Math Puzzles: Try to solve this problem

angle DEC its EDC did u think ??? D is on the BC, E is on AB is it possible DEC ???? 😉

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

I'm still pretty sure that the angle is ~32.54 degrees.

Call X the intersection of AD and CE. The angle CXA and therefore also DXE must be 30 deg greater than angle CEA. Since we know EDX is 20 deg, then CEA must be 65 deg. And that makes DEC also 65 deg.

Originally posted by luskin
Call X the intersection of AD and CE. The angle CXA and therefore also DXE must be 30 deg greater than angle CEA. Since we know EDX is 20 deg...

OK so far...

...then CEA must be 65 deg. And that makes DEC also 65 deg.

Bit of a leap here.

Look at the diagram - does CEA look like 65 degrees?

Originally posted by serigado
...snip...

- line EA is 2 times bigger then line EF (because angle FDE is 10º and angle EDA is 20º )

...snip...

There's your problem. EA is not 2xEF. Life isn't that simple.

It does give you an approximate answer though - and 30 degrees is an approximation of ~32.54.

I'm not going to go looking for another solution until someone points out a flaw in the argument using the sine rule. But I'm quite happy to critique other attempts 🙂

Originally posted by mtthw
There's your problem. EA is not 2xEF. Life isn't that simple.

It does give you an approximate answer though - and 30 degrees is an approximation of ~32.54.

I'm not going to go looking for another solution until someone points out a flaw in the argument using the sine rule. But I'm quite happy to critique other attempts 🙂

damn... you're right... for it to be 2:1 proportion, the AB segment would have to be vertical...
The rule of angles is for segments inside circles...
I'll do it later with brute force with innerproducts to check your result.

Originally posted by serigado
damn... you're right... for it to be 2:1 proportion, the AB segment would have to be vertical...

I assume you mean perpendicular.

No.

Suppose the angle was 45 degrees. Doubling the angle gives 90 degrees. The segment's length does NOT double.

Originally posted by mtthw
There's your problem. EA is not 2xEF. Life isn't that simple.

It does give you an approximate answer though - and 30 degrees is an approximation of ~32.54.

I'm not going to go looking for another solution until someone points out a flaw in the argument using the sine rule. But I'm quite happy to critique other attempts 🙂

I'm pretty sure that the 32,54 answer is right. I've been looking at it for a few hours. I made a drawing as accurate as I could (being a math teacher that means something 😛 ) and the angle is definitely just OVER 30.

Originally posted by mtthw
OK so far...

[b]...then CEA must be 65 deg. And that makes DEC also 65 deg.

Bit of a leap here.

Look at the diagram - does CEA look like 65 degrees?[/b]

Much too big a leap! The sine rule solution must surely be correct; will be interesting to see the elegant way to get there.

Originally posted by luskin
Much too big a leap! The sine rule solution must surely be correct; will be interesting to see the elegant way to get there.

i'm fairly certain, since the original poster has failed to acknowledge that the sine rule solution is correct, and continues to purport that the solution is a round number (e.g. 10, 20, 30, 40, or 50 degrees), that the sine rule IS in fact the elegant solution, and there is no alternate method involving only triangle similarity and 180 degree angle sum. i'm with mtthw in that until the sine rule answer is disproven in some miraculous way, i'm not going to waste any more time trying to come up with a better solution. but i'll keep reading everyone else's 🙂

Can't be 40 either (again, Q is where AD intersects CE):

Suppose DEC is 40. Then AEQ is 90, as is CDQ. And we know AQE = CQD = 60. So now AQE ~ CQD.

Similar argument to before:

AQ/CQ = EQ/DQ => AQ/EQ = CQ/DQ

We know AQC = EQD = 120, so AQC ~ EQD.
But this doesn't work. AQC's angles are 30, 120, and 30; while EQD's angles are 40, 120, and 20.

If it's not 30, and it's not 40, how can it be a multiple of 10?

Originally posted by TheMaster37
I assume you mean perpendicular.

No.

Suppose the angle was 45 degrees. Doubling the angle gives 90 degrees. The segment's length does NOT double.

I real meant vertical (perpendicular to segment AC)
The answer is around 32 using brute mathematics, like it has been pointed before.
I really want to see the elegant solution...

EDIT: considering ED makes an angle 10º with horizontal , making the inner product ED with EC gives an angle for DEC = 31.5085º
(i used exact numbers during calculation)

Originally posted by Jirakon
How can there be an elegant solution with an angle of ~32.54 degrees?

http://s117.photobucket.com/albums/o51/Jirakon/?action=view&current=Triangle.png

(1) a/sinz = b/sin50
(2) a/sin(70-z) = b/sin60

(2)/(1) => sinz/sin(70-z) = sin50/sin60

z = 32.54

There's something I can't understand in your reasoning:
The "a" in a/sin(z) and the "a" in a/(sin(70-z) are equal in your reasoning. Why? I don't see how they are the same.

The "a" in a/sin(z) and the "a" in a/(sin(70-z) are equal in your reasoning. Why? I don't see how they are the same.

If you connect the midpoint of one side of a triangle to a point on another side of the triangle, and the created segement is parallel to the third side of the triangle, then the second point is the midpoint of the second side.

In this case, I'm constructing a segment DF such that DF || CE. Since D is the midpoint of BC, F must be the midpoint of BE.

http://s117.photobucket.com/albums/o51/Jirakon/?action=view&current=Triangle.png

Originally posted by Jirakon
The "a" in a/sin(z) and the "a" in a/(sin(70-z) are equal in your reasoning. Why? I don't see how they are the same.

If you connect the midpoint of one side of a triangle to a point on another side of the triangle, and the created segement is parallel to the third side of the triangle, then the second point is the midpoint of the second side.

In of BE.

http://s117.photobucket.com/albums/o51/Jirakon/?action=view&current=Triangle.png

It would be so if E was the middle point of AB, but it isn't... or am I wrong?

And if DF || CE, the angle 70-z (FDB) would be 30º ...