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## #1 2005-07-23 05:16:29

flagitious-t
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### Im beginning to think this is impossible....

I have this question for an online course I'm doing, I don't want/need the answer I just would like to be pointed in the right direction...

Here is the question.

A fenced, circular field has a radius of 30m A goat is tethered by a rope to a fence post at the edge of the field. If the goat can graze over exactly half the area of the field, how long is the rope?

thanks.

## #2 2005-07-23 14:36:37

MathsIsFun

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### Re: Im beginning to think this is impossible....

You wouldn't believe it, but I have a cicular lawn! No goats, though.

So, you basically have the intersection of two circles. The goat can only graze the circular field where the circle formed by his rope intersects it.

And we know that the first circle has a radius of 30m and we also know that the two circles are 30m apart.

Let's cut this problem through the centre, and only look at the top half. We can get close by just looking at this sector of a circle:

And if we now add the sector from the other circle, we find that they now include the area we want, but the triangle in the centre is doubled up:

(to be continued ...)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #3 2005-07-23 16:46:17

MathsIsFun

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### Re: Im beginning to think this is impossible....

Now, that triangle (the one that gets its area double counted) - we can solve it! Because we know all three sides (two sides are 30m and the other is the length of the goat's tether)

If we can solve the triangle for the angles of the two shaded sectors. Then use the formula for the area of a sector A = ½r² θ  (need it for both sectors), then subtract the area of that triangle.

Anyone want to have a go at the formulas, then ... ?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #4 2005-07-23 17:34:15

mathsyperson
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### Re: Im beginning to think this is impossible....

Once again, you amaze me with your ability to make a problem so much simpler. I saw this puzzle last night, but I thought that it was probably one of the ones where you get the answer quicker by doing something else and letting your subconsious mind do the work. Plus I wanted to watch Titanic. Great film, with excellent (epic, even) music and good dramatic effects and just brilliant all round, really.

I've just got a few problems with it:
• Just before they hit the iceberg, someone says "Hard a' starboard" and then the ship turns left. Maybe that's meant to happen, I don't really know.
• When the posh guy gets angry and starts shooting them, he manages to fire off seven shots before his pistol starts clicking. What kind of a strange pistol was he using?
• Not to do with the film this time, but I hate it when any piece of moving drama finishes, and then it goes to the credits playing dramatic music to reflect the mood (usually sad) and then the BBC smash the mood with a sledgehammer (wait, not a sledgehammer, a sledge. They're heavier.) repeatedly and then they pick up all the pieces and smash them a bit more.

They do this by squashing the credits up to the right-hand side, then saying "Well, that was Titanic there, hope you enjoyed it! Next up it's [another program that I do not wish to advertise] . As I've already said, this completely wrecks all the emotion you may have felt from the film and turns it into hate for the BBC people.

I'm sorry, I was going to help and it's turned into a mini film review. Oh well. I did try the puzzle during the news.
You can work out the triangle that you've got twice using the semiperimeter rule: A=√(s(s-a)(s-b)(s-c)), where a, b and c are sides of the triangle and s is half their total.

So, for this one, s would be (60+x)/2 and the area would be √((30+x/2)(x/2)²(30-x/2))
I've done a bit of simplifying, so hopefully you can see how I got that.

We can get the angles for the sectors by attacking the triangle with a sledge, I mean, using trigonometry. I'll do that once I've had something to drink.

Why did the vector cross the road?
It wanted to be normal.

## #5 2005-07-23 17:42:38

justlookingforthemoment
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### Re: Im beginning to think this is impossible....

I see why you like icebergs, Mathsyperson - but shouldn't you hate them now?

## #6 2005-07-23 17:47:21

mathsyperson
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### Re: Im beginning to think this is impossible....

I would have been quicker, but I heard the Jonathon Creek theme music and rushed to see where it was coming from. Just the radio...

Also, flagitious-t, if you only wanted a push in the right direction, you should probably stop reading now.
Right then, time for the cosine rule: a²=b²+c²-2bcCosA, meaning A=cos-¹((b²+c²-a²)/2bc)
For the length x sector, A=cos-¹(x/60), having been simplified.
For the length 30 sector, A=cos-¹(1-x²/1800), again, having been simplified.

This makes the total area equal to cos-¹(x/60) + cos-¹(1-x²/1800) - √((30+x/2)(x/2)²(30-x/2)).

The field's area is pi*30²=900pi. As we are only looking at half of this, and we only want the goat to eat half of that, we want the area to be 225pi. Now all we need to do is solve it!

Edit: To justlooking, I was thinking about saying that I liked ships, but then I realised that as a ship is built and an iceberg is natural, then icebergs are much more beautiful. I also tried doing word association from that, but then I got to lettuce and I'm not saying that I like that because that would be a complete lie. I tried again and went from iceberg to glacier to mint to polo to horses, but I'm allergic to their hair (fur?), so I couldn't say I liked them either. So, icebergs it is.

Last edited by mathsyperson (2005-07-23 17:51:06)

Why did the vector cross the road?
It wanted to be normal.

## #7 2005-07-23 17:50:08

MathsIsFun

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### Re: Im beginning to think this is impossible....

LOL ... oh I agree, they may as well interrupt all the good points in the movie that way, it would have the same effect. it is a bit like putting a speech bubble on the Mona Lisa: "Hungry? The Cafe is upstairs!"

Not to burst your bubble about my supposed brains, but I got some clues by searching the internet about intersecting circles, then I started drafting and found how neat it was, so I put those graphics up.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #8 2005-07-23 17:52:35

MathsIsFun

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### Re: Im beginning to think this is impossible....

