A vertical wall is $f$ meters away from a vertical fence of height $m$ meters. We want to place an $h$ meter ladder so that it rests against the wall, the ground, and the top of the fence. Determine the distance of the contact points from the lower edge of the wall, with an accuracy of cm.- Example: $m=f=5, h=15$.


1. The support is naturally understood to mean that both longitudinal parallel sides of the ladder rest on the ground, the fence, and the wall. Thus, the rungs are perpendicular to the wall's surface, the horizontal plane considered as the ground, and the intersection line of the wall and the ground; the problem simplifies, and our consideration can be carried out in the vertical plane passing through one of the rungs (rods).

1984-02-063-1.eps

Let the support points of the rod be $T, K$, and $F$, and the projections of the latter two on the ground be $K^{\prime}$ and $F^{\prime}$. Not forgetting the practical background of the problem, we consider the $T K^{\prime}=x$ segment as unknown, which determines the first support point of the ladder. Thus, $T F^{\prime}=x+f, F F^{\prime}=\sqrt{h^{2}-(x+f)^{2}}$, and based on the similarity of the triangles $T K K^{\prime}$ and $T F F^{\prime}$,

$$
\frac{T K^{\prime}}{K K^{\prime}}=\frac{T F^{\prime}}{F F^{\prime}}, \quad \frac{x}{m}=\frac{x+f}{\sqrt{h^{2}-(x+f)^{2}}}
$$

From this, by the usual rearrangement, we arrive at a quartic equation:

$$
x^{4}+2 f x^{3}-\left(h^{2}-f^{2}-m^{2}\right) x^{2}+2 f m^{2} x+f^{2} m^{2}=0
$$

2. A quartic equation can only be solved in special cases based on high school knowledge. Such a case arises from the equality $f=m$, as with the example measurements given. Then the equation simplifies as follows:

$$
x^{4}+2 m x^{3}-\left(h^{2}-2 m^{2}\right) x^{2}+2 m^{3} x+m^{4}=0
$$

Among the known methods for solving quartic equations, several rely on transforming the expression into the difference of two squares. The left side can now be transformed into the difference of the squares of two polynomials in $x$, and then into a product:

$$
\begin{gathered}
\left(x^{2}+m x+m^{2}\right)^{2}-\left(h^{2}+m^{2}\right) x^{2}=0 \\
\left(x^{2}+m x+m^{2}-\sqrt{h^{2}+m^{2}} \cdot x\right)\left(x^{2}+m x+m^{2}+\sqrt{h^{2}+m^{2}} \cdot x\right)=0
\end{gathered}
$$

Since in our specific problem we expect a positive value for $x$, only the vanishing of the first factor can provide a solution:

$$
x^{2}-\left(\sqrt{h^{2}+m^{2}}-m\right) x+m^{2}=0
$$

from which

$$
x=\frac{1}{2}\left(\sqrt{h^{2}+m^{2}}-m \pm \sqrt{h^{2}-2 m \sqrt{h^{2}+m^{2}}-2 m^{2}}\right)
$$

Substituting the numerical values,

$$
x=\frac{5}{2}(\sqrt{10}-1 \pm \sqrt{7-2 \sqrt{10}})
$$

$x_{1}=7.460 \ldots, x_{2}=3.351 \ldots$

Thus, the lower end of the ladder can be supported in two ways, 7.46 meters and 3.35 meters from $K^{\prime}$, and these solutions meet the requirement of accuracy to the nearest centimeter. The two solutions are naturally symmetric about the angle bisector of $F^{\prime} K$.

Finally, the ends of the ladder touch the lower edge of the wall at 12.46 and 8.35 meters, respectively. It is easy to verify that these results meet the practical question.

3. If we do not notice the possibility of reduction, we can seek an approximate solution by trial and error on a scaled-down model of the figure (e.g., 1:100), easily adjust the rod, and obtain approximate values of 7.5 and 3.3 for $x$. These can then be refined as needed. With our numerical data,

$$
f(x)=x^{4}+10 x^{3}-17 x^{2}+250 x+625=0
$$

from this, according to the 522nd operational schema of the School Function Tables,

$$
\begin{aligned}
f(x) & =\{[(x+10) x-175] x+250\} x+625 \\
f(7.5) & =+39.06 \ldots, \quad f(3.3)=+22.21 \ldots \\
f(7.4) & =-57.10 \ldots, \quad f(3.4)=-21.3 \ldots
\end{aligned}
$$

Based on these, estimating graphically - we expect the roots around 7.46 and 3.35, since $f(x)$ is continuous everywhere, so it also takes the value 0 between the positive and negative values at the points $x=7.5$ and $x=7.4$. To decide on the correctness of the rounding, we perform 2 more substitutions:

$$
\begin{aligned}
& f(7.455)=-5.1 \ldots0 \\
& f(7.465)=+4.5 \ldots>0, \quad f(3.355)=-1.7 \ldots<0
\end{aligned}
$$

Thus, the sought 0 points, rounded to the nearest centimeter, 0.01 meters, are 7.46 and 3.35, as in the above calculations.

Remark. If we choose the $T F^{\prime}$ segment as the unknown, the constant term of the quartic equation at (1) will be negative: $-f^{2} m^{2}$, or $-m^{4}$. It is somewhat more complicated, but the factorization into a product is also possible.