A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?
$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(D) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(E) \frac{9}{\pi^2}(\sqrt{3}+3) }$

The three arcs make up the entire circle, so the circumference of the circle is $3+4+5=12$ and the radius is $\frac{12}{2\pi}=\frac{6}{\pi}$. Also, the lengths of the arcs are proportional to their corresponding central angles. Thus, we can write the values of the arcs as $3\theta$, $4\theta$, and $5\theta$ for some $\theta$. By Circle Angle Sum, we obtain $3\theta+4\theta+5\theta=360$. Solving yields $\theta=30$. Thus, the angles of the triangle are $90$, $120$, and $150$. Using $[ABC]=\frac{1}{2}ab\sin{C}$, we obtain $\frac{r^2}{2}(\sin{90}+\sin{120}+\sin{150})$. Substituting $\frac{6}{\pi}$ for $r$ and evaluating yields $\boxed{\frac{9}{\pi^2}(\sqrt{3}+3)}$.
\frac{9}{\pi^2}(\sqrt{3}+3)
Level 5
Geometry
Problem: A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?
$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(D) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(E) \frac{9}{\pi^2}(\sqrt{3}+3) }$

Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the area of the triangle inscribed in the circle, we first need to determine the angles corresponding to the arcs of lengths 3, 4, and 5. The total circumference of the circle is the sum of the arc lengths, which is 3 + 4 + 5 = 12. The radius of the circle can be calculated using the formula for the circumference, C = 2πr, leading to r = 12/(2π) = 6/π.

Next, we convert the arc lengths into angles in radians. The angle corresponding to an arc length s in a circle of radius r is given by θ = s/r. Thus, the angles for the arcs are:
- For arc length 3: θ1 = 3/(6/π) = 3π/6 = π/2.
- For arc length 4: θ2 = 4/(6/π) = 4π/6 = 2π/3.
- For arc length 5: θ3 = 5/(6/π) = 5π/6.

The angles of the triangle at the center of the circle are π/2, 2π/3, and 5π/6. The area of the triangle can be calculated using the formula for the area of a triangle inscribed in a circle, which is given by:
Area = (1/2) * r^2 * (sin(θ1) + sin(θ2) + sin(θ3)).

Substituting the radius r = 6/π and the angles:
Area = (1/2) * (6/π)^2 * (sin(π/2) + sin(2π/3) + sin(5π/6)).

Calculating the sine values:
- sin(π/2) = 1.
- sin(2π/3) = √3/2.
- sin(5π/6) = 1/2.

Thus, the area becomes:
Area = (1/2) * (36/π^2) * (1 + √3/2 + 1/2) = (1/2) * (36/π^2) * (2 + √3).

Simplifying this expression gives:
Area = (18/π^2) * (2 + √3).

Further simplification leads to:
Area = (18/π^2) * (√3 + 3).

This matches the final answer of:
Area = (9/π^2)(√3 + 3).<|END_OF_LATENT|> The final answer is $\frac{9}{\pi^2}(\sqrt{3}+3)$.