IMO Shortlist 2010 - Problem G6

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IMO Shortlist 2010 - Problem G6

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#1 h Jul 17, 2011, 2:18 AM • 5 Y

Y by TFIRSTMGMEDALIST, nobodyknowswhoIam, Adventure10, Autistic_Turk, and 1 other user

The vertices $X, Y , Z$ of an equilateral triangle $XYZ$ lie respectively on the sides $BC, CA, AB$ of an acute-angled triangle $ABC.$ Prove that the incenter of triangle $ABC$ lies inside triangle $XYZ.$

Proposed by Nikolay Beluhov, Bulgaria

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Vinoth

#2 h Jul 18, 2011, 5:45 AM • 5 Y

Y by Amir Hossein, TFIRSTMGMEDALIST, Adventure10, Mango247, Autistic_Turk

I can't think of a Euclidean way of doing it, but I can outline a way of killing it quickly with co-ordinates:

Let $Z$ be the origin, let $XY=YZ=XZ=1$ , say $BC$ is the $x$ -axis, say $\angle XZB = \theta$ ; then one can work out the equation of $XZ$ in terms of $\theta$ . It clearly suffices to prove that the incentre $I$ and $Y$ lie on the same side of the line $XZ$ . Then calculate the gradient of $XB$ , which gives the gradient of $IB$ (using say, the $\tan 2\theta$ formula). Similarly calculate the gradient of $YC$ , which gives the gradient of $IC$ . Now we know the equations of $IB$ and $IC$ , so solve for the co-ordinates of $I$ . Finally, check that $I$ and $Y$ lie on the same side of $XZ$ .

I can post the calculations if anyone is interested..

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349 posts

#3 h Mar 30, 2012, 4:54 AM • 5 Y

Y by Amir Hossein, oty, Adventure10, Mango247, and 1 other user

Can you post the calculations please? ( In your original post, I think there must be some typo: you say $Z$ the origin and $BC$ x-axis, but $BC$ does not pass through $Z$ . )

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#4 h Jan 10, 2014, 2:29 PM • 2 Y

Y by Ashutoshmaths, Adventure10

Due to miquel theorem we have the circumcircles of $BXZ,CXY,AZY$ passes through a common point namely $M$ ,notice that if the bisector of $B$ meet the circumcircle of $BXZ$ at $F$ then $FZ=FX$ ,so $F,Y$ lies perpendicular bisector of $ZX$ ,but $B$ is acute so $\angle ZFX>60$ ,similar analysis with other two triangles, then we get $I$ must be inside the triangle $XYZ$ .

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#5 h Jan 10, 2014, 3:54 PM • 2 Y

Y by Adventure10, Mango247

vanu1996 wrote:

... so $\angle ZFX>60$ ,similar analysis with other two triangles, then we get $I$ must be inside the triangle $XYZ$ .

Why?

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#6 h Apr 18, 2014, 3:01 PM • 2 Y

Y by Adventure10, Mango247

Trigonometry Another not so creative approach but can be done withouth much problems. I will only explain what are the cases to be dealt with and what to use.
Denote $d(P,l)$ distance of $P$ from line $l$
Let $XY=YZ=ZX=a$
Clearly if the bisector of $\angle A$ cuts $ZY$ at $D$ we need $d(D,BC)>d(D,AB)$ now we fix $X,Y,Z,A$ and find the maximal $k$ for which $d(D,BC)>k$ and prove that $k\ge d(D,AB)=d(D,AC)$ . To do this try to minimize the angle $\angle (DX,BC)$ from both sides. When minimizing it to one side say $\angle BXD$ consider the cases $\angle BZX\ge 90$ (this one is easy), $\angle BZX<90$ and $\angle AZD\ge 90$ (also not hard) and when both $\angle BZX,\angle AZD$ are acute.
For the last case you will need to express $k,d(D,AB)$ in termes of $90-\angle DZA=x$ and $a$ and $r=DZ$ now you get some inequality that you need for $tg x$ but for this you will need that $sin \angle AZD\le sin \angle AZD/sin \angle AYD=r/(a-r)$ .

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#7 h Apr 29, 2017, 2:42 AM • 2 Y

Y by Quidditch, Adventure10

Warning: This is a terrible solution.

