Question 459746
{{{drawing(300,300,0,10,0,10,


triangle(2,2,8,2,5,2+3sqrt(3)),
locate(2,2,A),
locate(5,2+3sqrt(3),B),
locate(8,2,C),
line(2,2,13/2,(4+3sqrt(3))/2),
circle(5,5.1,1.1),
locate(5.5,5,M),
line(8,2,4, (8+6sqrt(3))/3),
locate(4, (8+6sqrt(3))/3, P),
locate(13/2, (4+3sqrt(3))/2, D)
)
}}}


First, we can find the values for AM and MD using basic properties of inscribed circles within triangles. If we let AM = x, MD = y, and Q,R be the intersections of the circle and sides BD, BA respectively, we can say that QD = y, BQ = z, etc. and set up a system of three equations in terms of x, y, and z (I won't go too in-detail with this). Solving the system yields *[tex AM = \frac{\sqrt{3} + 1}{4}] and *[tex MD = \frac{\sqrt{3} - 1}{4}].


Next, extend PC through P to point S such that angle SBC is a right angle:


{{{drawing(300,300,0,10,0,10,


triangle(2,2,8,2,5,2+3sqrt(3)),
locate(2,2,A),
locate(5,2+3sqrt(3),B),
locate(8,2,C),
line(2,2,13/2,(4+3sqrt(3))/2),
circle(5,5.1,1.1),
locate(5.5,5,M),
line(8,2,4, (8+6sqrt(3))/3),
locate(4, (8+6sqrt(3))/3, P),
locate(13/2, (4+3sqrt(3))/2, D),


line(5, 2+3sqrt(3), 3.5, 2+3sqrt(3) - (sqrt(3)/3)(1.2)),
line(4, (8+6sqrt(3))/3,  3.5, 2+3sqrt(3) - (sqrt(3)/3)(1.2)),
locate(3.5, 2+3sqrt(3) - (sqrt(3)/3)(1.2), S)
)
}}}

Here, we conclude that triangles CMD and CSB are similar (due to the fact that SB, MD are parallel), and they have a 1:2 ratio. Since we have found that *[tex MD = \frac{\sqrt{3}-1}{4}], then SB is twice that, or *[tex \frac{\sqrt{3}-1}{2}].


Now we note that triangles SBP and MAP are also similar, with the ratio unknown. However, we have found SB and MA (two corresponding sides) so we can find AP:PB. We have


*[tex \LARGE \frac{SB}{MA} = \frac{BP}{AP}]


*[tex \LARGE \frac{ \frac{\sqrt{3}-1}{2}} {\frac{\sqrt{3} + 1}{4}} = \frac{BP}{AP}]


We simplify the fraction and replace BP with 1 - AP, since BP + AP = 1.


*[tex \LARGE \frac{1-AP}{AP} = 4 - 2\sqrt{3}]


Solving for AP yields


*[tex \LARGE AP = \frac{5 + 2\sqrt{3}}{13}].


Now we can find the area of triangle APC. This is because the ratio of AP:AB is the ratio of [APC]:[ABC] ([x] denotes area). Here, the area of triangle APC is AP multiplied by the area of ABC (which is easy to find).


The area of equilateral triangle ABC is


*[tex \LARGE [ABC] = \frac{AB*BC*\sin(60)}{2} = \frac{\sin(60)}{2} = \frac{\sqrt{3}}{4}].


Hence, the area of triangle APC is


*[tex \LARGE [APC] = AP*[ABC] = \frac{5 + 2\sqrt{3}}{13}\frac{\sqrt{3}}{4} = \frac{6 + 5\sqrt{3}}{52}].


You might want to check my arithmetic and algebraic manipulation as I do make silly mistakes once in a while. :)