Problem 1779

1. Given that $R, 1, R + \frac{1}{2}$ are the sides of a triangle, we need to prove that $2R\theta$ is between 1 and $\pi$, where $\theta$ is the angle between $R$ and $R + \frac{1}{2}$.

2. Since $R, 1, R + \frac{1}{2}$ form a triangle, the triangle inequality must hold:
$$R + \left( R + \frac{1}{2} \right) > 1 \implies 2R + \frac{1}{2} > 1 \implies 4R > 1 \implies R > \frac{1}{4}$$
Let $u = 4R$. Then $u > 1$.

3. The semiperimeter $s$ of the triangle is:
$$s = \frac{1}{2} \left( R + 1 + \left( R + \frac{1}{2} \right) \right) = R + \frac{3}{4} = \frac{1}{4}u + \frac{3}{4} = \frac{1}{4}(u + 3)$$

4. Using the half-angle formula for sine, we have:
$$\sin^2 \frac{\theta}{2} = \frac{(s - R)(s - (R + \frac{1}{2}))}{R(R + \frac{1}{2})}$$
Substituting $s = \frac{1}{4}(u + 3)$ and $R = \frac{1}{4}u$:
$$\sin^2 \frac{\theta}{2} = \frac{\left( \frac{1}{4}(u + 3) - \frac{1}{4}u \right) \left( \frac{1}{4}(u + 3) - \left( \frac{1}{4}u + \frac{1}{2} \right) \right)}{\frac{1}{4}u \left( \frac{1}{4}u + \frac{1}{2} \right)}$$
Simplifying:
$$\sin^2 \frac{\theta}{2} = \frac{\left( \frac{3}{4} \right) \left( \frac{1}{4}(u + 3) - \frac{1}{4}u - \frac{1}{2} \right)}{\frac{1}{4}u \left( \frac{1}{4}u + \frac{1}{2} \right)} = \frac{\left( \frac{3}{4} \right) \left( \frac{3}{4} - \frac{1}{2} \right)}{\frac{1}{4}u \left( \frac{1}{4}u + \frac{1}{2} \right)} = \frac{\left( \frac{3}{4} \right) \left( \frac{1}{4} \right)}{\frac{1}{4}u \left( \frac{1}{4}u + \frac{1}{2} \right)}$$
$$\sin^2 \frac{\theta}{2} = \frac{3}{u(u + 2)}$$

5. To show $1 < 2R\theta$, we need:
$$\frac{\theta}{2} > \frac{1}{4R} \implies \frac{\theta}{2} > \frac{1}{u}$$
Since $\sin^2 x$ is increasing on $[0, \frac{\pi}{2}]$, we need:
$$\sin^2 \frac{\theta}{2} > \sin^2 \frac{1}{u}$$
We will show:
$$\sin^2 \frac{\theta}{2} > \left( \frac{1}{u} \right)^2$$
Since $\sin x < x$ for $x > 0$:
$$\sin^2 \frac{\theta}{2} = \frac{3}{u(u + 2)} > \left( \frac{1}{u} \right)^2$$
Simplifying:
$$\frac{3}{u(u + 2)} > \left( \frac{1}{u} \right)^2 \implies 3u > u + 2 \implies 2u > 2 \implies u > 1$$
This is true since $u > 1$. Thus, $1 < 2R\theta$.

6. To show $2R\theta < \pi$, we need:
$$\frac{\theta}{2} < \frac{\pi}{4R} \implies \frac{\theta}{2} < \frac{\pi}{u}$$
Since $\sin^2 x$ is increasing on $[0, \frac{\pi}{2}]$, we need:
$$\sin^2 \frac{\theta}{2} < \sin^2 \frac{\pi}{u}$$
We will show:
$$\frac{3}{u(u + 2)} < \sin^2 \frac{\pi}{u}$$
Taking the reciprocal:
$$\frac{u(u + 2)}{3} > \csc^2 \frac{\pi}{u}$$
Subtracting 1:
$$\frac{u(u + 2)}{3} - 1 > \csc^2 \frac{\pi}{u} - 1 \implies \frac{(u - 1)(u + 3)}{3} > \cot^2 \frac{\pi}{u}$$
Taking the reciprocal:
$$\frac{3}{(u - 1)(u + 3)} < \tan^2 \frac{\pi}{u}$$
We will show:
$$\frac{3}{(u - 1)(u + 3)} < \left( \frac{3}{u} \right)^2$$
Simplifying:
$$\frac{3}{(u - 1)(u + 3)} < \left( \frac{3}{u} \right)^2 \implies 3u^2 < 9(u - 1)(u + 3) \implies 3u^2 < 9(u^2 + 2u - 3) \implies 3u^2 < 9u^2 + 18u - 27$$
Simplifying further:
$$0 < 6u^2 + 18u - 27 \implies 2u^2 + 6u - 9 > 0$$
The roots of $2u^2 + 6u - 9$ are:
$$u = \frac{-6 \pm \sqrt{36 + 72}}{4} = \frac{-6 \pm 6\sqrt{3}}{4} = \frac{-3 \pm 3\sqrt{3}}{2}$$
Since $u > 2$, $2u^2 + 6u - 9 > 0$. Thus, $2R\theta < \pi$.

The final answer is $\boxed{ 1 < 2R\theta < \pi }$