Another triangle?

Geometry Level 3

Consider an equilateral triangle:

Now pick any coplanar point outside the triangle, and call the distances from that point to each of the vertices, a a , b b , and c c .

Is it possible to pick a point such that a , b a, b , and c c , can't be the distances of another triangle? i.e. If a a is the largest distance, then a > b + c a > b + c ?


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

Inspiration

No Yes

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

David Vreken
Jan 11, 2019

Let the coordinates of the equilateral triangle be ( 2 , 0 ) (2, 0) , ( 1 , 3 ) (-1, \sqrt{3}) , and ( 1 , 3 ) (-1, -\sqrt{3}) , and let the picked point be ( p , q ) (p, q) .

Using the distance formula, a 2 = ( p 2 ) 2 + q 2 a^2 = (p - 2)^2 + q^2 , b 2 = ( p + 1 ) 2 + ( q 3 ) 2 b^2 = (p + 1)^2 + (q - \sqrt{3})^2 , and c 2 = ( p + 1 ) 2 + ( q + 3 ) 2 c^2 = (p + 1)^2 + (q + \sqrt{3})^2 .

Using Heron's formula , the area of the triangle simplifies to A = 1 4 3 ( p 2 + q 2 4 ) 2 A = \frac{1}{4}\sqrt{3(p^2 + q^2 - 4)^2} , so no matter the values of p p and q q , A 0 A \geq 0 , which means no triangle can be formed such that a > b + c a > b + c .

Note: A zero area triangle is formed such that a = b + c a = b + c when p 2 + q 2 = 4 p^2 + q^2 = 4 ; in other words, when the point ( p , q ) (p, q) is on the triangle's circumcircle.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...