Track / Stage 6 / 105 of 400 #1105 of 1964

Problem 1105

National Olympiad, first round

Geometry Difficulty 6.0 Prove it Ukrainian National Mathematical Olympiad, Third Round, First Tour · Ukraine

There are $n \ge 3$ segments, their lengths in centimeters are distinct positive integers. It's known that it's possible to form a nondegenerate triangle from any three of these $n$ segments. Suppose that among these segments there are segments with lengths $5$ cm and $12$ cm. What's the largest value $n$ can attain?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Reorder the segments by their lengths, so that $a_1 < a_2 < \dots < a_n$ . Clearly, any three segments form a triangle if and only if the sum of the lengths of the smallest two segments is larger than the length of the longest segment. So, the smallest segment except from the given two can't have a length smaller than $8$ , as in that case we wouldn't be able to form a triangle from segments $5$ , $a \le 7$ and $12$ , as $5 + 7 \le 12$ .

So, $a_1 = 5$ and $a_2 \ge 8$ . Clearly, $a_2 \le 12$ and the number of segments can't exceed the number of elements of the set $\{5, a_2, a_2 + 1, a_2 + 2, a_2 + 3, a_2 + 4\}$ , so $n \le 6$ .

Note that the set of segments with lengths $\{5, 8, 9, 10, 11, 12\}$ satisfies the conditions. Therefore, the answer is $n = 6$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.