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 The sum of the squares of the lengths from a point on unit circle to vertices of its inscribed equilateral triangle
04-21-2010, 05:40 PM
Post: #1
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
The sum of the squares of the lengths from a point on unit circle to vertices of its inscribed equilateral triangle
[float=right]     [/float]Let $$P$$ be a point on the circle $$\odot O$$ while $$A,B,C$$
are vertices of $$\triangle ABC$$, Find $$\overline{PA}^2+\overline{PB}^2+\overline{PC}^2$$
12-19-2010, 12:53 AM
Post: #2
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
Solution: The sum of the squares...
Let $\beta = \frac{\pi}{6}-\theta \in (0,\frac{\Pi}{6})$, then $\overline{PB}=2\cos\beta$. Now by the law cosine we get
$\overline{PA}^2+\overline{PB}^2+\overline{PC}^2 = \displaystyle{3 \overline{PB}^2 + 2(\sqrt{3})^2-2\overline{PB}\sqrt{3}(\cos(\frac{\pi}{6}-\beta)+\cos(\frac{\pi}{6}+\beta))}=$
$\displaystyle{=6+12\cos^2 \beta -8\sqrt{3}\cos^2\beta \cos \frac{\pi}{6}=6}$
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