[ Coordinate Method ] Complex [ Ratio of Areas of Triangles with a Common Base or Common Height ] Class

In quadrilateral $A B C D$, find a point $E$ such that the ratio of the areas of triangles $E A B$ and $E C D$ is 1:2, and the ratio of the areas of triangles $E A D$ and $E B C$ is 3:4, given the coordinates of all its vertices: $A(-2 ;-4)$, $B(-2 ; 3)$, $C(4 ; 6)$, $D(4 ;-1)$.
Since

$$
\overrightarrow{A B}=\overrightarrow{(-2-(-2): 3-(-4))}=\overrightarrow{(0: r)} \overrightarrow{D C}=\overrightarrow{(4-4: 6-(-1))}=\overrightarrow{(0: r)}=\overrightarrow{A B}
$$

then the quadrilateral $A B C D$ is a parallelogram, with sides $A B$ and $C D$ parallel to the y-axis.

Let $E$ be the desired point. The triangles $E A B$ and $E C D$ have equal and parallel bases $A B$ and $C D$, so the equality $\frac{S_{\triangle E, A B}}{S_{\triangle E C D}}=\frac{1}{2}$ is equivalent to the fact that their common vertex $E$ lies on a line parallel to the bases and at a distance from them in the ratio $\frac{1}{2}$, i.e., on the y-axis.

Let the line $B C$ intersect the y-axis at point $P$, and the line $A D$ at point $Q$. If $O$ is the origin, then $O P=3$ and $O Q=4$.

The triangles $E A D$ and $E B C$ have equal and parallel bases $A D$ and $B C$, so the equality $\frac{S_{\Delta E \cdot A D}=\frac{3}{4}}{S_{\triangle E B C}}$

is equivalent to the fact that their common vertex $E$ lies on a line parallel to the bases and at a distance from them in the ratio $\frac{3}{4}=\frac{O P}{O Q}$, i.e., on the line passing through point $O$.

Therefore, the point $E$ is the origin.

## Answer

$E(0 ; 0)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-01.jpg?height=140&width=2041&top_left_y=1571&top_left_x=0)

Aveo: : Puuvooov B.B.

On the side $A B$ of triangle $A B C$, a point $P$ is taken such that $A P=2 P B$, and on the side $A C$ - its midpoint, point $Q$. It is known that $C P=2 P Q$.

Prove that triangle $A B C$ is a right triangle.

## Solution

Extend the side $A B$ and mark a segment $B D=P B$. Then $P Q$ is the midline of triangle $A C D$.

Therefore, $C D=2 P Q=C P$, so triangle $P C D$ is isosceles. $C B$ is its median, and thus also its altitude.

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-01.jpg?height=191&width=2041&top_left_y=2191&top_left_x=0)

Author: Frreninin 5. .

Petya marked several (more than two) points on a plane, all distances between which are different. A pair of marked points ($A, B$) is called unusual if $A$ is the farthest marked point from $B$, and $B$ is the nearest marked point to $A$ (excluding the point $A$ itself). What is the maximum possible number of unusual pairs that Petya could have?

## Solution

Let $(A, B)$ be an unusual pair I, and $K$ be another marked point. Then $B K_{\mathrm{I}} B C>_{\text {II }} C D>_{\text {II }} A D>_{\mathrm{I}} A B$. Contradiction.

An example with one unusual pair $(A, B)$ - the vertices of triangle $A B C$, where $A C>A B>B C$.

## Answer

One pair