Trigonometry! #127
If A 0 , A 1 , A 2 , A 3 , A 4 and A 5 are the consecutive vertices of a regular hexagon inscribed in a unit circle, then find the product of the lengths of A 0 A 1 , A 0 A 2 and A 0 A 4 .
This problem is part of the set Trigonometry .
The answer is 3.
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1 solution
Why will the triangles be equilateral? And A 0 , A 1 , ... , A 5 are points.
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Sorry dude, I misunderstood the question as if you wanted the sum of A 0 A 1 , A 0 A 2 and A 0 A 4 .
Just let me know after you read this comment, I'll delete the solution as soon as possible . Can't let others see a wrong solution :)
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Yeah, delete it :P but why will the triangles formed be equilateral?
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@Omkar Kulkarni – Each triangle formed by joining the vertices to the centre will subtend an angle of 60 degree at the centre , ok .
Now two of it's sides are equal , yes the radii , therefore their opposite angles will be equal , so θ + θ + 6 0 = 1 8 0 ⇒ θ = 6 0
Hence equilateral .
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@A Former Brilliant Member – Oh I missed that! Delete this. Please. :P
Since this is a Regular Hexagon, there are 6 Equilateral Triangles that can be formed if you join all the Vertices to the Centre of the circle .
Since the Triangles are equilateral , all the sides are of equal length , i.e. equal to 1 cm (the radius of the circle) .
Therefore the product evaluates to be 1 ⋅ 1 + 1 ⋅ 1 + 1 ⋅ 1 = 3