2023 Canadian Senior
Mathematics Contest
Solutions

Wednesday, November 15, 2023
(in North America and South America)

Thursday, November 16, 2023
(outside of North American and South America)

Part A

Since $p+q$ is odd (because $31$ is odd) and $p$ and $q$ are integers, then one of $p$ and $q$ is even and the other is odd. (If both were even or both were odd, their sum would be even.)
Since $p$ and $q$ are both prime numbers and one of them is even, then one of them must be $2$, since $2$ is the only even prime number.
Since their sum is $31$, the second number must be $29$, which is prime.
Therefore, $pq = 2 \cdot 29 = 58$.

Answer: $58$

The integers between $100$ to $999$, inclusive, are exactly the three-digit positive integers.
Consider three-digit integers of the form $abc$ where the digit $a$ is even, the digit $b$ is even, and the digit $c$ is odd.
There are $4$ possibilities for $a$: $2, 4, 6, 8$. (We note that $a$ cannot equal $0$.)
There are $5$ possibilities for $b$: $0, 2, 4, 6, 8$.
There are $5$ possibilities for $c$: $1, 3, 5, 7, 9$.
Each choice of digits from these lists gives a distinct integer that satisfies the conditions.
Therefore, the number of such integers is $4 \cdot 5 \cdot 5 = 100$.

Answer: $100$

Solution 1

Since the distance from $(0,0)$ to $(x,y)$ is $17$, then $x^2 + y^2 = 17^2$.
Since the distance from $(16,0)$ to $(x,y)$ is $17$, then $(x-16)^2 + y^2 = 17^2$.
Subtracting the second of these equations from the first, we obtain $x^2 - (x-16)^2 = 0$ which gives $x^2 - (x^2 - 32x + 256) = 0$ and so $32x = 256$ or $x = 8$.
Since $x=8$ and $x^2 + y^2 = 17^2$, then $64 + y^2 = 289$ which gives $y^2 = 225$, from which we get $y = 15$ or $y = -15$.
Therefore, the two possible pairs of coordinates for $P$ are $(8,15)$ and $(8,-15)$.

Solution 2

The point $P$ is equidistant from $O$ and $A$ since $OP=PA=17$.
Suppose that $M$ is the midpoint of $OA$.
Since $O$ has coordinates $(0,0)$ and $A$ has coordinates $(16,0)$, then $M$ has coordinates $(8,0)$.
Since $OP = PA$, then $\triangle OPA$ is isosceles.
This means that median $PM$ in $\triangle OPA$ is also an altitude; in other words, $PM$ is perpendicular to $OA$.
Since $OA$ is horizontal, $PM$ is vertical, and so $P$ lies on the vertical line with equation $x = 8$.
Since $OM = 8$ and $OP = 17$ and $\triangle PMO$ is right-angled at $M$, then by the Pythagorean Theorem, $PM = \sqrt{OP^2 - OM^2} = \sqrt{17^2 - 8^2} = \sqrt{225} = 15$.
Since $PM$ is vertical and $M$ is on the $x$-axis, then $P$ is a distance of $15$ units vertically from the $x$-axis.
Since $P$ has $x$-coordinate $8$ and is $15$ units away from the $x$-axis, then the two possible pairs of coordinates for $P$ are $(8,15)$ and $(8,-15)$.

Answer: $(8,15)$, $(8,-15)$

The store sold $x$ shirts for $\$10$ each, $y$ water bottles for $\$5$ each, and $z$ chocolate bars for $\$1$ each.
Since the total revenue was $\$120$, then $10x + 5y + z = 120$.
Since $z = 120 - 10x - 5y$ and each term on the right side is a multiple of $5$, then $z$ is a multiple of $5$.
Set $z = 5t$ for some integer $t > 0$.
This gives $10x + 5y + 5t = 120$. Dividing by $5$, we obtain $2x + y + t = 24$.
Since $x > 0$ and $x$ is an integer, then $x \geq 1$.
Since $y > 0$ and $t > 0$, then $y + t \geq 2$ (since $y$ and $t$ are integers).
This means that $2x = 24 - y - t \leq 22$ and so $x \leq 11$.
If $x = 1$, then $y + t = 22$. There are $21$ pairs $(y,t)$ that satisfy this equation, namely the pairs $(y,t)=(1,21),(2,20),(3,19),\ldots,(20,2),(21,1)$.
If $x = 2$, then $y + t = 20$. There are $19$ pairs $(y,t)$ that satisfy this equation, namely the pairs $(y,t)=(1,19),(2,18),(3,17),\ldots,(18,2),(19,1)$.
For each value of $x$ with $1 \leq x \leq 11$, we obtain $y + t = 24 - 2x$.
Since $y \geq 1$, then $t \leq 23 - 2x$.
Since $t \geq 1$, then $y \leq 23 - 2x$.
In other words, $1 \leq y \leq 23 - 2x$ and $1 \leq t \leq 23 - 2x$.
Furthermore, picking any integer $y$ satisfying $1 \leq y \leq 23-2x$ gives a positive value of $t$, and so there are $23-2x$ pairs $(y,t)$ that are solutions.
Therefore, as $x$ ranges from $1$ to $11$, there are \[21 + 19 + 17 + 15 + 13 + 11 + 9 + 7 + 5 + 3 + 1\] pairs $(y,t)$, which means that there are this number of triples $(x,y,z)$.
This sum can be re-written as \[21 + (19+1) + (17+3) + (15+5) + (13+7) + (11+9)\] or $21 + 5 \cdot 20$, which means that the number of triples is $121$.