Abuse of Notation. The term ``segment'' will always denote a line segment (as opposed to a circular segment).

Note. Given a line $\ell$ , define its argument to be the angle between $\ell$ and the $x$ -axis, starting from the $x$ -axis and going counterclockwise. Arguments are always in the range $[0, 180^{\circ})$ .

Let $I$ be the incenter and let $\omega$ be the incircle of triangle $ABC$ . We will use the points $X,Y,Z$ and circle $\omega$ to try to locate $A,B,C$ .

Claim 1. Note that $X,Y,Z$ each lie on the boundary of triangle $ABC$ . Since $\omega$ lies within triangle $ABC$ , we have that $X,Y,Z$ lie outside or on the boundary of $\omega$ . Furthermore, if $\omega$ intersects line $XY$ , all of the intersection points must lie within segment $XY$ . Otherwise, part of $\omega$ would lie outside triangle $ABC$ , contradiction.

Claim 2. Let $D,E,F$ be the tangency points of $\omega$ with sides $BC,CA,AB$ respectively. Suppose none of the lines $XY,YZ,ZX$ intersect $\omega$ . If $X$ lies on segment $CD$ , then it is impossible to find $Z$ on segment $AB$ such that line $XZ$ does not intersect $\omega$ . Similarly, if $X$ lies on segment $BD$ , then it is impossible to find $Y$ on segment $AC$ such that line $XY$ does not intersect $\omega$ . Thus we have a contradiction. We conclude that at least one of the lines $XY, YZ, ZX$ must intersect $\omega$ .

We return to the main problem. Assume for the sake of contradiction that $I$ lies outside triangle $XYZ$ ; our goal is now to show that either $ABC$ cannot be acute, $\omega$ cannot be the incircle of triangle $ABC$ , or $X,Y,Z$ cannot be contained within segments $BC,CA,AB$ respectively.

We can assume without loss of generality that $I, X$ are on opposite sides of line $YZ$ . Furthermore, we can assume that line $YZ$ is ``horizontal'' and $X,I$ are ``below'' and ``above'' line $YZ$ respectively. (See Figure 1.)

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.928819800000007, xmax = 39.76706740000004, ymin = -19.4182868, ymax = 7.799332199999999; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.883420800000009,-6.0188436)--(15.061844000000018,-6.244315)); draw((15.061844000000018,-6.244315)--(9.777368439773172,-14.946352361668911)); draw((9.777368439773172,-14.946352361668911)--(4.883420800000009,-6.0188436)); /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (10.037052800000014,-1.4449952), NE * labelscalefactor); dot((4.883420800000009,-6.0188436),dotstyle); label("$Y$", (5.012261600000009,-5.6967416), NE * labelscalefactor); dot((15.061844000000018,-6.244315),dotstyle); label("$Z$", (15.19068480000002,-5.922213), NE * labelscalefactor); dot((9.777368439773172,-14.946352361668911),linewidth(3.pt) + dotstyle); label("$X$", (9.908212000000013,-14.7478078), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Figure 1. $I,X$ are on opposite sides of line $YZ$ .

We see that if segment $YZ$ does not intersect $\omega$ , we must have line $YZ$ does not intersect $\omega$ , so line $YZ$ must be ``below'' $\omega$ . This means that segments $ZX,XY$ must not intersect $\omega$ , contradicting Claim 2. Thus we must have that segment $YZ$ intersects $\omega$ .

We see that there are two tangents from $X$ to $\omega$ ; line $BC$ must be one of these tangents. Let $X_1, X_2$ be the intersections of the two tangents with $\omega$ .

Case 1. $\omega$ does not intersect lines $XY, XZ$ .

We have that $\omega$ is contained within the smaller region $\mathcal{R}$ of the plane bounded by rays $XY, XZ$ . Thus the rays $XX_1, XX_2$ must also be contained within $\mathcal{R}$ ; thus line $BC$ intersects $\mathcal{R}$ . However, line $BC$ does not intersect triangle $XYZ$ , contradiction.

Case 2. $\omega$ intersects line $XY$ but does not intersect line $XZ$ . (See Figure 2.)

