# Triangles: Similarity, angles, sides

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

On a unit circle $K$ there are twelve points $P_1, \ldots , P_{12}$ in that order and each two successive points have the same distance (as the number $1$ to $12$ on a clock). Let $g$ be the line that passes through $P_1$ and $P_8$ and let $h$ be the line that passes through $P_3$ and $P_{11}$. Let $S$ be the intersection of $g$ and $h$.

1. Show that the triangles $P_{11}P_8S$ and $P_1P_3S$ are similar.

2. Calculate the two acute angles that are formed by $g$ and $h$.

3. Calculate the side lengths of the triangle $P_1P_3S$.

For question 1 we have that at both triangles the angle of $S$ are similar, aren't they? What can we say about the other angles? the arcs are not equal.

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Here's a picture showing your problem statement.
\begin{tikzpicture}[scale=4]
\usetikzlibrary{quotes,angles}
\coordinate (o) at (0,0);
\coordinate[label=$P_1$] (p1) at (60:1);
\coordinate[label=right:$P_3$] (p3) at (0:1);
\coordinate[label=left:$P_8$] (p8) at (-150:1);
\coordinate[label=$P_{11}$] (p11) at (120:1);
\coordinate[label=below:$S$] (s) at (75:.518);
\draw[gray] (0,0) circle (1);
\draw[thick] (p1) -- (p8);
\draw[thick] (p11) -- (p3);
\draw[thick] (p8) -- (p11);
\draw[thick] (p1) -- (p3);
\filldraw (s) circle (.01);
\filldraw[gray] (o) circle (.01);
\draw (p3) -- (p8);

\draw[help lines] (p3) -- (o) -- (p8);
\pic["$\theta$",draw,angle radius=1cm] {angle=p8--p11--p3};
\pic["$\theta$",draw,angle radius=1cm] {angle=p8--p1--p3};
\draw[help lines] pic["$2\theta$",draw,angle radius=.5cm] {angle=p8--o--p3};
\end{tikzpicture}

The Inscribed angle theorem says that the angles marked $\theta$ are the same.
You already had that the acute angles at S are the same, so the triangles are similar.

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#### mathmari

##### Well-known member
MHB Site Helper
Here's a picture showing your problem statement.
\begin{tikzpicture}[scale=4]
\usetikzlibrary{quotes,angles}
\coordinate (o) at (0,0);
\coordinate[label=$P_1$] (p1) at (60:1);
\coordinate[label=right:$P_3$] (p3) at (0:1);
\coordinate[label=left:$P_8$] (p8) at (-150:1);
\coordinate[label=$P_{11}$] (p11) at (120:1);
\coordinate[label=below:$S$] (s) at (75:.518);
\draw[gray] (0,0) circle (1);
\draw[thick] (p1) -- (p8);
\draw[thick] (p11) -- (p3);
\draw[thick] (p8) -- (p11);
\draw[thick] (p1) -- (p3);
\filldraw (s) circle (.01);
\filldraw[gray] (o) circle (.01);
\draw (p3) -- (p8);

\draw[help lines] (p3) -- (o) -- (p8);
\pic["$\theta$",draw,angle radius=1cm] {angle=p8--p11--p3};
\pic["$\theta$",draw,angle radius=1cm] {angle=p8--p1--p3};
\draw[help lines] pic["$2\theta$",draw,angle radius=.5cm] {angle=p8--o--p3};
\end{tikzpicture}

The Inscribed angle theorem says that the angles marked $\theta$ are the same.
You already had that the acute angles at S are the same, so the triangles are similar.
Ahh I see!!

(Btw I could see the picture, I had to copied to the online editor.)

As for the question 2, do we use the formula of the arc length?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
As for the question 2, do we use the formula of the arc length?
Which formula for arc length?

We can use that for instance $\angle P_1OP_3$ is $\frac 2{12}$ of $360^\circ$...

