Obtuse, scalene, isosceles, equilateral?

Geometry Level 3

3 distinct points are chosen on a plane.

It turns out that it is impossible to chose a different point on the same plane such that their distances to the original three points $a, b$ , and $c$ can't be those found in a different triangle.

i.e. It is impossible to choose the fourth point such that if $a$ is the largest distance, then $a > b + c$ .

What type(s) of triangle(s) can the original three points form?

Note : The three distances are allowed to form a degenerate (or zero area) triangle where $a = b + c$ .

Isosceles only Any type Equilateral only Obtuse only Acute only Scalene only

1 solution

Geoff Pilling
Jan 11, 2019

Not a solution but a convincing demo...

Plug the following equations into desmos :

$\sqrt{\left(x-x_1\right)^{2\ }+\left(y-y_1\right)^2}\ >\ \sqrt{\left(x-x_2\right)^{2\ \ }+\ \left(y-y_2\right)^2}\ +\ \sqrt{\left(x-x_3\right)^{2\ }+\left(y-y_3\right)^2}$

$\sqrt{\left(x-x_2\right)^{2\ }+\left(y-y_2\right)^2}\ >\ \sqrt{\left(x-x_3\right)^{2\ \ }+\ \left(y-y_3\right)^2}\ +\ \sqrt{\left(x-x_1\right)^{2\ }+\left(y-y_1\right)^2}$

$\sqrt{\left(x-x_3\right)^{2\ }+\left(y-y_3\right)^2}\ >\ \sqrt{\left(x-x_1\right)^{2\ \ }+\ \left(y-y_1\right)^2}\ +\ \sqrt{\left(x-x_2\right)^{2\ }+\left(y-y_2\right)^2}$

Then play around with the $x_n$ and $y_n$ and you will see that the only time you get the empty set is when all three points are equidistant! (An equilateral triangle)

Here is a link!

Fun to play with the slider bars as well! :-)

Stay tuned for a more mathematically rigorous solution...

Obtuse, scalene, isosceles, equilateral?

1 solution

0 pending reports