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.928819800000007, xmax = 39.76706740000004, ymin = -19.4182868, ymax = 7.799332199999999; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.207006600000009,-2.1536196)--(16.70456420000002,-2.1214094)); draw((16.70456420000002,-2.1214094)--(10.48368025146099,-12.960716866859293)); draw((10.48368025146099,-12.960716866859293)--(4.207006600000009,-2.1536196)); /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (10.037052800000014,-1.4449952), NE * labelscalefactor); dot((4.207006600000009,-2.1536196),dotstyle); label("$Y$", (4.3358474000000085,-1.8315176), NE * labelscalefactor); dot((16.70456420000002,-2.1214094),dotstyle); label("$Z$", (16.83340500000002,-1.7993074), NE * labelscalefactor); dot((10.48368025146099,-12.960716866859293),linewidth(3.pt) + dotstyle); label("$X$", (10.616836400000015,-12.7829856), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Figure 2. $\omega$ intersects line $XY$ but does not intersect line $XZ$ .

Let $P,Q$ be the intersection points of line $XY$ with $\omega$ . (We could possibly have $P = Q$ .)

Let $t_y, t_z$ be the rays from $X$ tangent to $\omega$ such that $t_y$ is outside $\mathcal{R}$ and $t_z$ is inside $\mathcal{R}$ . By the above reasoning, we cannot have $t_z$ coincide with line $BC$ , so we must have that $t_y$ coincides with line $BC$ .

Now one of the tangent lines from $Y$ to $\omega$ must be line $AC$ ; let this tangent line be $u$ . Then the intersection of $u$ and $t_y$ must be point $C$ , and we can determine $\angle BCA$ by looking at the angle between $u$ and $t_y$ .

Sub-case 2.1. $u$ , $t_y$ are tangent to the same arc $\overarc{PQ}$ of $\omega$ .
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.7191574814425151, xmax = 24.134336132381687, ymin = -12.285790672877546, ymax = 0.6819027906836906; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.434862052742311,-1.9326727164537993)--(16.185425538842992,-1.947699012471831)); draw((16.185425538842992,-1.947699012471831)--(10.297130641716253,-12.11647235220784)); draw((10.297130641716253,-12.11647235220784)--(4.434862052742311,-1.9326727164537993)); draw((xmin, -1.5220791046650244*xmin + 3.5565750355544212)--(xmax, -1.5220791046650244*xmax + 3.5565750355544212)); /* line */ draw((10.297130641716253,-12.11647235220784)--(14.602673637027614,-4.389093254195133)); draw((xmin, -4.030405913580408*xmin + 15.941621526832158)--(xmax, -4.030405913580408*xmax + 15.941621526832158)); /* line */ /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (9.979565283395956,-1.6321467960931695), NE * labelscalefactor); dot((4.434862052742311,-1.9326727164537993),dotstyle); label("$Y$", (4.494967236814436,-1.782409756273485), NW * labelscalefactor); dot((16.185425538842992,-1.947699012471831),dotstyle); label("$Z$", (16.245530722915117,-1.7974360522915165), NE * labelscalefactor); dot((10.297130641716253,-12.11647235220784),linewidth(3.pt) + dotstyle); label("$X$", (10.355222683846746,-12.030343640571008), NE * labelscalefactor); label("$t_y$", (2.5265224584523023,-0.09946460225395042), NE * labelscalefactor); label("$u$", (4.089257244327585,-0.11449089827198197), NE * labelscalefactor); dot((4.937572906073237,-3.9588015125398126),linewidth(3.pt) + dotstyle); label("$C$", (4.990835005409479,-3.871064902779872), SW * labelscalefactor); dot((6.89408181391885,-6.204773342346854),linewidth(3.pt) + dotstyle); label("$P$", (6.959279783771613,-6.109983009466574), NE * labelscalefactor); dot((4.574048444430357,-2.1744641498317665),linewidth(3.pt) + dotstyle); label("$Q$", (4.630203900976721,-2.0829356766341163), SE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Note that the horizontal and vertical lines through $C$ are completely contained within angle $BCA$ . This is a contradiction.

