Problem:
Let β³ABC be an acute triangle with circumcircle Ο, and let H be the intersection of the altitudes of β³ABC. Suppose the tangent to the circumcircle of β³HBC at H intersects Ο at points X and Y with HA=3,HX=2, and HY=6. The area of β³ABC can be written as mnβ, where m and n are positive integers, and n is not divisible by the square of any prime. Find m+n.
Solution:
Extend AH to hit Ο at P.
Notice that HPβ HA=HXβ HYβHP=4.
Now notice how β CBH=β CAP=β CBP=90ββ C.
This yields β CBH=β CBP.
This yields that BC bisects HP.
This is useful as the length of the altitude from A to BC is AH+2HPβ=3+24β=5.
Notice that if two triangles are congruent, they must have the same circumradii, and we notice that similarly to the way β CBH=β CBP that β BCH=β BCP.
This combined with HP being bisected by BC shows that β³BCH is congruent to β³BCP meaning that their circumcircles have the same radii.
Notice that HM=XOβXH=2XYββXH=2. Similarly notice how the two circles are congruent, and "located" around the segment AH. This corresponding segment in the new circle is the line between the two circles circumcenters and that the distance between these two circumcwenters is the same as AH which is 3.
This is because the centers both lie through the perpendicular bisector of BC and the fact that they are "shifted" buy the same amount as AH.
Now, let's label some more sidelengths.
Let M denote the midpoint of XY and O the circumcenter of the triangle ABC.
OLD SOLUTION
Extend AH to intersect Ο again at P. The Power of a Point Theorem yields HP=HAHXβ HYβ=4. Because β CAP=β CBP, and β CAP and β CBH are both complements to β C, it follows that β CBP=β CBH, implying that BC bisects HP, so the length of the altitude from A to BC is haβ=AH+21βHP=5.
Let the circumcircle of β³BCH be Οβ². Because β³BCHβ β³BCP, the two triangles must have the same circumradius. Because the circumcircle of β³BCP is Ο, the circles Ο and Οβ² have the same radius R. Denote the centers of Ο and Οβ² by O and Oβ², respectively, and let M be the midpoint of XY. Note that trapezoid HMOOβ² has β H=β M=90β. Also HM=XMβXH=21ββ XYβHX=2 and HOβ²=R. Because Ο is a translation of Οβ² in the direction of AH, it follows that OOβ²=AH=3. Finally, the Pythagorean Theorem applied to β³XMO yields MO=R2β16β. Let T be the projection of O onto HOβ². Then TOβ²=RβMO, so the Pythagorean Theorem applied to β³TOOβ² yields
RβR2β16β=32β22β=5β
Solving for R gives R=25β21β. It follows from properties of the orthocenter H that
