Problem:
In triangle ABC,AB=10,BC=14, and CA=16. Let D be a point in the interior of BC. Let IB and IC denote the incenters of triangles ABD and ACD, respectively. The circumcircles of triangles BIBD and CICD meet at distinct points P and D. The maximum possible area of △BPC can be expressed in the form a−bc, where a,b and c are positive integers and c is not divisible by the square of any prime. Find a+b+c.
Solution:
Note that
∠BIBD=∠IBBA+∠BAD+∠ADIB=∠BAD+2∠DBA+2∠ADB
and
∠CICD=∠ICDA+∠DAC+∠ACIC=∠DAC+2∠CDA+2∠ACD
Because ∠BAD+∠DAC=∠BAC and ∠ADB+∠CDA=180∘, it follows that
∠BIBD+∠CICD=∠BAC+2∠ACD+2∠DBA+90∘=180∘+2∠BAC.
The points P and IB must lie on opposite sides of BC, and BIBDP and CICDP are convex cyclic quadrilaterals. If P and IB were on the same side, then both BIBPD and CICPD would be convex. It would then follow by (1) and the fact that quadrilaterals BIBPD and CICPD are cyclic that
Therefore, ∠BPC is constant, and so P lies on the arc of a circle passing through B and C.
The Law of Cosines yields cos∠BAC=2⋅10⋅16102+162−142=21, and so ∠BAC=60∘. Hence ∠BPC=150∘, and the minor arc subtended by the chord BC measures 60∘. Thus the radius of the circle is equal to BC=14. The maximum area of △BPC occurs when BP=PC. Applying the Law of Cosines to △BPC with BP=PC=x yields 142=2x2+x23, so x2=2+3196=196(2−3). The area of this triangle is 21x2sin150∘=98−493, and so a+b+c=150.
