- #1

maxkor

- 90

- 0

I've tried:

BC || EF

How to find angle GDC? I think GDC=7x but why?

I have an answer but how to solve this?

Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- MHB
- Thread starter maxkor
- Start date

- #1

maxkor

- 90

- 0

I've tried:

BC || EF

How to find angle GDC? I think GDC=7x but why?

I have an answer but how to solve this?

Last edited:

- #2

- 1,802

- 733

There's another approach. Look at angle BAC. It's 180 - 7x - 7x. (ABC is isosceles, but we don't need this.) So we can get angle DAC in terms of x.https://www.physicsforums.com/attachments/11871

View attachment 11873

I've tried:

BC || EF

How to find angle GDC? I think GDC=7x but why?

I have an answer but how to solve this?

Now find the angles ADB, BDC, and ADC. They sum to 180...

-Dan

- #3

maxkor

- 90

- 0

BAC=180-15x

ADB=180-3x

BDC=180-8x

ADC=11x

What next?

ADB=180-3x

BDC=180-8x

ADC=11x

What next?

Last edited:

- #4

- 1,802

- 733

Check that again. BAC = 180 - 14x.

ADB + BDC + ADC= 180. Why?

-Dan

ADB + BDC + ADC= 180. Why?

-Dan

- #5

maxkor

- 90

- 0

ok, BAC = 180 - 14x

Can you tel why ADB + BDC + ADC= 180? I don't see it.

Can you tel why ADB + BDC + ADC= 180? I don't see it.

- #6

- 1,802

- 733

Sorry, that was a typo. They sum to 360. Notice that all three angles sum to form a circle.ok, BAC = 180 - 14x

Can you tel why ADB + BDC + ADC= 180? I don't see it.

-Dan

- #7

maxkor

- 90

- 0

ADB=180-3x

BDC=180-8x

ADC=11x

ADB + BDC + ADC=180-3x+180-8x+11x=360

And what next??

BDC=180-8x

ADC=11x

ADB + BDC + ADC=180-3x+180-8x+11x=360

And what next??

- #8

- 1,802

- 733

Well, it made a lot more sense when I had the typo! :)ADB=180-3x

BDC=180-8x

ADC=11x

ADB + BDC + ADC=180-3x+180-8x+11x=360

And what next??

I'll have to get back to you on this. Sorry about that!

-Dan

- #9

I like Serena

Homework Helper

MHB

- 16,350

- 256

Picture to scale with $x=0.5^\circ$ where the angles are correct except for $\angle ACD$.

\begin{tikzpicture}

\def\x{0.5}

\def\d{4}

\def\b{\d * sin(3*\x) / sin(2*\x)}

\def\a{(\b) * sin(14*\x) / sin(7*\x)}

\def\e{(\b) * sin(\x) / sin(3 * \x)}

\coordinate[label=left:B] (B) at (0,0);

\coordinate[label=A] (A) at ({7*(\x)}:{\b});

\coordinate[label=D] (D) at ({5*\x}:{\e});

\coordinate[label=right:C] (C) at ({\a},0);

\draw (A) -- (B) -- (C) -- cycle (B) -- (D) -- (C) (D) -- (A);

\end{tikzpicture}

Note that $\angle ACD=4x$ is too big in the picture.

Btw, this is TikZ code in which we can change the angle $x$ if we want to play with it.

**Solution**

The angles at B and C are both $7x$. Therefore the triangle is isosceles, and

$$AB=AC$$

From the law of sines:

$$\triangle ABD:\quad\frac{AB}{\sin(180^\circ-3x)}=\frac{AD}{\sin(2x)}$$

$$\triangle ACD:\quad\frac{AC}{\sin(11x)}=\frac{AD}{\sin(4x)}$$

Combine to find:

$$\frac{AB}{AD}=\frac{\sin(3x)}{\sin(2x)}=\frac{\sin(11x)}{\sin(4x)} \implies \sin(3x)\sin(4x)=\sin(2x)\sin(11x)\tag 4$$

Wolfram seems to say that $x=0$.

**EDIT**: with some twiddling, $x=10^\circ$ rolls out as the only solution in which we actually have a triangle.

\begin{tikzpicture}

\def\x{0.5}

\def\d{4}

\def\b{\d * sin(3*\x) / sin(2*\x)}

\def\a{(\b) * sin(14*\x) / sin(7*\x)}

\def\e{(\b) * sin(\x) / sin(3 * \x)}

\coordinate[label=left:B] (B) at (0,0);

\coordinate[label=A] (A) at ({7*(\x)}:{\b});

\coordinate[label=D] (D) at ({5*\x}:{\e});

\coordinate[label=right:C] (C) at ({\a},0);

\draw (A) -- (B) -- (C) -- cycle (B) -- (D) -- (C) (D) -- (A);

\end{tikzpicture}

Note that $\angle ACD=4x$ is too big in the picture.

