Talk:2009 AIME I Problems/Problem 15

Sorry, I don't know LaTeX but I do think I still have this problem's solution, which hasn't yet been posted. Maybe someone can validate / compare, and format things nicely afterward.

Lemma: Angle BPC = 150deg regardless of choice of D

Proof:

• Call angles:

             IbBD = ABIb = a1, 
             BDIb = IbDA = a2, 
             IcDC = ADIc = a3, 
             DCIc = IcCA = a4.

(the pairs of angles are equal due to Ib, Ic being the incenters, implying that segments BIb and CIc are angle bisectors of angles ABD and DCA, respectively)

1) We can then write that angle BAC = (180-2a1-2a2) + (180-2a3-2a4) = 360-2(a1+a2+a3+a4) 2) angle BAC = 60deg using law of cosines on the original triangle

1) and 2) mean that (a1+a2+a3+a4) = 150deg

• Quads BPDIb and DPCIc are both cyclical, by problem definition. The property of cyclical quads that will be used is the fact that opposite angles are supplementary.

3) angle BPD = 180 - angle BIbD 4) angle BIbD = 180-a1-a2

3) and 4) mean that angle BPD = a1+a2

Similarly, angle DPC = a3+a4

Finally, angle BPC = angle BPD + angle DPC = a1+a2+a3+a4 = 150deg

The problem then reduces to finding the maximum area of a triangle with base 14 and opposite angle 150deg, which can be proven any number of ways to be when the triangle is isosceles.

This gives an area of 98-49sqrt(3), and a problem answer of (98+49+3) = 150