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

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?

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?

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!

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

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.
😞

my head hurts!

Originally posted by Jirakon

This one could qualify as having an alternate, elegant solution