Triangular Pick's Theorem
Pick’s theorem states that, given a polygon with vertices on a unit square lattice, the relationship between the area A S of the polygon, the number of boundary points B of the polygon, and the number of interior points I of the polygon is A S = I + 2 1 B − 1 .
There exists a similar relationship for a polygon with vertices on an equilateral triangle lattice in which the area of the smallest possible triangle is 1.
The area of the shaded triangle is 1.
If A T is the area of the polygon, B is the number of boundary points of the polygon, and I is the number of interior points of the polygon, then the relationship is A T = a I + b B + c .
What is ∣ a ∣ + ∣ b ∣ + ∣ c ∣ ?
The answer is 5.000.
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5 solutions
this solution makes all others look like brute force XD
They gave that the form of the equation would be linear, so I just chose 3 triangles that I knew what their area was. First was just the regular unit triangle, then a triangle with doubled side length, and finally a triangle with 3 side lengths. Solving that system gave a =2, b=1, and c= -2
0*a+ 3b + c = 1
0*a+ 6b + c =4
b=1 c=-2
a + 9* b + c = 9
a+9-2= 9
a = 2
2+2+1=5
*5 *
Three simple cases give three independent linear equations for a , b , c .
First case: the small triangle. Here I = 0 , B = 3 and A = 1 .
Second case: parallelogram made of two triangles. Now I = 0 , B = 4 and A = 2 .
Third case: regular hexagon made of six triangles. We have I = 1 , B = 6 and A = 6 .
Formally, we solve the equation ⎝ ⎛ 0 0 1 3 4 6 1 1 1 ⎠ ⎞ ⎝ ⎛ a b c ⎠ ⎞ = ⎝ ⎛ 1 2 6 ⎠ ⎞ , ⎝ ⎛ a b c ⎠ ⎞ = ⎝ ⎛ 2 1 − 2 ⎠ ⎞ . Or, note that between the first and second case, adding one to B also adds one to A , so that b = 1 ; from either case follows c = − 2 and the third case then gives a = 2 .
The solution is ∣ a ∣ + ∣ b ∣ + ∣ c ∣ = 2 + 1 + 2 = 5 .
Nicely done!
Note: This solution assumes that the linear relationship A T = a I + b B + c holds. David's solution explains why this assumption is valid.
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That is true, although he has to appeal to the usual Pick's Theorem to do so.
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Indeed. One way of proving Pick's theorem is by showing
- It is true for rectangles whose sides are parallel to the axis (obvious)
- It is true for right triangles whose base are parallel to the axis. (double it to form the rectangle)
- It is true for an arbitrary triangle (add triangles to form a rectangle)
- It is true for any polygon (by triangulating)
For a proof of the Pick's theorem in an equilateral triangle lattice, we can also proceed in a similar manner, using corresponding triangles/parallelograms with sides parallel to the axis.
Lmao I guessed 5 and I actually got it right
It's super late right now and I'm too tired to do this logically. I guessed a random number and got it right, so that's cool.
That's illegal
A unit square lattice grid can be stretched and skewed to a unit equilateral triangle lattice grid, where each unit square would transform to a parallelogram composed of 2 unit equilateral triangles and have an area of 2 .
Since the area is doubled for the same number of boundary and interior points, the formula for Pick’s Theorem is also doubled and becomes A T = 2 I + B − 2 , making a = 2 , b = 1 , and c = − 2 , and the sum of ∣ a ∣ , ∣ b ∣ , and ∣ c ∣ is ∣ a ∣ + ∣ b ∣ + ∣ c ∣ = 2 + 1 + 2 = 5 .