Four brothers inherited a circular meadow. Three of them have one goat each. One of the brothers suggests that they tie the 3 goats with equally long chains and secure the ends of the chains to suitable points on the perimeter of the meadow, so that the goats can graze one-quarter of the meadow each, as allowed by the chain, and finally the fourth brother will mow what the goats cannot reach. However, another brother is concerned that the areas accessible to the individual goats will overlap.

a) Show that this situation can be avoided.

b) Determine by trial and error what percentage of the meadow's radius the chains should be. (It is sufficient to determine 2 significant figures.)
a) The sections of the meadow that can be grazed by individual goats do not overlap if, for example, the three tethering points of the chain are chosen as the vertices of an equilateral triangle \(ABC\) inscribed in a circle \(k\) that encloses the meadow, and of course, the length of the chains is such that each goat can only reach \(1/4\) of the meadow. We will show that circles centered at \(A\), \(B\), and \(C\) that are tangent to each other and have the same radius - or \(AB/2\) radius - cover more than \(1/4\) of the area of \(k\). Knowing this, it is obvious that by keeping the points \(A\), \(B\), and \(C\) unchanged and reducing the radius to less than \(AB/2\), a radius \(\varrho\) can be found such that the circle centered at \(A\) (and similarly at \(B\) and \(C\)) with radius \(\varrho\) covers \(1/4\) of the meadow, and there is no overlap.

![](https://cdn.mathpix.com/cropped/2024_05_02_bd6b240496d6b0a63396g-1.jpg?height=589&width=512&top_left_y=459&top_left_x=793)

Figure 1

Let the circle centered at \(A\) with radius \(AB/2\) intersect \(k\) at point \(D\) on the shorter arc \(AB\). This point is not on the chord \(AB\), so it lies on the side of the perpendicular bisector of \(AB\) that contains \(A\). Therefore, if \(F\) is the midpoint of the arc \(AB\) (Figure 1),

\[
\angle DAB > \angle FAB = 30^\circ
\]

Thus, the central angle corresponding to the arc of \(k_1\) that lies inside \(k\) is greater than \(120^\circ\), so the area \(t\) of the part of \(k_1\) that lies inside \(k\) is greater than one-third of the area of \(k_1\):

\[
t > \frac{1}{3} \pi \left(\frac{AB}{2}\right)^2
\]

The right-hand side is exactly one-fourth of the area of \(k\):

\[
\frac{\pi}{12} AB^2 = \frac{\pi}{12} \cdot 3R^2 = \frac{\pi R^2}{4}
\]

where \(R\) is the radius of \(k\); indeed, by a simple calculation, \(AB = R \sqrt{3}\).

b) Let the length of the chain in question be \(\varrho\), the two intersection points of the circle centered at \(A\) with radius \(\varrho\) and \(k\) be \(M\) and \(N\), the center of \(k\) be \(O\), the midpoint of \(MN\) be \(G\), and \(\angle AOM = x\) (measured in degrees; Figure 2), which we will approximate.

![](https://cdn.mathpix.com/cropped/2024_05_02_bd6b240496d6b0a63396g-1.jpg?height=502&width=702&top_left_y=1904&top_left_x=687)

Figure 2

The part of the meadow that the goat can reach is divided into the segments cut off by \(AM\) and \(AN\) from \(k\) and the sector \(AMN\), and the former are obtained by removing the area of the kite \(OMAN\) from the sector \(OMN\). Clearly, \(\angle MAN = 180^\circ - x\), so the requirement, using symmetry, is

\[
\varrho^2 \pi \cdot \frac{180^\circ - x}{360^\circ} + R^2 \pi \cdot \frac{2x}{360^\circ} - OA \cdot MG = \frac{1}{4} R^2 \pi
\]

where \(MG = R \sin x\), and \(\varrho = 2R \sin \frac{x}{2}\). Simplification in the first term is given by the following transformation using the right triangle \(AMG\):

\[
\varrho^2 = 2R \varrho \sin \frac{x}{2} = 2R \cdot AG = 2R(OA - OG) = 2R^2(1 - \cos x)
\]

since by the inscribed angle theorem, \(\angle AMG = \angle AMN = \angle AON / 2\). Our equation thus becomes:

\[
\left(1 - \frac{x}{180^\circ}\right) \cos x + \frac{\sin x}{\pi} = \frac{3}{4}
\]

and since \(\varrho < R \sqrt{3} / 2\), therefore

\[
\sin \frac{x}{2} = \frac{\varrho}{2R} < \frac{\sqrt{3}}{4} < 0.4331, \quad \frac{x}{2} < 25^\circ 40', \quad x < 51^\circ 20'
\]

The value of the left-hand side at \(x = 45^\circ\) is 0.7554, which is greater than the right-hand side, and at \(x = 46^\circ\) it is 0.7461, which is already smaller. The sought value of \(x\) is expected around \(45^\circ 35'\) based on the excess of 0.0054 and the deficit of 0.0039. The values of \(\varrho / R\) corresponding to the chosen angles are 0.7654 and 0.7814, so only 77% and 78% are feasible.

At \(x = 45^\circ 35'\), the left-hand side of the equation is 0.7500, so this is the sought value of the angle, and to this \(\varrho / R = 0.7748\), so to three significant figures, \(\varrho\) is 77.5% of \(R\).

János Kuhár (Budapest, Berzsenyi D. Gymnasium, 1st year)