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Geometry Difficulty 8.2 Shortlist Prove it

Let OO be a point (in the plane) and TT be an infinite set of points such that P1P22012|P_1P_2| \le 2012 for every two distinct points P1,P2TP_1,P_2\in T. Let S(T)S(T) be the set of points QQ in the plane satisfying QP2013|QP| \le 2013 for at least one point PTP\in T.

Now let LL be the set of lines containing exactly one point of S(T)S(T). Call a line 0\ell_0 passing through OO [i]bad[/i] if there does not exist a line L\ell\in L parallel to (or coinciding with) 0\ell_0.

(a) Prove that LL is nonempty.

(b) Prove that one can assign a line (i)\ell(i) to each positive integer ii so that for every bad line 0\ell_0 passing through OO, there exists a positive integer nn with (n)=0\ell(n) = \ell_0.

[i]Proposed by David Yang[/i]

Solution

### Part (a)

1. **Define the function d(X) d(X) for XS(T) X \in S(T) **:
- Choose an arbitrary point T0T T_0 \in T .
- Define d(X) d(X) as the distance from X X to T0 T_0 .

2. **Boundedness of S(T) S(T) **:
- Notice that S(T) S(T) is bounded by a circle of radius 2012+2013=4025 2012 + 2013 = 4025 centered at T0 T_0 .
- This is because any point QS(T) Q \in S(T) satisfies QP2013 |QP| \le 2013 for some PT P \in T , and P1P22012 |P_1P_2| \le 2012 for any P1,P2T P_1, P_2 \in T .

3. **Closedness of S(T) S(T) **:
- S(T) S(T) is closed. For any convergent sequence of points in S(T) S(T) whose limit has a distance strictly more than 2013 2013 from all points in T T , we can find some point far enough along in the sequence that it too has a distance strictly more than 2013 2013 from all points in T T , contradicting its membership in S(T) S(T) .

4. **Compactness of S(T) S(T) **:
- Since S(T) S(T) is both bounded and closed, it is a compact set.

5. **Existence of a maximum for d d **:
- The continuous function d d on S(T) S(T) attains a maximum at some point S0 S_0 .

6. **Constructing a line in L L **:
- The line through S0 S_0 perpendicular to T0S0 T_0S_0 will intersect S(T) S(T) at exactly one point, S0 S_0 .
- Therefore, L L is nonempty.

### Part (b)

1. Definition of a bad line:
- A line 0 \ell_0 passing through O O is bad if there does not exist a line L \ell \in L parallel to (or coinciding with) 0 \ell_0 .

2. **Sweeping lines parallel to 0 \ell_0 **:
- If 0 \ell_0 is bad, then sweeping lines parallel to 0 \ell_0 across the plane means each one that intersects S(T) S(T) intersects it multiple times.
- This implies it intersects the convex envelope of S(T) S(T) multiple times.

3. Convex envelope and line segments:
- The convex envelope of S(T) S(T) must have a finite number of line segments of positive length.
- Suppose, for the sake of contradiction, there were an uncountably infinite number of (positive length) line segments.

4. Lemma: Sum of an uncountably infinite set of strictly positive numbers is infinite:
- Argue by contradiction and assume it were finite and equal to some M M .
- For each nN n \in \mathbb{N} , the set {xSx>1/n} \{x \in S | x > 1/n \} is finite (in particular, has size at most Mn Mn ).
- The union of these sets is S S , but the countable union of finite sets is countable, contradicting the uncountability of S S .

5. Contradiction:
- It follows that the convex envelope of S(T) S(T) must have infinite perimeter, but finite area (from the boundedness of S(T) S(T) ).
- Consider the set S(T) S'(T) of points which are within some ϵ>0 \epsilon > 0 of some point in S(T) S(T) ; clearly this figure has bounded area as well.
- By convexity, [S(T)][S(T)]Pϵ[S'(T)] - [S(T)] \ge P\epsilon where P P is the perimeter of the convex envelope of S(T) S(T) .
- This contradicts the fact that S(T) S(T) has infinite perimeter.

6. Countability of bad lines:
- Therefore, the set of bad lines is countable.
- We can assign a line (i) \ell(i) to each positive integer i i so that for every bad line 0 \ell_0 passing through O O , there exists a positive integer n n with (n)=0 \ell(n) = \ell_0 .

\blacksquare

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.