# Help !!!

An equilateral triangle ABC and there is a point out of the triangle O AO = 3 , BO = 8 , CO = 5 . What's the length of AB ?!!

Note by Mohab Mahdy
3 years, 1 month ago

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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. $x^2 = 8^2 + 5^2 - 2 \times 5 \times 8 \times cos(60) = 49$

- 3 years, 1 month ago

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

- 3 years, 1 month 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.

- 3 years, 1 month ago

which area ??

- 3 years, 1 month ago

Can you explain it with a design

- 3 years, 1 month ago

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

- 2 years, 1 month 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 $$\angle ABC = 60$$.

- 3 years, 1 month ago