Sub-case 2.2. $u$ , $t_y$ are tangent to different arcs $\overarc{PQ}$ of $\omega$ .
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.7191574814425151, xmax = 24.134336132381687, ymin = -12.285790672877546, ymax = 0.6819027906836906; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.434862052742311,-1.9326727164537993)--(16.185425538842992,-1.947699012471831)); draw((16.185425538842992,-1.947699012471831)--(10.297130641716253,-12.11647235220784)); draw((10.297130641716253,-12.11647235220784)--(4.434862052742311,-1.9326727164537993)); draw((xmin, -1.5220791046650244*xmin + 3.5565750355544212)--(xmax, -1.5220791046650244*xmax + 3.5565750355544212)); /* line */ draw((10.297130641716253,-12.11647235220784)--(14.602673637027614,-4.389093254195133)); draw((xmin, 5.269265571976335*xmin-25.30113864743315)--(xmax, 5.269265571976335*xmax-25.30113864743315)); /* line */ /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (9.979565283395956,-1.6321467960931695), NE * labelscalefactor); dot((4.434862052742311,-1.9326727164537993),dotstyle); label("$Y$", (4.494967236814436,-1.782409756273485), NE * labelscalefactor); dot((16.185425538842992,-1.947699012471831),dotstyle); label("$Z$", (16.245530722915117,-1.7974360522915165), NE * labelscalefactor); dot((10.297130641716253,-12.11647235220784),linewidth(3.pt) + dotstyle); label("$X$", (10.355222683846746,-12.030343640571008), NE * labelscalefactor); label("$t_y$", (2.5265224584523023,-0.09946460225395042), NE * labelscalefactor); label("$u$", (4.570098716904595,0.005719469872270494), NE * labelscalefactor); dot((6.89408181391885,-6.204773342346854),linewidth(3.pt) + dotstyle); label("$P$", (6.959279783771613,-6.109983009466574), NE * labelscalefactor); dot((4.574048444430357,-2.1744641498317665),linewidth(3.pt) + dotstyle); label("$Q$", (4.630203900976721,-2.0829356766341163), SE * labelscalefactor); dot((4.2491899700280085,-2.9110282295774117),linewidth(3.pt) + dotstyle); label("$C$", (4.314651684598058,-2.8192241815176624), SW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

We can do a few things to decrease $\angle BCA$ . We can dilate triangle $XYZ$ with respect to $X$ until $Y,Z,I$ are collinear; this will decrease $\angle BCA$ . Afterwards, we can shrink triangle $XYZ$ with respect to $Y$ until $Z$ lies on $\omega$ . Now we wish to show that in the new diagram, $\angle BCA \ge 90^{\circ}$ ; this will give us the desired contradiction.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.7191574814425151, xmax = 24.134336132381687, ymin = -12.285790672877546, ymax = 0.6819027906836906; /* image dimensions */ /* draw figures */ draw(circle((9.92,-1.78), 5.360485052679468)); draw((4.194441316453806,-1.7974360522915165)--(15.283847777761098,-1.782409756273485)); draw((15.283847777761098,-1.782409756273485)--(9.752157701183853,-11.393630612665907)); draw((9.752157701183853,-11.393630612665907)--(4.194441316453806,-1.7974360522915165)); draw((xmin, -1.5467491152852606*xmin + 3.690510683762557)--(xmax, -1.5467491152852606*xmax + 3.690510683762557)); /* line */ draw((xmin, 2.689327019123712*xmin-13.077660414759567)--(xmax, 2.689327019123712*xmax-13.077660414759567)); /* line */ /* dots and labels */ dot((9.92,-1.78),dotstyle); label("$I$", (9.979565283395956,-1.6321467960931695), NE * labelscalefactor); dot((4.194441316453806,-1.7974360522915165),dotstyle); label("$Y$", (4.254546500525932,-1.6471730921112009), NW * labelscalefactor); dot((15.283847777761098,-1.782409756273485),dotstyle); label("$Z$", (15.343952961833224,-1.6321467960931695), NE * labelscalefactor); dot((9.752157701183853,-11.393630612665907),linewidth(3.pt) + dotstyle); label("$X$", (9.814276027197609,-11.309081431705494), NE * labelscalefactor); label("$t_y$", (2.5866276425244283,-0.09946460225395042), NE * labelscalefactor); label("$u$", (4.585125012922626,-0.08443830623591886), NE * labelscalefactor); dot((6.639805421625977,-6.01970795315027),linewidth(3.pt) + dotstyle); label("$P$", (6.703832751465076,-5.929667457250195), NE * labelscalefactor); dot((4.610181019221642,-2.5152702845660646),linewidth(3.pt) + dotstyle); label("$Q$", (4.675282789030816,-2.428540485048842), E * labelscalefactor); dot((3.9584206153229724,-2.4321729009151873),linewidth(3.pt) + dotstyle); label("$C$", (4.014125764237427,-2.3383827089406526), W * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

