Sector Problem: Determine the radius of the circle

[Ilikebugs]

View attachment 6249 Is there a way other than guess and check to figure it out?

 

[MarkFL]

What kind of triangle is $\triangle MON$?

 

[Euge]

Hi Ilikebugs,

Yes, you can solve this systematically. Let $X$ be the center of the circle which passes through points $M$, $O$, and $P$. Show that $\triangle MXO\cong \triangle OXN$. Argue that $m\angle XON = 30º$. Draw an altitude of $\triangle XON$ from vertex $X$ to side $ON$. Say $Y$ is the point of intersection of the altidtude with side $ON$. Then consider the right triangle $\triangle OXY$ and solve the hypotenuse $OX$, which is the radius you want.

 

[MarkFL]

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 $r$?

 

[Ilikebugs]

I'm not sure

 

[MarkFL]

Well, what are $\sin\left(30^{\circ}\right)$ and $\sin\left(60^{\circ}\right)$?

 

[Ilikebugs]

0.5 and 0.86602540378

 

[topsquark]

0.5 and 0.86602540378

Hint: What are the exact values...

-Dan

 

[Ilikebugs]

1/2 and 43301270189/50000000000?

 

[topsquark]

1/2 and 43301270189/50000000000?

Put away the calculator! This will help in the long run... Memorize the sine and cosine functions for the angles 0, 30, 45, 60, and 90 degrees. It will save you a lot of time and grief.

sin(30) = 1/2 and sin(60) = sqrt(3) / 2.

Put these into your equation for r. Can you solve it now?

-Dan

 

[MarkFL]

Imagine we've take an equilateral triangle, bisected it and oriented one of the halves like so:

\begin{tikzpicture}
\draw[blue,thick] (0,0) -- (5.196,0);
\draw[blue,thick] (5.196,0) -- (5.196,3);
\draw[blue,thick] (5.196,3) -- (0,0);
\draw[gray,thin] (5.196,0.25) -- (4.946,0.25);
\draw[gray,thin] (4.946,0.25) -- (4.946,0);
\node[below=5pt of {(2.598,0)}] {\large $x$};
\node[right=5pt of {(5.196,1.5)}] {\large $\dfrac{1}{2}$};
\node[above=5pt of {(2.598,1.5)}] {\large $1$};
\end{tikzpicture}

Can you use the Pythagorean theorem to find $x$? Once you know $x$, then you can find the sine and cosine of 30 and 60 degrees...:D

 

[Ilikebugs]

I cant

 

[MarkFL]

I cant

The Pythagorrean theorem tells us that the square of the hypotenuse (the longest side in a right triangle, the one opposite the $90^{\circ}$ angle) is equal to the sum of the squares of the other two sides (the two shorter sides are called "legs"). So, using that with the triangle I gave in post #11, stated mathematically, this is:

\(\displaystyle x^2+\left(\frac{1}{2}\right)^2=1^2\)

\(\displaystyle x^2+\frac{1}{4}=1\)

\(\displaystyle x^2=1-\frac{1}{4}=\frac{3}{4}\)

Since $x$ represents a linear measure, we take the positive root:

\(\displaystyle x=\frac{\sqrt{3}}{2}\)

Okay, so we now have:

\begin{tikzpicture}
\draw[blue,thick] (0,0) -- (5.196,0);
\draw[blue,thick] (5.196,0) -- (5.196,3);
\draw[blue,thick] (5.196,3) -- (0,0);
\draw[gray,thin] (5.196,0.25) -- (4.946,0.25);
\draw[gray,thin] (4.946,0.25) -- (4.946,0);
\node[below=5pt of {(2.598,0)}] {\large $\dfrac{\sqrt{3}}{2}$};
\node[right=5pt of {(5.196,1.5)}] {\large $\dfrac{1}{2}$};
\node[above=5pt of {(2.598,1.5)}] {\large $1$};
\node[below=10pt of {(5.196,3)},xshift=-10pt] {\large $60^{\circ}$};
\node[right=18pt of {(0,0)},yshift=7pt] {\large $30^{\circ}$};
\end{tikzpicture}

Now, the sine of an angle in a right triangle is defined as the ratio of the side opposite the angle to the hypotenuse. Hence we have:

\(\displaystyle \sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\dfrac{1}{2}}{1}=\frac{1}{2}\)

\(\displaystyle \sin(60^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\dfrac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2}\)

So, returning to the equation:

\(\displaystyle 3\cdot\frac{1}{2}Rr\sin\left(30^{\circ}\right)=\frac{1}{2}R^2\sin\left(60^{\circ}\right)\)

All you need to do now is plug in the values for the sine functions, and solve for $r$. :D

 

[Ilikebugs]

I don't know how to solve for r

 

[MarkFL]

I don't know how to solve for r

Well, as I said, the first step is to substitute for the sine functions:

\(\displaystyle 3\cdot\frac{1}{2}Rr\cdot\frac{1}{2}=\frac{1}{2}R^2\cdot\frac{\sqrt{3}}{2}\)

Next, we may multiply through by \(\displaystyle \frac{4}{\sqrt{3}R}\) to obtain:

\(\displaystyle \sqrt{3}r=R\)

Hence:

\(\displaystyle r=\frac{R}{\sqrt{3}}\)

Now you need to plug in the given $R=6\text{ cm}$ and simplify to finish the problem.

 

[Ilikebugs]

r=2√3?

 

[MarkFL]

r=2√3?

$r=2\sqrt{3}\text{ cm}$

 

[Greg]

Having found that the triangle is equilateral one can determine its height (altitude) with the Pythagorean theorem:

$$\sqrt{6^2-3^2}=\sqrt{36-9}=\sqrt{27}=3\sqrt3$$

There is a point on a median (the line that emanates from a vertex and terminates on the midpoint of the opposite side) of the triangle, called the centroid, that is concurrent with the circumcenter in the case of an equilateral triangle and one of its properties is that it divides a median (in this case an altitude, which is concurrent with a median in the case of an equilateral triangle) in the ratio 2:1. Hence the radius we seek (the radius of the circumcircle of $\triangle{MON}$) is

$$\frac23\cdot3\sqrt3=2\sqrt3$$

The three medians of a triangle intersect at the centroid.