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Today’s Putnam problem – Nov. 27, 2007

By Rod Carvalho

From Harvard’s Math Department webpage, here’s today’s Putnam problem:

Given a plane k, a point P in the plane and a point Q not in the plane, find all points R in k such that the ratio \frac{QP + PR}{QR} is a maximum.

I like this problem. I will think about it and post my ideas later on.

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This entry was posted on November 27, 2007 at 1:37 pm and is filed under Brain Teasers, Mathematics, Recreational Math. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

7 Responses to “Today’s Putnam problem – Nov. 27, 2007”

  1. Ramzi Says:
    November 27, 2007 at 11:20 pm | Reply

    For the eq to be a maximum
    then QR must be a minimum
    QR approx -> 0

  2. rod. Says:
    November 27, 2007 at 11:41 pm | Reply

    Ramzi,

    You did not read the problem statement carefully enough. We cannot have QR \rightarrow 0 because point Q does not lie on plane k, while R must lie on plane k. We cannot make R = Q because we can not take R off the plane!

    My guess is that point R is the one in plane k which is nearest to point Q. That means that the line segment QR is perpendicular to plane k. Please note that this is a guess. I have tried some algebraic manipulation, but haven’t gotten anywhere yet… I thus opted for an “educated” guess.

  3. eldila Says:
    November 28, 2007 at 5:08 pm | Reply

    I attempted to solve the problem. I made a post on my own blog:
    http://www.jkwiens.com/2007/11/28/analysis-of-todays-putman-problem/

    Also, it would be nice if you allowed pings.

  4. rod. Says:
    November 28, 2007 at 5:19 pm | Reply

    Yup,

    I had already found and read your post a few hours ago, but thanks for notifying :-) WordPress does not allow pings from posts outside WordPress, so it’s not really my fault. Fortunately, there are many other tools at my disposal to find out who’s linking to my blog, and that’s how I happened to find your (very cool) blog.

    Cheers,
    rod.

  5. eldila Says:
    November 28, 2007 at 5:32 pm | Reply

    Thanks for the response. I am hosting wordpress on my own server, so I just assumed that wordpress.com would allow it as well.

    Anyways, my solution is not rigorous but I don’t think there is any fatal mistakes. Let me know if you find any problems with it. :D

  6. jkwiens.com: Analysis of Today’s Putman Problem Says:
    November 29, 2007 at 3:49 am | Reply

    [...] found an interesting problem on the blog Reasonable [...]

  7. kreso bilan Says:
    December 1, 2007 at 1:18 pm | Reply

    The problem is slightly ill-posed. There is only one point in the plane k which has maximum ratio (QP+PR)/QR. The R’ point lies directly beneath point Q.
    So, let’s find other points R’ lying on other rays in the plane k.
    As the case with QPR’, the other points R”, R”’,… must form rectangular triangle with P and Q, so that QPR”, QPR”’ triangles are rectangular.
    But this triangles don’t lie in the k plane… Nevertheless, if we project QPR”, QPR”’ to the k plane, they still form rectangular triangles R’PR”, R’PR”’,… So, the trace of R’, R”, R”,… must be part of circle with diameter R’P. The part of the circle is a one quarter of the circle (so that every ray through P has only one point R””).
    QED
    (It seems a bit too easy, I hope I am interpreting the problem correctly. Now I am going to check the solution of the other commenter).

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