Maths Olympiad Prep

Library / /12 of 15

Geometry Difficulty 5.8 AIME, harder Prove it Philippines

Problem:
A triangle has sides of lengths 2020 and 1919. If the triangle is not acute, how many possible integer lengths can the third side have?

Solution

Solution:
We begin by determining the possible integer lengths of the third side ss. By the triangle inequality, 2s382 \leq s \leq 38.

Now, we determine when the triangle is not acute. This means that the triangle is either a right triangle or an obtuse triangle. Since we are only considering integer lengths, notice that the third side will always be the shortest or the longest side. These are precisely the two cases to consider.

We use the Pythagorean Theorem to determine when the triangle is right or obtuse. If the two shorter sides of the triangle are aa and bb and the longest side is cc, then the triangle is right or obtuse when a2+b2c2a^{2}+b^{2} \leq c^{2}.

If the third side is the shortest, then by the above inequality we have that s2202192s^{2} \leq 20^{2}-19^{2}. This gives us s239s^{2} \leq 39. This in turn means that 2s62 \leq s \leq 6 for this case, a total of 55 possible lengths of the third side.

If the third side is the longest, then by the above inequality we have that 192+202s219^{2}+20^{2} \leq s^{2}. This gives us s2761s^{2} \geq 761. This in turn means that 28s3828 \leq s \leq 38 for this case, or a total of 1111 possible lengths of the third side.

Thus, there are 5+11=165+11=16 possible lengths of the third side.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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