Quadrectic

Probability Level 4

The grid above has width $6p-2$ and height $p+4$ .

If this grid contains $K$ rectangles of dimensions $3p\times2p$ with $p, K\in \mathbb Z$ , find the greatest value of $K$ .

Details and Assumptions:

Dimensions are expressed as width $\times$ height . Thus a $3p\times2p$ rectangle $($ that is, a rectangle of width $3p$ and height $2p)$ is considered to be different from a $2p\times3p$ $($ width $2p$ , height $3p)$ rectangle. In this problem, you only consider the $3p\times2p$ rectangles, not the $2p\times3p$ ones.

This is one part of Quadrilatorics .

The answer is 16.

1 solution

Kenneth Tan
Dec 26, 2016

From this note , we know that the number of $x\times y$ rectangles in an $a\times b$ grid given that $x\leqslant a$ and $y\leqslant b$ is $(a-x+1)(b-y+1)$

Here, we have $a=6p-2$ , $b=p+4$ , $x=3p$ , $y=2p$ , substitute them into the equation above we would get $\begin{aligned} K&=(6p-2-3p+1)(p+4-2p+1) \\&=(3p-1)(5-p) \\&=-3p^2+16p-5 \\&=-3(p-\frac83)^2+\frac{49}3 \end{aligned}$ As $p$ is an integer, and $\frac83$ rounded off to the nearest integer is 3, thus when $p=3$ , $K$ achieves its maximum value. $\therefore K\leqslant -3(3-\frac83)^2+\frac{49}3=16$

Quadrectic

This is one part of Quadrilatorics .

The answer is 16.

1 solution

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