Math Puzzles: Try to solve this problem

Originally posted by smaia
Let ABC be a triangle where AB = BC = AC.
From A, draw a line that intersects BC at a point D.
From the AD segment, draw a 20 degrees line that intersects AB at a point E. Determine the angle DEC.
This is not an easy problem.
Hint: Try to see invisible things.
Good luck. And consider yourself a genius if you resolve it in less than 30 minutes. Don't feel bad about yourself if you don't resolve it at all.

180 deg. minus 20 degress is my ans

Originally posted by LemonJello
I quickly got the same answer as Jirakon (~32.54 deg) just using a couple applications of simple law of sines. I fail to see how the problem is either all that difficult or all that interesting.

Originally posted by serigado
You are a genius.

Originally posted by TheGambit
Let him have his moment. 😉

You probably are, I can only just remember SOHCAHTOA let alone how to apply it.
All I want is an admission of err from the topic creator, many men have devoted their lives to finding a non-existent solution of elegance, we need to make sure it does not continue.


Disclaimer: If there is indeed an elegant solution I tip my hat to you smaia and offer a humble apology

I personally got 25.

Originally posted by TheGambit
You probably are, I can only just remember SOHCAHTOA let alone how to apply it.

Originally posted by TheMaster37
At first I didn't understand what you were referrig to :p

SOS CAS TOA here 😉

Originally posted by TheMaster37
At first I didn't understand what you were referrig to :p

SOS CAS TOA here 😉

Originally posted by LemonJello
Serigado was pointing out that I am a genius according to smaia-ian standards.

Originally posted by serigado
Yes, he said whomever solved in less then 30 min was a genius... 🙂

For whomever gets a result different then ~32.54 :
The jirakon solution is the right, and ELEGANT one. You only have to imagine a paralell segment and use sine rule... can't be simpler.

Here's the exact solution:

{x} = the square root of x ; *x* = the cube root of x

sinz/sin(70 - z) = sin50/sin60
sinz/(sin70cosz - sinzcos70) = sin50/sin60
sinz = sin50/sin60 (sin70cosz - sinzcos70)
sinz = sin50/sin60 (sin70cosz) - sin50/sin60 (sinzcos70)
(sin50cos70/sin60 + 1)sinz = sin50sin70/sin60 {1 - sin^2 z} ; p = sin50cos70/sin60 ; q = sin50sin70/sin60
(p + 1)sinz = q{1 - sin^2 z}
(p + 1)^2 sin^2 z = q^2 (1 - sin^2 z)
((p + 1)^2 + q^2)sin^2 z = q^2
sin^2 z = q^2/((p + 1)^2 + q^2)
z = arcsin{q^2/((p + 1)^2 + q^2)}

If you want it even more exact (that is, no trig, just radicals):
I'm not going to prove the first two of the following equations. You can show them by setting up sin3x = 3sinx - 4sin^3 x, plugging in 1 for x, and solving the cubic equation (you can get sin3 using sin(18 - 15)). The same applies to cos3x = 4cos^3 x - 3cosx.

sin1 = a = (*{s^2 - 1} - s* - *{s^2 - 1} + s*)/2 ; s = sin3 = {8 - {3} - {15} - {10 - 2{5}}/4
cos1 = b = (*{c^2 - 1} + c* - *{c^2 - 1} - c*)/2 ; c = cos3 = {8 + {3} + {15} + {10 - 2{5}}/4

sin2 = d = 2sin1cos1 = 2ab
cos2 = e = cos^2(1) - sin^2(1) = b^2 - a^2

sin4 = f = 2sin2cos2 = 2(2ab)(b^2 - a^2) = 4ab^3 - 4a^3b
cos4 = g = cos^2(2) - sin^2(2) = (b^2 - a^2)^2 - 4(a^2)(b^2)

sin54 = h = ({5} + 1)/4
cos54 = j = {10 - 2{5}}/4

sin72 = k = {10 + 2{5}}/4
cos72 = m = ({5} - 1)/4

sin50 = n = sin(54 - 4) = sin54cos4 - sin4cos54 = hf - gj
sin60 = r = {3}/2
sin70 = u = sin(72 - 2) = sin72cos2 - sin2cos72 = ke - dm
cos70 = v = cos(72 - 2) = cos72cos2 + sin72sin2 = me - kd

p = nv/r ; q = nu/r

z = arcsin{q^2/((p + 1)^2 + q^2)}

Now isn't that elegant?

Originally posted by wolfgang59
32.54 is NOT the EXACT answer.

If it is an approximation it is certainly NOT elegant.

The problem was flawed.

Originally posted by serigado
I didn't say it was the exact answer. I told jirakon did the right way, and that the exact result was around 32.54
It's the arcsin of something .

edit: Well... jirakon got it with great detail in prev post.