Consider the figure bounded by the positive halves of the coordinate axes and the arc of the curve $y=\cos x$ over the interval $(0, \pi / 2)$.

a) Divide the figure into two equal areas with a line parallel to the $y$-axis,

b) Divide the figure into two equal areas with a line parallel to the $x$-axis,

c) Divide the figure into two equal areas with a line passing through the origin,

d) Divide the figure into two equal areas with a line starting from the point $(0 ; 1)$,

e) Divide the figure into two equal areas with a line starting from the point $(\pi / 2,0)$.

(For cases where an exact solution is not possible, provide an approximate solution and give an upper bound for the error of the result.)
The area of the $OAB$ sector to be bisected

$$
\int_{0}^{\pi / 2} \cos x \, dx = 1
$$

According to this, the line bisecting the area must divide it into two parts, each with an area of $1/2$ square units. The equation of the bisecting line can be precisely determined under conditions a), d), and e), while in cases b) and c) it can only be approximated.

In case a), let the equation of the sought line be $x = a$; from the condition

$$
\int_{0}^{a} \cos x \, dx = \sin a = \frac{1}{2} \text{, we get } a = \frac{\pi}{6}
$$

In case d), the part of the $OAB$ sector below the line is a right triangle with a height of 1, so its base is also 1, and the line intersects the x-axis at the point $(1; 0)$, with the equation $y = 1 - x$.

Similarly, in case e), the base of the right triangle is $\pi / 2$, so its height is $2 / \pi$, which is also the y-intercept of the line, and the equation is:

$$
\frac{x}{\pi / 2} + \frac{y}{2 / \pi} = 1, \quad y = \frac{2}{\pi} - \frac{4}{\pi^2} x, \quad \text{or } 4x + \pi^2 y - 2\pi = 0
$$

In case b), we seek the equation of the area-bisecting line in the form $y = b$ and denote the arc (angle) for which $\cos \beta = b$ by $\beta$.

For the area above the sought line in the $OAB$ sector, the condition is

$$
\int_{0}^{\beta} (\cos x - b) \, dx = [\sin x - bx]_{0}^{\beta} = \sin \beta - \beta \cos \beta = \frac{1}{2}
$$

Thus, we need to solve the following equation:

$$
f(\beta) = \sin \beta - \beta \cos \beta - \frac{1}{2} = 0
$$

We will return to this after setting up the equation for the remaining case c).

Finally, in case c), we seek the area-bisecting line in the form $y = cx$ and denote the x-coordinate of its intersection with the curve $y = \cos x$ by $\gamma$. Thus,

$$
c \gamma = \cos \gamma, \quad c = \frac{\cos \gamma}{\gamma}
$$

For the area above the line in the $OAB$ sector, we have

$$
\int_{0}^{\gamma} (\cos x - cx) \, dx = \sin \gamma - \frac{c}{2} \gamma^2 = \sin \gamma - \frac{\gamma}{2} \cos \gamma = \frac{1}{2}
$$

Thus, we need to solve the following equation, similar to (1):

$$
g(\gamma) = \sin \gamma - \frac{\gamma}{2} \cos \gamma - \frac{1}{2} = 0
$$

The value of the left-hand side of (1) at $\beta = \pi / 2 \approx 1.571$ is 0.5; at $\beta = \pi / 6$ it is $-\pi \sqrt{3} / 12 \approx -0.45$; and at the midpoint of these two values, $\beta = \pi / 3 \approx 1.047$, it is $(3 \sqrt{3} - \pi - 3) / 6 \approx -0.134$. Since the left-hand side is a continuous function of $\beta$, we expect its graph to cross the x-axis from below to above between 1.047 and 1.571, much closer to the former. Using the 7th table of the school Function Table (trigonometric functions for angles measured in radians), we find:

$$
\begin{array}{llll}
\beta = 1.20 \text{ gives: } & \sin \beta = 0.9320, \quad \cos \beta = 0.3624, & f(1.20) = -0.00290
\end{array}
$$

We see that the ratio of $|f(1.21)|$ to $|f(1.20)|$ is approximately 3, so by linear interpolation,

$$
\beta = 1.20 + \frac{10^{-2}}{4}
$$

is an approximate root of (1), which cannot be refined further using our table.

![](https://cdn.mathpix.com/cropped/2024_05_02_c0d994c29e677fef56a1g-2.jpg?height=776&width=1111&top_left_y=127&top_left_x=496)

Thus, the equation of the area-bisecting line in case b) is

$$
y = b = \cos \beta = 0.360
$$

and we can accept that the result has 3 significant digits. (Since $b$ was obtained by performing multiple operations on data with 4 significant digits, we cannot expect greater precision.)

Essentially the same method gives us, from the values $g(0.90) = +0.0036 (>0)$ and $g(0.89) = -0.0030 (<0)$ in (2),

$$
\gamma = 0.89 + \frac{30 \cdot 10^{-4}}{(30 + 36) 10^{-4}} \cdot 10^{-2} = 0.89 + \frac{0.05}{11}
$$

from which $\cos \gamma = 0.6260$ and $c = 0.700$, again with 3 significant digits, so the equation of the bisecting line is $y = 0.700 x$.

## József Kópházi (Tatabánya Árpád Gimn., III. o. t.)

Remarks. 1. Most of the solutions used the tables of arcs, sines, and cosines for angles measured in degrees (the 5th and 8th tables of the school Function Table). This makes the calculations more difficult and less precise.

2. From the results of a) and b), we can read the further bisecting of the $OAB$ sector in an analogous manner, in the sense that the line $x = a$ can be considered a translation of the boundary segment $OB$ along $OA$, and the line $y = b$ a translation of $OA$ along $OB$. We can ask: by how much must the arc $AB$ be translated, holding it at $A$ along $OA$, and holding it at $B$ along $OB$, so that in its new position it divides $OAB$ into two equal areas.

Since half of the area of $OAB$ lies to the left of the line $x = \pi / 6$ and half to the right, the latter part must be translated to the origin by holding it at the point $(\pi / 6; 0)$, so the translation is $-a = -\pi / 6$. Similarly, the translation along the y-axis by $b = -0.360$ is the solution to the other question.