, 2018

Solution 1

Among a group of $n$ people, there are $\dfrac{n(n-1)}{2}$ ways of choosing a pair of these people:

There are $n$ people that can be chosen first.

For each of these $n$ people, there are $n-1$ people that can be chosen second.

This gives $n(n-1)$ orderings of two people.

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 $\dfrac{n(n-1)}{2}$.

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 $\dfrac{8\cdot 7}{2}$ pairs. This leaves 6 people for the remaining 3 canoes.

Next, we choose 2 people to put in X. There are $\dfrac{6\cdot 5}{2}$ pairs. This leaves 4 people for the remaining 2 canoes.

Next, we choose 2 people to put in Y. There are $\dfrac{4\cdot 3}{2}$ pairs. This leaves 2 people for the remaining canoe.

There is now $1$ way to put the remaining people in Z.

Therefore, there are $$\dfrac{8 \cdot 7}{2} \cdot \dfrac{6 \cdot 5}{2} \cdot \dfrac{4 \cdot 3}{2} = \dfrac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{2^3} = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3$$ ways to put the 8 people in the 4 canoes.

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 $4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 \cdot 3$ ways in which the 8 people can be put in 4 canoes so that no two of Barry, Carrie and Mary are in the same canoe.

Therefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $\dfrac{4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 \cdot 3}{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3} = \dfrac{4 \cdot 3 \cdot 2}{7 \cdot 6} = \dfrac{24}{42} = \dfrac{4}{7}$.

Solution 2

Let $p$ be the probability that two of Barry, Carrie and Mary are in the same canoe.

The answer to the original problem will be $1-p$.

Let $q$ be the probability that Barry and Carrie are in the same canoe.

By symmetry, the probability that Barry and Mary are in the same canoe also equals $q$ as does the probability that Carrie and Mary are in the same canoe.

This means that $p=3q$.

So we calculate $q$.

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 $\frac{1}{7}$. The other 6 people can be put in the canoes in any way.

This means that the probability that Barry and Carrie are in the same canoe is $q = \frac{1}{7}$.

Therefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is $1 - 3\cdot\frac{1}{7}$ or $\frac{4}{7}$.
Solution 1

Consider square $WXYZ$ in the first quadrant. Suppose that $WY$ makes an angle of $\theta$ with the horizontal.

[[IMAGE0]]

Since the slope of $WY$ is 2, then $\tan \theta = 2$, since the tangent of an angle equals the slope of a line that makes this angle with the horizontal.

Since $\tan \theta = 2 > 1 = \tan 45^\circ$, then $\theta > 45^\circ$.

Now $WY$ bisects $\angle ZWX$, which is a right-angle.

Therefore, $\angle ZWY = \angle YWX = 45^\circ$.

Therefore, $WX$ makes an angle of $\theta + 45^\circ$ with the horizontal and $WZ$ makes an angle of $\theta - 45^\circ$ with the horizontal. Since $\theta > 45^\circ$, then $\theta - 45^\circ > 0$ and $\theta + 45^\circ > 90^\circ$.

We note that since $WZ$ and $XY$ are parallel, then the slope of $XY$ equals the slope of $WZ$.

To calculate the slopes of $WX$ and $WZ$, we can calculate $\tan(\theta + 45^\circ)$ and $\tan(\theta - 45^\circ)$.

Using the facts that $\tan(A+ B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$ and $\tan(A- B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$, we obtain: $$\begin{aligned} \tan(\theta + 45^\circ) & = \dfrac{\tan \theta + \tan 45^\circ}{1 - \tan\theta \tan 45^\circ} = \dfrac{2+1}{1-(2)(1)} = -3 \\ \tan(\theta - 45^\circ) & = \dfrac{\tan \theta - \tan 45^\circ}{1 - \tan\theta \tan 45^\circ} = \dfrac{2-1}{1+(2)(1)} = \dfrac{1}{3}\end{aligned}$$ Therefore, the sum of the slopes of $WX$ and $XY$ is $-3 + \dfrac{1}{3} = -\dfrac{8}{3}$.

Solution 2

Consider a square $WXYZ$ whose diagonal $WY$ has slope $2$.

Translate this square so that $W$ is at the origin $(0,0)$. Translating a shape in the plane does not affect the slopes of any line segments.

Let the coordinates of $Y$ be $(2a,2b)$ for some non-zero numbers $a$ and $b$.

Since the slope of $WY$ is $2$, then $\dfrac{2b-0}{2a-0}=2$ and so $2b = 4a$ or $b=2a$.

Thus, the coordinates of $Y$ can be written as $(2a,4a)$.

Let $C$ be the centre of square $WXYZ$.

Then $C$ is the midpoint of $WY$, so $C$ has coordinates $(a,2a)$.

We find the slopes of $WX$ and $XY$ by finding the coordinates of $X$.

Consider the segment $XC$.

Since the diagonals of a square are perpendicular, then $XC$ is perpendicular to $WC$.

Since the slope of $WC$ is $2$, then the slopes of $XC$ and $ZC$ are $-\frac{1}{2}$.

Since the diagonals of a square are equal in length and $C$ is the midpoint of both diagonals, then $XC = WC$.

Since $WC$ and $XC$ are perpendicular and equal in length, then the “rise/run triangle” above $XC$ will be a $90^\circ$ rotation of the “rise/run triangle” below $WC$.

[[IMAGE1]]

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 $X$ is to the left of $W$ and $Z$ is to the right of $W$. Since the slopes of parallel sides are equal, it does not matter which vertex is labelled $X$ and which is labelled $Z$. We would obtain the same two slopes, but in a different order.

To get from $W(0,0)$ to $C(a,2a)$, we go up $2a$ and right $a$.

Thus, to get from $C(a,2a)$ to $X$, we go left $2a$ and up $a$.

Therefore, the coordinates of $X$ are $(a-2a,2a+a)$ or $(-a,3a)$.

Thus, the slope of $WX$ is $\dfrac{3a-0}{-a-0}=-3$.

Since $XY$ is perpendicular to $WX$, then its slope is the negative reciprocal of $-3$, which is $\dfrac{1}{3}$.

The sum of the slopes of $WX$ and $XY$ is $-3 + \dfrac{1}{3} = -\dfrac{8}{3}$.