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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

diagram in post 24.

The diagram shows a triangle ABC

with ABC = 80 and ACB = 80

D lies on AC so that DBC = 60

and E lies on AB so that ECB = 50.

To find (by Euclidean geometry) x = EDB

*Last edited by bob bundy (2010-07-03 00:47:52)*

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

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

whoops. I've got BBCode on but my image tag hasn't worked.

What no FAQs on how to post images?

But you can follow the link to my site for the diagram

EDIT No you cannot. See post 24.

*Last edited by bob bundy (2010-07-03 00:37:16)*

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: 86,366

Hi Bob;

Try this:

Image removed. see post 24.

But that is not making it any easier to solve it. Does the diagram have everything you need?

**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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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Well, there are certinaly enough restrictions to make the question well-defined and only have one answer.

I'm not quite there though. I've managed to get all the angles that aren't partially made by line DE.

I have a feeling the last step will be more complicated than the rest of it.

Why did the vector cross the road?

It wanted to be normal.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

That is one of (the so called "easier" one) the "toughest easy geometry problems" of this world. I don't know who composed it but the solution does take a lot of constructions. I couldn't solve it on my own!

If two or more thoughts intersect with each other, then there has to be a point.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

Thanks for inserting my diagram.

All the information you need is there.

Now lets see if I can get a picture on this post.

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: 86,366

Hi;

bobbym wrote:

R. I. P. O. S. T. P.

This is all I will post so others can work on it.

**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: 6,380

Hi,

Euclid wants the proof too.

That's why he wrote all those books!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Sorry, bob, Euclid is dead but this is what I did. My command of compass and ruler is limited so forgive my ugly diagram. Also Point A is not shown at the top of the diagram.

**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: 6,380

Hi bobbym

Euclid may be dead but his spirit lives on ... and it's strong in you. Constructing a certain point was an excellent move and things fall nicely into place then.

Mathsyperson said that there was enough information for the problem to be well defined so now I'm wondering whether all such well defined geometry problems can be solved by Euclidean methods. Trouble with finding one that can't is that it might just be we haven't found the solution yet.

Bob

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi bob bundy;

Constructing a certain point was an excellent move and things fall nicely into place then.

Yes, it is an excellent move but it is not mine. I think the problem originates with G.H. Hardy and actually there are many solutions to it ( about 15 ). I just had to look at my notebooks, there I found 2 solutions to it. So I just had to try to understand what had been done before. Took me a couple of days just to do that!

I had a original idea that I really liked but that idea just kept falling apart.

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: 6,380

Hi bobbym

To find out what the angle was I first used the sine rule. Then I spent ages trying to establish connections between the sines of angles so that I could get the values without a calculator. Eventually did it.

Then I thought: the sine rule is usually proved using angle properties of circles, so there ought to be a circle (circles) that lead to a solution. Trouble is there's loads of possibilities but eventually I hit on your point as the centre of a very helpful circle and that led to a solution very similar to yours.

But then I thought: all well defined problems like this can be solved using the sine rule (I think) so maybe they're all solvable if you can find the right circle(s). That's as far as I've got so far.

Bob

ps. If you've got a book with more like this, perhaps you would post one (just one please, I've got things to do!)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi bob;

You are right on the money on all your points. One solution uses the law of sines and is perhaps the best way because it doesn't need to add anything to the diagram. Another solution does use a very helpful circle, so you are right again. Lastly, you are correct in surmising that research solved this problem for me.

bob bundy wrote:

But then I thought: all well defined problems like this can be solved using the sine rule (I think) so maybe they're all solvable if you can find the right circle(s). That's as far as I've got so far.

There is another solution where he draws a helpful circle and inscribes an eighteen sided polygon in it to solve the problem.

bob bundy wrote:

ps. If you've got a book with more like this, perhaps you would post one (just one please, I've got things to do!)

Take a look at post #7.

R. I. P. O. S. T. P. stands for Research Is Part Of Solving The Problem. You are correct in assuming I have a book ( 3 ) on this problem and its variants.

(Problem moved to exercises)

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: 6,380

Arrhhh. Is that what it means?

I thought it might be Rest In Peace Or Start Trying Properly. (in other words Do it or die trying)

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi bob bundy;

Only advantage of being hundreds of years older than the average poster in here ( age 13 - 25 ). Experience! Knowledge of what went before.

Once you have seen a proof, it is very difficult to forget it. Some of the particulars might fade but you know you have seen it before. Eventually you will piece it together.

Let's examine the chronology of this thread. I thought that you might really need the answer ( I didn't know you were posing a problem) . I just put the answer there ( post #7) because I knew it, I said that was all I would post. I hoped it wouldn't prevent others from working on the problem. When you asked for the proof ( post #8 ), I again reasoned that you needed the answer for some practical purpose. In such a case quoting someone else's proof is okay. I was only trying to help.

In post #11 I immediately set the record straight that it was not my proof. To not do so would be dishonest. It now became clear for the first time that you were not seeking help but were posing a problem that you already knew the answer to. Perhaps that is why it was in puzzles and games, something I missed. In any event, I believe research is an invaluable tool in not rediscovering America. Solving a problem using research or memory in no way demeans the problem or the solver, just as long as proper credit is given.

bob bundy wrote:

I thought it might be Rest In Peace Or Start Trying Properly. (in other words Do it or die trying)

I understand this attitude. I have a problem that I have been working on for more than 10 years. If I were to take and dedicate a single machine to it I could have the answer in about a year of computing time. I want a mathematical solution, so 10+ years...

Point is I didn't think you wanted to wait 10 years...

I post problems here too. I usually don't get the answer or the method I am looking for. Usually when I do not want a known solution, I state that in the problem. Anyway, I enjoyed the problem as I did have to review some material to even post the solution.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zoe11221121****Member**- Registered: 2012-10-17
- Posts: 5

bobbym: why is angle DBF equal to 40 degrees??

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi;

The diagram is gone, and I remember little about this problem. But a straight line is 180 degrees. I suppose the other two angles are given or can be deduced. 180 ° - (80 ° +60 °) = 40 °

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zoe11221121****Member**- Registered: 2012-10-17
- Posts: 5

your diagram is still up there!

But angles EBD DBF and FBC are not on a straight line.....

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

The original drawing is gone.

Line segment DFC is a straight line because it is on the side of triangle.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zoe11221121****Member**- Registered: 2012-10-17
- Posts: 5

oh..... Then can you post the original one/send it to me?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi;

I am sorry but that problem was a long time ago and I have misplaced both the problem and the solutions. Maybe the OP can provide it.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zoe11221121****Member**- Registered: 2012-10-17
- Posts: 5

urm....what is the OP?

and..... Can you calculate it again? i don't think it is a difficult question to you

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

I will try to recreate the problem and draw a diagram.

bob bundy wrote:

The diagram shows a triangle ABC

with ABC = 80 and ACB = 80

D lies on AC so that DBC = 60

and E lies on AB so that ECB = 50.

To find (by Euclidean geometry) x = EDB

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: 6,380

hi zoe11221121

Welcome to the forum.

I was a new member when I posted that question.

I think I was having trouble uploading images, so I linked it into the post.

More recently I had to remove the image to save on-line space.

Anyway, here it is again (below)

The question is "Find x"

Bob

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**zoe11221121****Member**- Registered: 2012-10-17
- Posts: 5

hi bob and bobbym

I have just registered recently and i still don't know how to upload images.

So..... How can this question be solved? Thanks.

Zoe

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