#### mathmari

##### Well-known member
MHB Site Helper
We can use that for instance $\angle P_1OP_3$ is $\frac 2{12}$ of $360^\circ$...
By $O$ you mean the center of the circle?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
By $O$ you mean the center of the circle?
Yep.
It's a unit circle after all.

Perhaps this picture will help:
\begin{tikzpicture}[scale=4]
\usetikzlibrary{quotes,angles}
\coordinate (o) at (0,0);
\coordinate[label=$P_1$] (p1) at (60:1);
\coordinate[label=right:$P_3$] (p3) at (0:1);
\coordinate[label=left:$P_8$] (p8) at (-150:1);
\coordinate[label=$P_{11}$] (p11) at (120:1);
\coordinate[label=below:$S$] (s) at (75:.518);
\draw[gray] (0,0) circle (1);
\filldraw[gray] (o) circle (.01);
\filldraw[gray] foreach \i in {1,...,12} { ({\i*30}:1) circle (.01) };
\draw[thick] (p1) -- (p8);
\draw[thick] (p11) -- (p3);
\draw[thick] (p8) -- (p11);
\draw[thick] (p1) -- (p3);
\filldraw (s) circle (.01);
\draw (p3) -- (p8);

\draw[help lines] (p3) -- (o) -- (p8);
\pic["$\theta$",draw,angle radius=1cm] {angle=p8--p11--p3};
\pic["$\theta$",draw,angle radius=1cm] {angle=p8--p1--p3};
\draw[help lines] pic["$150^\circ$",draw,angle radius=.6cm] {angle=p8--o--p3};

\draw[help lines] (p3) -- (o) -- (p8) (p1) -- (o) -- (p11);
\draw[help lines, rotate=120] rectangle (.1,.1);
\draw[help lines] pic["$60^\circ$",draw,angle radius=1.1cm] {angle=p3--o--p1};
\draw[help lines] pic["$60^\circ$",angle radius=1.1cm] {angle=p1--o--p11};
\end{tikzpicture}

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#### mathmari

##### Well-known member
MHB Site Helper
Yep.
It's a unit circle after all.

Perhaps this picture will help:
\begin{tikzpicture}[scale=4]
\usetikzlibrary{quotes,angles}
\coordinate (o) at (0,0);
\coordinate[label=$P_1$] (p1) at (60:1);
\coordinate[label=right:$P_3$] (p3) at (0:1);
\coordinate[label=left:$P_8$] (p8) at (-150:1);
\coordinate[label=$P_{11}$] (p11) at (120:1);
\coordinate[label=below:$S$] (s) at (75:.518);
\draw[gray] (0,0) circle (1);
\filldraw[gray] (o) circle (.01);
\filldraw[gray] foreach \i in {1,...,12} { ({\i*30}:1) circle (.01) };
\draw[thick] (p1) -- (p8);
\draw[thick] (p11) -- (p3);
\draw[thick] (p8) -- (p11);
\draw[thick] (p1) -- (p3);
\filldraw (s) circle (.01);
\draw (p3) -- (p8);

\draw[help lines] (p3) -- (o) -- (p8);
\pic["$\theta$",draw,angle radius=1cm] {angle=p8--p11--p3};
\pic["$\theta$",draw,angle radius=1cm] {angle=p8--p1--p3};
\draw[help lines] pic["$150^\circ$",draw,angle radius=.6cm] {angle=p8--o--p3};

\draw[help lines] (p3) -- (o) -- (p8) (p1) -- (o) -- (p11);
\draw[help lines, rotate=120] rectangle (.1,.1);
\draw[help lines] pic["$60^\circ$",draw,angle radius=1.1cm] {angle=p3--o--p1};
\draw[help lines] pic["$60^\circ$",angle radius=1.1cm] {angle=p1--o--p11};
\end{tikzpicture}
But how is the angle of $S$ related to the angle of $O$ ? They have the same arc, do we use that?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
But how is the angle of $S$ related to the angle of $O$ ? They have the same arc, do we use that?
We can deduce the other angles.

