# Math Puzzles: Try to solve this problem

smaia
Posers and Puzzles 17 Feb '08 03:24
1. 17 Feb '08 03:24
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.
2. 17 Feb '08 08:29
80° is my try after 2 mins
3. wolfgang59
17 Feb '08 13:55
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.
Doesnt seem to be enough info?

Is that ANY point D on BC or midpoint?

And E could be anywhere depending on where the 20degree angle on AD is taken from. ... or is EDA 20?
4. 17 Feb '08 15:48
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 segment (having AD as reference) that starts at D and intersects AB at a point E, i.e. the angle ADE is 20 degrees.

4- Determine angle DEC

I hope this helps
5. 17 Feb '08 16:00
It is not 80. Take at least 15 minutes and see what you get.
6. wolfgang59
17 Feb '08 16:19
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.

4- Determine angle DEC

I hope this helps
I understand the question now! Whether that helps or not .....
7. wolfgang59
17 Feb '08 16:42
Originally posted by wolfgang59
I understand the question now! Whether that helps or not .....
Good Puzzle. Cant do it!
8. wolfgang59
17 Feb '08 16:42
Originally posted by wolfgang59
I understand the question now! Whether that helps or not .....
Good Puzzle. Cant do it!
9. 17 Feb '08 16:551 edit
21º?

Edit: 32º
10. 17 Feb '08 17:21
21º?

Edit: 32º
Not 21 degrees.
Of course you are not alowed to measure the angle, or to use the measured the angle to proof backwards. I will post the solution in about 2 weeks.
11. 17 Feb '08 17:24
Originally posted by wolfgang59
Good Puzzle. Cant do it!
Try more. It should take more than a quick look to find the solution. No one that I know resolved it on the spot.
Good luck.
12. 17 Feb '08 18:412 edits
DEC = 32.55º

I gave the equilateral triangle sides with 9 units.

AE = 3.48
EC = 7.86
ACE = 22.55º
ECD = 37.45º

First I worked out AD. With AD you can solve AE. With AE you can work out EC. With EC you can work out angle ACE.

Angle ECD = 37.45º (60 - 22.5). Angle EDC (110) plus angle ECD (37.45) = 147.45º. The sum of the angles of a triangle = 180º -147.45º = 32.55º.

Is this correct?
13. 17 Feb '08 19:46
DEC = 32.55º

I gave the equilateral triangle sides with 9 units.

AE = 3.48
EC = 7.86
ACE = 22.55º
ECD = 37.45º

First I worked out AD. With AD you can solve AE. With AE you can work out EC. With EC you can work out angle ACE.

Angle ECD = 37.45º (60 - 22.5). Angle EDC (110) plus angle ECD (37.45) = 147.45º. The sum of the angles of a triangle = 180º -147.45º = 32.55º.

Is this correct?
Nice try, but the answer is not 32.55
If you want to follow your approach, you need to set AB = x where x is any real number.
Good luck.
14. 17 Feb '08 20:32
Originally posted by smaia
Nice try, but the answer is not 32.55
If you want to follow your approach, you need to set AB = x where x is any real number.
Good luck.
Ok, I've checked my numbers. Final attempt (with no rounding off) - 32.725229402º
15. 17 Feb '08 20:52