Task B-3.4. A group of children found a wooden board in the shape of a quadrilateral and decided to use it for the game "pick-a-point". Since the target was of an unusual shape, they had to adapt the game rules. They asked for advice from Mark's older brother, a good mathematician. He measured, calculated, and wrote something... and came to an important conclusion. He established that there is a point $O$ equally distant from all sides of the quadrilateral. Then he divided the board into areas by drawing lines from point $O$ to all vertices $A, B, C, D$. He assigned points to the areas inversely proportional to their area. Which area yields the most points and why? Mark's brother added a condition that points are only awarded if the distance from the hit point $T$ to point $O$ is not greater than the distance from point $O$ to any side of the board. Determine the ratio of the area of the part of the board that does not yield points to the area of the part that does yield points.

Mark's brother measured these values $|A B|=5 \mathrm{dm},|B C|=3 \mathrm{dm},|A D|=4 \mathrm{dm}$, $|\angle B A D|=60^{\circ}$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_2595713ee0b414c2653fg-13.jpg?height=745&width=1022&top_left_y=1015&top_left_x=519)

If the point $O$ is equidistant from the sides of the quadrilateral, then a circle can be inscribed in that quadrilateral.

In a tangential quadrilateral, $a+c=b+d$ holds. We conclude that $c=2 \mathrm{dm}$.

The tile is divided into 4 regions, triangles $A B O, B C O, C D O, A D O$. The triangle with the smallest area is triangle $C D O$ because it has the shortest side on which $r$ is the height, so this region scores the most points.

Points are earned by hitting a point inside the inscribed circle of the quadrilateral.

Let's calculate the radius of the inscribed circle, i.e., the distance from point $O$ to the sides of the quadrilateral.

From $\triangle A B D$ we get

$$
\begin{aligned}
|B D|^{2} & =16+25-2 \cdot 4 \cdot 5 \cdot \cos 60^{\circ}=21 \\
|B D| & =\sqrt{21} \mathrm{dm}
\end{aligned}
$$

From $\triangle B C D$ we get

$$
\cos \gamma=\frac{9+4-21}{2 \cdot 2 \cdot 3}=\frac{-2}{3}, \text { and } \sin \gamma=\frac{\sqrt{5}}{3}
$$

Consider the area of the quadrilateral.

$$
P(A B C D)=r \cdot \frac{a+b+c+d}{2}=r \cdot s=7 r
$$

Similarly,

$$
P(A B C D)=P(A B D)+P(B C D)=\frac{3 \cdot 2 \sin \gamma}{2}+\frac{4 \cdot 5 \sin 60^{\circ}}{2}
$$

so

$$
\begin{aligned}
7 r & =\sqrt{5}+5 \sqrt{3} \\
r & =\frac{\sqrt{5}+5 \sqrt{3}}{7} \mathrm{dm}
\end{aligned}
$$

The ratio of the desired areas is

$$
\frac{7 r-r^{2} \pi}{r^{2} \pi}=\frac{7}{r \pi}-1=\frac{7}{\pi} \cdot \frac{7}{\sqrt{5}+5 \sqrt{3}}-1=\frac{7}{10 \pi}(5 \sqrt{3}-\sqrt{5})-1
$$

Note. The radius can also be calculated without using areas.

From $\triangle A E O$ we have $\operatorname{tg} 30^{\circ}=\frac{r}{x}, x=r \sqrt{3}$,

$$
\begin{aligned}
y & =5-r \sqrt{3} \\
\operatorname{tg} \frac{\gamma}{2} & =\frac{\sin \frac{\gamma}{2}}{\cos \frac{\gamma}{2}}=\sqrt{\frac{1-\cos \gamma}{1+\cos \gamma}}=\sqrt{5}
\end{aligned}
$$

From $\triangle O G C$ we have $\operatorname{tg} \frac{\gamma}{2}=\frac{r}{z}, z=\frac{r \sqrt{5}}{5}$.

$$
\begin{aligned}
z & =3-y=3-(5-r \sqrt{3})=r \sqrt{3}-2 \\
\frac{r \sqrt{5}}{5} & =r \sqrt{3}-2 \\
r & =\frac{10}{5 \sqrt{3}-\sqrt{5}}=\frac{5 \sqrt{3}+\sqrt{5}}{7} \mathrm{dm}
\end{aligned}
$$