Brain Exercise...

[joo]

The values are not specified. It should be solved without values ?

Yes. The answer should be given in terms of a, b and c.

And there is no such angle called AEF in the figure. Please correct it.

Oh sorry. Fixed; although I can't edit my post, so you might need to bypass your browser's cache to see it. (The angle AEF does exist, by the way, but it obviously isn't a right angle).

[Snickie]

I know how to work out this problem, I'm just having trouble finding the answer in terms of a, b, and c. Does it /have/ to be just in a, b, and c? Or can I use other segments and angles too?

[joo]

I know how to work out this problem, I'm just having trouble finding the answer in terms of a, b, and c. Does it /have/ to be just in a, b, and c? Or can I use other segments and angles too?

It can be expressed in terms of other angles and segments, as long as those angles and segments are expressed in terms of a, b or c.

On another note, I just noticed another error on the diagram which would change the answer from what I originally intended, but a solution exists in both cases (although the solution for the erroneous labeling is a bit horrible), so I will consider solutions in terms of both the old labeling and the new labeling. The corrected diagram is:

[Snickie]

I know how to work out this problem, I'm just having trouble finding the answer in terms of a, b, and c. Does it /have/ to be just in a, b, and c? Or can I use other segments and angles too?

It can be expressed in terms of other angles and segments, as long as those angles and segments are expressed in terms of a, b or c.

Well that's even more than less than unhelpful. Thanks, I guess.

It's too bad we can't prove that |AF|=|CD|. That would make things easier, but then again, that's not the point of the puzzle, is it.

[joo]

Well that's even more than less than unhelpful. Thanks, I guess.

It's too bad we can't prove that |AF|=|CD|. That would make things easier, but then again, that's not the point of the puzzle, is it.

It happens that every length and angle in the diagram is uniquely determined by a, b and c, which is why it's possible to express x in terms of them.

[Snickie]

I know that. Otherwise you wouldn't've given the problem.

h-trig wouldn't happen to be associated with this, would it? (ex: sinh of some angle) Because if it is, then I can't do it because my trig teacher never got that far last year (courtesy of people being all like "What? Huh? I don't get it! Can we spend another week on it?" when we were still in our Algebra 2 review).

[joo]

I know that. Otherwise you wouldn't've given the problem.

h-trig wouldn't happen to be associated with this, would it? (ex: sinh of some angle) Because if it is, then I can't do it because my trig teacher never got that far last year (courtesy of people being all like "What? Huh? I don't get it! Can we spend another week on it?" when we were still in our Algebra 2 review).

It is doable with just ordinary circular trigonometry.