ARO 2011 10-4

Art of Problem Solving

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68 posts

sartt

#1 h Apr 28, 2011, 5:35 PM • 4 Y

Y by Adventure10, Mango247, NO_SQUARES, Mogmog8

Perimeter of triangle $ABC$ is $4$ . Point $X$ is marked at ray $AB$ and point $Y$ is marked at ray $AC$ such that $AX=AY=1$ . Line segments $BC$ and $XY$ intersectat point $M$ . Prove that perimeter of one of triangles $ABM$ or $ACM$ is $2$ .

(V. Shmarov).

This post has been edited 1 time. Last edited by v_Enhance, May 31, 2024, 2:48 PM
Reason: LaTeX

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801 posts

#2 h May 14, 2011, 1:49 AM • 5 Y

Y by El_Ectric, thczarif, khina, rayfish, Adventure10

Let $\omega$ denote the $A$ -excircle of $\triangle{ABC}$ . Let $\omega$ be tangent to $BC$ , $AC$ and $AB$ at $D$ , $E$ and $F$ , respectively. By power of a point, $AE=AF$ and $AE+AF=AC+CD+AB+BE=4$ which implies that $AE=AF=2$ . Hence $X$ and $Y$ are the midpoints of $AE$ and $AF$ , respectively and $XY$ is the radical axis betwen $\omega$ and the point $A$ . This implies that $AM=MD$ . Since $M \in (BC)$ , we may assume without the loss of generality that $M \in (BD]$ . This implies that the perimeter of $\triangle{AMB}$ is $AM+MB+AB=AM+(BD-MD)+AB=AB+BD=AF=2$ .

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MBGO

#3 h Dec 22, 2012, 7:46 PM • 2 Y

Y by Adventure10, Mango247

sartt wrote:

Perimeter of triangle ABC is 4. Point X is marked at ray AB and point Y is marked at ray AC such that AX=AY=1. BC intersects XY at point M. Prove that perimeter of one of triangles ABM or ACM is 2.
(V. Shmarov).

is the problem correct? also since WLOG $M$ is on the external of $BC$ nearer to $B$ than $C$ so we get:
$AM+MB+AB=MD+MB+AB=2MB+BD+AB=AF+2MB>AF=2$ so we get (maybe some kinda) contradiction here...

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3091 posts

#4 h Jun 21, 2014, 2:09 AM • 2 Y

Y by Adventure10, Mango247

MBGO wrote:

is the problem correct? also since WLOG $M$ is on the external of $BC$ nearer to $B$ than $C$ so we get:
$AM+MB+AB=MD+MB+AB=2MB+BD+AB=AF+2MB>AF=2$ so we get (maybe some kinda) contradiction here...

If $D$ is between $B$ and $M$ then $\triangle ACM$ will have perimeter 2, not $\triangle ABM$ .

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Stens

#5 h Mar 29, 2016, 3:20 PM • 2 Y

Y by Adventure10, Mango247

In the triangle below, $(AB,BC,CA)=(1.5\bar{3},1.3\bar{3},1.1\bar{3})$ , and all the conditions in the problem are satisfied. But clearly the perimeters of both $ABM$ and $ACM$ are $>2$ . Have I overlooked something or is the problem actually incorrectly stated?

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62861

#6 h Mar 29, 2016, 3:28 PM • 5 Y

Y by Stens, Adventure10, Mango247, NO_SQUARES, v_Enhance

Stens wrote:

In the triangle below, $(AB,BC,CA)=(1.5\bar{3},1.3\bar{3},1.1\bar{3})$ , and all the conditions in the problem are satisfied. But clearly the perimeters of both $ABM$ and $ACM$ are $>2$ . Have I overlooked something or is the problem actually incorrectly stated?

The problem should say that segments BC and XY intersect at M.

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1980 posts

#7 h Dec 18, 2016, 3:14 PM • 2 Y

Y by Adventure10, Mango247

Very elegant!

