Math Puzzles: Try to solve this problem

Originally posted by smaia
let me clarify a bit -
Going clockwise, mark points A, B, C such that AB = BC = CA. (i.e , the triangle is equilateral).
1- take A: The side opposed to A is BC. Draw a perpendicular line from A to BC. This line will intersect BC at point D - Note that because the line is perpendicular to BC => BD = DC.

2- Now, take the segment AD. Draw a 20 degrees segme ...[text shortened]... .e. the angle ADE is 20 degrees.

3- Link E to C.
4- Determine angle DEC

I hope this helps

That's pretty easy. It's 30 degrees...

I am not sure how to get there very well. I think it would be best to look at possible combinations of numbers that equal to 130 that make both triangles BEC and ACE true.

Originally posted by TheMaster37
As I understand it there is no single answer.


The quotation above has no restrictions on choosing the point E.

You can manage it so that point E is point B, hence the requested angle is 0.

What am I missing?

That's obviously not true. There is only one triangle. This is not an obtuse case of the law of sines, it's pretty simple if you guess and check, because I am too lazy to plug in the things into internal sums of 180 and then find properties of similar triangles.

Originally posted by smaia
The question is: what is the angle?
Assume you have a multiple choice test:
A- 10
B- 20
C- 30
D- 40
E- 50

Without using a calculator what is the correct answer?

Obviously C from those choices, DEC is lesser than ECD but more than ACE, and ACE is would work out as 20, so if ECD is 40 than DEC must be 30.

Originally posted by ParShooter
50 degrees.

The angle ADE is 20 degrees so its complement, EDC, along the right angle ADC, is 70 degrees. That's one corner of the triangle EDC. The second corner is DCE which, by definition of an equilateral triangle, is 60 degrees. 60 plus 70 is 130. That leaves 180 - 130 for the third corner of the triangle EDC which equals 50 degrees.

This is the fruit of my 10th grade geometry. I am 47 years old now. Thank you Miss Hurst!

Miss Hurst would be ashamed! You did not get that right!

That's pretty easy. It's 30 degrees...

(@_@)

I wonder if you're all doing it on purpose now, just to annoy me. Again, I refer you to post 47 or 105.

If I see another post declaring that the angle is 30 (or 40 for that matter) without refuting my argument of why it can't possibly be 30 or 40... ... ... I don't know what I'll do, but I won't stand for it!

It's ~32.54! Let's put this to rest already!

Originally posted by alexdino
first of all CZA and CZE are not oposite angles!
and second if they were oposite angles (as are CZA and DZE) you can't say that the other angles are equal too!!!



P.S.: smaia if you're reading this please be kind enough to stop this nonsens

Sorry for taking so long to reply. I apologize, there is no "elegant" solution for the data I provided. Something is wrong with the angles...But this is a famous problem that was once proposed in the math olympics (IMO) and nobody managed to resolve. But the angles are not 30 and 20 degrees.
I wonder if someone can determine what angles would result in DEC being exactly 30.
Once again I apologize for making so many people waste their times with a flawed problem. It's a pity because the real problem is really beautiful.
😞

I wonder if someone can determine what angles would result in DEC being exactly 30.

Same method, but now QDE is g:

(1) a/sin30 = b/sin(30+g)
(2) a/sin(60-g) = b/sin60

(2)/(1) => 2sin(60-g) = sin60/sin(30+g)

=> g = QDE = 30

This one could qualify as having an alternate, elegant solution.

Originally posted by smaia
Sorry for taking so long to reply. I apologize, there is no "elegant" solution for the data I provided. Something is wrong with the angles...But this is a famous problem that was once proposed in the math olympics (IMO) and nobody managed to resolve. But the angles are not 30 and 20 degrees.
I wonder if someone can determine what angles would result in DEC be ...[text shortened]... ir times with a flawed problem. It's a pity because the real problem is really beautiful.
😞

Yep it is 32.54381074 degrees, Autocad says so too.
I didn't calculate it but drafted it so it must be true.

By the way thanks a lot! I went to look up your famous matholympics problem and got to http:\\library.thinkquest.org there I clicked "take the special triangles test" (yes I was bored) and now I am stuck with an infinite succession of little popups on my screen asking me annoying intransparent and possibly personal (it is hard to determine) questions about my hypotenuses and trigoniometric ratios. No matter where I click, the little buggers keep coming back.

Doing math always makes me feel stupid.

my head hurts!

Originally posted by Jirakon

This one could qualify as having an alternate, elegant solution

Yes, and also seems much easier than we were led to believe.