Draw a diameter in a circle and connect the midpoint of the resulting semicircle with the two one-third points of the other semicircle. The sum of the legs and the base of the isosceles triangle formed by the three lines drawn is approximately equal to one quarter of the circumference of the circle. Determine the accuracy of this approximate construction.
Let's choose the radius of the circle as the unit, and let the center of the circle be $O$, the endpoints of the diameter be $A$ and $B$, the midpoint of one semicircle be $C$, and the points dividing the other semicircle into thirds be $D$ and $E$. Finally, let the points where the lines $C D$ and $C E$ intersect $A B$ be $F$ and $G$.

![](https://cdn.mathpix.com/cropped/2024_05_02_ec4308190d059be9a6f9g-1.jpg?height=531&width=555&top_left_y=293&top_left_x=774)

The triangle $O D E$ is equilateral, and its sides are of unit length. Similarly, the height of the triangle $C F G$ from $C$ to $O$ is also of unit length. Due to the similarity of triangles $C D E$ and $C F G$, the ratio of the sides $D E$ and $F G$ is the same as the ratio of the heights of these triangles:

$$
\begin{gathered}
D E: F G=\left(1+\frac{\sqrt{3}}{2}\right): 1 \\
F G=\frac{D E}{1+\frac{\sqrt{3}}{2}}=\frac{2}{2+\sqrt{3}}=2(2-\sqrt{3})
\end{gathered}
$$

In the right triangle $C O F$, by the Pythagorean theorem,

$$
C F=\sqrt{F O^{2}+O C^{2}}=\sqrt{(2-\sqrt{3})^{2}+1}=\sqrt{8-4 \sqrt{3}}=\sqrt{6-2 \sqrt{6} \sqrt{2}+2}=\sqrt{6}-\sqrt{2}
$$

The sum of the base and one leg of the triangle $C G F$ is therefore

$$
q=C F+F G=\sqrt{6}-\sqrt{2}+4-2 \sqrt{3}
$$

According to the problem, this value $q$ should approximate the quarter of the circumference of the circle, $\pi / 2$. Based on our table,

$$
\begin{aligned}
q & =2.449-1.414+4-3.464=1.571 \\
\pi / 2 & \approx 1.571
\end{aligned}
$$

Thus, $q$ and $\pi / 2$ appear to be equal to three decimal places. However, this does not mean that the error is less than the rounding error, which is $5 \cdot 10^{-4}$. In the determination of $\left(\frac{\pi}{2}-q\right)$, four values are involved, each rounded to three decimal places. In the worst case, the rounding errors add up, so we can only say that

$$
\left|q-\frac{\pi}{2}\right|<4 \cdot 5 \cdot 10^{-4}=0.002
$$

Calculating $q$ and $\pi / 2$ more precisely,

$$
q=1.571 \quad 17, \quad \frac{\pi}{2}=1.57080
$$

Thus, the error is

$$
0<q-\frac{\pi}{2}<0.0004
$$

which means $q$ is slightly larger than the length of the quarter arc, but the excess is less than 0.03% of the arc length.