We will use Cartesian coordinates. Assume that the radius of $\omega$ equals 1. Let $I = (0,0), Z = (1,0), Y = (-x, 0)$ for some $x \ge 1$ . Since $XY$ needs to intersect $\omega$ , we have $1 \le x \le \frac{2}{\sqrt{3}}.$ We have $X = (\frac{-x+1}{2}, \frac{-x-1}{2}\sqrt{3})$ . Let $\theta$ be the argument of $u$ and let $\phi_1$ be the argument of line $XI$ . Let $\phi_2$ be the angle between $t_y$ and ray $XI$ . In particular, $\phi_1 + \phi_2$ is the argument of $t_y$ . It is clear that $\angle BCA = 180^{\circ} - \phi_1 - \phi_2 + \theta$ . We wish to show that $\angle BCA \ge 90^{\circ}$ ; this is equivalent to showing that $\phi_1 + \phi_2 - \theta \le 90^{\circ}$ , or $90^{\circ} + \theta \ge \phi_1 + \phi_2.$
Note that $\tan \theta = \frac{1}{\sqrt{x^2 - 1}}$ , so $\tan (90^{\circ} + \theta) = -\sqrt{x^2 - 1}.$ We have $\tan \phi_1 = \frac{(x+1)\sqrt{3}}{x-1}.$ Letting $d = XI$ , we see that $\tan \phi_2 = \frac{1}{\sqrt{d^2 - 1}}$ , so $\tan \phi_2 = \frac{1}{\sqrt{x^2+x}}.$ This yields $\tan (\phi_1 + \phi_2) = \frac{(x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}}{(x-1)\sqrt{x} - \sqrt{x+1}\sqrt{3}}.$
I claim that the denominator $(x-1)\sqrt{x} - \sqrt{x+1}\sqrt{3}$ is always negative. This is equivalent to showing that $(x-1)^2x < 3(x+1)$ holds for all $x$ in our interval. Expanding, we get $f(x) = x^3 - 2x^2 - 2x - 3 < 0.$ To prove this, we note that $f(1) = -6 < 0$ . Furthermore, $f'(x) = 3x^2 - 4x - 2 = 3(x-r_1)(x-r_2)$ where $r_1 = \frac{2-\sqrt{10}}{3}, r_2 = \frac{2+\sqrt{10}}{3}.$ Note that $r_1 < 0$ and that $r_2 > \frac{5}{3} > \frac{2}{\sqrt{3}}$ , so $f'(x) < 0$ for $x$ in our interval. This means that $f(x) < f(1) < 0$ for $x$ in our interval.

Since $90^{\circ} + \theta, \phi_1 + \phi_2$ are both in the interval $(90^{\circ}, 180^{\circ}]$ , $90^{\circ} + \theta \ge \phi_1 + \phi_2$ is equivalent to $\tan (90^{\circ} + \theta) \ge \tan (\phi_1 + \phi_2).$ This in turn is equivalent to $-\sqrt{x^2 - 1} \ge \frac{(x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}}{(x-1)\sqrt{x} - \sqrt{x+1}\sqrt{3}}.$ This is equivalent to $\sqrt{x^2 - 1} \le \frac{(x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}}{-(x-1)\sqrt{x} + \sqrt{x+1}\sqrt{3}}$ and after clearing denominators, we get $(x+1)\sqrt{3}\sqrt{x-1} - (x-1)\sqrt{x}\sqrt{x^2-1} \le (x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}.$ This is true because $(x+1)\sqrt{3}\sqrt{x-1} - (x-1)\sqrt{x}\sqrt{x^2-1} \le (x+1)\sqrt{3}\sqrt{x-1} < (x+1)\sqrt{3}\sqrt{x} \le (x+1)\sqrt{3}\sqrt{x} + \frac{x-1}{\sqrt{x+1}}$ . We have our desired contradiction.

