Math Puzzles: Try to solve this problem

i don't know whether everyone's sine calculations were right... i didn't check... but heuristically speaking.... try this

because CDE is 110 and i fill CED as 30 which meens ECD is 40 leaving ACE as 20

so now you just ask yourself, would the angle ADE and the angle ACE be similar in an equilateral triangle when a line is perpindicularly dropped to point D (a midpoint) and the triangle created is 50 60 70 ?

it would be very very close, as lines CA and DE is NEAR parallel...but not quite.

I agree with 32 degrees...because heuristically, 30 degrees calculates rather nearly.

I look forward to seeing a simple solution.

Originally posted by Checkman
I look forward to seeing a simple solution.

a month has pased and smaia still hasn't a simple solution..............
so much for the ''genius'' part!!!

Originally posted by Adorea
i don't know whether everyone's sine calculations were right... i didn't check... but heuristically speaking.... try this

because CDE is 110 and i fill CED as 30 which meens ECD is 40 leaving ACE as 20

so now you just ask yourself, would the angle ADE and the angle ACE be similar in an equilateral triangle when a line is perpindicularly dropped to poin ...[text shortened]... ite.

I agree with 32 degrees...because heuristically, 30 degrees calculates rather nearly.

Originally posted by alexdino
take ADF = 30º => CFD = 30º
smaia miget have thought to split the 10º in 2 and get CED=35 and after checking with sin he realised that he made a mistake..

but at least he could appologise for his error and put an end to this quest for a big NOTHING! 🙄

The first clue to this puzzle is that the angle DEC is invariant.

It is always best to simplify a puzzle before attempting to solve it, and the best way to simplify such a puzzle is to introduce symmetry. The angle between the lines DA and DE is fixed at 20 degrees. However, the solver has a choice, when selecting the point D, of determining the angle between the line AB and the line AD.

One type of symmetry is that of identity (i.e., making both angles identical). Another type of symmetry is that of complementarity (i.e., selecting angles which together add to some useful total).

Question: Isn't it possible to select a point D such that the line DE is the same length as the line DC?

Also note that "elegant solution" needn't imply an integral answer. For example, a solution that required one to divide an odd whole number by two wouldn't require a calculator but wouldn't have an integral answer either. Certain other divisions yield easily calculable fractions also.

Also note that "elegant solution" needn't imply an integral answer. For example, a solution that required one to divide an odd whole number by two wouldn't require a calculator but wouldn't have an integral answer either. Certain other divisions yield easily calculable fractions also.

I already posted the exact solution (with no calculator, and nothing but radicals). It was not very elegant. If one actually substitutes all those letters I used in the final solution, it would be a huge radical mess. It's not a simple division yielding a terminal or repeating decimal. It's an irrational nightmare. In fact, now that I look again, I see that I didn't even keep it completely trig-free. There's still an arcsin in the final solution, which, if it were even possible to eliminate, would yield an even less elegant solution.

Originally posted by Jirakon
Also note that "elegant solution" needn't imply an integral answer. For example, a solution that required one to divide an odd whole number by two wouldn't require a calculator but wouldn't have an integral answer either. Certain other divisions yield easily calculable fractions also.

I already posted the exact solution (with no calculator, and not ...[text shortened]... hich, if it were even possible to eliminate, would yield an even less elegant solution.

I will post the solution next week.

The solution comes down to the principles of similar triangles. If you mark the point where lines CE and AD intersect as Z, you have have two similar triangles. The angle CZA and CZE are identical since opposite angles of two intersecting lines are equal. We also know that CAD is 30 and ADE is 20. Since the two triangles have the same angles, ADE is the same as ACE. Angles CAD and DEC are also the same. Therefore, since CAD is 30 and the equal to DEC, DEC is 30.

Originally posted by cysan
The solution comes down to the principles of similar triangles. If you mark the point where lines CE and AD intersect as Z, you have have two similar triangles. The angle CZA and CZE are identical since opposite angles of two intersecting lines are equal. We also know that CAD is 30 and ADE is 20. Since the two triangles have the same angles, ADE is the same a ...[text shortened]... ngles CAD and DEC are also the same. Therefore, since CAD is 30 and the equal to DEC, DEC is 30.

Therefore, since CAD is 30 and the equal to DEC, DEC is 30.

Remember this?:

There's no way it can be 30. (Let Q be the intersection of AD and CE).

If DEC were 30, then ACQ ~ EDQ.
=> AQ/EQ = CQ/DQ
=> AQ/CQ = EQ/DQ

We know that AQE = CQD
=> AQE ~ CQD

But this can't be true, since EAQ = 30 and QCD = 180-110-30 = 40 =/= 30.
Also, QDC = 90 and QEA = 180-50-30 = 100 =/= 90

If anyone still adheres to an answer of exactly 30 degrees, then tell me what I did wrong here.