Problem:
Let △ABC be an acute triangle with circumcenter O and centroid G. Let X be the intersection of the line tangent to the circumcircle of △ABC at A and the line perpendicular to GO at G. Let Y be the intersection of lines XG and BC. Given that the measures of ∠ABC,∠BCA, and ∠XOY are in the ratio 13:2:17, the degree measure of ∠BAC can be written as nm, where m and n are relatively prime positive integers. Find m+n.
Solution:
In this solution, all angle measures are in degrees. Let M be the midpoint of BC so that OM⊥BC and A,G,M are collinear. Let ∠ABC=13k,∠BCA=2k and ∠XOY=17k.
Note that:
Since ∠OGX=∠OAX=90, quadrilateral OGAX is cyclic by the Converse of the Inscribed Angle Theorem. It follows that ∠OAG=∠OXG, as they share the same intercepted arc OG.
Since ∠OGY=∠OMY=90, quadrilateral OGYM is cyclic by the supplementary opposite angles. It follows that ∠OMG=∠OYG, as they share the same intercepted arc OG.
Together, we conclude that △OAM∼△OXY by AA, so ∠AOM=∠XOY=17k.
Next, we express ∠BAC in terms of k. By angle addition, we have
∠AOM=∠AOB+∠BOM=2∠BCA+21∠BOC by Inscribed Angle Theorem and Perpendicular Bisector Property =2∠BCA+∠BAC. by Inscribed Angle Theorem
Substituting back gives 17k=2(2k)+∠BAC, from which ∠BAC=13k.
For the sum of the interior angles of △ABC, we get
∠ABC+∠BCA+∠BAC=18013k+2k+13k=18028k=180k=745
Finally, we obtain ∠BAC=13k=7585, from which the answer is 585+7=592.
