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Topic review (newest first)

bob bundy
2012-02-19 07:13:26


I kept this problem on 'subscription', as I was hopeful of coming up with a grade 9 solution using Pythagoras.

Here it is.

Set up a spreadsheet with 'w' as a variable.

use formulas to compute

y = root(30*30 - w*w)
x = root(40*40-y*y)
z = root(50*50-x*x)

Compute w + x - y - z.  For a square this value should be zero.

Use goal seek to 'find' a value of zero by varying w.

The screen shot from Excel shows the formula set up and the goal seek command with the finished values.

I have used goal seek with this age of child so I think it is a valid method.

Sorry it has taken so long to dream it up.


2010-12-29 21:46:19

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.

bob bundy
2010-12-29 20:28:54


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!


2010-12-29 12:49:23

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

2010-12-29 11:44:50

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,

2010-12-28 22:00:53

Hi Bob;

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

bob bundy
2010-12-28 21:48:59

hi bobbym

someone  wrote:


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

Happy New Year!


2010-12-28 05:28:30

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!

bob bundy
2010-12-28 01:18:25

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.


2010-12-28 01:15:26

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.

bob bundy
2010-12-28 01:10:35

hi bobbym and Tonia,

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

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!


2010-12-27 22:01:03

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..

2010-12-27 21:48:18

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,

2010-12-27 18:37:52

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!

bob bundy
2010-12-27 03:07:05

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)


Now use cosine rule on triangle AEB



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