Triangle $\triangle OP_1P_3$ must be an equilateral triangle with sides $1$ and angles $60^\circ$.
Triangle $\triangle OP_3P_{11}$ must be an isoscales triangle. One angle is $120^\circ$, so the remaining angles must be $30^\circ$.
Triangles $\triangle OP_3P_8$ and $\triangle OP_1P_8$ must be an isoscales triangles as well. One angle is $150^\circ$ so the remaining angles must be $15^\circ$ in both cases.

Can we find $\theta$ now? And subsequently the angle at $S$?

#### mathmari

##### Well-known member
MHB Site Helper
We can deduce the other angles.

Triangle $\triangle OP_1P_3$ must be an equilateral triangle with sides $1$ and angles $60^\circ$.
Triangle $\triangle OP_3P_{11}$ must be an isoscales triangle. One angle is $120^\circ$, so the remaining angles must be $30^\circ$.
Triangles $\triangle OP_3P_8$ and $\triangle OP_1P_8$ must be an isoscales triangles as well. One angle is $150^\circ$ so the remaining angles must be $15^\circ$ in both cases.

Can we find $\theta$ now? And subsequently the angle at $S$?
We have that $\theta=75^{\circ}$.

To find $S$ do we have to consider for example $P_1SP_3$ ? Or how is $\theta$ else related to $S$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
We have that $\theta=75^{\circ}$.

To find $S$ do we have to consider for example $P_1SP_3$ ? Or how is $\theta$ else related to $S$ ?
Yes and yes.

Can we find $\angle P_1 P_3 P_{11}$?

#### mathmari

##### Well-known member
MHB Site Helper
Yes and yes.

Can we find $\angle P_1 P_3 P_{11}$?
I don't really see how we could that.... Could you give me a hint?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I don't really see how we could that.... Could you give me a hint?
$\triangle OP_3P_{11}$ is isoscales, so $\angle P_{11} P_3 O = 30^\circ$.
$\triangle OP_3P_1$ is equilateral, so $\angle P_1 P_3 O = 60^\circ$.

#### mathmari

##### Well-known member
MHB Site Helper
$\triangle OP_3P_{11}$ is isoscales, so $\angle P_{11} P_3 O = 30^\circ$.
$\triangle OP_3P_1$ is equilateral, so $\angle P_1 P_3 O = 60^\circ$.
We have the following:
\begin{align*}&\angle P_1SP_3+\angle SP_3P_1+\angle P_3P_1S=180^{\circ} \\ & \Rightarrow \angle P_1SP_3+\left (\angle P_1 P_3 O-\angle P_{11} P_3 O\right )+\theta=180^{\circ} \\ & \Rightarrow \angle P_1SP_3+\left (60^\circ-30^\circ\right )+75^{\circ}=180^{\circ} \\ & \Rightarrow \angle P_1SP_3+30^\circ+75^{\circ}=180^{\circ} \\ & \Rightarrow \angle P_1SP_3+105^{\circ}=180^{\circ} \\ & \Rightarrow \angle P_1SP_3=75^{\circ}\end{align*} right?

So we know now all three agles of the triangle $\triangle P_1P_3S$, which is isosceles, which means that $P_3P_1=P_3S$.

To find the length of the sides do we apply the sine rule?

$$\frac{P_3S}{\sin 75^{\circ}}=\frac{P_1S}{\sin (180^{\circ}-2\cdot 75^{\circ})} \Rightarrow \frac{P_3S}{\sin 75^{\circ}}=\frac{P_1S}{\sin (30^{\circ})} \Rightarrow \frac{P_3S}{0.965}=\frac{P_1S}{0.5}$$

How could we continue?

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
How could we continue?
$\triangle OP_1P_3$ is equilateral and 2 sides have the length of the radius. The radius is 1 because we have a unit circle. Therefore $P_1 P_3=P_3S=1$.

#### mathmari

##### Well-known member
MHB Site Helper
$\triangle OP_1P_3$ is equilateral and 2 sides have the length of the radius. The radius is 1 because we have a unit circle. Therefore $P_1 P_3=P_3S=1$.
Ahh yes! Thanks a lot!!