MathDB
I got the horses in the BAC

Source: 2021 AIME II #14

March 19, 2021
AIMEAIME II

Problem Statement

Let ABC\triangle ABC be an acute triangle with circumcenter OO and centroid GG. Let XX be the intersection of the line tangent to the circumcircle of ABC\triangle ABC at AA and the line perpendicular to GOGO at GG. Let YY be the intersection of lines XGXG and BCBC. Given that the measures of ABC,BCA,\angle ABC, \angle BCA, and XOY\angle XOY are in the ratio 13:2:17,13 : 2 : 17, the degree measure of BAC\angle BAC can be written as mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+nm+n. [asy] unitsize(5mm); pair A,B,C,X,G,O,Y; A = (2,8); B = (0,0); C = (15,0); dot(A,5+black); dot(B,5+black); dot(C,5+black); draw(A--B--C--A,linewidth(1.3)); draw(circumcircle(A,B,C)); O = circumcenter(A,B,C); G = (A+B+C)/3; dot(O,5+black); dot(G,5+black); pair D = bisectorpoint(O,2*A-O); pair E = bisectorpoint(O,2*G-O); draw(A+(A-D)*6--intersectionpoint(G--G+(E-G)*15,A+(A-D)--A+(D-A)*10)); draw(intersectionpoint(G--G+(G-E)*10,B--C)--intersectionpoint(G--G+(E-G)*15,A+(A-D)--A+(D-A)*10)); X = intersectionpoint(G--G+(E-G)*15,A+(A-D)--A+(D-A)*10); Y = intersectionpoint(G--G+(G-E)*10,B--C); dot(Y,5+black); dot(X,5+black); label("AA",A,NW); label("BB",B,SW); label("CC",C,SE); label("OO",O,ESE); label("GG",G,W); label("XX",X,dir(0)); label("YY",Y,NW); draw(O--G--O--X--O--Y); markscalefactor = 0.07; draw(rightanglemark(X,G,O)); [/asy]