Wednesday, April 11, 2018
(in North America and South America)
Thursday, April 12, 2018
(outside of North American and South America)
©2018 University of Waterloo
When
We multiply the equation
Solving, we get
Solution 1:
Since the cost of one chocolate bar is $1.00 more than that of a pack of gum, then if we replace a pack of gum with a chocolate bar, then the price increases by $1.00.
Starting with one chocolate bar and two packs of gum, we replace the two packs of gum with two chocolate bars.
This increases the price by $2.00 from $4.15 to $6.15.
In other words, three chocolate bars cost $6.15, and so one chocolate bar costs
Solution 2:
Let the cost of one chocolate bar be $
Let the cost of one pack of gum be $
Since the cost of one chocolate bar and two packs of gum is $4.15, then
Since one chocolate bar costs $1.00 more than one pack of gum, then
Since
Since
Solving, we obtain
In other words, the cost of one chocolate bar is $2.05.
Suppose that the five-digit integer has digits
The digits
Since
Since
Since
So far, the integer is
Since the two-digit integer
This means that the the five-digit integer is
By the Pythagorean Theorem in
By the Pythagorean Theorem in
Therefore,
Solution 1:
The area of the shaded region equals the area of square
Since square
Also,
Since the equation of the line is
Since the slope of the line is
Since
Thus, the area of
Finally, the area of shaded region must be
Solution 2:
Since square
Since the slope of the line is
Since
Since
The shaded region is a trapezoid with parallel sides
Therefore, the area of the shaded region is
Calculating,
Since
Therefore,
Since
We make a table listing the possible values of
Since the line with equation
Since the parabola with equation
To find the coordinates of
Since
Since
In summary, (i)
Since
For an integer
Since
This leaves the list
There are
We start with
Since the hexagon has perimeter 12 and has 6 sides, then each side has length 2.
Since equilateral
Consider equilateral triangles with side length
Six of these triangles can be combined to form a regular hexagon with side length 2 and four of these can be combined to form an equilateral triangle with side length 4.
Note that the six equilateral triangles around the centre of the hexagon give a total central angle of
Also, the length of each side of the hexagon is
Since the triangle is made from four identical smaller triangles and the hexagon is made from six of these smaller triangles, the ratio of the area of the triangle to the hexagon is
Since sector
For the line
We determine the length of
Since sector
Drop a perpendicular from
The area of
Since
For the area of
(Alternatively, we could have used the fact that the area of
Let
The given inequalities become
When
This means that we can can multiply both sides by
Therefore,
Note that
Therefore, the original inequality is true exactly when
Note that
When
From
From
From
Therefore, the original inequality is true exactly when
The integers
Suppose that Karuna and Jorge meet for the first time after
When they meet for the first time, Karuna has run partway from
At this instant, the sum of the distances that they have run equals the total distance from
Since Karuna runs at 6 m/s for these
Since Jorge runs at 5 m/s for these
Therefore,
When they meet for the second time, Karuna has run from
Since they each finish running after 99 seconds, then each has
At this instant, the sum of the distances that they have left to run equals the total distance from
Since Karuna runs at 6 m/s for these
Since Jorge runs at 7.5 m/s for these
Therefore,
Alternatively, to calculate the value of
This means that they have each run the full distance from
Karuna has run at 6 m/s for
Jorge ran the first 297 m at 5 m/s, which took
Therefore,
Solution 1:
Among a group of
There are
For each of these
This gives
Each pair is counted twice (given two people A and B, we have counted both the pair AB and the pair BA), so the total number of pairs is
We label the four canoes W, X, Y, and Z.
First, we determine the total number of ways to put the 8 people in the 4 canoes.
We choose 2 people to put in W. There are
Next, we choose 2 people to put in X. There are
Next, we choose 2 people to put in Y. There are
There is now
Therefore, there are
Now, we determine the number of ways in which no two of Barry, Carrie and Mary will be in the same canoe.
There are 4 possible canoes in which Barry can go.
There are then 3 possible canoes in which Carrie can go, because she cannot go in the same canoe as Barry.
There are then 2 possible canoes in which Mary can go, because she cannot go in the same canoe as Barry or Carrie.
This leaves 5 people left to put in the canoes.
There are 5 choices of the person that can go with Barry, and then 4 choices of the person that can go with Carrie, and then 3 choices of the person that can go with Mary.
