Problem: In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$ . Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$ $\angle{BAD}=\angle{CAD}$ , and $BM=CM$ . Point $N$ is the midpoint of the segment $HM$ , and point $P$ is on ray $AD$ such that $PN\perp{BC}$ . Then $AP^2=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: [asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm);  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 = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471;  /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882);   draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq);   /* draw figures */ draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr);  draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */ draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */ draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */ draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr);  draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */ draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr);  draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr);  draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr);  draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */ draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr);  draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr);  draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq);  draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq);  draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq);  draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq);  draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq);   /* dots and labels */ dot((-1.4934334172297545,2.6953043701763835),dotstyle);  label("$A$", (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor);  dot((1.1286284157632023,-6.954814372303504),dotstyle);  label("$B$", (1.2093379191072373,-6.719031552166216), NE * labelscalefactor);  dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle);  label("$C$", (8.292420110475998,-6.741880204396438), NE * labelscalefactor);  dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle);  label("$H$", (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor);  dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle);  label("$E$", (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor);  dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle);  label("$L$", (4.705181710331174,6.441792132441429), NE * labelscalefactor);  dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle);  label("$O$", (4.728030362561396,-0.6412900589272691), NE * labelscalefactor);  dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle);  label("$D$", (4.2025113612662945,-6.764728856626659), NE * labelscalefactor);  dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle);  label("$F$", (4.750879014791618,-7.701523598065745), NE * labelscalefactor);  dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle);  label("$G$", (4.750879014791618,-3.2231877609423107), NE * labelscalefactor);  dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle);  label("$K$", (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor);  dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle);  label("$P$", (1.6891596159418953,-2.4234849328845542), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy] Draw the $45-45-90 \triangle AHC$ . Now, take the perpendicular bisector of $BC$ to intersect the circumcircle of $\triangle ABC$ and $AC$ at $F, L, G$ as shown, and denote $O$ to be the circumcenter of $\triangle ABC$ . It is not difficult to see by angle chasing that $AHBGO$ is cyclic, namely with diameter $AB$ . Then, by symmetry, $EH = HB$ and as $HB, OG$ are both subtended by equal arcs they are equal. Hence, $EH = GO$ . Now, draw line $HL$ and intersect it at $AC$ at point $K$ in the diagram. It is not hard to use angle chase to arrive at $AEOL$ a parallelogram, and from our length condition derived earlier, $AL \parallel HG$ . From here, it is clear that $AK = KG$ ; that is, $P$ is just the intersection of the perpendicular from $K$ down to $BC$ and $AD$ ! After this point, note that $AP = PF$ . It is easily derived that the circumradius of $\triangle ABC$ is $\frac{10}{\sqrt{2}}$ . Now, $APO$ is a $30-60-90$ triangle, and from here it is easy to arrive at the final answer of $\boxed{077}$ . ~awang11's sol

Answer:  077