There is a formula that can be derived for the side length of the equilateral triangle given distances from a point inside to the vertices. The formula is
\[\sqrt{(4 \sqrt{3} A + a^2 + b^2 + c^2)/2}\]
where A is area of a triangle formed by those distances. In our case the point is outside of equilateral triangle, and the distances form degenerate triangle of area 0. \[AB=\sqrt{(3^2 + 5^2 + 8^2)/2} = \boxed{7}\]
I think formula for the side length when the point is outside would be:
\[ x = \sqrt{(a^2 + b^2 + c^2-4 \sqrt{3} A)/2}\]
Degenerate triangle in our case is a segment of length 8. When we construct equilateral triangles on one end of the segment, one with side length 3 the other with side length 5,connect the vertices to the other end of the segment, we will get the side length AB. \(\angle AOB = \angle BOC = 60\). Using cosine law would get the result. \[x^2 = 8^2 + 5^2 - 2 \times 5 \times 8 \times cos(60) = 49\]

A is this formula is an area of a triangle formed by AO, BO and CO - the formula for triangle area when just side lengths are given is called Heron's formula. In this case the triangle would be just a segment.

@Mohab Mahdy
–
I can not post an image for some reason.
One way to do it would be to construct equilateral triangles on both sides of the 8 segment - BO. Then you can prove that when points A and C lie on the sides of these new triangles ABC is equilateral. The reason is that \(\triangle ABO and \triangle BCO2 \) are congruent and \(\angle ABC = 60\).

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThere is a formula that can be derived for the side length of the equilateral triangle given distances from a point inside to the vertices. The formula is \[\sqrt{(4 \sqrt{3} A + a^2 + b^2 + c^2)/2}\] where A is area of a triangle formed by those distances. In our case the point is outside of equilateral triangle, and the distances form degenerate triangle of area 0. \[AB=\sqrt{(3^2 + 5^2 + 8^2)/2} = \boxed{7}\] I think formula for the side length when the point is outside would be: \[ x = \sqrt{(a^2 + b^2 + c^2-4 \sqrt{3} A)/2}\] Degenerate triangle in our case is a segment of length 8. When we construct equilateral triangles on one end of the segment, one with side length 3 the other with side length 5,connect the vertices to the other end of the segment, we will get the side length AB. \(\angle AOB = \angle BOC = 60\). Using cosine law would get the result. \[x^2 = 8^2 + 5^2 - 2 \times 5 \times 8 \times cos(60) = 49\]

Log in to reply

sorry but i didn't understand what's the A ?!!

Log in to reply

A is this formula is an area of a triangle formed by AO, BO and CO - the formula for triangle area when just side lengths are given is called Heron's formula. In this case the triangle would be just a segment.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply