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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
(43A+a2+b2+c2)/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=(32+52+82)/2=7
I think formula for the side length when the point is outside would be:
x=(a2+b2+c2−43A)/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. ∠AOB=∠BOC=60. Using cosine law would get the result. x2=82+52−2×5×8×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
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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 △ABOand△BCO2 are congruent and ∠ABC=60.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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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
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 (43A+a2+b2+c2)/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=(32+52+82)/2=7 I think formula for the side length when the point is outside would be: x=(a2+b2+c2−43A)/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. ∠AOB=∠BOC=60. Using cosine law would get the result. x2=82+52−2×5×8×cos(60)=49
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sorry but i didn't understand what's the A ?!!
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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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△ABOand△BCO2 are congruent and ∠ABC=60.
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