25.19 In the coordinate plane, construct a right triangle such that its two legs are parallel to the $x$ and $y$ axes, and the medians to these legs lie on the lines $y=3x+1$ and $y=mx+2$, respectively. The number of constants $m$ for which such a triangle exists is 
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) More than 3.
(37th American High School Mathematics Examination, 1986)
[Solution 1] From the following image, it can be seen that: there are two $m$ values that satisfy the problem, i.e., in the line bundle passing through $(0, 2)$, there are two lines that meet the requirements.

Specific calculations can be as follows:
Given that $B C \parallel x$-axis, $A C \parallel y$-axis, $A D$ is the median of $B C$, and $B E$ is the median of $A C$,
and let $B C=a, A C=b$, then
$$
k_{B E}=\frac{E C}{B C}=\frac{\frac{b}{2}}{a}=\frac{b}{2 a}, \quad k_{A D}=\frac{A C}{C D}=\frac{b}{\frac{a}{2}}=\frac{2 b}{a},
$$

Thus,
$$
k_{B E}: k_{A D}=1: 4:
$$

That is, the ratio of the slopes of the two medians is 4. Therefore, $\frac{3}{m}=4$, or $\frac{m}{3}=4$, yielding $m=12$ or $\frac{3}{4}$. Hence, $m$ has two values.
Therefore, the answer is (C).

[Solution 2] Let the coordinates of points $A, B, C$ be (since they lie on the medians given in the problem):
$$
\begin{array}{l}
A(b, m b+2), B(a, 3 a+1), \\
C(b, 3 a+1).
\end{array}
$$

Then the coordinates of point $E$ are $\left(b, \frac{3 a+m b+3}{2}\right)$.
So the coordinates of point $G$ are
$$
x=\frac{a+2 b}{3}, \quad y=\frac{6 a+m b+4}{3},
$$

Since $G$ lies on $B E$ and $A D$, we also have
$$
y=\frac{a+2 b}{3} \cdot 2+1
$$

and $\quad y=\frac{a+2 b}{3} m+2$,
From (1) and (2), we get $6 a+m b+4=3 a+6 b+3$,
which simplifies to $(m-6) b+3 a+1=0$.
Similarly, from (1) and (3), we have
$$
(m-6) a+m b+2=0.
$$

From (4) and (3), we can get $m=12$.
Similarly, by setting up as shown in the right figure, we can solve for $m=\frac{3}{4}$.
In summary, $m$ has two values.
Therefore, the answer is (C).