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Math Help - Pigeon-hole Principle

  1. #1
    MHF Contributor alexmahone's Avatar
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    Pigeon-hole Principle

    Let T be a triangle with angles of 30, 60 and 90 degrees whose hypotenuse is of length 1. We choose ten points inside T at random. Prove that there will be four points among them that can be covered by a half-circle of radius 0.42.

    My solution: Let us divide our triangle into three smaller right triangles as shown in the figure. By the Pigeon-hole Principle, one of these triangles will contain four of our points. The circumradii of each of these triangles is less than 0.42, so we are done.

    I'm not sure that my proof is right because the statement in the question could apparently have been made stronger by restricting the radius of the half-circle to 0.375 (as the longest hypotenuse has length 3/4=2*0.375). Could someone please check my solution? Thanks.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Pigeon-hole Principle

    Quote Originally Posted by alexmahone View Post
    Let T be a triangle with angles of 30, 60 and 90 degrees whose hypotenuse is of length 1. We choose ten points inside T at random. Prove that there will be four points among them that can be covered by a half-circle of radius 0.42.

    My solution: Let us divide our triangle into three smaller right triangles as shown in the figure. By the Pigeon-hole Principle, one of these triangles will contain four of our points. The circumradii of each of these triangles is less than 0.42, so we are done.

    I'm not sure that my proof is right because the statement in the question could apparently have been made stronger by restricting the radius of the half-circle to 0.375 (as the longest hypotenuse has length 3/4=2*0.375). Could someone please check my solution? Thanks.

    Correct!
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