Each triangle has 6 basic determinants: 3 line segments and 3 angles. Is it possible to have two triangles, which have 5 of the basic determinants in common, but are not congruent?

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    $\begingroup$ By "line segment" do you mean the length of a side? $\endgroup$ Dec 2 '19 at 5:44
  • $\begingroup$ How is this not just a basic math problem that is frequently flagged and closed around here? $\endgroup$ Dec 2 '19 at 16:01
  • $\begingroup$ @ThomasMarkov Because it takes some creativity to come up with the solution. $\endgroup$ Dec 9 '19 at 1:02

Yes. A triangle with sidelengths 4,6,9 and one with sidelengths 6,9,13.5 have this property. (I know their angles are congruent because their side, taken pairwise in increasing order, are in equal proportions. And I know such a triangle exists by the converse of the triangle inequality.)


Any triangle with sides $a,ak,ak^2$ with $a+ak\gt ak^2$ is similar to a triangle with sides $ak,ak^2, ak^3$. An interesting thing about this is the value of $k$. We have $k^2-k-1\lt0$, so $k=\frac{1\pm\sqrt{5}}{2}$ (see the Golden ratio) gives the roots, $k$ must be positive to avoid negative lengths, and so $0\lt k \lt\frac{1+\sqrt{5}}{2}$.


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