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?

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!

Hi;

bobbym wrote:

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

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

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.

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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.

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)

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.

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 °

The original drawing is gone.

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

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.

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

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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