The remaining 2 people are put in the remaining empty canoe.
This means that there are
Therefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is
Solution 2:
Let
The answer to the original problem will be
Let
By symmetry, the probability that Barry and Mary are in the same canoe also equals
This means that
So we calculate
To do this, we put Barry in a canoe. Since there are 7 possible people who can go in the canoe with him, then the probability that Carrie is in the canoe with him equals
This means that the probability that Barry and Carrie are in the same canoe is
Therefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is
Solution 1:
Consider square
Since the slope of
Since
Now
Therefore,
Therefore,
We note that since
To calculate the slopes of
Using the facts that
Solution 2:
Consider a square
Translate this square so that
Let the coordinates of
Since the slope of
Thus, the coordinates of
Let
Then
We find the slopes of
Consider the segment
Since the diagonals of a square are perpendicular, then
Since the slope of
Since the diagonals of a square are equal in length and
Since
This is because these triangles are congruent (each is right-angled, their hypotenuses are of equal length, and their remaining angles are equal) and their hypotenuses are perpendicular.
In this diagram, we have assumed that
To get from
Thus, to get from
Therefore, the coordinates of
Thus, the slope of
Since
The sum of the slopes of
Since the base of a logarithm must be positive and cannot equal
This tells us that
We note further that
Using logarithm rules, the following equations are equivalent:
Rectangle
Let
Since
Since
Since
Note that
Let
Since
Since
Since
Since
Continuing in the same way, we can show that
Since
Similarly,
In particular, this means that
Since
Thus,
Also, by the Pythagorean Theorem in
By the Pythagorean Theorem in
Since
Therefore, substituting into
Finally,
Here is a tiling of a
There are many other tilings.
First, we note that it is possible to tile each of a
Next, we note that it is not possible to tile a
Finally, we show that it is possible to tile a
To do this, we show that such a
Case 1:
Suppose that
We build a
Each
Therefore, each such rectangle can be tiled.
Case 2:
Suppose that
We build a
The
Each
Therefore, each such rectangle can be tiled.
Thus, a
Suppose that
Since the area of each triomino is 3, then the area of any rectangle that can be tiled must be a multiple of 3 since it is completely covered by triominos with area 3.
Since the area of an
Since a rectangle that is
We show that if
Case 1:
Here,
Since
By stacking
Since each
Case 2:
Suppose that
To tile a
Therefore, every rectangle in this case can be tiled. (Note that in this case the fact that
Case 3:
Since
(There are also other ways to tile this rectangle.)
Since
Since
Thus, the
We break this rectangle into three rectangles – one that is
(If
The
If
If
Therefore, the
Through these three cases, we have shown that any
Since the roles of
We draw part of the array using the information that
Thus,
Since
Since
So far, this gives
We draw part of the array:
Proof of statement (P)
Suppose that all of the entries in the array are positive integers.
Assume that not all of the entries in the array are equal.
Since all of the entries are positive integers, there must be a minimum entry. Let
Choose an entry in the array equal to
If not all of the entries
If all of the entries
In other words, since not all of the entries in the array are equal, then there exists an integer
But
Since not all of
This means that
Therefore our assumption that not all of the entries are equal must be false, which means that all of the entries are equal, which proves statement (P).
Proof of statement (Q)
Suppose that all of the entries are positive real numbers.
Assume that not all of the entries in the array are equal.
As in (b), define
Also, define
Step 1: Prove that the numbers
From (b),
Re-arranging, we see
Since this is true for all integers
Step 2: Prove that
Suppose that
Since the terms
If the sequence of terms
Note that
Since
Thus, the sequence cannot be increasing.
If the sequence of terms
Note that
Since
Thus, the sequence cannot be decreasing.
Therefore, since all of the entries are positive and the sequence
Step 3: Determine range of possible values for
We note that
Since
Similarly,
Therefore,
Since
In other words,
Step 4:
Using a similar approach to our solution to (b),
Step 5: Final contradiction
We want to show that
This will show that
Since
Suppose that
We may assume that
Thus,
We may assume that
Suppose that
Then
Similarly,
Continuing in this way, we see that
The other possibility is that
Here, we re-arrange
Thus,
Extending this using a similar method, we see that
Therefore, a contradiction is obtained in all cases and so it cannot be the case that
Since