#### mathsyperson wrote:

cos-¹(x/60) + cos-¹(1-x²/1800) - √((30+x/2)(x/2)²(30-x/2)) = 225pi.

Time for that mathematical toy - Excel!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #9 2005-07-23 17:56:25

mathsyperson
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### Re: Im beginning to think this is impossible....

To the best of my knowledge, Excel can't do inverse cos functions.
Also, is it possible to put pi up on the symbols list?

Why did the vector cross the road?
It wanted to be normal.

## #10 2005-07-23 18:00:54

MathsIsFun

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### Re: Im beginning to think this is impossible....

ACOS() ?

Pi is there, it just looks weird: Π

By the way, I tried that formula in excel with "30" for x and got -387.617, have we somehow created anti-grass?

Formula: =ACOS(B8/60) + ACOS(1-B8*B8/1800) - SQRT((30+B8/2) * (B8/2)* (B8/2) * (30-B8/2))

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #11 2005-07-23 18:05:14

mathsyperson
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### Re: Im beginning to think this is impossible....

Sorry, I've given you a duff formula. The two sectors 'areas' are actually just their angles. *Feels stupid*

The real area: 900Π*cos-¹(x/60)/360 + x²Π*cos-¹(1-x²/1800)/360 - √((30+x/2)(x/2)²(30-x/2))
Sorry again!

Why did the vector cross the road?
It wanted to be normal.

## #12 2005-07-23 18:20:02

MathsIsFun

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### Re: Im beginning to think this is impossible....

"Sorry"? Try "Haha"!

( Gee, if we all have to be worried about making mistakes then we may as well take the "Fun" off the forum name. )

Still not quite there ... maybe i am not getting my formula right.

=900*PI()*ACOS(B8/60) + B8*B8*PI() *ACOS(1-B8*B8/1800) - SQRT((30+B8/2) * (B8/2)* (B8/2) * (30-B8/2))

If you have a chance, see if you can get it working, then just delete the last two or three posts and show the world how neat and perfect we are!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #13 2005-07-23 18:34:10

mathsyperson
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### Re: Im beginning to think this is impossible....

And then it would just be strange: "Math Is      Forum"

All I can see is that you haven't divided the sector areas by the number of whatever you're measuring by. Excel works with radians by default so try this:
=900*PI()*ACOS(B8/60)/(2*pi()) + B8*B8*PI() *ACOS(1-B8*B8/1800)/(2*pi()) - SQRT((30+B8/2) * (B8/2)* (B8/2) * (30-B8/2))

Why did the vector cross the road?
It wanted to be normal.

## #14 2005-07-23 18:51:09

MathsIsFun

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### Re: Im beginning to think this is impossible....

See? My turn to be foolish. I have already had my flash of brilliance for today - it is all downhill now

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #15 2005-07-23 19:03:54

mathsyperson
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### Re: Im beginning to think this is impossible....

We still haven't cracked it though. When the goat's leash has length 0m, it can apparantly eat around 700m² of grass. It must have a long neck...

Why did the vector cross the road?
It wanted to be normal.

## #16 2005-07-23 19:14:34

MathsIsFun

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### Re: Im beginning to think this is impossible....

Oh, rats ... I mean goats. Well, the answer is somewhere in there. Maybe flagitious can take it from here.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #17 2005-07-23 19:39:24

mathsyperson
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### Re: Im beginning to think this is impossible....

My fault again! "Haha!"

I got the sectors the wrong way around.
Length x sector's angle: cos-¹(1-x²/1800)
Length 30 sector's angle: cos-¹(x/60)

So, the latest formula is =900*PI()*ACOS(1-B8*B8/1800)/(2*PI())+B8*B8*PI()*ACOS(B8/60)/(2*PI())-SQRT((30+B8/2)*(B8/2)*(B8/2)*(30-B8/2))

Edit: And it works perfectly! Cookies for everyone!
To the nearest cm, the goat's leash is 34.76m long.

Last edited by mathsyperson (2005-07-23 19:45:56)

Why did the vector cross the road?
It wanted to be normal.

## #18 2005-07-23 19:58:25

ganesh
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### Re: Im beginning to think this is impossible....

My head is reeling

Character is who you are when no one is looking.

## #19 2005-07-23 20:21:34

MathsIsFun

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### Re: Im beginning to think this is impossible....

These look particularly yummy.

Mathsy gets first choice.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #20 2005-07-23 20:29:02

ganesh
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### Re: Im beginning to think this is impossible....

Thanks, Mathsisfun,
I am on the verge of post # 200
and this an occassion for me to celebrate,
I shall take a big chunk of thhat

Character is who you are when no one is looking.

## #21 2005-07-24 03:57:14

flagitious-t
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### Re: Im beginning to think this is impossible....

thanks!

wow, really complicated.

## #22 2005-07-24 08:01:15

MathsIsFun

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### Re: Im beginning to think this is impossible....

#### flagitious-t wrote:

wow, really complicated.

Titanic or the Cookies?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

## #23 2005-07-24 08:23:17

Zach
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### Re: Im beginning to think this is impossible....

Cookies. They're really complex.

Boy let me tell you what:
I bet you didn't know it, but I'm a fiddle player too.
And if you'd care to take a dare, I'll make a bet with you.

## #24 2005-07-24 08:39:45

flagitious-t
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### Re: Im beginning to think this is impossible....

uhh... both?

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