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. \[x^2 = 8^2 + 5^2 - 2 \times 5 \times 8 \times cos(60) = 49\]
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Maria Kozlowska
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1 year, 12 months ago

@Mohab Mahdy
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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.
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Maria Kozlowska
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1 year, 12 months ago

@Mohab Mahdy
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A triangle formed with sides 3, 5, and 8 is a straight line that has 0 area.
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Niranjan Khanderia
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11 months, 2 weeks ago

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@Mohab Mahdy
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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 \(\angle ABC = 60\).
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Maria Kozlowska
·
1 year, 12 months ago

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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. \[x^2 = 8^2 + 5^2 - 2 \times 5 \times 8 \times cos(60) = 49\] – Maria Kozlowska · 1 year, 12 months ago

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– Mohab Mahdy · 1 year, 12 months ago

sorry but i didn't understand what's the A ?!!Log in to reply

– Maria Kozlowska · 1 year, 12 months ago

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

– Mohab Mahdy · 1 year, 12 months ago

which area ??Log in to reply

– Mohab Mahdy · 1 year, 12 months ago

Can you explain it with a designLog in to reply

– Niranjan Khanderia · 11 months, 2 weeks ago

A triangle formed with sides 3, 5, and 8 is a straight line that has 0 area.Log in to reply

– Maria Kozlowska · 1 year, 12 months ago

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 \(\angle ABC = 60\).Log in to reply