We have a homogeneous material, square-shaped, thin plate with a side length of $20 \mathrm{~cm}$. Find a point on its perimeter such that when the plate is suspended from this point, the nearest corner will be the highest relative to the suspension point. Determine the distance between the suspension point and the nearest corner in this situation with an accuracy of one-tenth of a millimeter.
Let the side of the disk be $2a$, and the distance of the suspension point $P$ from the midpoint $F$ of the side it is on be $x$, so its distance from the nearest vertex $A$ is $a-x$.

![](https://cdn.mathpix.com/cropped/2024_05_02_23b3d8cc88f7703786f6g-1.jpg?height=800&width=702&top_left_y=256&top_left_x=706)

The center of gravity of the homogeneous disk $S$ is at the center of the square, so $PS$ will be in a vertical position, and the projection of $A$ on this is denoted by $Q$. We are looking for the maximum of the height difference $PQ=y$ depending on $FP=x$.

The right-angled triangles $PAQ$ and $PSF$ are similar because the angles at $P$ are vertical angles, so $\angle PSF = \angle PAQ$. Let's denote this angle by $\alpha$, then in the triangle $PSF$ we have $x = a \tan \alpha$ and in the triangle $PAQ$ we have

$$
y = PQ = AP \sin \alpha = (a-x) \sin \alpha = a(1-\tan \alpha) \sin \alpha
$$

Since $0 \leq x \leq a$, we have $0 \leq \alpha \leq 45^\circ$. The value of $y$ is 0 at the endpoints and positive in the interior of the domain, so $y$ can take its maximum in the interior of the domain. If there is an $\alpha$ value for which $y$ is maximal, then the derivative of $y$ with respect to $\alpha$ is 0:

$$
\frac{dy}{d\alpha} = -\frac{a}{\cos^2 \alpha} \sin \alpha + a(1-\tan \alpha) \cos \alpha = 0
$$

Since $a > 0$ and $\cos \alpha > 0$, we can divide by them, and also by $(-1)$:

$$
\frac{\sin \alpha}{\cos^3 \alpha} + \tan \alpha - 1 = 0
$$

Substituting $\sin \alpha$ with $\sin \alpha (\sin^2 \alpha + \cos^2 \alpha)$, we get a cubic equation for the unknown $x/a = \tan \alpha$:

$$
\tan^3 \alpha + 2 \tan \alpha - 1 = 0
$$

One of the roots of this equation, according to Cardano's formula, is:

$$
\frac{x}{a} = \sqrt[3]{\frac{1}{2} + \sqrt{\frac{1}{4} + \frac{8}{27}}} + \sqrt[3]{\frac{1}{2} - \sqrt{\frac{1}{4} + \frac{8}{27}}} = \sqrt[3]{\frac{1}{2} + \frac{\sqrt{177}}{18}} + \sqrt[3]{\frac{1}{2} - \frac{\sqrt{177}}{18}}
$$

The required 0.1 mm error limit for the data $a = 1 \text{dm}$ is $1/10^3$ of it, and we stay within this if we do not exceed the $5/10^4$ error in either of the cube roots.

According to our table, the value of $\sqrt{177}$ is between 13.30 and 13.31, and its 18th part is between 0.7390 and 0.7395, so the two cube roots are between 1.2390 and 1.2395, and -0.2395 and -0.2390, respectively.

The cube roots, calculated using logarithms, have values between 1.0740 and 1.0742 for the first, and between -0.6207 and -0.6206 for the second, so the value of $x/a$ is between 0.4533 and 0.4536, so we can take either 0.453 or 0.454, and the error committed is less than $10^{-3}$.

Equation (2) has no other real roots because the derivative of its left side,

$$
3\left(\frac{x}{a}\right)^2 + 2
$$

is positive everywhere, and the left side is strictly monotonically increasing. Thus, the only extremum can be at $x/a = 0.453$. Here there is indeed a maximum, because the sign of $\frac{dy}{d\alpha}$ is the opposite of the sign of the expression on the left side of (2), and since the derivative of the left side of (2) is positive, the left side of (2) is monotonically increasing in $\alpha$, so the expression on the left side of (2) is negative before the root and positive after it. Accordingly, $y$ as a function of $\alpha$ is monotonically increasing before the root and monotonically decreasing after it.

Considering $a = 10 \text{cm}$, the sought distance $PA = 10 - 4.53 = 5.47 \text{cm}$. (The maximum height difference is then $22.6 \text{mm}$.

Simon Júlia (Györ, Kazinczy F. Gimn.)

[^0]: ${ }^{1}$ See e.g. Hack Frigyes-Kugler Sándorné: Függvénytáblázatok, matematikai és fizikai összefüggések, Tankönyvkiadó, Budapest, 1968, $65-66$. o.