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**Tonia96TT****Member**- Registered: 2010-12-23
- Posts: 5

Hello;

I am trying to solve what must be a ridiculously simple problem involving Pythagoras theorem and some simple geometry but I seem to be missing something. The problem is this:

A farmer has a square field and has set up a drinking trough at a point in the middle of the field that is 50m from one corner of the field, 30m from another corner, and 40m from a third corner (the diagram shows the configuration). The goal is to find the size of the field (either the area of the length of one side it is a square field).

Note that the 30m and 50m lines do not make the diagonal line of the square field.

I have constructed several right angled triangles and looked for common sides to try and solve simultaneous equations but I end up with 3 equations and 4 unknowns and am thus unable to solve this.

Any help (especially in the form of a detailed explanation) would be greatly appreciated. By the way, I am not a student with a homework assignment this is a problem I came across and am embarrassed that at my age I cannot figure it out!

Kind regards,

Tonia

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi Tonia96TT;

Tonia96TT wrote:

I came across and am embarrassed that at my age I cannot figure it out!

Welcome to the forum. With the method I used and the solution I got, you should not be embarassed if you were 179.

Checks out! Now I am taking you on your honor that this is not a contest problem or a homework assignment.

Merry Christmas and a Happy New Year!

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,418

OK I'm impressed.

Are you going to reveal your method?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi Bob,

Sorry did not see you. I filled in the important parts. The computational problem is tough, I am trying to work that down.

The z variable can be eliminated but it is not necessary.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,418

Thanks bobbym

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Tonia96TT****Member**- Registered: 2010-12-23
- Posts: 5

Hi Bobbym;

Thanks for your reply and your solution. Your answer is correct but I still cannot figure out how you arrived at it. Your simultaneous equations all make sense (they are in fact the same ones I derived), but when I tried to solve them I got stuck. I could not figure out what you meant by Solve by using the equation that only has x's in it first (equation number 3 in your list), as when I solve it all I get is either x^2 = x^2 or 50=50. Also, by The z variable can be expressed in terms of x also I assume you mean z=x+y-w but I did not see how that helps.

I really appreciate your taking the time to respond and I feel I am close but I am still not quite there. Any more details would be greatly appreciated.

Kind regards,

Tonia

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,418

hi Tonia and bobbym

There is a lesser known formula for the area of any triangle with sides a, b and c.

Let 2s = a + b + c

then

So I let the side of square be 2x. The table below shows the formula creation.

By using the formula and also the area of half the square (=2.x.x)

I came up with this formula:

I put this into Excel and used goalseek to home in on the vaue x = 28.2695196043031, which gives the length of a side as 2x = 56.5390392086062

Merry Christmas and a Happy New Year!

Got to go a peel some sprouts now

Bob

*Last edited by bob bundy (2010-12-25 00:05:23)*

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi Tonia96TT;

I considered that problem solved with the formation of the simultaneous set of nonlinear equations. I then used mathematica to solve them.

This is perfectly acceptable, you would not take the square root of 123612563.87614523 without using calculator or a computer would you?

But let us see how far we can get with some hand methods. Save me some sprouts , Bob!

When you try to solve the equation with only x's in it you should come up with no solution. So just eliminate that equation from the set. This is only natural since we had 5 equations and only 4 unknowns.You should be left with this.

You can eliminate x and after some algebra.

Now you have a 4th degree equation so a numerical method involving a computer is maybe necessary. We will see.

We can eliminate w:

Can you finish now with the top equation? Solving for y exactly is possible. Once you have y you can get z and then x. Remembering that x + y = a side!

See a better solution in post #15.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**Tonia96TT****Member**- Registered: 2010-12-23
- Posts: 5

Hi Bobbym and Bob Bundy;

Thank you both for your help and patience in this. Unfortunately I am still unable to solve the problem, even with the help and advice proffered.

[By the way, the background to this problem is that it was presented in a math book aimed at grade 8 students, so I am somewhat surprised at the complexity of the answer and the need for such heavy computation]

Bobbym, I can understand your math and logic up to the point where you write "You can eliminate x and after some algebra", as it is from here on in that I am lost. How does one eliminate x, and what is the algebra involved (I cannot generate the next 3 equations you list). Also, I was unsure what you meant by "Can you finish now with the top equation?" Which is the top equation (the one with the y^4 term in it or the one starting with 800z=)?

Mr. Bundy, I'm sorry but I don't even understand your basic premise (eg "let 2s = a + b + c"; why 2s?) and I could not decipher your response (even though your answer is correct).

I'm sorry if I appear thick but both your answers do not make sense to me. Perhaps this is beyond me (although I am still surprised that someone obviously thought a grade 8 student could solve this without the help of a computer), and I wouldn't be offended if either of you gave up on me.

