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ABC is equilateral triangle inscribed in circle with Radius R. D point on the circle. I want to proof that: $DA^2+DB^2+DC^2=6R^2$ (in three ways- the proof can be in anyway not just geometry)

enter image description here

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closed as off-topic by Namaste, Qwerty, Daniel W. Farlow, MathOverview, Watson Dec 29 '16 at 21:38

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  • $\begingroup$ please any help $\endgroup$ – ShiraMathmatics Dec 29 '16 at 16:08
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You can solve this by using the coordinates of your points.

Without any loss of generality, you have:

$$ A(0, R), \ B(\frac{\sqrt{3}}{2}R, -\frac{1}{2}R), \ C(\frac{-\sqrt{3}}{2}R, -\frac{1}{2}R) \textrm{ and } D(x,y)$$ with $x^2+y^2 = R^2$.

Then $$DA^2 = x^2 + (y - R)^2 = 2R^2-2Ry$$ $$DB^2 = (x- \frac{\sqrt{3}}{2}R)^2+(y+\frac{1}{2}R)^2 = 2R^2+Ry-\sqrt{3}xR$$ $$DC^2 = (x+ \frac{\sqrt{3}}{2}R)^2+(y+\frac{1}{2}R)^2 = 2R^2+Ry+\sqrt{3}xR$$ Finally you get $$DA^2+DB^2+DC^2=6R^2$$

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  • $\begingroup$ Why not $x^2+y^2=R^2$? $\endgroup$ – Guacho Perez Dec 29 '16 at 16:32
  • $\begingroup$ @GuachoPerez I edited my post, I first thought of a circle of radius 1 $\endgroup$ – fonfonx Dec 29 '16 at 16:33
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You can use the law of sines:

$${DB\over\sin(60-\angle DAC)}={DC\over\sin(\angle DBC)}={DA\over\sin(60+\angle DBC)}=2R$$

Now we know that $\angle DBC=\angle DAC=\beta$ (both of them are over the same arc $\overset{\frown}{DC}$), so we have:

$$DB=2R\sin(60-\beta)=2R\left({\sqrt{3}\over2}\cos\beta-{\sin\beta\over2}\right)$$ $$DC=2R\sin\beta$$ $$DA=2R\sin(60+\beta)=2R\left({\sqrt{3}\over2}\cos\beta+{\sin\beta\over2}\right)$$

so we get:

$$DB^2+DC^2+DA^2=\\4R^2\left({3\over4}\cos^2\beta+{\sin^2\beta\over4}-{\sqrt3\over2}\sin\beta\cos\beta+\sin^2\beta+{3\over4}\cos^2\beta+{\sin^2\beta\over4}+{\sqrt3\over2}\sin\beta\cos\beta\right)=4R^2{3\over2}=6R^2$$

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Without loss of generality, let the circle have unit radius and rotate the image so $A$ $B$ and $C$ are the cube roots of unity, and let $D=e^{i\theta}$

Then the expression on the left hand side is $$|e^{i\theta}-1|^2+|e^{i\theta}-\omega|^2+|e^{i\theta}-\omega^2|^2$$

Noting that $1+\omega+\omega^2=0$ and that $|z|^2=zz^*$ you can expand this expression and it quickly reduces to $6$

I leave the details to you.

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WLOG, assume D is located on the shorter circular arc between $B$ and $C$ (as in the picture). Let $\theta=\angle CBD$. If $O$ is the center of the circle, one can show that $\angle BOD=\frac{2\pi}{3}-2\theta$, $\angle AOD=\frac{2\pi}{3}+2\theta$, and $\angle COD=2\theta$.

From the law of cosines on triangles $BOD$, $AOD$, and $COD$, we get:

$$DA^2=2R^2-2R^2\cos(\frac{2\pi}{3}+2\theta)$$ $$DB^2=2R^2-2R^2\cos(\frac{2\pi}{3}-2\theta)$$ $$DC^2=2R^2-2R^2\cos(2\theta)$$ However, note that

$$\cos(\frac{2\pi}{3}+2\theta)+\cos(\frac{2\pi}{3}-2\theta)+\cos(2\theta)=0$$

So, adding the three equations above, we get:

$$DA^2+DB^2+DC^2=6R^2$$

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Using triangles $AOD, BOD, COD$ and angle names $A=DBA, B=BAD, C=CAD$ we have $A=60^{\circ}+C$ and $B=60^{\circ}-C$ (noting that angles $DBC$ and $DAC$ are equal, because they are on the same arc) and using the cosine formula in triangles $OAD, OBD, OCD$

$$DA^2+DB^2+DC^2=2R^2-2R^2\cos2A+2R^2-2R^2\cos 2B+2R^2-2R^2\cos 2C$$

Now,using $\cos (P+Q)=\cos P \cos Q-\sin P\sin Q$, we have $$\cos 2A+\cos 2B+\cos 2C= \cos (120^{\circ}+2C)+\cos (120^{\circ}-2C)+\cos 2C =$$$$=(2\cos120^{\circ}+1)\cos 2C -\sin 120^{\circ} (\sin 2C-\sin 2C)$$

and the terms in brackets reduce to zero.

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