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In an acute-angled triangle ABC, point D lies on AC such that AD = 1, DC = 2, and BD is the altitude from B to AC. A circle of radius 2 passes through A and D and is externally tangent to the circumcircle of triangle BDC at point D. If the area of triangle ABC equals p * sqrt(q) where p is prime and q is a natural number, find the value of (q - p) / 2.

1. 4
2. 5
3. 6
4. 7

Correct answer: 6

Solution

Set D=(0,0), A=(-1,0), C=(2,0), B=(0,h). Circumcircle of BDC has center O1=(1, h/2), radius R1=sqrt(1+h²/4). Circle through A,D has center O2=(-1/2, k) with (1/4+k²)=4, so k=+-sqrt(15)/2. Tangency at D requires O1, O2, D collinear. Direction O1->D: (-1,-h/2); direction O2->D: (1/2, -k). Proportionality gives -2 = -h/(2k). With k=-sqrt(15)/2: h=2*sqrt(15). Check: R1=sqrt(1+15)=4, R2=2, |O1O2|=6=R1+R2 (external tangency confirmed). Area = (1/2)*AC*BD = (1/2)*3*2*sqrt(15) = 3*sqrt(15). So p=3, q=15, (q-p)/2=6.

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