I did want to express my thanks and appreciation for your help thus far, and I am still interested in how to get the final answer if this is possible without extensive computing power!

Kind regards,

Tonia

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi;

Tonia96TT wrote:

By the way, the background to this problem is that it was presented in a math book aimed at grade 8 students,

There is the problem, if it is for 8 th graders I am outclassed, but I will try anyway. There may be an easier solution to this but you did say find by Pythagoras so...

Tonia96TT wrote:

I'm sorry if I appear thick but both your answers do not make sense to me. Perhaps this is beyond me (although I am still surprised that someone obviously thought a grade 8 student could solve this without the help of a computer), and I wouldn't be offended if either of you gave up on me.

I am too stupid to ever give up. Back in the old country if you give up your whole family is sauteed in onion juice.

I am famous for my ugly solutions so here we go!

Tonia96TT wrote:

I did want to express my thanks and appreciation for your help thus far, and I am still interested in how to get the final answer if this is possible without extensive computing power!

I am showing you how this can be done without heavy computing but it is going to be pure drudgery. I know you do not understand how it was arrived at but forget that for now. First we will get one answer, y.

You solve the top equation like this:

As a fourth degree polynomial it is difficult but using the substitution y = x^2 you get this poly that you can do easily.

This is a simple quadratic can you now solve this? If you cannot then I will try to teach you how to solve a quadratic. Please give it a try.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,418

hi Tonia,

If I may use the analogy that we are at war with this problem, then I have a racial characteristic that " We may lose the first battle, but we go on to win the war!". If that was good enough for Winston Churchill, it's good enough for me. In terms of helping you, that means I'm not ready to give up yet.

If I've understood correctly what 'grade 8' means then I'm very surprised. It is far from a 'ridiculously simple problem'. Bobbym's method uses Pythagoras as you first specified, but requires a lot of advanced algebra to get to an answer. The trick of substituting y = x^2 to solve a quartic (highest power x^4) is taught to mathematics students in England at age 17/18 (=Advanced level), not earlier. No simultaneous equations with more than 2 variables would occur before this level too. Maybe we, all three, are missing a direct and simple method, but I doubt it. And no one else on the forum has jumped in with an answer either, so I think the person writing this book may have set a problem that is just too hard for grade 8. Nor does it lend itself to an easy scale drawing type solution. I have a program called 'Sketchpad' that allows an accurate construction, but even with that, it took me three attempts before I got one that worked.

Now to the methods.

bobbym wrote:

You can make x the subject of D and substitute this into C. ***

Then eliminate w and z using A and B to leave an equation in y.

This is a quartic but can be converted to a quadratic that can be solved using the formula. (at grade 8 ?)

(bobbym, now I look more closely at this I'm stuck also as to how the step *** works as there are some terms with w and z rather than w^2 and z^2. This approach is tough for those of us who don't have Mathematica.) These are for you:

So I thought I'd try another approach. The formula

is one I learnt in 1968 when I was doing A level maths. This is the first time I've actually used it since!

It is derived from the cosine formula and double angle formulas, so again is an Advanced level topic. 's' is called the semi-perimeter and is introduced only to make the final version of the formula neat to remember. Otherwise it is

I used it to get the area of triangles, AEB, BEC and AEC. Then added them together and set this equal to the area of half the square. Advantage: only one unknown; I used x as half the length of a side because this kept lots of ½s out of the equation. The result was again an equation that wouldn't yield to analytic solutions so I resorted to Excel. I constructed formulas for the left hand side and the right hand side of the equation and then trialled to make them equal. Excel has a macro that does trials for you where you specify the target value and what you want to vary and the procedure 'homes in' on the solution.

I will post the proof for the formula and the spreadsheet if you want.

Bob

*Last edited by bob bundy (2010-12-25 21:48:10)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,418

Hi Tonia,

How about this method:

See diagram below.

The square ABCD is rotated 90 around B, to give ABA'D'.

E' is the rotated point for E.

Join EE'.

EB = E'B = 40. and angle EBE' = 90

So EBE' is isosceles with angle E'EB = 45 and E'E = 56.6 (by Pythag)

AE' = CE = 30 so use cosine rule to calculate angle AEE' (dash removed in next two lines as it gives an error)

so

Now use cosine rule on triangle AEB

so

Bob

*Last edited by bob bundy (2010-12-26 04:08:07)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi All;

Sorry, for the no reply but my internet provider has been going out for large portions of the last 3 days. They are claiming that a bunch of nasty aliens have vandalized their satellite. I think they just have junky lines.

