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Posers and Puzzles

Posers and Puzzles

  1. Standard member wolfgang59
    Infidel
    21 Oct '07 10:54
    Not sure how hard/easy this is but I found a neat solution.


    Consider two touching circles. Let their common tangent be the line T. Take any point on T, lets call it P.

    Now take a line from P through the centre of the first circle so that this line cuts the circle at points A & B.

    Similarly for the second circle to get points C & D.

    Prove that triangle PAC is similar to triangle PDB.
  2. Standard member celticcountry
    Copyright ©2001-2006
    22 Oct '07 15:40
    anyone can see they are, why do you need proof?
  3. Standard member Mitsurugi
    Not the dead
    22 Oct '07 15:41
    Originally posted by wolfgang59
    Not sure how hard/easy this is but I found a neat solution.


    Consider two touching circles. Let their common tangent be the line T. Take any point on T, lets call it P.

    Now take a line from P through the centre of the first circle so that this line cuts the circle at points A & B.

    Similarly for the second circle to get points C & D.

    Prove that triangle PAC is similar to triangle PDB.
    I'm not getting the english language nomenclature but I'm assuming "similar" is "its angles are the same as the other", right?
  4. 22 Oct '07 16:29
    OK. A few more letters: X is the centre of the first circle, radius x. Y is the centre of the second circle, radius y. Q is the point where the circles meet.
    For convenience, I'll say PA = a, PC = c and PQ = q.

    Since they have a common angle at P, one way of showing PAC and PDB are similar is to show that PA/PC = PD/PB

    PQX and PQY are right-angled triangles. So Pythagoras gives us:
    (a + x)^2 = x^2 + q^2
    and
    (c + y)^2 = y^2 + q^2

    => (a + x)^2 - x^2 = (c + y)^2 - y^2
    => a(a + 2x) = c(c + 2y)
    => a/c = (c + 2y)/(a + 2x)
    => PA/PC = PD/PB
    QED
  5. Standard member wolfgang59
    Infidel
    23 Oct '07 07:29
    Originally posted by celticcountry
    anyone can see they are, why do you need proof?
    thats what I thought at first but then I thought "I need proof" so I got four pieces of string and cut them to the lengths of the triangle sides. I then used a tape measure and carefully measured them (I wanted my proof to be be vigourous so I measured to the millimetre!). I then compared the ratios using my slde rule. The result was close enough to pursuade me I had similar triangles.

    btw Well done mtthw for an alternative method!
  6. 23 Oct '07 09:58
    Originally posted by wolfgang59
    Not sure how hard/easy this is but I found a neat solution.


    Consider two touching circles. Let their common tangent be the line T. Take any point on T, lets call it P.

    Now take a line from P through the centre of the first circle so that this line cuts the circle at points A & B.

    Similarly for the second circle to get points C & D.

    Prove that triangle PAC is similar to triangle PDB.
    The line through the centres of both circles is perpendicular to T. Therefore, at the start the half-planes on both sides of T are mirror images. Since you only take mirror-image actions, at the end both sides are still mirror images. In particular, PAB is a mirror image of PCD. Mirror images are similar.

    Richard
  7. Standard member wolfgang59
    Infidel
    23 Oct '07 10:26
    Originally posted by Shallow Blue
    The line through the centres of both circles is perpendicular to T. Therefore, at the start the half-planes on both sides of T are mirror images. Since you only take mirror-image actions, at the end both sides are still mirror images. In particular, PAB is a mirror image of PCD. Mirror images are similar.

    Richard
    You're looking at different triangles. The triangles I'm talking about are not the same size.
  8. 23 Oct '07 10:42
    Originally posted by wolfgang59
    You're looking at different triangles. The triangles I'm talking about are not the same size.
    Egh. PAB and PCD aren't triangles at all, or if you wish, they are degenerate triangles, and all degenerate triangles are of necessity similar, being all similar to a linear section.

    Ok, another argument, also based on symmetry:
    The angle at A is similar to that at C, because (see above) the whole setup is symmetric, and therefore angle A is the mirror image of angle C; and mirrored angles are identical. Therefore, PAC is an isosceles triangle. According to the same argument, so is PBD.
    But PAC and PDB are two isosceles triangles with the top angle identical (because the top angle in both triangles is formed by the same two lines); and therefore, they are similar isosceles triangles.

    Richard
  9. 23 Oct '07 10:49 / 1 edit
    Originally posted by Shallow Blue
    Ok, another argument, also based on symmetry:
    The angle at A is similar to that at C, because (see above) the whole setup is symmetric
    The circles aren't necessarily the same size, so the diagram isn't symmetric about T. If it was, it's a pretty trivial problem.

    The angle at A turns out to match that at D, and the angle at C to match that at B.
  10. 23 Oct '07 11:07
    Originally posted by mtthw
    The circles aren't necessarily the same size, so the diagram isn't symmetric about T.
    Ok, here's another proof:

    I need more caffeine. Q.E.D.

    Richard, off to get a cuppa