6. On a rectangular sheet of paper, a circle is drawn, inside which Misha mentally selects $n$ points, and Kolya tries to guess them. In one attempt, Kolya indicates one point on the sheet (inside or outside the circle), and Misha tells Kolya the distance from it to the nearest unguessed point. If this distance turns out to be zero, the indicated point is then considered guessed. Kolya can mark points on the sheet, measure distances, and perform constructions with a compass and ruler. Can Kolya definitely guess all the selected points in fewer than $(n+1)^{2}$ attempts?

## Variant 6
6. Let's first show how to guess one point in $2 n+1$ attempts. Suppose the line $l$ does not intersect the circle given in the problem's condition; draw a segment of this line and mark $n+1$ points $X_{0}, \ldots, X_{n}$ on it. Find out the distance from each of these points to the nearest hidden point. By the Pigeonhole Principle (see Fact 1), at least two (say, $X_{i}$ and $X_{j}$, $i<j$) of the marked $n+1$ points will have the same distance to the same hidden point $Y$.

Lemma. For all $k$ such that $i<k<j$, Misha will also name the distance to point $Y$.

Proof. Let $\omega_{i}$ and $\omega_{j}$ be the circles centered at points $X_{i}$ and $X_{j}$, respectively, passing through point $Y$. Let $Y^{\prime}$ be the second point of intersection of these circles (the case of tangent circles is left to the reader). Let $\omega$ be the circle centered at point $X_{k}$, passing through points $Y$ and $Y^{\prime}$ (Figure 192). It is sufficient to prove that the circle $\omega$ is contained within the union of the circles bounded by $\omega_{i}$ and $\omega_{j}$.

First, note that $\omega$ intersects $\omega_{i}$ and $\omega_{j}$ only at points $Y$ and $Y^{\prime}$ (since circles intersect in at most two points). Therefore, each of the arcs into which circle $\omega$ is divided by points $Y$ and $Y^{\prime}$ either lies entirely inside the union of the circles or entirely outside. To prove that the second case is impossible, it is enough to check that the points of intersection of $\omega$ and line $l$ lie inside the union of the circles bounded by $\omega_{i}$ and $\omega_{j}$.

Indeed, let $Z$ be one of the points of intersection of $\omega$ and line $l$. Without loss of generality, we can assume that it lies on the extension of segment $X_{k} X_{i}$ beyond point $X_{i}$. Then $X_{i} Z = X_{k} Y - X_{k} X_{i} < X_{i} Y$, so $Z$ lies inside circle $\omega_{i}$. The lemma is proved.

By the lemma, if we draw circles centered at the marked $n+1$ points with radii equal to the distances named by Misha for these points, then one of the intersection points of the circles corresponding to adjacent points will be the hidden point. Moreover, of the two intersection points of the circles, the one that lies on the same side as the circle given in the problem's condition should be chosen. Therefore, if Kolya sequentially names all such intersection points of the circles ($n$ attempts), then at least once the distance named by Misha will be zero.

Thus, we have guessed one point in $2 n+1$ attempts. Acting similarly, we will guess all points in

$$
(2 n+1) + (2 n-1) + \ldots + 3 = (n+1)^{2} - 1
$$