circleseuclidean-geometrygeometry
Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of the equilateral triangle?
My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ I will manipulate the formula afterwards,,,
By construction you have $$1=\sqrt 3r+r+r+\sqrt 3r\Rightarrow r=\frac{\sqrt3-1}{4}\approx 0.183012701$$ Later add an explanatory figure.
Circle with the center $O$ is inscribed in the equilateral triangle $ABC$ with the height $CF$ and inradius $OF$, circle with the center $R$ is inscribed in the equilateral triangle $DEC$ with the height $CG$ and inradius $RG$,
\begin{align}
|CF|&=3|CG|
,
\end{align}
hence, the radius
\begin{align} |OF|&=3\cdot|RG|=3\cdot4=12 , \end{align}
and the side length of $\triangle ABC$ \begin{align} |AC|&= \frac{|CF|}{\tfrac {\sqrt3}2} =\frac{3\cdot|OF|}{\tfrac {\sqrt3}2} =2\sqrt3\cdot|OF|=24\sqrt3 . \end{align}
Best Answer
Using the construction that @Michael Rozenberg suggested
I will leave the following exercise for you (which isn't that hard)
Thus $\angle BDC=180°-\angle CAB=120°$. In virtue of the law of Cosines $$\begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·\cos(\angle BDC)\\ &=1+4-2·1·2·(-0.5)\\ &=5+2=7 \end{array}$$