Case 3. $\omega$ intersects both lines $XY, XZ$ .

One of the tangents from $X$ to $\omega$ must be line $BC$ ; proceed as in the previous case. $\blacksquare$

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#8 h Dec 18, 2017, 8:38 PM • 2 Y

Y by Adventure10, Mango247

wrong solution... ignore

This post has been edited 7 times. Last edited by Ghoshadi, Dec 20, 2017, 5:21 PM

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#9 h Nov 4, 2018, 5:25 AM • 4 Y

Y by Amir Hossein, BobaFett101, Adventure10, Mango247

Here's a very clean and mostly synthetic solution

By Miquel's Theorem, $(AYZ), (BZX), (CXY)$ meet at some point $P$ . Clearly $\angle ABP + \angle ACP = \angle ZXP+\angle YXP = 60^{\circ}$ and similarly for the other vertices. This implies two things: Firstly, $\angle BPC = 180^{\circ} - \angle PBC-\angle PCB = 180^{\circ} - (180^{\circ} - \angle A -60^{\circ}) = 60^{\circ} + \angle A$ and similarly for $\angle CPA, \angle APB$ , meaning that $P$ is fixed. Secondly, if $Q$ is the image of $P$ under $\sqrt{bc}$ -inversion, then $Q$ lies on the opposite side of $BC$ as $A$ with $\triangle QBC$ equilateral.

We'll show $I$ is outside $\triangle AYZ$ ; similar arguments combined with the fact that $I$ lies inside $\triangle ABC$ will that $I$ is inside $\triangle XYZ$ . By performing a $\sqrt{bc}$ inversion, $Y$ and $Z$ map to points $E,F$ on rays $AC,AB$ extended past $C,B$ with $E,Q,F$ collinear; instead of showing $I$ is outside $\triangle AYZ$ , we're then left with showing the $A$ -excenter $I_A$ is inside $(AEF)$ (because $I,I_A$ are inverses in $\sqrt{bc}$ -inversion).

Consider any point $J$ on the internal angle bisector of $J$ and any points $U,V$ on $AB,AC$ with $AUJV$ cyclic. For any fixed $J$ , I claim $AU+AV$ is always fixed; indeed, let $J_1,J_2$ be the projections of $J$ onto $AB,AC$ . It then follows that $\triangle JJ_1U\cong \triangle JJ_2V$ , so $J_1U=J_2V\implies AU+AV=AJ_1+AJ_2$ . Furthermore, as $J$ moves towards $A$ along the angle bisector, this value of $AJ_1+AJ_2$ obviously decreases. When $J=I_A$ , we have $AJ_1+AJ_2 = a+b+c$ , so to prove that $I_A$ lies inside $(AEF)$ , it's enough to show $AE+AF > a+b+c$ or $CE+BF>a$ .

Now we're almost done. Let $\angle E=y, \angle F=z, \angle CQE=w, \angle BQF=x$ . Since $E,F$ are not within segments $AC,AB$ , we must have $0<w,x<120^{\circ}$ . Then clearly $w+x=120^{\circ}$ . Meanwhile, we have $w+y=\angle ACQ = \angle C+ 60^{\circ}$ , so since $ABC$ is acute we have $60^{\circ} < w+y < 150^{\circ}$ and similarly for $x+z$ . By the Law of Sines in $\triangle CEQ, \triangle BQF$ , we get $CE= \frac{\sin w}{\sin y}CQ,BF=\frac{\sin x}{\sin z}BQ$ . Conveniently we have $CQ=BQ=a$ , so we just need to show $\frac{\sin w}{\sin y}+\frac{\sin x}{\sin z}>1$ .

Case 1: $15^{\circ} \le w,x\le 105^{\circ}$ . Then $\frac{\sin w}{\sin y}+\frac{\sin x}{\sin z} \ge \sin w + \sin x$ . Since $\sin$ is concave on $[0,\pi]$ and $w+x$ is fixed, we have by Karamata that $\sin w +\sin x \ge \sin 15^{\circ} + \sin 105^{\circ} =\frac{\sqrt{6}}{2}>1$ , so we're done.

