In the picture, the nodes of a square grid are marked, two of which are named $A$ and $B$. Let $C$ be one of the remaining nodes. Find all possible positions of point $C$ such that triangle $A B C$ has an area of 4.5 squares.

(E. Novotná)

Hint. The area of any triangle can be expressed using the area (at most) of two right triangles.
By randomly testing nodal points, we will soon realize that we need to discuss three qualitatively different cases when the triangle $ABC$ is a) right-angled with the right angle at vertex $A$ or $B$, b) acute-angled, or c) obtuse-angled.

a) Assume that the triangle $ABC$ is right-angled with the right angle at vertex $A$ or $B$. We will consider the first possibility; the solution in the second case will be symmetrical.

Such a triangle forms exactly half of a rectangle, one side of which is $AB$ and the other is $AC$. The area of this rectangle should be equal to 9 square units. The length of $AB$ is 3 units, so point $C$ must be $9: 3=3$ units from $A$. This point is denoted as $C_{1}$, and the symmetrical solution to the left of point $B$ is denoted as $C_{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_22cb57ce78fb8d5dd5c3g-2.jpg?height=320&width=277&top_left_y=794&top_left_x=889)

b) Now assume that point $C$ is some nodal point in the strip between the lines $AC_{1}$ and $BC_{2}$. The triangle $ABC$ will be divided by the altitude to side $AB$ into two smaller right-angled triangles. Each of these triangles is half of some rectangle, see the figure. These two rectangles form a larger rectangle, whose area is twice the area of triangle $ABC$. One side of this rectangle is $AB$ and the other is equal to the altitude $CP$. We are looking for point $C$ such that the area of such a rectangle is 9 square units. By trial or reasoning as above, we will find two solutions, which are marked as $C_{3}$ and $C_{4}$:

![](https://cdn.mathpix.com/cropped/2024_04_17_22cb57ce78fb8d5dd5c3g-2.jpg?height=300&width=263&top_left_y=1620&top_left_x=902)

c) Finally, we must discuss the possibilities where point $C$ is some nodal point outside the strip defined by the lines $AC_{1}$ and $BC_{2}$. The triangle $ABC$ is obtuse in this case, and the altitude to side $AB$ lies outside the triangle; the foot of this altitude is again denoted as $P$. We will consider only triangles with an obtuse angle at vertex $A$, the remaining solutions are symmetrical.

The area of triangle $ABC$ is now the difference in the areas of the right-angled triangles $CPB$ and $CPA$. Each of these triangles is half of a suitable rectangle, see the figure. The difference in the areas of these rectangles is therefore twice the area of triangle $ABC$ and is the same as the area of the rectangle outlined with a solid line. One side of this rectangle is $AB$ and the other is equal to the altitude $CP$. We are looking for point $C$ such that the area of such a rectangle is 9 square units. By trial or reasoning as above, we will find two solutions, which are marked as $C_{5}$ and $C_{6}$. The symmetrical solutions above the line $BC_{2}$ are $C_{7}$ and $C_{8}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_22cb57ce78fb8d5dd5c3g-3.jpg?height=406&width=277&top_left_y=428&top_left_x=889)

The problem has a total of 8 solutions, which we have sequentially marked as $C_{1}, \ldots, C_{8}$.

Notes. The idea of a sum rectangle in paragraph b), or a difference rectangle in paragraph c), is not necessary for successfully solving the problem. It is sufficient for the solver to realize that the area of triangle $ABC$ is the sum, or difference, of the areas of the right-angled triangles $CPB$ and $CPA$. The justification that, for example, the marked point $C_{5}$ is a solution, can look like this:

$$
S_{ABC_{5}}=\frac{3 \cdot 5}{2}-\frac{3 \cdot 2}{2}=\frac{9}{2}.
$$

Notice that for any point $C$ on the line $C_{1}C_{2}$, the area of triangle $ABC$ will also be 4.5 square units according to the previous reasoning. In fact, we have thus proven that the area of any triangle is equal to half of a rectangle that has one side in common with the triangle and the other equal to the altitude to that side.

If the student already knows this fact, the solution is particularly easy: it is enough to find one suitable point, and all the others are precisely the nodal points lying on a line parallel to $AB$ that passes through this suitable point.