Given the curve $y=x^{2}+1$. Let's determine points $P_{1}, P_{2}, \ldots, P_{n}$ on it such that their projections on the $x$-axis are denoted by $Q_{1}, Q_{2}, \ldots, Q_{n}$, respectively, and the tangent to the parabola at $P_{i}$ passes through $Q_{i+1}$ if $1 \leqq i \leqq n-1$ and through $Q_{1}$ if $i=n$. Find the number of different point $n$-tuples for $n=6$.
a) The equation of the tangent line to the parabola at the point $P\left(t, t^{2}+1\right)$ is

$$
y-\left(t^{2}+1\right)=2 t(x-t)
$$

According to this, the tangent line intersects the $x$-axis at the point $Q$ with abscissa

$$
x=\frac{t^{2}-1}{2 t}
$$

(t=0 can be excluded because the tangent at (0, 1) does not intersect the $x$-axis and thus cannot be a $P_{i}$). Therefore, the problem requires that for the $n$-tuple of points $\left(P_{1}, P_{2}, \ldots, P_{n}\right)$, the $n$-tuple of abscissas $T\left(t_{1}, t_{2}, \ldots, t_{n}\right)$ satisfies

$$
t_{i+1}=\frac{t_{i}^{2}-1}{2 t_{i}}
$$

for $i=1,2, \ldots, n-1$, and the $t_{n+1}$ derived from this should be equal to $t_{1}$.

The basis of our solution is the (fortunate) observation that if $\alpha_{i}$ is an angle such that ctg $\alpha_{i}=t_{i}$, then

$$
t_{i+1}=\frac{\operatorname{ctg}^{2} \alpha_{i}-1}{2 \operatorname{ctg} \alpha_{i}}=\operatorname{ctg} 2 \alpha_{i}
$$

and furthermore, for every real number $t_{1}$, there exists a unique $\alpha_{1}$ such that $0<\alpha_{1}<\pi$ and $\operatorname{ctg} \alpha_{1}=t_{1}$. Then the numbers $t_{i}$ can be expressed using $\alpha_{1}$ as follows:

$$
t_{i}=\operatorname{ctg} 2^{i-1} \alpha_{1}
$$

and the requirement $t_{n+1}=t_{1}$ becomes:

$$
\operatorname{ctg} 2^{n} \alpha_{1}=\operatorname{ctg} \alpha_{1}
$$

Accordingly, the difference between $2^{n} \alpha_{1}$ and $\alpha_{1}$ is an integer multiple of $\pi$:

$$
\left(2^{n}-1\right) \alpha_{1}=k \pi, \quad \alpha_{1}=k \cdot \frac{\pi}{2^{n}-1}
$$

where, considering the requirement $0<\alpha_{1}<\pi$,

$$
0<k<2^{n}-1
$$

Thus, every $n$-tuple of points $\left(P_{i}, Q_{i}\right)$ that satisfies the problem's requirements corresponds to an initial value

$$
t_{1}=\operatorname{ctg} \frac{k \pi}{2^{n}-1} \quad\left(k=1,2, \ldots, 2^{n}-2\right)
$$

and the $n$-tuple $T$ derived from the recurrence (1). The number of such $n$-tuples is less than $2^{n}-2$, since each $n$-tuple can be obtained from at least two different points, and it is not possible that $P_{i+1} \equiv P_{i}$, and thus $Q_{i+1} \equiv Q_{i}$ either; however, as we saw in problem 1660, for $n=2$ there is a solution [and this applies to every $n=2 m$ (even number)].
b) For $n=6$ - as can be easily seen - the values

$$
k=1,3,5,7,11,15,19,23,31
$$

yield 6 different points; for the value

$$
k=21
$$

$P_{3} \equiv P_{5} \equiv P_{1}$ and $P_{4} \equiv P_{6} \equiv P_{2}$ (this is the solution of problem 1660, $\left.21 /\left(2^{6}-1\right)=1 / 3\right)$; finally, for the values

$$
k=9, \quad 27
$$

$P_{1}, P_{2}, P_{3}$ are different from each other, $P_{4} \equiv P_{1}$, and from here on the points repeat. Therefore, the number of different 6-tuples of points is 12. (It is a matter of interpretation whether to accept or exclude the latter two types of point systems. The problem did not exclude the possibility that the same point can appear more than once in the system.) The diagram shows the point system obtained for $k=11$, with $x$-axis stretching[^0]

![](https://cdn.mathpix.com/cropped/2024_05_02_e2e53c49bf79ea4c06f0g-2.jpg?height=821&width=1268&top_left_y=126&top_left_x=426)


[^0]:    ${ }^{1}$ See the solution in K. M. L. 39. (1969) 128.