## Task B-3.5.

Grandpa Ivo, a retired mathematician, turned every game with his grandson into a math problem. So, in response to the question of how to make a paper dragon, he gave a very unusual answer. The dragon has the shape of a kite, whose diagonals are determined by the vectors

$\overrightarrow{A C}=(5 a+4) \vec{i}-5 a \vec{j}$ and $\overrightarrow{B D}=4 a \vec{i}+(2 a+4) \vec{j}, a \in \mathbb{R}, a \neq 0$. If two vertices of the kite are at points $A(-1,2), D(1,4)$, where would the vertices $B$ and $C$ be located? Can the dragon be cut out of a circular piece of paper so that all vertices of the dragon are on the edge of the paper? Explain.
## Solution.

The diagonals of the deltoid are perpendicular to each other, so we have:

$$
\begin{aligned}
& \overrightarrow{A C} \cdot \overrightarrow{B D}=0 \Leftrightarrow \\
& (5 a+4) \cdot 4 a-5 a(2 a+4)=0 \Leftrightarrow \\
& 10 a^{2}-4 a=0 \\
& a \neq 0 \Rightarrow a=\frac{2}{5}
\end{aligned}
$$

Then, $\overrightarrow{A C}=6 \vec{i}-2 \vec{j}, \overrightarrow{B D}=\frac{8}{5} \vec{i}+\frac{24}{5} \vec{j}$. If $A(-1,2), D(1,4)$, it follows that $x_{C}+1=6, y_{C}-2=-2, \quad C(5,0)$

$1-x_{B}=\frac{8}{5}, 4-y_{B}=\frac{24}{5}, \quad B\left(-\frac{3}{5},-\frac{4}{5}\right)$.

Next, we need to check if the given deltoid is a cyclic quadrilateral, i.e., whether a circle can be circumscribed around the deltoid.

$\overrightarrow{B A}=-\frac{2}{5} \vec{i}+\frac{14}{5} \vec{j}$

$\overrightarrow{B C}=\frac{28}{5} \vec{i}+\frac{4}{5} \vec{j}$

$\overrightarrow{D A}=-2 \vec{i}-2 \vec{j}$

$\overrightarrow{D C}=4 \vec{i}-4 \vec{j}$

$\overrightarrow{D A} \cdot \overrightarrow{D C}=-8+8=0$ and $\overrightarrow{B A} \cdot \overrightarrow{B C}=-\frac{56}{25}+\frac{56}{25}=0$

$\Rightarrow \angle A D C=90^{\circ}=\angle A B C \Rightarrow$

$\angle A D C+\angle A B C=180^{\circ} \Longrightarrow$

Quadrilateral $A B C D$ is a cyclic quadrilateral, i.e., a circle can be circumscribed around it, and in this case, the minimum waste is achieved.

Note: The proof can also be done by first determining the equation of the circumscribed circle, for example, through points $A, C, D$, and then checking whether point $B$ belongs to it.

$S(2,1), r^{2}=\frac{36+4}{4}=10$

The equation of the circle is $(x-2)^{2}+(y-1)^{2}=10$.

## NATIONAL MATHEMATICS COMPETITION

## 4th grade - high school - B variant

Primošten, April 4, 2017.