All-Russian MO 2011/10/4 wrote:

Perimeter of triangle $ABC$ is $4$ . Point $X$ is marked at ray $AB$ and point $Y$ is marked at ray $AC$ such that $AX=AY=1$ . Line segments $BC$ and $XY$ intersect at point $M$ . Prove that perimeter of one of triangles $ABM$ or $ACM$ is $2$ .

Proposed by V.Shmarov

Point $E$ is the touching point of the $A$ -excircle $\omega$ with line $\overline{BC}$ . WLOG, point $E$ lie on segment $CM$ . It is clear that $AB+BE=2$ . We will show that perimeter of $\triangle ABM$ is $2$ .

The main idea is to note that line $\overline{XY}$ is the radical axis of circle $\omega$ and "point-size circle" $A,$ as $X,Y$ bisect the tangents drawn from $A$ to $\omega$ (since $AX=AY=1$ ).

So, $MA=ME \Longrightarrow AB+BM+MA=AB+(BM+ME)=AB+BE=2. \, \square$

This post has been edited 2 times. Last edited by v_Enhance, May 31, 2024, 2:48 PM
Reason: missing space

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#8 h Feb 1, 2022, 10:08 PM

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Let $Y$ lie on segment $AC.$ Denote by $U,V,Z$ touch-points of $A-\text{excircle }\omega$ with $AB,AC,BC$ respectively.
Note that $X,Y$ are midpoints of $AU,AV$ and hence $XY$ is radical axis of $A,\omega.$ Thus $|AB|+|BM|+|AM|=|AB|+|BZ|=|AU|=2.$

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6877 posts

#9 h May 31, 2024, 2:47 PM • 2 Y

Y by szpolska, OronSH

The solution from my textbook, which I apparently never posted (note that as written in post #6, the problem only works if $M$ is on segments $XY$ and $BC$ ):

Let $I_A$ be the center of the $A$ -excircle, tangent to $\overline{BC}$ at $T$ , and to the extensions of $\overline{AB}$ and $\overline{AC}$ at $U$ and $V$ . We see that $AU = AV = s = 2$ . Then $\overline{XY}$ is the radical axis of the $A$ -excircle and the circle of radius zero at $A$ . Therefore $AM = MT$ .

$[asy] size(12cm); pair A = dir(130); pair B = dir(190); pair C = dir(-10); pair I = incenter(A, B, C); pair L = dir(-90); pair I_A = 2*L-I;; pair T = foot(I_A, B, C); pair U = foot(I_A, A, B); pair V = foot(I_A, A, C); pair X = midpoint(A--U); pair Y = midpoint(A--V); pair M = extension(B, C, X, Y); filldraw(A--B--C--cycle, invisible, blue); draw(B--U, blue); draw(C--V, blue); draw(X--Y, blue); draw(A--M, deepgreen); draw(arc(I_A, abs(I_A-T), -10, 190), grey); dot("$A$", A, dir(A)); dot("$B$", B, dir(150)); dot("$C$", C, dir(30)); dot("$I_A$", I_A, dir(I_A)); dot("$T$", T, dir(90)); dot("$U$", U, dir(135)); dot("$V$", V, dir(45)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$M$", M, dir(270)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); A = dir 130 B 150 = dir 190 C 30 = dir -10 I := incenter A B C L := dir -90 I_A = 2*L-I; T 90 = foot I_A B C U 135 = foot I_A A B V 45 = foot I_A A C X = midpoint A--U Y = midpoint A--V M 270 = extension B C X Y A--B--C--cycle / 0.1 palecyan / blue B--U / blue C--V / blue X--Y / blue A--M / deepgreen !draw(arc(I_A, abs(I_A-T), -10, 190), grey); */ [/asy]$
Assume without loss of generality that $T$ lies on segment $MC$ , as opposed to segment $MB$ . Then $AB + BM + MA = AB + BM + MT = AB + BT = AB + BU = AU = 2$ as desired.

This post has been edited 1 time. Last edited by v_Enhance, May 31, 2024, 2:49 PM

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