3. Inside the rectangle $A B C D$ there is a point $T$. If point $T$ is connected to the vertices of the rectangle, the rectangle will be divided into four triangles. The areas of three of these triangles are $15 \mathrm{~cm}^{2}$, $54 \mathrm{~cm}^{2}$ and $75 \mathrm{~cm}^{2}$. Determine all possible areas of the rectangle $A B C D$.
3. Inside the rectangle $A B C D$ there is a point $T$. If point $T$ is connected to the vertices of the rectangle, the rectangle will be divided into four triangles. The areas of three of these triangles are $15 \mathrm{~cm}^{2}, 54 \mathrm{~cm}^{2}$, and 75 $\mathrm{cm}^{2}$. Determine all possible areas of the rectangle $A B C D$.

## Solution.

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_ca549eb60a6f3c3b6299g-24.jpg?height=614&width=853&top_left_y=590&top_left_x=316)

Let's denote the areas of triangles $\triangle A T D$ as $P_{1}$, $\triangle A B T$ as $P_{2}$, $\triangle B C T$ as $P_{3}$, and $\triangle C D T$ as $P_{4}$.

Draw the heights from vertex $T$ in $\triangle A T D$ and $\triangle B C T$, and denote them as $x$ and $y$, respectively.

Then, $a = x + y$.

1 POINT

Furthermore,

$P_{1} = \frac{1}{2} b x, P_{3} = \frac{1}{2} b y$.

1 POINT

By adding the areas $P_{1}$ and $P_{3}$, we get the following equation:

$P_{1} + P_{3} = \frac{1}{2} b x + \frac{1}{2} b y$

1 POINT

$= \frac{1}{2} b (x + y)$

$= \frac{1}{2} b a$

$= \frac{1}{2} P_{A B C D}$

1 POINT

We can conclude that $P_{2} + P_{4} = \frac{1}{2} P_{A B C D}$.

1 POINT

Let's denote the fourth unknown area of one of the triangles as $Q$. Then we have 3 possibilities:
I. $15 + 54 = 75 + Q$

II. $15 + 75 = 54 + Q$

III. $54 + 75 = 15 + Q$

The first case is invalid because $15 + 54 < 75$, and thus $15 + 54 < 75 + Q$. 1 POINT

From the second case, we have $Q = 36 \mathrm{~cm}^{2}$. 1 POINT

From the third case, we have $Q = 114 \mathrm{~cm}^{2}$. 1 POINT

Therefore, the problem has two solutions, meaning there are two rectangles that satisfy the given conditions.

The areas of these rectangles are $15 + 54 + 75 + 36 = 180 \mathrm{~cm}^{2}$ and $15 + 54 + 75 + 114 = 258 \mathrm{~cm}^{2}$. 2 POINTS

TOTAL 10 POINTS

Note: A student who writes only one correct solution should, according to the scoring, receive 7 POINTS. A student who does not show that one solution is not possible should receive 9 POINTS.