Given a certain number of points on a plane, the straight line connecting any two of them does not pass through any other given point. Each point is marked with either red, white, or green. The number of lines connecting pairs of points of different colors is 213, and the number of lines connecting pairs of points of the same color is 112. Among the triangles determined by the points, the number of those with each vertex of a different color is 540, and the number of triangles with 2 red vertices is 612. What can we say about the number of triangles with 2 white vertices?
Based on the data, denoting the number of red, white, and green points by $p$, $f$, and $z$ respectively, we can set up the following equations (1)-(4) for these three unknowns. For equations (2) and (4), it is necessary to know how many ways a pair can be chosen from $n$ elements. Arranging the elements in a sequence, the first, second, third, ..., $(n-1)$-th element can form a previously uncounted pair with the following $n-1, n-2, n-3, \ldots, 1$ elements, respectively, so the number of possible pairs is:

\[
\begin{aligned}
(n-1)+(n-2)+(n-3)+\ldots+1 & =\frac{n(n-1)}{2} \\
p f + p z + f z & = 213 \\
\frac{p(p-1)}{2} + \frac{f(f-1)}{2} + \frac{z(z-1)}{2} & = 112 \\
p f z & = 540 \\
\frac{p(p-1)}{2} \cdot (f + z) & = 612
\end{aligned}
\]

Based on these, we need to say something about the value of the expression \( F = \frac{f(f-1)}{2} \cdot (p + z) \). It is cautionary that we have more equations than unknowns.

Notice that under (1) and (2), we counted every line segment connecting any two points exactly once, since any pair of points is either of two different colors or of the same color. Based on this, if we denote the total number of points by $x$, the number of line segments is

\[
\frac{x(x-1)}{2} = 213 + 112 = 325
\]

From this, we can only use the positive root: \( x = p + f + z = 26 \).

Using this in (4), we get a single-variable cubic equation for $p$:

\[
p(p-1)(26-p) = 1224
\]

Each factor on the left side is a natural number, so we can look for the solution by factoring the right side into different bases. \( 1224 = 2^3 \cdot 3^2 \cdot 17 \), so one of the factors must be divisible by 17. Moreover, we see that \( p < 26 \), so the factor must be equal to 17. For \( p = 17 \) and \( p - 1 = 17 \), the value on the left side of (5) is not 1224, so the only value that fits for us is \( 26 - p = 17 \), which gives \( p = 9 \).

Now we have \( f + z = 17 \), and combining this with (3), we can calculate $f$ and $z$, and if these are natural numbers, we can also calculate $F$. However, we can see in advance that our answer cannot be unique, because the roots of the resulting quadratic equation, if integers, are different, since their sum is odd, and each root can be taken as $f$.

From (3), \( f z = f(17 - f) = 60 \), so \( f = 5 \) or \( f = 12 \), and thus \( z = 12 \) or \( z = 5 \). These, along with \( p = 9 \), satisfy all the equations (1)-(4). In the first case, the number of triangles with two white vertices is \( F = 10 \cdot 21 = 210 \), and in the second case, it is 924.

Farkas György (Budapest, Landler J. Gép- és Hír. ip. Techn., III. o. t.) Kóczy László (Budapest, Fazekas M. gyak. gimn. II. o. t.)

Remarks. The problem can be solved in several other ways. For example, to get the result \( p + f + z = 26 \), adding (1) and (3) gives that for \( p \), \( f \), and \( z \) each, the following cubic equation holds:

\[
y^3 - 26 y^2 + 213 y - 540 = 0
\]

The roots of this equation are 5, 9, and 12 (one of them can be found among the divisors of 540 by trial, and the other two can be obtained from a quadratic equation after removing this one), but from (4), only \( p = 9 \) gives integer values for \( f + z \).

Several others, based on the factorization \( 540 = 2^2 \cdot 3^3 \cdot 5 \) in (3), tried, and based on (1), added that among \( p \), \( f \), and \( z \), two are odd, so the third is divisible by \( 2^2 \).