What is the maximum number of positive integers among which any three are sides of an obtuse triangle?

I can find four, $11,11,16,20$. Is it possible to get five or more? We need $a^2+b^2<c^2$ and $a+b>c$ for all $a,b,c$.

  • $\begingroup$ The smallest possible quadruple is: $6,6,9,11$. The first case where all four numbers are distinct (and hence no two of the four obtuse triangles are congruent): $7,8,11,14$ $\endgroup$ – Jeppe Stig Nielsen May 2 '16 at 10:14
  • $\begingroup$ @JeppeStigNielsen: How did you get those solutions? $\endgroup$ – Rory Daulton May 2 '16 at 11:01
  • $\begingroup$ @RoryDaulton I used brute force, a computer program. I ran through a lot of integer quadruples $(a,b,c,d)$ where (WLOG) $0<a\le b\le c\le d$. I checked a bunch of inequalities to make sure all four combinations were in fact triangles, and to check that those triangles were in fact obtuse. Upon thinking a bit more, when $0<a\le b\le c\le d$ is known, we need to check only $a+b>d$ to make sure all combinations are triangles (triangle inequality). And we need to check only $b^2+c^2<d^2$ and $a^2+b^2<c^2$ to make sure every triangle is obtuse (from law of cosines, generalizing Pythagoras). $\endgroup$ – Jeppe Stig Nielsen May 2 '16 at 12:49

Suppose we have five positive integers $a_1\le a_2\le a_3\le a_4\le a_5$ such that any three are the sides of an obtuse triangle. Then by repeated substitution of $a_n^2+a_{n+1}^2<a_{n+2}^2$, $$a_5^2>a_4^2+a_3^2>2a_3^2+a_2^2>3a_2^2+2a_1^2$$ But $a_1+a_2>a_5$, so $$(a_1+a_2)^2>a_5^2>3a_2^2+2a_1^2=(a_2+a_1)^2+a_2^2+(a_2-a_1)^2$$which is impossible.

  • $\begingroup$ I guess this works for five positive real numbers as well. $\endgroup$ – Jeppe Stig Nielsen May 2 '16 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.