We have a square grid with 77 nodal points. Two of them are marked as $A$ and $B$ as shown in the figure. Let $C$ be one of the remaining nodal points. Find all possible positions of point $C$ such that triangle $ABC$ has an area of 6 square units.

(E. Novotná)

![](https://cdn.mathpix.com/cropped/2024_04_17_47cbfa10f45526792f91g-6.jpg?height=562&width=376&top_left_y=1087&top_left_x=840)

Hint. Try points first such that one side of the triangle lies on one of the lines forming the square grid.
If we are looking for a solution by trial, we will most likely start testing the nodal points as indicated by the hint. Let's first consider the nodal points on the horizontal line passing through point $A$; the position of point $C$ is calculated from $A$ to the left:

1. the triangle is right-angled and its area is clearly 2 squares, which is too small,
2. the height from point $B$ divides the triangle into two (congruent) right-angled triangles, the area is $2+2=4$ squares, which is still too small,
3. the height from point $B$ divides the triangle into two (non-congruent) right-angled triangles, the area is $4+2=6$ squares, and we have our first valid solution, which we denote as $C_{1}$.

The same point can also be found without trial if we realize in time that the area of each discussed triangle is equal to half the area of a rectangle, one side of which is $A C$ and the other is 4 units (the height from point $B$), see the figure. We are looking for a nodal point on the considered line such that the area of the corresponding rectangle is

12 squares. Point $C$ must therefore be $12: 4=3$ units from $A$, and to stay within the marked area, we must go to the left.

![](https://cdn.mathpix.com/cropped/2024_04_17_47cbfa10f45526792f91g-7.jpg?height=566&width=445&top_left_y=471&top_left_x=811)

In a very similar way, we can justify the other solution on the horizontal line passing through point $B$, which we denote as $C_{2}$. (The symmetric point with respect to $B$ again falls outside the marked area.) The line $C_{1} C_{2}$ is parallel to $A B$, so every triangle $A B C$ whose vertex $C$ lies on this line has the same height to the side $A B$, and thus the same area. We are therefore looking for nodal points that simultaneously lie on the line $C_{1} C_{2}$. This way we find the point marked as $C_{3}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_47cbfa10f45526792f91g-7.jpg?height=585&width=460&top_left_y=1595&top_left_x=798)

By similar reasoning in the opposite half-plane defined by the line $A B$, we find that the remaining solutions are precisely those nodal points that simultaneously lie on the marked unnamed line. This way we find the last valid point, which is marked as $C_{4}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_47cbfa10f45526792f91g-8.jpg?height=582&width=462&top_left_y=246&top_left_x=797)

The problem has a total of 4 solutions, which we have sequentially marked as $C_{1}, C_{2}, C_{3}$, and $C_{4}$.

Notes. In the given solution, we assume knowledge of the fact that the area of any triangle is equal to half the area of a rectangle that has one side in common with the triangle and the other side equal to the height to this side. A simple justification of this statement is essentially shown in the solution to problem Z6-I-2.

Even without this knowledge, the problem can be solved by trial, as we indicated at the beginning. The area of any triangle $A B C$ with vertices at nodal points can always be expressed as follows:

- we circumscribe a rectangle around the triangle $A B C$, whose sides lie on the lines forming the square grid,
- if necessary, we divide the supplementary areas to the triangle in the circumscribed rectangle into right-angled triangles, or right-angled rectangles,
- we express the area of the triangle as the difference between the area of the circumscribed rectangle and the areas of the individual supplementary parts from the previous step.

The calculation of the areas of some triangles would look like this according to this guide (see the figures above):

$$
\begin{aligned}
& S_{A B C_{1}}=3 \cdot 4-\frac{2 \cdot 4}{2}-\frac{1 \cdot 4}{2}=12-6=6, \\
& S_{A B C_{2}}=4 \cdot 4-\frac{4 \cdot 4}{2}-\frac{1 \cdot 4}{2}=16-10=6, \\
& S_{A B C_{3}}=5 \cdot 8-\frac{5 \cdot 8}{2}-\frac{4 \cdot 4}{2}-1 \cdot 4-\frac{1 \cdot 4}{2}=40-34=6, \\
& S_{A B C_{4}}=2 \cdot 8-\frac{1 \cdot 4}{2}-\frac{2 \cdot 4}{2}-\frac{1 \cdot 8}{2}=16-10=6 .
\end{aligned}
$$

Note that even with this method, it is not necessary to exhaust all possibilities: for example, if we have found that point $C_{2}$ is a solution, we certainly do not need to consider such nodal points where the corresponding triangle either contains triangle $A B C_{2}$ or is part of it (in the first case, the resulting triangle would have a larger than required area, in the second case a smaller area).