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In an equilateral triangle what is sum of distance from vertices to any arbitrary point inside the triangle? enter image description here

What is the relation between $a$ and $x + y +z$. The special condition is that the interior point cannot be considered to be a special point like centroid or circumcenter,etc.

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    $\begingroup$ It doesnt look like there is any "nice" formula, but if you use law of cosines on the smaller triangles and the relations we get from angles in the corners summing to $60^\circ$ you can get a messy formula relating them $\endgroup$
    – N. Owad
    Oct 31 '13 at 14:04
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It seems that the minimal value of $s=x+y+z$ is $a\sqrt{3}$, attained at the centroid. The maximal value of $s$ is $2a$, attained at a vertex.

By continuity of $s$, all intermediate values between $a\sqrt{3}$ and $2a$ are possible too.

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  • $\begingroup$ The values of s remain a√3 for any point that is in the interior. $\endgroup$ Oct 30 '13 at 6:09
  • $\begingroup$ @shauryagupta It seems the following. Let the fixed point be the centroid of the triangle. Then $x=y=z=\frac 23m$, where $m$ is length of the median. But $m=a\frac{\sqrt{3}}2$. Then $s=3x=2m=a\sqrt{3}\ne 2a$. $\endgroup$ Oct 30 '13 at 18:44
  • $\begingroup$ yes i know that it is a√3 but what if the point is not a centroid...What if it were to be just a random point inside the triangle? $\endgroup$ Oct 31 '13 at 13:02
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    $\begingroup$ It seems that, as I arleady wrote, all intermediate values of $s$ between $a\sqrt{3}$ and $2a$ are possible, and this value depends on the chosen point. $\endgroup$ Oct 31 '13 at 18:29
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I have not even tried to simplify this, but it is a relation.

$$ \frac{a^2+x^2-y^2}{2xa} = \frac{\sqrt{3}}{2}\frac{a^2+x^2-z^2}{2xa}+\frac{1}{2} \sqrt{1- \Big(\frac{a^2+x^2-z^2}{2xa}\Big)^2} $$

Perhaps it is what you were looking for. Of course, if you require your point to be inside the equilateral, you need to have restraints on your $x,y,$ and $z$.

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The sum $x+y+z$ is not a constant but varies from $\sqrt3 a$ (at the centroid) to $2a$ (at a vertex).

Probably the neatest relation is $$(x^2+y^2+z^2+a^2)^2=3(x^4+y^4+z^4+a^4)$$ as quoted in wikipedia : Equilateral Triangle - Other Properties.

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