Consider an equilateral triangle:
Now pick any coplanar point outside the triangle, and call the distances from that point to each of the vertices, , , and .
Is it possible to pick a point such that , and , can't be the distances of another triangle? i.e. If is the largest distance, then ?
Note : The three distances are allowed to form a degenerate (or zero area) triangle where .
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Let the coordinates of the equilateral triangle be ( 2 , 0 ) , ( − 1 , 3 ) , and ( − 1 , − 3 ) , and let the picked point be ( p , q ) .
Using the distance formula, a 2 = ( p − 2 ) 2 + q 2 , b 2 = ( p + 1 ) 2 + ( q − 3 ) 2 , and c 2 = ( p + 1 ) 2 + ( q + 3 ) 2 .
Using Heron's formula , the area of the triangle simplifies to A = 4 1 3 ( p 2 + q 2 − 4 ) 2 , so no matter the values of p and q , A ≥ 0 , which means no triangle can be formed such that a > b + c .
Note: A zero area triangle is formed such that a = b + c when p 2 + q 2 = 4 ; in other words, when the point ( p , q ) is on the triangle's circumcircle.