25 points, square of side 3, perimeter <=5 2008 Clock-Tower School Junior p4

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25 points, square of side 3, perimeter <=5 2008 Clock-Tower School Junior p4

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#1 h Oct 18, 2020, 8:25 PM

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Consider any $25$ points, three by three non-collinear, in the interior of a square of side length $3$ . Show that there exist, four among them that form a quadrilateral perimeter less than $5$ .

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#2 h Dec 22, 2020, 4:43 PM • 2 Y

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Obviously the problem shall be solved by application of Pigeonhole Theorem. Pigeonhole Theorem is simple and "Pigeon" might be easy to identify. The key point is to figure out how to construct the "Pigeonhole" and by how many.
We know that the "Pigeon" shall be the " $25$ points". We want to have at least four points falling within a "Pigeonhole". Therefore, we shall construct $8$ "Pigeonholes" in the interior of a square of side length $3$ . (See attachment).
By application of Pigeonhole Theorem, for $25$ points and $8$ trapezoids, at least $\lfloor {\frac {25}{8}} \rfloor+1=4$ points shall fall within a trapezoid $ABCD$ with a perimeter of $3+\frac {\sqrt{10}}{2}$ which is less than $5$ , and we are done. $\blacksquare$ .
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This post has been edited 2 times. Last edited by BackToSchool, Dec 22, 2020, 4:49 PM

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#3 h Dec 22, 2020, 5:17 PM

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Correct me if I am wrong, but couldn't you just use $.75\times 1.5$ rectangles?

nice solution btw.

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#4 h Dec 22, 2020, 5:46 PM • 1 Y

Y by Mango247

natmath wrote:

Correct me if I am wrong, but couldn't you just use $.75\times 1.5$ rectangles?

nice solution btw.

Definitely you could. It is easier to calculate the perimeter of $.75\times 1.5$ rectangles.

We just need to have $8$ identical quadrilaterals with perimeters less than $5$ . The reason I came up with trapezoids is that I explored the triangles at my first try and then turned to trapezoids. Anyway, I will leave my solution as is.

This post has been edited 2 times. Last edited by BackToSchool, Dec 22, 2020, 6:04 PM

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