April 2021
(in North America and South America)
April 2021
(outside of North American and South America)
©2021 University of Waterloo
Each business card has dimensions
Each single page with dimensions
The entire page is used with no waste and so the number of business cards that can be printed without overlap is
We begin by first considering the portrait page layout.
The width of each card is 5 cm and the width of the page is 19 cm. Since 3 adjacent cards give a combined width of 15 cm (less than 19 cm) and 4 adjacent cards give a combined width of 20 cm (greater than 19 cm), then the greatest number of adjacent cards that can be printed horizontally across the page in this layout is 3.
The height of each card is 9 cm and the height of the page is 29 cm. Since 3 adjacent cards give a combined height of 27 cm (less than 29 cm) and 4 adjacent cards give a combined height of 36 cm (greater than 29 cm), then the greatest number of adjacent cards that can be printed vertically on the page in this layout is 3.
Thus the portrait page layout allows a maximum of
Next, we consider the landscape page layout.
The width of each card is 9 cm and the width of the page is 19 cm. Since 2 adjacent cards give a combined width of 18 cm (less than 19 cm) and 3 adjacent cards give a combined width of 27 cm (greater than 19 cm), then the greatest number of adjacent cards that can be printed horizontally across the page in this layout is 2.
The height of each card is 5 cm and the height of the page is 29 cm. Since 5 adjacent cards give a combined height of 25 cm (less than 29 cm) and 6 adjacent cards give a combined height of 30 cm (greater than 29 cm), then the greatest number of adjacent cards that can be printed vertically on the page in this layout is 5.
Thus the landscape page layout allows a maximum of
Throughout the solution to this problem, all coordinates represent lengths in metres.
Let the point
On Monday, Franklin walks from school straight home (a distance of 260 m) at a constant speed of 80 m/min. Since time equals distance divided by speed, it takes Franklin
Points
Thus, it takes Franklin
Since Giizhig leaves the school at the same time as Franklin, and they meet halfway between the two homes, then it also takes Giizhig
As in part (a),
Therefore, the distance along the straight path from the school to Giizhig’s home to the halfway point between the two homes is
Since Giizhig’s speed,
When
The two Reverse Operations change the order of the first four numbers in the list,
After performing
Performing
Thus, the two Reverse Operations are
(It is worth noting that there are no other ways to do this using exactly two Reverse Operations.)
Each Reverse Operation, with the exception of
That is, the number 3 moves to the last position only when
Is it possible to perform a Reverse Operation on the original list that moves 3 to the first position of the list?
If
If
Thus, performing
Since in (i) we found a way to achieve the desired result using 2 Reverse Operations, we need only to explain why performing 1 Reverse Operation is not possible.
As described in (i), the number 3 moves to the last position only when
Together, (i) and (ii) show that the minimum number of Reverse Operations that need to be performed on the list
Performing the Reverse Operations
Thus, 3 Reverse Operations can achieve a list of the desired form.
Can the desired result be achieved by performing fewer than 3 Reverse Operations?
As similarly demonstrated in (c), performing the Reverse Operations
This is the minimum number of Reverse Operations needed to move 4 to the last position in the list and further, these are the only 2 Reverse Operations that move 4 to the last position in the list. (Can you see why?)
However, performing
Since it is clearly not possible to achieve the desired result in 1 Reverse Operation, then the minimum number of Reverse Operations is 3.
The 3 Reverse Operations
The
We begin by labelling the vertices as shown.
Beginning at
At this point, each edge from
Thus, it must be that if an
From
Proceeding up to
Proceeding from
Finally, proceeding down from
Therefore, there is no
We define the “middle column” by labelling the points
Each
Each
Each
Each
There are exactly 3
There are exactly 3
There are exactly 4
There are exactly 3
In each of these cases, the left endpoint (
This leaves exactly 18 squares on each side of the middle column through which the path travels.
In the first 18 squares, an
These choices of top or bottom edges uniquely determine the path since it is forced to proceed along vertical edges exactly when the path through two adjacent squares has different horizontal segments (one follows a top edge and the other a bottom).
The diagram below shows where these vertical edges must be added to the previous example. There is no choice in the selection of vertical edges once the horizontal edges have been chosen.
It is worth noting that a path cannot proceed along both the top and bottom edges of a given square since such a path would visit a vertex more than once.
Two choices (top or bottom edge) in each of the first 18 squares gives
Let
The
How many of these 2
Each of the paths arriving at
Thus, there are exactly
Similarly, there are
Similar arguments show that there are 2
From our earlier analysis, there are exactly 3
There are
Thus, there is a total of