Pessimist Singularitarian
#4
December 3rd, 2016,
15:05
Let's begin by constructing the line segment $\overline{MN}$. since $\overline{OM}=\overline{ON}=R$, we know $\triangle MON$ is isosceles, and so we can label $\angle OMN=\angle ONM=\theta$:
\begin{tikzpicture}
\draw [blue,thick,domain=60:120] plot ({6*cos(\x)}, {6*sin(\x)});
\draw [blue,thick](0,0) -- ++(60:6cm)
(0,0) -- ++(120:6cm);
\draw[blue,thick] (-3,5.196) -- (3,5.196);
\node[below=5pt of {(0,0)}] {\large O};
\node[left=5pt of {(-3,5.196)}] {\large M};
\node[right=5pt of {(3,5.196)}] {\large N};
\node[above=15pt of {(0,0)}] {\large 60$^{\circ}$};
\node[left=5pt of {(-1.5,2.598)}] {\large $R$};
\node[right=5pt of {(1.5,2.598)}] {\large $R$};
\node[right=8pt of {(-3,5.196)},yshift=-8pt] {\large $\theta$};
\node[left=8pt of {(3,5.196)},yshift=-8pt] {\large $\theta$};
\end{tikzpicture}
Given that the sum of the interior angles in a triangle is $180^{\circ}$, we may state:
$ \displaystyle 2\theta+60^{\circ}=180^{\circ}$
Solving this, we find:
$ \displaystyle \theta=60^{\circ}$
And so we know $\triangle MON$ is equilateral:
\begin{tikzpicture}
\draw [blue,thick,domain=60:120] plot ({6*cos(\x)}, {6*sin(\x)});
\draw [blue,thick](0,0) -- ++(60:6cm)
(0,0) -- ++(120:6cm);
\draw[blue,thick] (-3,5.196) -- (3,5.196);
\node[below=5pt of {(0,0)}] {\large O};
\node[left=5pt of {(-3,5.196)}] {\large M};
\node[right=5pt of {(3,5.196)}] {\large N};
\node[left=5pt of {(-1.5,2.598)}] {\large $R$};
\node[right=5pt of {(1.5,2.598)}] {\large $R$};
\node[above=5pt of {(0,5.196)}] {\large $R$};
\end{tikzpicture}
Next, let's decompose $\triangle MON$ into 3 congruent isosceles triangles sharing a central common vertex:
\begin{tikzpicture}
\draw [blue,thick,domain=60:120] plot ({6*cos(\x)}, {6*sin(\x)});
\draw [blue,thick](0,0) -- ++(60:6cm)
(0,0) -- ++(120:6cm);
\draw[blue,thick] (-3,5.196) -- (3,5.196);
\draw[red,thick] (0,0) -- (0,3.464);
\draw[red,thick] (-3,5.196) -- (0,3.464);
\draw[red,thick] (3,5.196) -- (0,3.464);
\node[below=5pt of {(0,0)}] {\large O};
\node[left=5pt of {(-3,5.196)}] {\large M};
\node[right=5pt of {(3,5.196)}] {\large N};
\node[left=5pt of {(-1.5,2.598)}] {\large $R$};
\node[right=5pt of {(1.5,2.598)}] {\large $R$};
\node[above=5pt of {(0,5.196)}] {\large $R$};
\node[right=3pt of {(0,1.732)}] {\large $r$};
\node[above=5pt of {(-1.5,3.464)}] {\large $r$};
\node[above=5pt of {(1.5,3.464)}] {\large $r$};
\end{tikzpicture}
Now, using the formula for the area of a triangle, we may state:
$ \displaystyle 3\cdot\frac{1}{2}Rr\sin\left(30^{\circ}\right)=\frac{1}{2}R^2\sin\left(60^{\circ}\right)$
Can you now solve for $r$?