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Triangles: Similarity, angles, sides

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,274
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.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,012

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,012
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. 🤔
 
Last edited:

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,274
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? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,012
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 $360^\circ$... :unsure:
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,274

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,012
By $O$ you mean the center of the circle?
Yep. (Nod)
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}
🤔
 
Last edited:

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,274
Yep. (Nod)
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? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,012
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 $S$? :unsure:
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,274
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 $S$? :unsure:
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 $S$ ? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,012
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. (Nod)

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

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,274

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,012
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
Apr 14, 2013
4,274
$\triangle OP_3P_{11}$ 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? :unsure:

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? :unsure:
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,012
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
Apr 14, 2013
4,274
$\triangle OP_1P_3$ 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!! (Sun)