Problem:
Let β³ABC have incenter I, circumcenter O, inradius 6, and circumradius 13. Suppose that IAβ₯OI. Find ABβ AC.
Solution:
Extend AI to meet the circumcircle at M.
Because AIβ₯OI, it follows that I is the midpoint of AM. Note that BM=CM because line AM is a bisector of angle β A. In addition, β BIM=β IBM=2β A+β Bβ, and thus BM=CM=IM=AI. Applying Ptolemy's Theorem to cyclic quadrilateral ABMC and using BM=CM=21ββ AM gives AMβ BC=ABβ CM+ACβ BM=21ββ AMβ (AB+AC), so BC=2AB+ACβ.
Expressing the area of triangle ΞABC in two different ways gives
Area(ΞABC)=21β(AB+AC+BC)r=23ββ BCβ r, and
Area(ΞABC)=4RABβ ACβ BCβ.
It follows that ABβ AC=6β Rβ r=6β 6β 13=468β.
OR
Because the inradius is r=6 and the circumradius is R=AO=13, it follows from Euler's Formula for the distance from the incenter to the circumcenter that OI2=R2β2β rβ R=132β2β 6β 13=13. Then applying the Pythagorean Theorem to ΞAIO gives AI=AO2βOI2β=132β13β=156β=239β.
Let F be the projection of I onto AB and note that IF=r=6, so applying the Pythagorean Theorem to ΞAIF gives AF=AI2βIF2β=156β62β=120β=230β. Then
so by the Extended Law of Sines, it follows that BC=2Rsin(β BAC)=430β.
Using properties of the incircle gives AF=2AB+ACβBCβ, so AB+AC=2AF+BC=830β, and the semiperimeter of ΞABC is 630β. The area of ΞABC is its semiperimeter times its inradius. Thus
