In triangle △ABC, the reflection of its incenter over side AB is its circumcenter.
Find the measure of ∠C in degrees.
Let point I be the incenter and point O be the circumcenter.
I is always inside the triangle, so for O to be the reflection of I, then O must be outside the triangle which means that triangle ABC is an obtuse triangle with angle C being the obtuse angle.
Draw the perpendicular bisector of side AB and call the intersection M. Trivially, angles AMI and BMI are congruent and AM=BM.
O must lie on the bisector line; because I and O are reflections then I is also on this line.
Draw AI and BI. Then with angle AMI=BMI, AM=BM, and MI=MI triangles AMI and BMI are congruent. This means that angles IAM and IBM are congruent.
Lines AI and BI are the angle bisectors of angles CAB and CBA, respectively due to I being the incenter of ABC. Angles AMI and BMI being congruent then implies angles CAB and CBA are congruent.
Thus, ABC is an isosceles triangle with vertex angle C. The perpendicular bisector of AB must then be the altitude of ABC from vertex C and then the entire figure is symmetric over the bisector line (which contains points C, I, M, and O).
Let the inradius have unit length and the circumradius be x. Then CM=1, OA=OB=OC=x, MC=OC-CM=x-1.
Triangle AMO is a right triangle, then by the Pythagorean theorem AM=sqrt[x^2-1], similarly BM=sqrt[x^2-1].
The formula for the distance between the incenter and circumcenter is d = sqrt[R*(R-2r)] where R is the circumradius and r is the inradius. In this case d=2, R=x, and r=1. Then 2 = sqrt[x*(x-2)]. This equation has one positive root at x=1+sqrt[5].
Then CM = sqrt[5] and AM=sqrt[5+2*sqrt[5]]. Angle ACB = 2* angle ACM = 2* arctan(AM/CM) = 2*arctan(sqrt[sqrt[5]+2]). Then by calculator ACB = 2*arctan(2.0582) = 128.173 degrees.
Solution
