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1300
Incenter Distance — Editorial
geometrytrianglesincenter
The semiperimeter is s=(13+14+15)/2=21.s = (13+14+15)/2 = 21. By Heron's formula, K=21⋅8⋅7⋅6=84,K = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84, so the inradius is r=K/s=4.r = K/s = 4. Let A=∠BAC.A = \angle BAC. By the Law of Cosines,
cos⁡A=169+225−196390=3365,\cos A = \frac{169 + 225 - 196}{390} = \frac{33}{65},
hence
sin⁡A2=1−cos⁡A2=1665=465.\sin\frac{A}{2} = \sqrt{\frac{1-\cos A}{2}} = \sqrt{\frac{16}{65}} = \frac{4}{\sqrt{65}}.
Since AI=rcsc⁡A2,AI = r\csc\tfrac{A}{2},
AI=4⋅654=65,AI2=65.AI = 4 \cdot \frac{\sqrt{65}}{4} = \sqrt{65}, \quad AI^2 = 65.