Symmetric axis-parallel integral hexagon

Geometry Level 4

A polygon is called a symmetric axis-parallel integral hexagon if:

  • It has six sides (a hexagon). It is also not degenerate (no vertex has an angle of 0 ∘ 0^\circ or 18 0 ∘ 180^\circ ) and it doesn't self-intersect.
  • Its side lengths are (positive) integers.
  • Each of its sides is parallel to either the x-axis or the y-axis.
  • It has a line of symmetry: upon reflection on this line, the reflection coincides exactly with the original.

Consider a symmetric axis-parallel integral hexagon H H . Suppose its perimeter is P ( H ) P(H) units of length and its area is A ( H ) A(H) units of length squared (that is, P ( H ) , A ( H ) P(H), A(H) are its perimeter and area respectively without the dimension). Let f ( H ) = ∣ P ( H ) − A ( H ) ∣ f(H) = |P(H)-A(H)| . Determine the minimum value of f f .


The answer is 1.

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Ivan Koswara by Ivan Koswara , 25, Indonesia

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1 solution

Daniel Liu
Feb 28, 2015

Note that this hexagon must be square-shaped with one corner having a square cut-out. (easy to see just by testing some hexagons out)

Let a a be the side length of the longest side (side length of the square) and let b b be the side length of the shortest side (side length of the square cut-out).

We see that the perimeter is P ( H ) = 4 a P(H)=4a and the area is A ( H ) = a 2 − b 2 A(H)=a^2-b^2

To minimize f ( H ) = ∣ P ( H ) − A ( H ) ∣ f(H)=|P(H)-A(H)| , ideally we would want P ( H ) = A ( H ) P(H)=A(H) so let's see if that is possible.

4 a = a 2 − b 2 4 a = ( a + b ) ( a − b ) Let a = 2 a ′ , b = 2 b ′ 2 a ′ = ( a ′ + b ′ ) ( a ′ − b ′ ) Let a ′ = 2 a ′ ′ , b ′ = 2 b ′ ′ a ′ ′ = ( a ′ ′ + b ′ ′ ) ( a ′ ′ − b ′ ′ ) \begin{aligned} 4a&=a^2-b^2\\ 4a&=(a+b)(a-b)\\ \text{Let }&a=2a', b=2b'\\ 2a'&=(a'+b')(a'-b')\\ \text{Let }&a'=2a'',b'=2b''\\ a''&=(a''+b'')(a''-b'') \end{aligned} which is clearly impossible.

So, the next best bet is ∣ P ( H ) − A ( H ) ∣ = 1 |P(H)-A(H)|=1 . Fortunately, ( a , b ) = ( 4 , 1 ) (a,b)=(4,1) does the trick, since P ( H ) = 16 P(H)=16 and A ( H ) = 15 A(H)=15 .

Thus, our answer is 1 \boxed{1}

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Note that this hexagon must be square-shaped with one corner having a square cut-out. (easy to see just by testing some hexagons out)

This is the most difficult part to prove rigorously.

Ivan Koswara - 6 years, 3 months ago

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Just do a construction argument by drawing each segment one by one, and considering cases of going left/right and up/down. This gives that the hexagon must be a rectangle with a little rectangle cut out from the corner. Then, clearly the big rectangle must be a square since any symmetry lines parallel to the x and y axis will not work because of the little rectangle cut out. Also, one of the diagonal symmetry lines will not work because of the rectangle cut out, so the symmetry line must go through the rectangle cut-out. But then, this means the little rectangle cut-out must actually be square because rectangles do not have diagonal symmetry lines.

Daniel Liu - 6 years, 3 months ago

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