1. H. T. & E. hte
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    06 Oct '07 17:121 edit
    Here is a beautiful puzzle from geometry for all RHP members.
    ABC is an isosceles triangle with AB = AC. A perpendicular PQ is drawn on AB at its mid-point P (i.e. AP =PB), and this perpendicular line cuts AC at Q. It turns out that QC = BC. What is the the vertex angle /_BAC? The solution is to be obtained analytically and exactly. No trigonometrical tables..No numerical solutions. I assure you it is analytically solvable. 😳😲
  2. Standard memberwolfgang59
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    06 Oct '07 17:29
    I have succeeded.
  3. H. T. & E. hte
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    06 Oct '07 18:261 edit
    Originally posted by wolfgang59
    I have succeeded.
    Well ..plz give your solution..
  4. Standard memberwolfgang59
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    06 Oct '07 19:27
    No solution.

    Just succeeded at trying to crack it. (As post challenged!) 😉


    Seriously though I did try and as with most of these triangle puzzles I just went around in circles and proved the triangle had 180degrees!

    Good Luck to anyone else who tries.
  5. Standard memberwolfgang59
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    06 Oct '07 20:54
    45/7 degrees

    Obvious 😀
  6. H. T. & E. hte
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    07 Oct '07 06:58
    Originally posted by wolfgang59
    45/7 degrees

    Obvious 😀
    Nope that's not correct..
  7. Standard memberwolfgang59
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    07 Oct '07 09:51
    This has been driving me crazy! All I have succeeded in doing is
    * proving that the angle equals itself
    * proving that there are 180 degress in the triangle
    * designing a new Star Trek logo

    You will have to PM me answer if I dont get it in next 24 hours!
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    07 Oct '07 11:081 edit
    180/7.

    I'll come back and give the reasoning later, if it's right.
  9. Standard memberwolfgang59
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    07 Oct '07 11:34
    B@St*** beat me!

    180/7 is correct.

    Call the angle in question 'a'

    The final equation to solve is 7a/2 = 90 and in my drunkeness last night I made that a=45/7 instead of 180/7.

    Proof

    The isoceles triangle ABC has angle at B & C of (90-a/2)


    Now consider the isoceles triangle QBC. It has angle at C of (90-a/2)

    The angle at Q is {180 -[90-a] - [90-a]} = 2a

    The angle at B is {[90-a/2]-a} = {90-3a/2}

    But the angle at B must equal the angle at Q

    Theefore 90-3a/2 = 2a

    90 = 7a/2

    180 = 7a

    180/7=a

    QED
  10. Joined
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    07 Oct '07 12:25
    Originally posted by wolfgang59
    B@St*** beat me!
    🙂

    Deceptively simple, isn't it? I started off looking for something much more complicated than that, using the sine rule etc.
  11. Standard memberwolfgang59
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    07 Oct '07 12:42
    Interestingly this is the same triangle from another (much harder) post.

    http://www.redhotpawn.com/board/showthread.php?threadid=65139
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    07 Oct '07 15:20
    Originally posted by wolfgang59
    Interestingly this is the same triangle from another (much harder) post.

    http://www.redhotpawn.com/board/showthread.php?threadid=65139
    It's the same answer. Not exactly the same triangle.
    Nice solution!
  13. H. T. & E. hte
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    07 Oct '07 15:23
    Originally posted by wolfgang59
    B@St*** beat me!

    180/7 is correct.

    Call the angle in question 'a'

    The final equation to solve is 7a/2 = 90 and in my drunkeness last night I made that a=45/7 instead of 180/7.

    Proof

    The isoceles triangle ABC has angle at B & C of (90-a/2)


    Now consider the isoceles triangle QBC. It has angle at C of (90-a/2)

    The angle at Q is {180 - ...[text shortened]... must equal the angle at Q

    Theefore 90-3a/2 = 2a

    90 = 7a/2

    180 = 7a

    180/7=a

    QED
    Right you are..nice and neat
  14. H. T. & E. hte
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    07 Oct '07 15:281 edit
    Originally posted by luskin
    It's the same answer. Not exactly the same triangle.
    Nice solution!
    puzlles are different . Solutions are the same though...This puzzle is simpler than the one in the thread quoted.
  15. Standard memberwolfgang59
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    07 Oct '07 17:02
    Originally posted by luskin
    It's the same answer. Not exactly the same triangle.
    Nice solution!
    "It's the same answer. Not exactly the same triangle. "



    Two isoceles triangles with the same angle at the apex? That makes them the same in my book!
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