Solutions
(Grade 11)
Wednesday, April 10, 2019 (in North America and South America)
Thursday, April 11, 2019 (outside of North American and South America)
©2019 University of Waterloo
The radius of each hole is 2 cm, and so the diameter of each hole is 4 cm.
Since there are 4 holes, then their diameters combine for a total distance of
Let the radius of each hole be
Since there are 4 holes, then their diameters combine for a total distance of
The distance along the midline between adjacent holes is equal to the radius,
Solution 1
As in part (b), if the diameter of each hole is
If the distance along the midline between adjacent holes is 5 cm, then the 5 equal spaces along the midline combine for a total distance of
The total length of the midline is 91 cm, and so we get
However, the vertical distance from the midline to each edge of the metal is 8 cm, and since the holes must be circles, the radius of each hole cannot be 8.25 cm, and so the distance between adjacent holes cannot be 5 cm.
Solution 2
The minimum possible distance between adjacent holes is determined by maximizing the radius of each of the circles.
Since the holes must be circles, and the vertical distance from the midline to each edge of the metal is 8 cm, the maximum radius is 8 cm.
Since there are 4 holes, then their diameters combine for a maximum total distance of
The total length of the midline is 91 cm, and so the 5 equal spaces combine for a minimum total length of
To add a bump, the line segment of length 21 is first broken into three segments, each having length
The middle segment of these three segments is removed, and two new segments each having length 7 are added.
Thus, after a bump is added to a segment of length 21, the new path will have length
A path with exactly one bump has four line segments of equal length.
If such a path has length 240, then each of the four line segments has length
To add the first bump, the line segment of length 36 is broken into three segments, each having length
The middle segment of these three segments is removed, and two new segments each having length 12 are added.
Thus, after the first bump is added to a segment of length 36, the new path will have length
Next, a bump is added to each of the four segments having length 12.
Consider adding a bump to one of these four segments.
The segment of length 12 is broken into three segments, each having length
The middle segment is removed, two new segments each having length 4 are added, and so the new length is
To add a bump, the line segment of length
The middle segment of these three segments is removed, and two new segments each having length
Thus, after a bump is added to a segment of length
To create Path 2, a bump is added to each of the four segments having length
Consider adding a bump to one of these four segments.
The segment of length
The middle segment is removed, two new segments each having length
There are four such segments to which this process happens, and so the total length of Path 2 is
To summarize, when a bump is added to a line segment of length
When bumps are then added to Path 1, the length of the resulting Path 2 is
This process will continue with each new path having a total length that is
That is, Path 3 will have length
If the length of Path 5 is an integer, then
The smallest integer
The arithmetic mean of 36 and 64 is
If the arithmetic mean of two positive real numbers
If the geometric mean of two positive real numbers
Multiplying the first equation by 2 gives
Substituting
Factoring the left side of this equation, we get
When
We are required to solve the equation
Simplifying first, we get
This occurs exactly when
Since
1 | 2 | 8 |
2 | 8 | 18 |
3 | 18 | 32 |
4 | 32 | 50 |
The pairs
We begin by multiplying the first equation by
By subtracting the second equation from the first, we get
Similar to part (a), we will first solve for
Multiplying the first equation by
We can subtract
From the first equation, we get
We need to find values of
Since
For this to happen, we need
However, for
Since any divisor of
The divisors of
Therefore, the possible values of
In order to simplify things, we will first show that, regardless of the values of
To see this, we begin by multiplying the first equation by
We now add the two equations.
When doing this, the
Observe that
Suppose
We are interested in finding positive integer values of
The first equation simplifies to
Dividing equation (1) by
We want
Assuming
Notice that
To summarize, we want to understand integer pairs
By rearranging and factoring, this is the same as finding integer pairs
Therefore, we need to find integer pairs
Using the same reasoning as before, since
Notice that
Putting all of this together, given that
Phrasing the question in these terms, we are looking for a positive integer
Beginning with
When
The factors of
When
When
When
When
When
The factors of
There are exactly
Therefore, if