Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  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. Standard member wolfgang59
    Infidel
    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.

    3- Link E to C.
    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. Standard member wolfgang59
    Infidel
    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.

    3- Link E to C.
    4- Determine angle DEC

    I hope this helps
    I understand the question now! Whether that helps or not .....
  7. Standard member wolfgang59
    Infidel
    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. Standard member wolfgang59
    Infidel
    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 16:55 / 1 edit
    21º?

    Edit: 32º
  10. 17 Feb '08 17:21
    Originally posted by Green Paladin
    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 18:41 / 2 edits
    DEC = 32.55º

    I gave the equilateral triangle sides with 9 units.

    AD = 7.8
    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
    Originally posted by Green Paladin
    DEC = 32.55º

    I gave the equilateral triangle sides with 9 units.

    AD = 7.8
    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
    Originally posted by Green Paladin
    Ok, I've checked my numbers. Final attempt (with no rounding off) - 32.725229402º
    In two weeks I will post the solution here.
    You will be amazed how beautiful and simple the solution is.
    No need to use calculators or perform complex computations.
    It can be resolved mentally if you discover a few hidden things.