If $p$ and $q$ are positive integers, $\max (p, q)$ is the maximum of $p$ and $q$ and $\min (p, q)$ is the minimum of $p$ and $q$. For example, $\max (30,40)=40$ and $\min (30,40)=30$. Also, $\max (30,30)=30$ and $\min (30,30)=30$.

Determine the number of ordered pairs $(x, y)$ that satisfy the equation

$$
\max (60, \min (x, y))=\min (\max (60, x), y)
$$

where $x$ and $y$ are positive integers with $x \leq 100$ and $y \leq 100$.

## PART B

For each question in Part B, your solution must be well organized and contain words of explanation or justification when appropriate. Marks are awarded for completeness, clarity, and style of presentation. A correct solution, poorly presented, will not earn full marks.
Solution 1

We consider two cases: $y<60$ and $y \geq 60$.

We use the facts that $\min (a, b) \leq a, \min (a, b) \leq b, \max (a, b) \geq a$, and $\max (a, b) \geq b$. In other words, the minimum of two numbers is less than or equal to both numbers and the maximum of two number is greater than or equal to both numbers.

We also use the fact that if $a \leq b$, then $\min (a, b)=a$ and $\max (a, b)=b$. Note that the case of $a=b$ does not affect these equations.

Case 1: $y<60$

If $y<60$, then $\min (x, y) \leq y$ and $y<60$, so $\min (x, y)<60$.

Since $\min (x, y)<60$, then LS $=\max (60, \min (x, y))=60$.

Also, $\max (60, x) \geq 60$, so since $y<60$, then $\mathrm{RS}=\min (\max (60, x), y)=y$.

Therefore, in the case $y<60$, the equation is satisfied whenever $60=y$, which is never true.

Therefore, there are no ordered pairs $(x, y)$ with $y<60$ that satisfy the equation.

Case 2: $y \geq 60$

If $x<y$, then $\min (x, y)=x$ and so LS $=\max (60, x)$.

If $x<y$, then both 60 and $x$ are no larger than $y$, so the largest of 60 and $x$ is no larger than $y$. In other words, $\max (60, x) \leq y$ and so $\mathrm{RS}=\min (\max (60, x), y)=\max (60, x)$.

If $x<y$, the equation becomes $\max (60, x)=\max (60, x)$ and so is true for every ordered pair $(x, y)$.

If $x \geq y$, then $\min (x, y)=y$ and so $\mathrm{LS}=\max (60, y)=y$ since $y \geq 60$.

If $x \geq y$, then $x \geq 60$ so $\max (60, x)=x$ and so $\mathrm{RS}=\min (x, y)=y$.

If $x \geq y$, the equation becomes $y=y$ and so is true for every pair $(x, y)$.

Therefore, the given equation is satisfied by all pairs $(x, y)$ with $y \geq 60$.

Since $1 \leq x \leq 100$, there are 100 possible values for $x$.

Since $60 \leq y \leq 100$, there are 41 possible values for $y$.

Therefore, there are $100 \times 41=4100$ ordered pairs that satisfy the equation.

## Solution 2

In this solution, we use the facts that if $a \leq b$, then $\min (a, b)=a$ and $\max (a, b)=b$.

We also label as $(*)$ the original equation $\max (60, \min (x, y))=\min (\max (60, x), y)$.

We consider the possible arrangements of the integers $60, x$ and $y$.

There are six possible arrangements. We examine the various pieces of the left and right sides of the given equation for each of these orders, moving from the innermost function out on each side:

$$
\begin{array}{|c||c|c||c|c|}
\text { Case } & \min (x, y) & \mathrm{LS}=\max (60, \min (x, y)) & \max (60, x) & \mathrm{RS}=\min (\max (60, x), y) \\
\hline 60 \leq x \leq y & =x & =\max (60, x)=x & =x & =\min (x, y)=x \\
60 \leq y \leq x & =y & =\max (60, y)=y & =x & =\min (x, y)=y \\
x \leq 60 \leq y & =x & =\max (60, x)=60 & =60 & =\min (60, y)=60 \\
x \leq y \leq 60 & =x & =\max (60, x)=60 & =60 & =\min (60, y)=y \\
y \leq 60 \leq x & =y & =\max (60, y)=60 & =x & =\min (x, y)=y \\
y \leq x \leq 60 & =y & =\max (60, y)=60 & =60 & =\min (60, y)=y
\end{array}
$$

If $60 \leq x \leq y$, then $(*)$ is equivalent to $x=x$, so all $(x, y)$ with $60 \leq x \leq y$ satisfy $(*)$. If $60 \leq y \leq x$, then $(*)$ is equivalent to $y=y$, so all $(x, y)$ with $60 \leq y \leq x$ satisfy $(*)$. If $x \leq 60 \leq y$, then $(*)$ is equivalent to $60=60$, so all $(x, y)$ with $x \leq 60 \leq y$ satisfy $(*)$. If $x \leq y \leq 60$, then $(*)$ is equivalent to $60=y$, so only $(x, y)$ with $y=60$ and $x \leq 60$ satisfy
$(*)$. These $(x, y)$ are already accounted for in the third case.

If $y \leq 60 \leq x$, then $(*)$ is equivalent to $60=y$, so only $(x, y)$ with $y=60$ and $x \geq 60$ satisfy $(*)$. These $(x, y)$ are included in the second case.

If $y \leq x \leq 60$, then $(*)$ is equivalent to $60=y$, so only $y=60$ and $60 \leq x \leq 60$, which implies $x=60$. In this case, only the pair $(60,60)$ satisfies $(*)$ and this pair is already included in the first case.

Therefore, we need to count the pairs of positive integers $(x, y)$ with $x \leq 100$ and $y \leq 100$ which satisfy one of $60 \leq x \leq y$ or $60 \leq y \leq x$ or $x \leq 60 \leq y$.

From these three ranges for $x$ and $y$ we see that $y \geq 60$ in all cases and $x$ can either be less than or equal to 60 or can be greater than or equal to 60 and either larger or smaller than $y$.

In other words, these three ranges are equivalent to the condition that $y \geq 60$.

Therefore, $(*)$ is satisfied by all pairs $(x, y)$ with $y \geq 60$.

Since $1 \leq x \leq 100$, there are 100 possible values for $x$.

Since $60 \leq y \leq 100$, there are 41 possible values for $y$.

Therefore, there are $100 \times 41=4100$ ordered pairs that satisfy the equation.

ANSWER: 4100

## Part B