Math Puzzles: Try to solve this problem

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!

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

Ok, second attempt. I drew it correctly this time. My answer is 40 degrees. It's always 60 minus the given angle, 20 degrees in this case because the angle EDA is always the same as ECA as point E slides up and down side AB.

If this is wrong, it's back to chess for me.

My answer is 40 degrees

I answered that one as well:

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.

Again, if anyone still thinks it's 40, you'll have to find a flaw in that reasoning.

20 degrees

By my reckoning it's 25 degrees. I could be wrong though...

Still nobody has tried to refute the ~32.5 answer. If you want to suggest a different answer, you really ought to try and refute that argument.

Originally posted by mtthw
Still nobody has tried to refute the ~32.5 answer. If you want to suggest a different answer, you really ought to try and refute that argument.

I just assumed the 30 degree answer was correct...
I can't, however, find the flaw in the resultant diagram.
...can someone point it out?

De ja vu...

See post 47 or 105.

And the next poster had better not tell me it's 40! (unless they can refute post 56 or 109)

Originally posted by Jirakon
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.

Side-angle-side similarity: When two triangles have corresponding angles that are congruent and corresponding sides with identical ratios, the triangles are similar.

Just because two opposing triangles of an intersection are similar (AQC and EQD) doesn't mean the other two have to be.

Yes it does. There are two possibilities. One is that ED || AC, which you stated. What you failed to mention is the possibiliy that the identical angles could be switched in both triangles; instead of E=C and A=D, it could be that E=A and C=D, which is the case here. Sure, this switches the sides that are proportional, but the fact remains that the two included sides of the other two triangles are proportional, and the angle between them is equal. They have to be similar, and this is where the contradiction arises.

You may want to see a diagram to help clear it up:

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

Originally posted by Jirakon
Side-angle-side similarity: When two triangles have corresponding angles that are congruent and corresponding sides with identical ratios, the triangles are similar.

Just because two opposing triangles of an intersection are similar (AQC and EQD) doesn't mean the other two have to be.

Yes it does. There are two possibilities. One is that ED || A ...[text shortened]... lear it up:

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

In that diagram, which triangles are you comparing?
...which angles are you referring to?

I think I made it clear in my original explanation. Suppose DEC were 30. We already know that CAD is 30, and the angles AQC (sorry about not labelling Q, but it wasn't needed in my solution) and EQD are vertical. That's two congruent angles, which automatically means that triangle AQC ~ triangle EQD. Now since they're similar, we know that their sides are in proportion. Namely,

AQ/EQ = CQ/DQ.

By multiplying both sides by EQ/CQ, we get that

AQ/CQ = EQ/DQ

Now let's ignore triangles AQC and EQD, and let's focus on triangles AQE and CQD. We know the vertical angles AQE and CQD are equal. We also know that the two sides forming the angles are in proportion by the previous equation. Therefore triangles AQE and CQD must must be similar if DEC is 30.

I don't see why AQE ~ CQD, since QD is perpedicular to CD, and QE is obviously not perpedicular to AE

This was the last part of my explanation, and that's why there's a contradiction. This is why DEC can not be 30.

It seems that no-one can refute the trig-bash
I suppose that if there were a 'nice' answer it would be possible to reverse reconstruct it?