Case 2: Suppose case 1 does not hold. WLOG $w<15^{\circ}, x>105^{\circ}$ . Then $x+z<150^{\circ}$ yields $z<45^{\circ}$ . Since $x<120^{\circ}$ , we have $\frac{\sin w}{\sin y}+ \frac{\sin x}{\sin z} > \frac{\sin x}{\sin z}\ge \frac{\sin 120^{\circ}}{\sin 45^{\circ}}>1$ , so once again we're done.

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MathStudent2002

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MathStudent2002

#10 h Feb 3, 2019, 6:16 AM • 1 Y

Y by Adventure10

Proceed as above until we want to show that $CE+BF > a$ . We will prove the more general result that if $BCQ$ is equilateral, and $\ell$ passes through $Q$ but not the interior of the triangle, and $E,F$ lie on $\ell$ so that $E,C$ and $B,F$ are on the same side of $BJ$ , then $CE+BF > a$ . Suppose $\angle BQC$ is $60$ measured counterclockwise (henceforth CCW). If the angle $\angle(CQ,\ell)$ measured CCW is greater than $90$ then $CE>a$ and we are done. Similarly if $\angle(\ell,QB)$ measured CCW is greater than $90$ then we are done.

Now we may assume that $E,F$ are the foot from $C,B$ to $\ell$ . Letting $\alpha = \angle(CQ,\ell)$ , and $\beta = \angle(\ell,QB)$ we want $\sin\alpha + \sin \beta > 1$ for acute $\alpha, \beta$ summing to $120$ . But this is clear from $\sin\alpha+\sin\beta = 2\sin((\alpha+\beta)/2)\cos((\alpha-\beta)/2) = \sqrt 3 \cos((\alpha-\beta)/2) > 1.5 > 1$ since $|\alpha-\beta| < 30$ , where all angle measures are in degrees.

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#11 h Nov 23, 2020, 11:49 AM

