81. A branch of diameter $d$ (it can have any shape; we neglect the thickness of the branch) is floating through a channel of width 1, which at a certain point turns $90^{\circ}$ relative to its initial direction (Fig. 33). What is the maximum $d$ for which the branch can have any shape and still pass the turn without getting stuck?

![](https://cdn.mathpix.com/cropped/2024_05_21_d4ffc236376258137647g-071.jpg?height=346&width=324&top_left_y=486&top_left_x=196)

Fig. 33.
81. First solution. Draw a circle through the "inner angle" M of the channel, touching its "outer" banks at points \(A_{0}\) and \(B_{0}\) (Fig. 142a); it is clear that the radius \(R = O A_{0} = O M = O B_{0}\) of this circle is determined from the relationship (note that the side of the square Oaмb is \(R-1\))

\[
2(R-1)^{2} = R^{2}, \quad \text{from which} \quad R = 2 + \sqrt{2} (\approx 3.41)
\]

and thus

\[
A_{0} B_{0} = R \sqrt{2} = 2 \sqrt{2} + 2 \quad (\approx 4.83)
\]

It is not difficult to understand that a branch in the form of the arc \(A_{0} B_{0}\) can pass through the channel bend: indeed, when rotated around point \(O\) by \(45^{\circ}\) in one or the other direction, \(A_{0} B_{0}\) takes the position \(A_{0}^{\prime} M\) or \(M B_{0}^{\prime}\) (Fig. 142a) - and in both these positions, the branch can freely move along one or the other bend of the channel.

Now let's prove that if the diameter of the branch \(A B = d > 2 \sqrt{2} + 2\), then regardless of its shape, it will not be able to pass through the channel bend. Indeed, if the branch is positioned in the straight section of the channel with the bank \(M A_{0}^{\prime}\) and if its end \(B\) is closer to the channel bend than the end \(A\), then (directed) segment \(\overline{A B}\) (whose length exceeds \(2 \sqrt{2} + 2\)) forms an obtuse angle with the "bisector" \(\overline{M N}\) of the channel (i.e., with the common bisector of the angles formed by both banks of the channel). Indeed, if we parallel shift the branch to the position \(A_{1} B_{1}\) such that the end \(B_{1}\) coincides with \(M\), then the end \(A_{1}\) of the segment \(B_{1} A_{1}\) will occupy a position on the arc \(P Q\) of the circle with radius \(d > 2 \sqrt{2} + 2\) centered at \(M\). But if \(A_{1} \in P Q\), then \(\angle A_{1} M A > 90^{\circ}\) (since \(d > 2 \sqrt{2} + 2 > \sqrt{2}\)), and thus \(\angle N M A_{1} > 90^{\circ}\). Similarly, when the branch, after the bend, moves along the second "bend" of the channel, it will take the position \(A_{2} B_{2}\), where the point \(A_{2}\) will be closer to the bend than \(B_{2}\) (since the branch with a diameter \(> \sqrt{2}\) cannot "flip" and continue moving with the other end forward); in this case, the angle between \(\overline{B_{2} A_{2}}\) and \(\overline{M N}\) will also be obtuse, and the angle between \(\overline{A_{2} B_{2}}\) and \(\widehat{M N}\) (note the direction of the segments!) will be acute. Since the angle between \(\overline{A B}\) and \(\overline{M N}\) changes continuously during the branch's movement along the river, the branch must pass through a position \(\bar{A} \bar{B}\) such that \(\angle(\bar{A} \bar{B}, M N) = 90^{\circ}\), i.e., \(\bar{A} \bar{B} \perp M N\).

Let \(C\) and \(D\) be the points of intersection of the segment \(\bar{A} \bar{B}\) with the banks \(M A_{0}^{\prime}\) and \(M B_{0}^{\prime}\) of the river (Fig. 142, \(\sigma\)). Since \(C \bar{A} \leqslant C A^{\prime} = \sqrt{2}\) and \(D \bar{B} = (2 \sqrt{2} + 2) - \sqrt{2} - \sqrt{2} = 2\),

it follows that if \(M P\) is the distance from \(M\) to \(\bar{A} \bar{B}\), then

\[
M P = \frac{1}{2} C D > 1
\]

and even more so, the distance \(F P\) from \(\bar{A} \bar{B}\) to the point \(F\) of intersection of the branch \(\bar{A} \bar{B}\) with the segment \(M N\) is greater than 1.

Now imagine the branch \(\bar{A} \bar{B}\) moving along one of the straight arms of the channel. In this case, all points \(\bar{A}, \bar{B}\), and \(F\) will be inside the channel (which can be understood simply as a strip of width 1 bounded by parallel lines); therefore, the entire \(\triangle \bar{A} \bar{B} F\) will be inside the channel (due to the convexity of the strip; see the definition on page 56). But this is impossible, because the width of the obtuse-angled triangle \(\bar{A} \bar{B} F\) (which is already \(\angle \bar{A} M \bar{B}\), which is less than \(\angle \bar{A} F \bar{B}\)), i.e., the width of the narrowest strip in which this triangle can be enclosed, is equal to its smallest height, i.e., \(F P > 1\). From this it follows that the branch \(\bar{A} \bar{B}\) will not be able to pass through the channel of width 1.

Second solution. This problem can also be solved similarly to the solution (similar in theme) of problem 49. The possibility of passing for a branch with a diameter of \(2 \sqrt{2} + 2\) is easily established (see the first solution); the main difficulty lies in establishing that no branch with a diameter \(d = A B > 2 \sqrt{2} + 2\) will be able to pass the channel bend without getting stuck. Imagine a rigidly connected to the branch \(A B\) and enclosing the branch inside a rectangle \(\Pi = A_{1} B_{1} B_{2} A_{2}\) of width 1, two sides of which lie on the banks of the channel (Fig. 142, b), and follow the movement of this rectangle during the movement of the branch. As in the first solution, it is shown that when the branch moves forward and turns, the segment \(A B\), and with it the rectangle \(\Pi\), will rotate more than \(45^{\circ}\). Focus now on the moment in the process of the branch's movement when the sides \(A_{1} B_{1}\) and \(A_{2} B_{2}\) of the rectangle \(\Pi\) are perpendicular to the "bisector of the channel" \(N M\).

Draw a line \(C^{\prime} D^{\prime} = l^{\prime} \perp M N\) through point \(M\) and on the other side of it, relative to point \(N\), draw a line \(C_{1}^{\prime} D_{1}^{\prime} = l_{1}^{\prime} \| l^{\prime}\), at a distance of 1 from \(l^{\prime}\). If the diameter \(A B\) of the branch is \(> 2 \sqrt{2} + 2\), then at this moment there will be a point of the branch located in the triangle \(C^{\prime} N D^{\prime}\); after all, the diameter of each of the rhombuses \(C_{1}^{\prime} c M C^{\prime}\) and \(D_{1}^{\prime} d M D^{\prime}\) depicted in Fig. 142, b is \(C_{1}^{\prime} M = D_{1}^{\prime} M = 6 > 2 \sqrt{2} + 2\) (why?), and the branch will not be able to fit into either of these