A box contains 5 rods whose lengths make triangles.

Question

A box contains five rods whose lengths are 1", 3", 6", 10", 15". How many different obtuse triangles can be made using only three rods at a time.

I determined that the answer is 1 because the triangle inequality nullified the others without using the Pythagorean inequalities. I feel like brute force and a bit of the triangle inequality is need the only way to solve this. Is there a more efficient way of solving it?

brute force will not be so brute since you have precise datas. — , Jul 24 '16 at 17:48
The version of this problem with $n$ lengths can be solved in time $\mathcal O(n)$ using the two pointer method. — Asinomás, Jul 24 '16 at 17:50
Even if you actually did, it sounds like you didn't check that the one possible triangle is obtuse. — Joffan, Jul 24 '16 at 17:57

score 3 · Answer 1 · answered Jul 24 '16 at 17:46

The smallest length can't be $1$ because all bigger lengths differ by more than $1$.

The smallest length can't be $3$ because all bigger lengths differ by more than $3$.

The only possibility is therefore $6,10,15$ and it works, since $6+10<15$ and $36+100<225$.

score 0 · Answer 2 · answered Jul 24 '16 at 17:51

You just need the triangle inequality (as you say) to get the one viable triangle and then check that $a^2 + b^2 < c^2$ to see if you have an obtuse triangle. You're choosing $3$ from $5$ so there are only ${3 \choose 5 } = 10$ options.

\begin{array}{c|c|c} a & b & c & TI & \text{obtuse}\\ 1 & \text{any} & \text{any} & \color{red} \times\\ 3 & 6/10 & 10/15 & \color{red} \times \\ 6 & 10 & 15 & Y & 36+100 < 225 \quad \color{blue} \checkmark\\ \end{array}

A box contains 5 rods whose lengths make triangles.

2 Answers2