Btw, this is TikZ code in which we can change the angle $x$ if we want to play with it.

The angles at B and C are both $7x$. Therefore the triangle is isosceles, and

$$AB=AC$$

From the law of sines:

$$\triangle ABD:\quad\frac{AB}{\sin(180^\circ-3x)}=\frac{AD}{\sin(2x)}$$

$$\triangle ACD:\quad\frac{AC}{\sin(11x)}=\frac{AD}{\sin(4x)}$$

Combine to find:

$$\frac{AB}{AD}=\frac{\sin(3x)}{\sin(2x)}=\frac{\sin(11x)}{\sin(4x)} \implies \sin(3x)\sin(4x)=\sin(2x)\sin(11x)\tag 4$$

Wolfram seems to say that $x=0$.

Last edited:

- #10

maxkor

- 90

- 0

https://hobbydocbox.com/Board_Games...ual-ksf-meeting-november-protaras-cyprus.htmlI've tried to find a trick solution.

- #11

I like Serena

Homework Helper

MHB

- 16,350

- 256

Interesting, Wolfram confirms that $x=10^\circ$ is also a solution of the same equation. It appears that I misinterpreted the output of Wolfram.Ok but answer is 10 not 0, so how solve this without wolfram.

I've tried to find a trick solution.

\begin{tikzpicture}

\def\x{10}

\def\d{4}

\def\b{\d * sin(3*\x) / sin(2*\x)}

\def\a{(\b) * sin(14*\x) / sin(7*\x)}

\def\e{(\b) * sin(\x) / sin(3 * \x)}

\coordinate[label=left:B] (B) at (0,0);

\coordinate[label=A] (A) at ({7*(\x)}:{\b});

\coordinate[label=D] (D) at ({5*\x}:{\e});

\coordinate[label=right:C] (C) at ({\a},0);

\draw (A) -- (B) -- (C) -- cycle (B) -- (D) -- (C) (D) -- (A);

\end{tikzpicture}

Last edited:

- #12

maxkor

- 90

- 0

But that equation is hard to solve.

- #13

maxkor

- 90

- 0

This is the problem with the 4 point kangaroo competition.

- #14

I like Serena

Homework Helper

MHB

- 16,350

- 256

There doesn't seem to be a problemThis is the problem with the 4 point kangaroo competition.

- #15

I like Serena

Homework Helper

MHB

- 16,350

- 256

I've tried to find a trick solution.

If only we could prove that $\triangle CGD$ is similar to $\triangle AGC$. 🤔

We have 1 angle that is the same.

We either need another angle,

Last edited:

- #16

maxkor

- 90

- 0

Look page 74.

- #17

maxkor

- 90

- 0

$\triangle CGD$ is not also isosceles.If only we could prove that $\triangle CGD$ is similar to $\triangle AGC$. 🤔

We have 1 angle that is the same.

We either need another angle, or we need to prove $\triangle CGD$ is also isosceles.

If only we could prove that $\triangle CGD$ is similar to $\triangle AGC$, if GDC=7x ??

- #18

I like Serena

Homework Helper

MHB

- 16,350

- 256

The solution $x=10^\circ$ is the only one such that all angles of the triangle are less than $180^\circ$.

The interesting part is that angles with an integer number of degrees are solutions.

For that to make sense, there must be angles hidden somewhere of a multiple of $x$ that is $90^\circ$ so that the cosine is zero, or $180^\circ$ so that the sine is zero.

Put otherwise, if we can find an angle of $9x$ somewhere (or $18x$), and if we can prove that it is a right angle (respectively straight angle), then the solution $x=10^\circ$ comes rolling out.

Unfortunately I haven't found an angle of $9x$ yet, although I do have $x,2x,3x,4x,5x,7x,8x,10x,11x,12x$ and their complements.

- #19

maxkor

- 90

- 0

Maybe you have GDC=7x? we at home :)

- #20

maxkor

- 90

- 0

Share:

- Last Post

- Replies
- 0

- Views
- 418

- Last Post

- Replies
- 0

- Views
- 470

- Last Post

- Replies
- 4

- Views
- 379

- Replies
- 8

- Views
- 332

- Replies
- 2

- Views
- 509

- Replies
- 6

- Views
- 541

- Last Post

- Replies
- 1

- Views
- 687

- Last Post

- Replies
- 5

- Views
- 371

- Last Post

- Replies
- 1

- Views
- 45

- Last Post

- Replies
- 1

- Views
- 527