#### 0rthodontist

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I was thinking about a problem on one of the old Putnam exams and as an aside started thinking about this: is it true that for any convex polygon of 5 or more sides, there exist 3 adjacent vertices A, B, C, such that if you "complete the parallelogram" for those 3 points, the fourth vertex of the parallelogram lies within the polygon?

I came up with a mostly pictorial proof for quadrilaterals, though instead of ensuring that the fourth vertex of the parallelogram lies strictly within the quadrilateral, it might only fall at an edge or vertex of the quadrilateral.

http://img226.imageshack.us/my.php?image=parall3qh.png

In the top diagram, I have placed vertices A, B, C of the quadrilateral and drawn with dotted lines the completed parallelogram for A, B, C, whose fourth vertex I denote p(A, B, C). The fourth point D of the quadrilateral cannot lie within the upper red region without allowing p(A, B, C) to lie within the quadrilateral. And D cannot lie within the lower red region, because then p(C, D, A) would lie within the quadrilateral. If D lies on the boundaries of these regions (exactly on a dotted line) then p(A, B, C) or p(C, D, A) will lie on the boundary of the quadrilateral. But assume that neither p(A, B, C) nor p(C, D, A) lies on either the boundary of the quadrilateral or within the quadrilateral, and continue.

WLOG, say that D lies on the left side of the red regions (if not, reflect the figure). Then as shown in the lower diagram, angle DAC equals angle u plus angle t, and angle ABC equals angle v plus angle u, with all of those angles greater than zero for reasons that can be seen by looking at parallel lines and the location of D. So angle u is strictly less than both angle DAC and angle ABC, and it follows that p(D, A, B) lies within the quadrilateral.

So in a quadrilateral, of all the fourth vertices of parallelograms completed from three adjacent vertices of the quadrilateral, there must exist at least one that either lies on the boundary of the quadrilateral or within it.

I don't know how or if I can use this to prove the case for polygons with more than 4 sides. Also, a lot of this proof sort of depends on geometric "eyeballing" which isn't as symbolic as I'd like, so maybe there is a clearer way to approach the problem. Any ideas?

I came up with a mostly pictorial proof for quadrilaterals, though instead of ensuring that the fourth vertex of the parallelogram lies strictly within the quadrilateral, it might only fall at an edge or vertex of the quadrilateral.

http://img226.imageshack.us/my.php?image=parall3qh.png

In the top diagram, I have placed vertices A, B, C of the quadrilateral and drawn with dotted lines the completed parallelogram for A, B, C, whose fourth vertex I denote p(A, B, C). The fourth point D of the quadrilateral cannot lie within the upper red region without allowing p(A, B, C) to lie within the quadrilateral. And D cannot lie within the lower red region, because then p(C, D, A) would lie within the quadrilateral. If D lies on the boundaries of these regions (exactly on a dotted line) then p(A, B, C) or p(C, D, A) will lie on the boundary of the quadrilateral. But assume that neither p(A, B, C) nor p(C, D, A) lies on either the boundary of the quadrilateral or within the quadrilateral, and continue.

WLOG, say that D lies on the left side of the red regions (if not, reflect the figure). Then as shown in the lower diagram, angle DAC equals angle u plus angle t, and angle ABC equals angle v plus angle u, with all of those angles greater than zero for reasons that can be seen by looking at parallel lines and the location of D. So angle u is strictly less than both angle DAC and angle ABC, and it follows that p(D, A, B) lies within the quadrilateral.

So in a quadrilateral, of all the fourth vertices of parallelograms completed from three adjacent vertices of the quadrilateral, there must exist at least one that either lies on the boundary of the quadrilateral or within it.

I don't know how or if I can use this to prove the case for polygons with more than 4 sides. Also, a lot of this proof sort of depends on geometric "eyeballing" which isn't as symbolic as I'd like, so maybe there is a clearer way to approach the problem. Any ideas?

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