Let's decide whether the point $P(15.2, 12.4)$ is inside the triangle formed by the lines $8x-15y-35=0$, $x-2y-2=0$, and $y=0$.
Preliminary remark. Based on the representation of the given points and lines, it is easy to decide the question. Below, we only consider solutions based solely on calculations. (The diagram is not to scale.)

![](https://cdn.mathpix.com/cropped/2024_05_02_79b28280a08bfc874b66g-1.jpg?height=453&width=681&top_left_y=267&top_left_x=711)

I. Solution. If a point $P$ is inside a triangle, then drawing any line parallel to one of the sides through $P$, the segment cut off by the other two sides will contain $P$. If, however, any of the segments cut off by the mentioned parallels does not contain $P$, then $P$ is not inside the triangle.

In the given case, let's first take the line parallel to the side $y=0$ and passing through $P$. The equation of this line is $y=12.4$, so the x-coordinates of the intersection points with the other two lines are 27.625 and 26.8. Both are greater than the x-coordinate of $P$, so we can immediately conclude that $P$ is not inside the triangle.

György Csákó (Sátoraljaújhely, Kossuth L. g. III. o. t.)

Remark. If the x-coordinate of $P$ were between the two x-coordinate values, this would not prove that $P$ is inside the triangle; it could also be in the angle formed by the extensions of the sides intersecting the x-axis beyond their common vertex.

II. Solution. If $P$ is inside or on the perimeter of triangle $ABC$ but does not coincide with $A$, then the intersection point of lines $PA$ and $BC$ lies between $B$ and $C$, or at one of the endpoints. Let's take the intersection point of the first two lines $(40; 19)$ as $A$. Thus, the equation of $PA$ is $6.6x - 24.8y + 207.2 = 0$, and the line $BC$ is the line $y=0$. The first two lines intersect at $(4.375; 0)$ and $(2; 0)$, while $AP$ intersects at $M(-207.2 / 6.6; 0)$. Without precise calculation of the latter, we can see that it lies outside the segment between the first two, so $P$ is not an interior point of the triangle.

Rezső Gáspár (Debrecen, Kossuth L. gyak. g. III. o. t.)

Remark. Similar to the above method: if $\angle PAB > \angle CAB$, then $P$ is an exterior point. Taking $A$, $B$, and $C$ as $(2; 0)$, $(4.375; 0)$, and $(40; 19)$ respectively, we have $\tan \angle PAB = 12.4 / 13.2 \approx 0.94$ and $\tan \angle CAB = 0.5$, so $P$ is an exterior point.

László Butor (Budapest, I. István g. IV. o. t.)

III. Solution. If $P$ is inside the triangle, then any line passing through $P$ will intersect the sides of the triangle at three points that enclose $P$. Therefore, if we find a line through $P$ for which this is not true, then $P$ is an exterior point. In the given case, the vertical line $x=15.2$ passing through $P$ is such a line, because the y-coordinates of the intersection points with the given lines are 5.773, 6.6, and 0; all are smaller than the y-coordinate of $P$, so $P$ is an exterior point.

Andrea Bede (Budapest, Szilágyi Erzsébet lg. III. o. t.)