P100.1.0.2011.7.9

Let $ABC$ be an equilateral triangle. A point $T$ is chosen on $AC$ and on arcs $AB$ and $BC$ of the circumcircle of $ABC$, $M$ and $N$ are chosen respectively, so that $MT$ is parallel to $BC$ and $NT$ is parallel to $AB$. Segments $AN$ and $MT$ intersect at point $X$, while $CM$ and $NT$ intersect in point $Y$. Prove that the perimeters of the polygons $AXYC$ and $XMBNY$ are the same.
Solution
 
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(16cm); 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 = -17.14111604398607, xmax = 3.895698381258108, ymin = -6.5237873096099985, ymax = 6.002679461785393; /* image dimensions */pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-8.972436308995295,-3.521212348883271)--(0.48756369100471453,-3.481212348883271)--(-4.277077325146665,4.691387970917528)--cycle, linewidth(2) + rvwvcq); /* draw figures */draw((-8.972436308995295,-3.521212348883271)--(0.48756369100471453,-3.481212348883271), linewidth(2) + rvwvcq); draw((0.48756369100471453,-3.481212348883271)--(-4.277077325146665,4.691387970917528), linewidth(2) + rvwvcq); draw((-4.277077325146665,4.691387970917528)--(-8.972436308995295,-3.521212348883271), linewidth(2) + rvwvcq); draw(circle((-4.2539833143790835,-0.7703455756163375), 5.461782370862713), linewidth(2) + wrwrwr); draw((xmin, -1.715260455530013*xmin-14.031567575887422)--(xmax, -1.715260455530013*xmax-14.031567575887422), linewidth(2) + wrwrwr); /* line */draw((xmin, 0.004228329809725158*xmin-1.0255502250204778)--(xmax, 0.004228329809725158*xmax-1.0255502250204778), linewidth(2) + wrwrwr); /* line */draw((-8.972436308995295,-3.521212348883271)--(1.2020688938546569,-1.0204674812832488), linewidth(2) + wrwrwr); draw((-4.277077325146665,4.691387970917528)--(-4.551825721347898,-6.224000915594996), linewidth(2) + wrwrwr); draw((-6.48411613999653,-2.9096195718874647)--(-4.421447455341107,-1.04424556309803), linewidth(2) + wrwrwr); draw((-4.551825721347898,-6.224000915594996)--(0.48756369100471453,-3.481212348883271), linewidth(2) + wrwrwr); draw((0.48756369100471453,-3.481212348883271)--(1.2020688938546569,-1.0204674812832488), linewidth(2) + wrwrwr); draw(circle((-8.995530319762882,1.940521197650598), 5.461782370862718), linewidth(2) + wrwrwr); draw(circle((-3.1863748570961805,0.25376697258226966), 4.569694956851869), linewidth(2) + wrwrwr); /* dots and labels */dot((-8.972436308995295,-3.521212348883271),dotstyle); label("$A$", (-8.917634041390617,-3.381925414930675), NE * labelscalefactor); dot((0.48756369100471453,-3.481212348883271),dotstyle); label("$B$", (0.5489324499692619,-3.3409446076087708), NE * labelscalefactor); dot((-4.277077325146665,4.691387970917528),dotstyle); label("$C$", (-4.218501468478931,4.827896318557471), NE * labelscalefactor); dot((-7.56388611647572,-1.0575328301641385),dotstyle); label("$T$", (-7.510626323338572,-0.9230769756164215), NE * labelscalefactor); dot((-4.551825721347898,-6.224000915594996),linewidth(4pt) + dotstyle); label("$M$", (-4.49170685062496,-6.113979236390957), NE * labelscalefactor); dot((1.2020688938546569,-1.0204674812832488),linewidth(4pt) + dotstyle); label("$N$", (1.2592664435489354,-0.9094167065091201), NE * labelscalefactor); dot((-6.48411613999653,-2.9096195718874647),linewidth(4pt) + dotstyle); label("$X$", (-6.4314650638617605,-2.7945338433167146), NE * labelscalefactor); dot((-4.421447455341107,-1.04424556309803),linewidth(4pt) + dotstyle); label("$Y$", (-4.368764428659247,-0.9367372447237229), NE * labelscalefactor); dot((-6.13455216274132,-3.5092128387511186),linewidth(4pt) + dotstyle); label("$D$", (-6.0762980670719235,-3.3955856840379766), NE * labelscalefactor); dot((-0.941770262729685,-1.0295323402962904),linewidth(4pt) + dotstyle); label("$E$", (-0.8853958062973866,-0.9230769756164215), NE * labelscalefactor); dot((-9.707725273060063,-1.06659768917718),linewidth(4pt) + dotstyle); label("$X'$", (-9.655288573184894,-0.9640577829383258), NE * labelscalefactor); dot((-9.146612557869142,1.657255246679742),linewidth(4pt) + dotstyle); label("$Y'$", (-9.095217539785535,1.767996038521956), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
We have $$60^{\circ}=\angle CBA=\angle CNA=\angle CNX=\angle ATX$$
so $CNXT$ is cyclic. Similarly, $AMYT$ is cyclic.

Now, we have $$\angle AXC=\angle AXT+\angle CXT=\angle TCN+\angle TNC=120^{\circ}$$Similarly, $\angle AYC=120^{\circ}$

Reflect $X$ and $Y$ over AC to get $X'$ and $Y'$, respectively. Note that $$\angle AX'C=\angle AY'C=120^{\circ}=180^\circ-\angle ABC$$so $X'$ and $Y'$ lie on the circumcircle of $ABC$.

Now, since $AB$ and $X'N$ are parallel and all points lie on the circumcircle, we can say $\triangle X'TA\cong \triangle NEB$

This means $$BN=AX'=AX \quad\text{and}\quad BM=CY'=CY$$
Define $D=AB\cup TM$ and $E=BC\cup TN$. Now, $$XM+YN=(XD+DM)+(YE+EN)=(XD+TY')+(YE+TX')=TX+XD+TY+YE=TD+TE=AC$$
Hence, we have $AC+CY+AX+XY=XM+YN+BM+BN+XY$, so the perimeters of the two figures are equal.