Problem: Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$ $B(0,12)$ $C(16,0)$ $A'(24,18)$ $B'(36,18)$ $C'(24,2)$ . A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$ , will transform $\triangle ABC$ to $\triangle A'B'C'$ . Find $m+x+y$

Solution: [asy]  /* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ 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 = -8.0451801958033, xmax = 47.246151591238494, ymin = -10.271454747548662, ymax = 21.426040258770957;  /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqffff = rgb(0,1,1);   draw((16,0)--(0,0)--(0,12)--cycle, linewidth(2) + zzttqq);  draw((24,2)--(24,18)--(36,18)--cycle, linewidth(2) + blue);  draw((16,0)--(21,-3)--(24,2)--cycle, linewidth(2) + qqwuqq);  draw((21.39134584768662,-2.3477569205223032)--(20.73910276820892,-1.9564110728356852)--(20.347756920522304,-2.608654152313382)--(21,-3)--cycle, linewidth(2) + qqffff);   /* draw figures */ draw((16,0)--(0,0), linewidth(2) + zzttqq);  draw((0,0)--(0,12), linewidth(2) + zzttqq);  draw((0,12)--(16,0), linewidth(2) + zzttqq);  draw((24,2)--(24,18), linewidth(2) + blue);  draw((24,18)--(36,18), linewidth(2) + blue);  draw((36,18)--(24,2), linewidth(2) + blue);  draw((16,0)--(24,2), linewidth(2));  draw((16,0)--(21,-3), linewidth(2) + qqwuqq);  draw((21,-3)--(24,2), linewidth(2) + qqwuqq);  draw((24,2)--(16,0), linewidth(2) + qqwuqq);  draw((21,-3)--(20,1), linewidth(2.8) + qqffff);   /* dots and labels */ dot((0,0),linewidth(4pt) + dotstyle);  label("$A$", (-0.6228029714727868,0.12704474547474198), NE * labelscalefactor);  dot((0,12),dotstyle);  label("$B$", (0.1301918194013232,12.354245873478124), NE * labelscalefactor);  dot((16,0),dotstyle);  label("$C$", (16.15822379657881,0.34218611429591583), NE * labelscalefactor);  dot((24,18),dotstyle);  label("$A'$", (24.154311337765787,18.342347305667463), NE * labelscalefactor);  dot((24,2),dotstyle);  label("$C'$", (23.186175178070503,1.95574638045472), NE * labelscalefactor);  dot((36,18),dotstyle);  label("$B'$", (36.13051420214449,18.342347305667463), NE * labelscalefactor);  dot((21,-3),dotstyle);  label("$P$", (21.35747354309052,-3.458644734878156), NE * labelscalefactor);  dot((20,1),linewidth(4pt) + dotstyle);  label("$D$", (20.13833911977053,1.2744653791876692), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]
We first draw a diagram with the correct Cartesian coordinates and a center of rotation $P$ . Note that $PC=PC'$ because $P$ lies on the perpendicular bisector of $CC'$ (it must be equidistant from $C$ and $C'$ by properties of a rotation).
Since $AB$ is vertical while $A'B'$ is horizontal, we have that the angle of rotation must be $90^{\circ}$ , and therefore $\angle P = 90^{\circ}$ . Therefore, $CPC'$ is a 45-45-90 right triangle, and $CD=DP$
We calculate $D$ to be $(20,1)$ . Since we translate $4$ right and $1$ up to get from point $C$ to point $D$ , we must translate $1$ right and $4$ down to get to point $P$ . This gives us $P(21,-3)$ . Our answer is then $90+21-3=\boxed{108}$ . ~Lopkiloinm & samrocksnature

Answer:  108