(bobbym, now I look more closely at this I'm stuck also as to how the step *** works as there are some terms with w and z rather than w^2 and z^2. This approach is tough for those of us who don't have Mathematica.)

( Original comments deleted see post below )

Thanks for the sprouts!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Tonia96TT****Member**- Registered: 2010-12-23
- Posts: 5

Hi Bobbym and Bob Bundy;

Mr. Bundy's latest solution is quite elegant and satisfying, and the one I like the best (and can understand).

Bobbym, regarding solving for "13 x^2-13120 x =-3200000", I can do this using the equation to solve for the roots of a quadratic (I get X=596.7 and X=412.5 as 2 solutions), but your offer to help me solve quadratics was lovely and very gracious. Thank you.

I can't tell you how much I appreciate the amount of time and effort you have both spent trying to help out with this. I think I agree with Mr. Bundy that this is far from a ridiculously simple problem (all the other examples in this section of the grade 8 math book were just that - ridiculously simple, and I thought I was simply missing something obvious for this problem). I too am too stupid to give up, but I must add that I am often too stupid to know when to give up as well (this was a problem that I thought I could solve in 2-3 minutes but has eaten up a good portion of the holidays. I even dreamed about it at one point!).

Yes, some problems cannot be easily solved with pen and paper, but I cannot understand why such an example was given in a grade 8 math book (especially as none of the other examples needed anything more than a calculator).

Thanks again to you both for all your help. It is hugely appreciated, and I wish you all the best in the New Year. As for sprouts, the less said the better I think.

With best wishes,

Tonia

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi Tonia;

Hold the presses! I did it!

Subtract the 2 nd equation from the first.

Subtract the 3 rd equation from the second.

Eliminate x and z in equations 6 and 7 using equations 4 and 5.

Solve for y and w in A and B.

That is the trick 2 variables are expressed in terms of S!

Use equation 3 because it uses w and y.

Times by 4S^2

Now you should know how to do this from an earlier post.

Just paper, pen and pythagoras, maybe too hard for an 8th grader. I would not take a chance and say too hard for every 8th grader..

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,418

hi bobbym and Tonia,

Very impressed by this, bobbym. That's got to be the best method so far. :):)

Tonia wrote:

I appreciate the amount of time and effort you have both spent trying to help out with this.

You are very welcome. I have enjoyed trying the problem and haven't given up on finding the 'grade 8' solution.

Watch this space!

Bob

*Last edited by bob bundy (2010-12-27 02:13:35)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Not really Bob, your idea is probably best. I wished she had not mentioned pythagoras, that took a whole day. There are undoubtedly more ways to do this.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,418

hi bobbym,

That's very gracious of you. I keep coming back to (30,40,50) as a Pythagorean Triple. Surely that triangle crops up somewhere, but I have't found it yet.

Oh, yes, Tonia; it's just occurred to me: Maths Is Fun apparently.

Bob

*Last edited by bob bundy (2010-12-27 02:20:44)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Bob Bundy wrote:

That's very gracious of you.

I have to quote that in contrast to the usual comments I get. I have been called boring, annoying, dishonorable, sesquipedalian, idiotic, sickening, rude, pathetic, cheesy, creepy, arrogant, a big blowhard, bad at math, messy, incomprehensible, bad at german, bad at chess, bad at computers, despicable, unimaginative, retarded, dishonest, a lizard and a lowdown punk. That is just from the girls.

Yes, Math is Fun!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,418

hi bobbym

someone wrote:

sesquipedalian

I had to look that up. Perhaps it was a very subtle joke.

Happy New Year!

Bob

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Hi Bob;

Yes, that person knew I would not know what it means.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**Tonia96TT****Member**- Registered: 2010-12-23
- Posts: 5

Thanks again to you both for your superb help and perseverance. I must admit I was very pleasantly surprised at the obvious enjoyment and passion the both of you feel for math (and for helping others appreciate the beauty of math), and want to let you know I think you are both a credit to the community.

Thanks again for all your help, and all the best in the New Year.

Kind regards,

Tonia

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,535

I second that! This has been a fun thread to watch.

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,418

Hi MIF,

Thank you for that comment. I still have the brain cells working on this. My Sketchpad files are up to version 7 now. I'm hoping there's a 'straight pythag. with 1 variable' solution lurking in there somewhere.

Happy New Year!

Bob

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,164

Tonia wrote:

I think you are both a credit to the community.

MIF wrote:

I second that!

I was going to rant on and on about how that comment does not apply to me. I always get more out of answering any question then I give to the OP. So actually, in my case it is total selfishness. It is impossible to do a selfless act. What was I going to say? Oh, yea, since MIF has seconded it I cannot rant about it.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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