The Scissors Are Back (Count 'em All 23!)

Let's first pretend that the grid is 'perfect': it has no holes, and no cut corners. Let's also patch the cut corners with imaginary golden squares.

For convenience sake, I'll label the two golden squares $A$ and $B$ , and the hole $C$ .

From this note , we know that the number of quadrilaterals in an $8\times8$ perfect grid is ${9\choose2}\times{9\choose2}=\frac{8\times8\times9\times9}{4}=1296$

Now we need to remove all 'invalid' quadrilaterals, invalid quadrilaterals consists of those which contain the hole, and those that turn into a pentagon or triangle when the golden corner squares are cut.

Suppose $n(\square)$ is the number of invalid quadrilaterals that contain at least the specified squares (for example, $n(A, C)$ is the number of invalid quadrilaterals that contain the corner $A$ and the hole $C$ , which might also contain the corner $B$ ), then by PIE , the total number of invalid quadrilaterals is $n(A)+n(B)+n(C)-n(A, B)-n(A, C)-n(B, C)+n(A, B, C)$

Let's first look at $n(A)$ . Note that we can construct quadrilaterals by choosing 2 squares in the grid as its opposite corner squares, with that in mind, if a quadrilateral's bottom-left corner square lies on $A$ , then for this quadrilateral to be invalid, its opposite corner square must either not lie on the same row or column as $A$ (since it will become a pentagon when $A$ is cut) or lie exactly on $A$ (since it would be just the corner square $A$ itself, it will be cut into a triangle). Hence, the number of possible locations for its top-right corner square for the quadrilateral to be invalid is $n(A)=(8-1)\times(8-1)+1=50$

With similar reasoning, we can also conclude that $n(B)=n(A)=50$ .

Now let's look at $n(C)$ . Another way to construct quadrilaterals is by choosing two vertical lines and two horizontal lines from the grid as its edges. Thus for a quadrilateral to contain the hole $C$ , its left edge must be on the left of $C$ , its right edge must be on the right of $C$ , its top edge must be above $C$ , and its bottom edge must be below $C$ . There are $4$ vertical lines on the left of $C$ , $5$ vertical lines on the right of $C$ , $5$ horizontal lines above $C$ , and $4$ horizontal lines below $C$ . Hence, $n(C)=4\times5\times5\times4=400$

Consider $n(A, C)$ . To construct quadrilaterals of this type, fix its bottom-left corner square on $A$ , and choose a location for its top-right corner square such that it is not on the left of $C$ and is not below $C$ . $\therefore n(A, C)=5\times5=25$

With similar reasoning, we can also conclude that $n(B, C)=4\times4=16$ .

Let's now consider $n(A, B)$ and $n(A, B, C)$ . It's easy to see that $n(A, B)=n(A, B, C)=1$ .

$\begin{aligned}\therefore n(A)+n(B)+n(C)-n(A, B)-n(A, C)-n(B, C)+n(A, B, C)&=50+50+400-25-16-1+1 \\ &=459 \end{aligned}$

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The number of invalid quadrilaterals is $459$ , therefore the number of quadrilaterals in the original grid is $1296-459=837$ .