## Task 4 - 270834

Let $\overparen{A B} M$ be a sector of a circle, where the length $r=\overline{M A}=\overline{M B}$ is given and the central angle $\angle A M B$ is $60^{\circ}$. A line $g$, which is parallel to $A B$ and intersects the segments $M A$ and $M B$ at points $C$ and $D$ respectively, is known to have the property that the perimeter $u_{1}$ of triangle $M C D$ is equal to the perimeter $u_{2}$ of the figure $\widehat{A B D C}$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0779.jpg?height=343&width=400&top_left_y=1422&top_left_x=1322)

a) Determine the length $x=\overline{M C}$ in dependence on $r$ under this condition!

b) The length $r$ is measured with an accuracy that gives $r=6.7 \mathrm{~cm}$ to one decimal place. Furthermore, for the calculation, it is used that $\pi=3.14$ to two decimal places.

Prove that from this, the length $x$ (in centimeters) can be determined to one decimal place by calculating bounds! What is this length $x$?
a) Since $\angle A M B=60^{\circ}=\frac{360^{\circ}}{6}$, the arc length of $\overparen{A B}$ is

$$
b=\frac{2 \pi r}{6}=\frac{1}{3} \pi \cdot r
$$

Since triangle $M A B$ is isosceles with $M A=M B$, the base angles $\angle M A B=\angle M B A$ are equal, and by the interior angle sum theorem, $\angle M A B+\angle M B A=180^{\circ}-60^{\circ}$; thus, $\angle M A B=\angle M B A=60^{\circ}$.

Since $C D \| A B$, it follows that $\angle M C D=\angle M A B=60^{\circ}$ (corresponding angles) and similarly $\angle M D C=60^{\circ}$. Therefore, triangle $M C D$ is equilateral, i.e.,

$$
M C=M D=C D=x
$$

From (2) it follows that $u_{1}=3 x$ and $C A=D B=r-x$; from this and (1) it follows that $u_{2}=x+2(r-x)+\frac{1}{3} \pi r$.

Thus, the condition $u_{1}=u_{2}$ implies

$$
\begin{aligned}
3 x & =x+2 r-2 x+\frac{1}{3} \pi r \\
4 x & =\left(2+\frac{\pi}{3}\right) r=\frac{6+\pi}{12} r \\
x & =\frac{6+\pi}{12} r
\end{aligned}
$$

b) From the precision requirements, it follows from (3) that the numerical value of $x$ has a lower bound of

$$
\frac{6+3.135}{12} \cdot 6.65=5.0623125
$$

and an upper bound of

$$
\frac{6+3.145}{12} \cdot 6.75=5.1440625
$$

Thus, it is proven that to one decimal place, $x=5.1 \mathrm{~cm}$.