B5. For a regular decagon $A B C D E F G H I J$ with sides of length 12, let $M$ be the center. Let $S$ be the intersection of $M C$ and $A D$.

Calculate the difference between the perimeter of quadrilateral $A B C D$ and the perimeter of triangle $D M S$.

## C-Problems

For the C-Problems, the answer is not the only thing that matters; there should also be a reasoning that shows your answer is correct. Each correctly worked-out C-Problem is worth 10 points. Partial solutions can also earn points. Therefore, write everything clearly and submit your scratch paper as well.

NOTE: Solve each C-Problem on a separate sheet of paper and also submit the corresponding scratch paper per problem.
B5. Due to symmetry, the line segments from $M$ to the vertices divide the decagon into ten congruent triangles. The angle at $M$ of such a triangle is equal to $\frac{360^{\circ}}{10}=36^{\circ}$ (indicated with $\circ$ in the figure) and the other two angles, which are equal, are $\frac{180^{\circ}-36^{\circ}}{2}=72^{\circ}$ (indicated with $x$ in the figure). From this, it follows that the angles of the original decagon are each $2 \cdot 72^{\circ}=144^{\circ}$. For the remainder, we will no longer look at the entire decagon, but only at the "diamond" given by the pentagon

![](https://cdn.mathpix.com/cropped/2024_04_17_cc6c26309fec00623354g-2.jpg?height=386&width=577&top_left_y=201&top_left_x=1251)
$A B C D M$. See the picture below.

First, we look at quadrilateral $A B C D$. The sum of the four angles is $360^{\circ}$. The angles at $B$ and $C$ are $144^{\circ}$ and the angles at $A$ and $D$, which are equal due to the symmetry of the decagon, are thus $36^{\circ}$. This means that the line segment $A D$ divides the angles $\angle M A B$ and $\angle M D C$ into exactly two parts of $36^{\circ}$. We also know that $\angle M A S=36^{\circ}$ and $\angle M D S=36^{\circ}$.

Now consider triangle $S D C$. Because the sum of the angles is $180^{\circ}$, we find that the angle $\angle C S D=180^{\circ}-\angle C D S-\angle M C D=180^{\circ}-36^{\circ}-72^{\circ}=72^{\circ}$ and the triangle is thus isosceles. It follows that $|S D|=|C D|=12$, the side of the decagon (marked with two lines in the figure).

Then consider triangle $M A S$: due to vertical angles, $\angle M S A=72^{\circ}$. We already saw that $\angle M A S=36^{\circ}$. Because of the angle sum of $180^{\circ}$ in triangle $M A S$, $\angle A M S=72^{\circ}$ and this triangle is also isosceles. It follows that $|S A|=|M A|$, the radius of the decagon (marked with three lines in the figure).

We examine triangle $M D S$. We already saw that $\angle M D S=36^{\circ}$, and at the beginning, we also saw that $\angle D M S=36^{\circ}$. Again, an isosceles triangle, so $|M S|=|D S|$ and we already saw that this is equal to 12.

Finally, we can calculate the difference between the perimeter of quadrilateral $A B C D$ and the perimeter of triangle $D M S$:

$$
\begin{aligned}
& (|A B|+|B C|+|C D|+|D A|)-(|D M|+|M S|+|S D|) \\
& =(12+12+12+(12+|S A|))-(|D M|+12+12) \\
& =24+|M A|-|M D| \\
& =24 .
\end{aligned}
$$

## C-problems