Wednesday, April 15, 2015
(in North America and South America)
Thursday, April 16, 2015
(outside of North American and South America)
©2015 University of Waterloo
Evaluating,
Alternatively, we could factor
Since
Therefore,
Alternatively, we could note that since
Thus,
Since
Therefore,
Alternatively, we could note that since
Since
Let the radius of the larger circle be
Since the radius of the smaller circle is 1, then its area is
Since the area between the circles is equal to the area of the smaller circle, then the area of the larger circle is
Thus,
Since 30 students had an average mark of 80, then the sum of the marks of these 30 students was
After 2 students dropped the class, there were 28 students left. Their average mark was 82.
Thus, the sum of the marks of the remaining 28 students was
Therefore, the sum of the marks of the 2 students who dropped the class was
Solution 1
Join
Since
Since
Since
Since
Each of these angles equals
Since
Therefore,
Solution 2
Join
Since
Since
By the cosine law in
Solution 3
Join
Since
Since
Since
Thus,
Since
Since
Each of these angles equals
But
Points
The corresponding height of
Since the area of
Since
Therefore,
Since
The sum of these two values is
(We could also have noted that, since the two values of
To find the
Therefore, the
To find the points of intersection of the graphs with equations
We want to find all values of
Solving, we obtain
Note that
Therefore, for there to be exactly two points of intersection between the two graphs, the quadratic equation
Setting the discriminant equal to 0 (to obtain a single root), we obtain
If
Therefore, the values of
(We can check that
Suppose that
Since
Since
By the Pythagorean Theorem in
Since
Since each list contains 6 consecutive positive integers and the smallest integers in the lists are
Note that
We first determine the pairs
The first bullet tells us that
Since
If 1 is in one of the lists, then either
If 49 is in the second list, then one of
Therefore, for 1 and 49 to appear in the two lists, then
Therefore, for 7 to appear in both lists, then, knowing that
Given that the possible values of
There is no multiple of 32 or 64 in these lists.
Thus, for a pair of integers from these lists to have a product that is a multiple of 64, one is a multiple of 4 and the other is a multiple of 16, or both are multiples of 8.
If
If
If
If
Therefore, after considering the first two bullets, the possible pairs
The third bullet tells us that there is at least one number in the third list that is larger than 75.
Given the possible pairs
The corresponding largest integers in the third list are the products of the largest integers in the two lists; these products are
Therefore, the remaining pairs
Having considered the three conditions, the possible pairs
We are told that when
This means that if
Starting with the given 2 and 3 and proceeding clockwise, we obtain
Since there are 36 terms in total, then the 6 terms repeat exactly
Therefore, the sum of the 36 numbers is
We consider two cases:
Case 1:
We take the given inequality
Thus,
From the first, we obtain
Since
From the second, we obtain
Since
Since
Case 2:
We take the given inequality
Thus,
From the first, we obtain
Since
From the second, we obtain
Since
Since
In summary, the values of
Join
Since
Since
Now
This means that
Now the diagonals of a rectangle partition it into four triangles of equal area. (Diagonal
Since
Since
Suppose that the arithmetic sequence
Then, for each positive integer
Since
Since
Substituting, we obtain
Thus,
For
Since
Checking, we note that
Therefore, the only possible value of
First, we note that since
Next, we note that the given parabola passes through the point
Therefore, for every positive integer
If
Now, the given circle with equation
This means that the
To find the other points of intersection, we re-write
(We note that since the two graphs intersect at
Therefore, for
This gives
Since
If
This gives the two points
Consider the three points
Now
The vertical distance from
Therefore, the area of
We now repeat these calculations for each of the other values of
|
Base | Height | Area of triangle | ||
---|---|---|---|---|---|
When
Finally, the values of
Suppose that
Suppose that the centre of the smaller circle is
Suppose that the smaller circle touches
Join
Since
Join
Now
Draw a perpendicular from
Consider
Each triangle is right-angled (at
Also,
Therefore,
Thus,
Now
Also,
Lastly,
Therefore,
Using logarithm rules
This transforms the system of equations to the equivalent system
In a similar way, we re-write the second and third of these equations to obtain the equivalent system
This transforms the original system of equations into the equivalent system
Suppose that
Then the last version of the system is
Since
Therefore,
Therefore, the solutions to the last version of the system are
Converting back to the original variables, we see that the solutions to the original system when
We consider the various possibilities for the product,
Case 1:
As in (a), we multiply the three equations together to obtain
Since the left side is at least 0 and the right side is negative, then there are no solutions to the system of equations in this case.
Case 2:
As in (a), we multiply the three equations together to obtain
Since
Since
Also, since
As we did in (a), we can solve to obtain
Case 3:
Suppose that exactly one of
Without loss of generality, suppose that
Since
In this case, either
Therefore, it cannot be the case that exactly one of
Suppose next that exactly two of
Without loss of generality, suppose that
Since
Since
In this case, any triple
Thus, when
Each of these solutions corresponds to a solution to the original system of equations in
Similarly, if
Finally, we must consider the case of
Here, we must solve the system of equations
Therefore, when
Therefore, the system of equations has an infinite number of solutions
The subsets of
Consider the Furoni family
Each of the following subsets of
Each of the following subsets of
Each of the following subsets of
Since a Furoni family of
This leaves the following subsets of
If
If any of
Thus,
Therefore, the two Furoni families of
Solution 1
Suppose that
Consider each element
Each
We use
Since there are
Since there are
Each possible
Each of these
We repeat this process for each of the elements
Since, for each
Is it possible that two elements
Suppose that two permutations
Suppose that
Either
Without loss of generality, suppose that
Then the length of
But
But the entries in
Therefore, each of the permutations generated by each of the subsets of
Therefore,
Solution 2
Suppose that
Consider
The probability that the intersection of
Note that since each element of
If
There are
subsets of that contain exactly integer. The probability that any particular one of these subsets is
equals . Since
of these subsets are in , then the probability that one of these subsets is equals . (Note that we use the convention that if
, then .)
The probability that any of the elements of
Therefore,
Set
Then
From (b),
Is it possible to find a Furoni family of size
Yes – the
We now prove the algebraic result above.
First, we note that
Therefore, if
Suppose first that
We prove that
Since
, then If for some non-negative integer , then Since the right side is the product of non-negative fractions, each of which is smaller than 1, then their product is smaller than 1. Thus,
if .
Suppose next that
We prove that
Since
, then If for some non-negative integer , then Since the right side is the product of non-negative fractions, each of which is smaller than 1, then their product is smaller than 1. Thus,
if .
This completes our proof.