof for this but it requires to consider the position of point $X$ on line $AA'$ , which gives a lot of case work. I hope someone can have a better proof for this.

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#6 h Oct 12, 2018, 3:37 PM • 2 Y

Y by Adventure10, Mango247

(redacted)

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#7 h Jul 7, 2019, 10:01 AM • 6 Y

Y by a_simple_guy, aops29, primesarespecial, Infinityfun, Adventure10, Mango247

Not as smart as the above solution, but a very harmonic one.
Let $D'$ be the anti-pole of $D$ . Then $A,D',A'$ are collinear. Let $G = D'F \cap ED$ , $H = DF \cap D'E$ , $R = EF \cap DD'$ . Then by Brokard's theorem, $HG$ is the polar line of $R$ , in particular $HG \parallel BC$ . Since the polar line $EF$ of $X$ passes through R, by La Hire's theorem, $X$ lies on $HG$ . Let $S = HG \cap DD'$ , $P = HG \cap EF$ , $M=XR \cap BC$ then
$-1 = (E,F; P,R) \stackrel{X}{=} (Y,Z; \infty, M)$ Therefore $M$ is the midpoint of $YZ$ . Moreover,
$-1 = (D',D; S,R) \stackrel{X}{=} (D,A'; \infty, M)$ Hence $M$ is the midpoint of $DA'$ as well. As a result, $XY+XZ = YD+ZD = DA'$ which is fixed.

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#8 h Jul 29, 2020, 4:00 PM

Y by

shinichiman wrote:

I think the problem is based on the following lemma.
Lemma. Triangle $ABC$ with $\omega$ is incircle. The A-excircle is tangent to $BC$ at $A'$ . $X$ is in line $AA'$ . Tangents from $X$ to $\omega$ intersect $BC$ at $Y,Z$ then $BY=CZ$ .

I have a proof for this but it requires to consider the position of point $X$ on line $AA'$ , which gives a lot of case work. I hope someone can have a better proof for this.

I like it. Let $D,E,F$ be points of tangency of the incircle with sides $BC,CA,AB$ respectively. Place the unit circle on Argand plane in such way that it's a unit circle centered at $0$ . Take $P,Q$ as two different points of incircle such that lines $PX,QX$ are tangent to $\omega$ . Denote wlog $Y,Z$ as intersections of line $BC$ with lines $PX,QX$ respectively. We have $x=\frac{2pq}{p+q},\ b=\frac{2df}{d+f},\ c=\frac{2de}{d+e},\ a=\frac{2ef}{e+f},\ y=\frac{2dp}{d+p},\ z= \frac{2dq}{d+q}.$ Let $g=-d$ . It is a known lemma following by homothethy centered at $A$ that $G\in AA'$ . Because $A\neq G, A'$ we have $X\in AA'\iff X\in AG\iff \frac{x-g}{a-g}=\overline{\left(\frac{x-g}{a-g}\right)}\iff d(2ef+de+df)\cdot\overline{x}=(2d+e+f)x+2(d^2-ef)\iff$ $d(2ef+de+df)\cdot \frac{2}{p+q}=(2d+e+f)\cdot \frac{2pq}{p+q}+2(d^2-ef)\iff q=\frac{d(2ef+de+df)+(ef-d^2)p}{(2d+e+f)p+d^2-ef}.$ Move on to the length equality.
$y-b=\frac{2d^2(p-f)}{(p+d)(d+f)},\ z-c=\frac{2d^2(q-e)}{(q+d)(d+e)}.$ We're left with proving $\left|\frac{p-f}{(p+d)(d+f)}\right|=\left|\frac{q-e}{(q+d)(d+e)}\right|$ Since $q-e=\frac{(d+e)^2(f-p)}{(2d+e+f)p+d^2-ef},\ q+d=\frac{(p+d)(d+e)(d+f)}{(2d+e+f)p+d^2-ef}$ we're done. QED
#1679

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#9 h Jul 10, 2021, 1:31 PM

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Denote by $D,E,F,U,V$ touch-points of $\omega$ with $BC,AC,AB,XY,XZ$ respectively. Let $AA'\cap \omega =\left\{S,R\right\}$ (where $S$ lies on segment $AR$ ), and let $M$ be an intersection of $BC$ with tangent to $\omega$ through $R$ . Note that $URVS,RESF$ are harmonic, and as well-known tangent to $\omega$ through $S$ is parallel to $BC$ , so by duality wrt $\omega$ we obtain $(YZM\infty)=(DU,DV,DR,DS)=-1=(DF,DE,DR,DS)=(BCM\infty),$ i.e. $M$ is midpoint of $YZ,BC$ and thus also is midpoint of $DA'$ . Hence $|XY|+|XZ|=|XU|\pm |UY|+|ZV|\mp |XV|=|YD|+|ZD|=|DA'|$ which not depends to $X$ .

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#10 h Aug 14, 2023, 1:31 PM

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solution

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#11 h Feb 9, 2024, 7:40 PM

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After so many misreads I finally solved this one...
Let $D'$ be the antipode of $D$ in $\omega$ let $XY \cap \omega = E, XZ \cap \omega = F$ , let $DD' \cap EF = K$ , the line through $X$ parallel to $BC$ meet $DD',EF,KZ$ at $R,P,G$ . Let $H = D'E \cap DF$ , we first claim $DD',EF,HY,GZ$ concur, this follows from repeated application of brocard's and la hire's, now $(EF;PK) = (YZ,M \infty )$ so $M$ is mid pt. of $YZ$ but it is also $DA'$ so we have $XY + XZ = DA' = AB-AC$ so done.
Funny Remark: I was trying to solve the case where the tangents are to excircle for 30+ min before realizing I misread, I somehow got some progress there also maybe some version of this is true.

This post has been edited 1 time. Last edited by math_comb01, Feb 9, 2024, 7:41 PM

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