Equilateral triangle with vertices on 3 concentric circles

Using the construction that @Michael Rozenberg suggested

enter image description here

I will leave the following exercise for you (which isn't that hard)

Prove that the quadrilateral $ABCD$ is cyclic.

Thus $\angle BDC=180°-\angle CAB=120°$. In virtue of the law of Cosines $$\begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·\cos(\angle BDC)\\ &=1+4-2·1·2·(-0.5)\\ &=5+2=7 \end{array}$$


The hint.

Take $A$ on the biggest circle and rotate the smallest circle by $60^{\circ}$ around $A$.

Now, take an intersection point $B$ with the middle circle.

Thus, $AB$ is a side of the needed triangle.

I took $A(-3,0)$ and got $AB=\sqrt7.$


Here's the image

I'll use the process of Dr. Mathva in a different way.

We'll first prove $ABCD$ is cyclic.

Let $AB=AC=BC=s$

$DC=1, DB=2, DA=3$

We see that $AB×DC+BD×AC=1s+2s=3s=AD×BC$

By Converse of Ptolemy's theorem, we conclude that $ABCD$ is cyclic.

After this you can find out a through pure trigonometric means. I got $s = \sqrt{7}$

Tags:

Geometry

Euclidean Geometry

Circles

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