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Discrete Geometry and Extremal Graph Theory - IMSA

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We prove the following result using a similar technique.<br />

Theorem 2. A graph G on n vertices without a “Θ-graph” contains at most c ′ n 3/2 edges<br />

for some constant c ′ > 0.<br />

Proof of Theorem 2 (sketch). As in the proof of Theorem 1, let d 1 , d 2 , . . . , d n be the degrees<br />

of the vertices v 1 , v 2 , . . . , v n . For a vertex v i , we count the pairs of the d i vertices<br />

connected to v i . As we continue this count for all the vertices of G, note that each pair is<br />

counted at most twice, or else G must have a “Θ-graph.” In other words, we arrive at the<br />

inequality<br />

( ) ( ) ( ) (<br />

Create PDF with GO2PDF for free, if you d1 wish to d2 remove this line, dn click here n to buy Virtual PDF Printer<br />

2)<br />

2<br />

+<br />

2<br />

+ · · ·<br />

Thus, as in the proof of Theorem 1, we see c ′ ≈ √ 2c will work for sufficiently large n. <br />

Finally, we prove our main result, showing an unexpected connection between Erdős’ problem<br />

on unit distances <strong>and</strong> graph theoretic results.<br />

Theorem 3. Let f(n) be the maximum number of unit distances among n points in the<br />

plane. Then, f(n) ≤ c ′ n 3/2 .<br />

Proof of Theorem 3. Consider the unit distance graph of the n points defined as follows: two<br />

points are joined by a segment if <strong>and</strong> only if they are unit distance apart. Now, note that<br />

no “Θ-graph” can appear in such a graph, for let us consider the two end vertices. The two<br />

unit circles centered at these vertices must each pass through three common points, but two<br />

circles can intersect in at most two points, contradiction. Thus, by Theorem 2, the result<br />

follows. <br />

2<br />

≤ 2<br />

.<br />

3

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