2013 Indonesia MO Problems/Problem 4

Problem

In a $4 \times 6$ grid, all edges and diagonals are drawn (see attachment). Determine the number of parallelograms in the grid that uses only the line segments drawn and none of its four angles are right.

$[asy] draw((0,0)--(6,0)--(6,4)--(0,4)--(0,0)); for (int i=1; i<6; ++i) { draw((i,0)--(i,4)); } for (int i=1; i<4; ++i) { draw((0,i)--(6,i)); } draw((0,1)--(1,0)); draw((0,2)--(2,0)); draw((0,3)--(3,0)); draw((0,4)--(4,0)); draw((1,4)--(5,0)); draw((2,4)--(6,0)); draw((3,4)--(6,1)); draw((4,4)--(6,2)); draw((5,4)--(6,3)); draw((0,3)--(1,4)); draw((0,2)--(2,4)); draw((0,1)--(3,4)); draw((0,0)--(4,4)); draw((1,0)--(5,4)); draw((2,0)--(6,4)); draw((3,0)--(6,3)); draw((4,0)--(6,2)); draw((5,0)--(6,1)); [/asy]$

Solution

In the grid, you can make a rectangle and construct a parallelogram, notice how you can always make a parallelogram so long as the rectangle that was chosen was not a square, the ammount of ways to pick a rectangle is $\binom{7}{2}\binom{5}{2}=210$ since there are 7 horizontal lines and you choose 1, and there are 5 vertical lines and you choose 2, and the total ammount of squares are $6\cdot 4+5\cdot 3+4\cdot 2+3\cdot 1=50$ , so the total ammount of rectangles you can make are $210-50=160$ , also for each rectangle there are 2 parallelograms you can make, one of them is fliped, so myltiply the total ammount of rectangles by 2 which is $160\cdot 2=\boxed{320}$

2013 Indonesia MO (Problems)
Preceded by First Problem	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 2
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2013 Indonesia MO Problems/Problem 4

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