# geometry question (triangle & circle) [closed]

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)

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Qwerty, Daniel W. Farlow, MathOverview, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

• please any help – ShiraMathmatics Dec 29 '16 at 16:08

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

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

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:

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

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.

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:

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

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.