Problem:
In โณABC with side lengths AB=13,AC=12, and BC=5, let O and I denote the circumcenter and incenter, respectively. A circle with center M is tangent to the legs AC and BC and to the circumcircle of โณABC. What is the area of โณMOI ?
Answer Choices:
A. 25โ
B. 411โ
C. 3
D. 413โ
E. 27โ Solution:
Place the figure on coordinate axes with coordinates A(12,0),B(0,5), and C(0,0). The center of the circumscribed circle is the midpoint of the hypotenuse of right triangle ABC, so the coordinates of O are (6,25โ). The radius r of the inscribed circle equals the area of the triangle divided by its semiperimeter, which here is 30รท15=2, so the center of the inscribed circle is I(2,2). Because the circle with center M is tangent to both coordinate axes, its center has coordinates (ฯ,ฯ), where ฯ is its radius. Let P be the point of tangency of this circle and the circumscribed circle. Then M,P, and O are collinear because MP and OP are perpendicular to the common tangent line at P. Thus MO=OPโMP=213โโฯ. By the distance formula, MO=(ฯโ6)2+(ฯโ25โ)2โ. Equating these expressions and solving for ฯ shows that ฯ=4. The area of โณMOI can now be computed using the shoelace formula:
Alternatively, the area can be computed as 21โ times MI, which by the distance formula is (4โ2)2+(4โ2)2โ=22โ, times the distance from point O to the line MI, whose equation is xโy+0=0. This last value is