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Lemma 1. If $0\leq a\leq b\leq 180^{\circ}$ , let $I=[a,b]$ then
$\underset{\theta\in I}{\min}\sin\theta=\min\{\sin a,\sin b\}$ Proof. Obvious. $\blacksquare$
Lemma 2. If $\triangle ZYX$ is an equilateral triangle and $A$ is a point lie inside $\angle ZXY$ but outside $\triangle ZYX$ , with $\angle AZY>30^{\circ}$ . Let $B_1$ be the projection of $X$ on $AZ$ . Let $A_1$ be the intersection of the angle bisector of $\angle ZAY$ with $ZY$ . Then
$ZB_1A'<\frac{1}{2}\angle ZB_1X=45^{\circ}$ Proof.
Let $\omega$ be the circle with diameter $ZX$ . Let $M,N$ be the intersection of the perpendicular bisector of $ZX$ and $\omega$ , such that $M$ lies between $M$ and $N$ .
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CASE I: $\angle AZY<75^{\circ}$
Then $B_1$ lies in arc $ZN$ of $\omega$ . We have
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CASE II: $\angle AZY\geq 75^{\circ}$ .
Let $D$ be the projection of $M$ onto $ZY$ . Then we have $\frac{ZA'}{ZY}=\frac{\sin\angle AYZ}{\sin\angle AYZ+\sin\angle AZY}\leq \frac{1}{1+\sin 75^{\circ}}<\frac{\sin75^{\circ}}{\sqrt{2}}=\frac{ZD}{ZY}$ Therefore $ZA'<ZD$ . From this, combine with the fact that $B_1$ lies in arc $NX$ of $\omega$ , we have
$\angle ZB_1A'<\angle ZB_1D<\angle ZB_1D=45^{\circ}$ $\blacksquare$
We will now show the equivalent problem.
CLAIM. Given a fixed point $A$ and an equilateral triangle $XYZ$ such that $A$ lies inside $\angle ZXY$ and that $A,Z$ lie on different side of $YZ$ . Let $B$ be a point on $AZ$ beyond $Z$ and that $BX$ intersect $AY$ at $C$ , where $Y$ lies between $A,C$ . Furthermore, $\triangle ABC$ is acute. Then the incenter of $\triangle ABC$ doesn't lie in $\triangle AZY$ .
Proof.
Let $A'$ be the intersection of the angle bisector of $\angle ZAY$ and $ZY$ . Suppose on the contrary that the incenter $I$ of $\triangle ABC$ lies inside $AZY$ , then it must lie in the segment $AA'$ . Therefore,
$\angle ABA'\geq \angle ABI=\frac{1}{2}\angle ABX$ Hence $\angle ABA'\geq \angle A'BX$ . Since $\angle ABA'$ and $\angle ABX$ are both acute we have
$\frac{\sin\angle ABA'}{\sin\angle A'BX}\geq 1 \quad(1)$ Now for any point $B$ on $AZ$ such that $ABC$ is well-defined and acute define the function
$f(B)=\frac{\sin\angle ABA'}{\sin\angle A'BX}$ We will show that this function is strictly less than $1$ , hence obtaining a contradiction with $(1)$ .
Now, applying Ceva's theorem we have
$f(B)=\frac{AA'}{\sin\angle BXA'}\frac{\sin\angle BAA'}{A'X}=\left(\frac{AA'}{\sin\angle ZAA'}\cdot{A'X}\right)\frac{1}{\sin\angle BXA'}$ Magically, the term in the bracket does not depend on $B$ , so it is a constant. We now further distinguish two cases. Let $\angle BXA'=\theta$
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CASE I: One of $\angle AZY,\angle AYZ$ is less than or equal to $30^{\circ}$ .
WLOG assume $\angle AZY<30^{\circ}$ , then $\angle AYZ>60^{\circ}$ since $A$ is acute. Let the projection from $X$ to $AY$ meet $AZ$ , $AY$ at $B_1,C_1$ respectively. Now we have $\theta\in[\angle BZX,\angle BB_1X]$ . Therefore, from Lemma 1 we have
$f(B)\leq \max\{f(Z),f(B_1)\}$ However, we have $f(Z)=\frac{\sin\angle AZY}{\sin 60^{\circ}}<1$ . Meanwhile, from Lemma 2, we have $\frac{\sin\angle ACA'}{\sin\angle B_1CA'}<1$ , hence by Ceva's theorem
$f(B_1)=\frac{\sin\angle AB_1A'}{\sin\angle C_1B_1A'}<1$ This shows that $f$ is strictly less than $1$ .
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CASE II: Both $\angle AZY,\angle AYZ$ is at least $30^{\circ}$ .
Suppose the perpendicular line from $X$ to $AZ$ intersect $AZ,AY$ at $B_1,C_1$ , and the perpendicular line from $X$ to $AY$ intersect $AZ,AY$ at $B_2,C_2$ . Notice that $B_1,B_2,C_1,C_2$ lies in the extension of either $AZ$ or $AY$ . Now, similar to the above case we have
$f(B)\leq \max\{f(B_1),f(B_2)\}$ Moreover, $f(B_1)<1, f(C_2)<1$ by Lemma 2. Therefore, by Ceva's theorem we have
$f(B_2)=\frac{\sin\angle AB_2A'}{\sin\angle C_2B_2A'}<1$ as well, this implies $f$ is strictly less than 1. $\blacksquare$

This contradiction completes the proof.

Motivational Remarks

This post has been edited 1 time. Last edited by mathaddiction, Nov 23, 2020, 11:54 AM

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#12 h Dec 25, 2024, 4:38 AM

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Amir Hossein wrote:

The vertices $X, Y , Z$ of an equilateral triangle $XYZ$ lie respectively on the sides $BC, CA, AB$ of an acute-angled triangle $ABC.$ Prove that the incenter of triangle $ABC$ lies inside triangle $XYZ.$

Proposed by Nikolay Beluhov, Bulgaria

Prove by contradiction: notice incenter is either in triangle AZY or BZX or CYX without using generality of problem we could assume I is in the triangle AZY
Now we choose BZX our angel of freedom now define T as intersection of angle bisector of ABC and segment ZY now we could easily calculate ZT/ZX now we define Q as intersection of angle bisector of ACB and segment ZY now we could easily calculate QY/ZX therfore we could calculate QY+ZT/ZX and by bit of calculation we know QY+ZT/ZX>1 therfore I is in the Quadrilateral BZYC with is contradiction

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