Answer: $121$

We consider $r^2 - r(p+6) + p^2 + 5p + 6 = 0$ to be a quadratic equation in $r$ with two coefficients that depend on the variable $p$.
For this quadratic equation to have real numbers $r$ that are solutions, its discriminant, $\Delta$, must be greater than or equal to $0$. A non-negative discriminant does not guarantee integer solutions, but may help us narrow the search.
By definition, \[\begin{align*} \Delta & = (-(p+6))^2 - 4 \cdot 1 \cdot (p^2 + 5p + 6) \\ & = p^2 + 12p + 36 - 4p^2 - 20p - 24 \\ & = -3p^2 - 8p + 12\end{align*}\] Thus, we would like to find all integer values of $p$ for which $-3p^2 - 8p + 12 \geq 0$. The set of integers $p$ that satisfy this inequality are the only possible values of $p$ which could be part of a solution pair $(r,p)$ of integers. We can visualize the left side of this inequality as a parabola opening downwards, so there will be a finite range of values of $p$ for which this is true.
By the quadratic formula, the solutions to the equation $-3p^2 - 8p + 12 = 0$ are \[p = \dfrac{8 \pm \sqrt{8^2 - 4(-3)(12)}}{2(-3)} = \dfrac{8 \pm \sqrt{208}}{-6} \approx 1.07, -3.74\] Since the roots of the equation $-3p^2 - 8p + 12 = 0$ are approximately $1.07$ and $-3.74$, then the integers $p$ for which $-3p^2 - 8p + 12 \geq 0$ are $p=-3,-2,-1,0,1$. (These values of $p$ are the only integers between the real solutions 1.07 and $-3.74$.)
It is these values of $p$ for which there are possibly integer values of $r$ that work.
We try them one by one:

When $p=1$, the original equation becomes $r^2 - 7r + 12 =0$, which gives $(r-3)(r-4)=0$, and so $r = 3$ or $r=4$.
When $p=0$, the original equation becomes $r^2 - 6r + 6 =0$. Using the quadratic formula, we can check that this equation does not have integer solutions.
When $p=-1$, the original equation becomes $r^2 - 5r + 2 =0$. Using the quadratic formula, we can check that this equation does not have integer solutions.
When $p=-2$, the original equation becomes $r^2 - 4r =0$, which factors as $r(r-4)=0$, and so $r = 0$ or $r=4$.
When $p=-3$, the original equation becomes $r^2 - 3r =0$, which factors as $r(r-3)=0$, and so $r = 0$ or $r=3$.

Therefore, the pairs of integers that solve the equation are \[(r,p) = (3,1), (4,1), (0,-2), (4,-2), (0,-3), (3,-3)\]

Answer: $(3,1), (4,1), (0,-2), (4,-2), (0,-3), (3,-3)$

We start by determining the heights above the bottom of the cube of the points of intersection of the edges of the pyramids.
For example, consider square $AFGB$ and edges $AG$ and $FP$. We call their point of intersection $X$.
We assign coordinates to the various points using the fact that the edge length of the cube is $6$: $F(0,0)$, $G(6,0)$, $B(6,6)$, $A(0,6)$, $P(6,3)$ ($P$ is the midpoint of $BG$).

The first quadrant of the coordinate plane with the origin, F, along with points A, B, and G plotted forming a square. Point P is plotted on side BG. Diagonal AG and segment FP intersect at point X inside the square.

Line segment $AG$ has slope $-1$, and so has equation $y = - x + 6$.
Line segment $FP$ has slope $\frac{3}{6} = \frac{1}{2}$ and so has equation $y = \frac{1}{2}x$.
To find the coordinates of $X$, we equate expressions in $y$ to obtain $-x + 6 = \frac{1}{2}x$ which gives $\frac{3}{2}x = 6$ or $x = 4$, and so $y = - 4 + 6 = 2$.
Therefore, point $X$ is a height of 2 above square $EFGH$.
Using a similar argument, the point of intersection between $PH$ and $GC$ is $2$ units above square $EFGH$.
To see why the point of intersection of $GD$ and $PE$ is also $2$ units above $EFGH$, we note that rectangle $DEGB$ has a height of $6$ (like square $AFGB$) and a width of $6\sqrt{2}$. As a result, we can think of obtaining rectangle $DEGB$ by stretching square $AFGB$ horizontally by a factor of $\sqrt{2}$. This horizontal stretch will not raise or lower the point of intersection between $GD$ and $PE$ and so this point is also two units above $EFGH$.
Now, imagine drawing a plane through the three points of intersection of the edges of the pyramids.
Since each of these points is $2$ units above $EFGH$, this plane must be horizontal and will also intersect $BG$ $2$ units above $G$, forming a square. (The points of intersection form a square because every horizontal cross-section of both pyramids is a square.) This square has side length $2$ because the $x$-coordinate of $X$ was $4$, which is $2$ units from $BG$ in that coordinate system.
This square divides the common three-dimensional region into two square-based pyramids.
One of these pyramids points upwards and has fifth vertex $P$. This pyramid has a square base with edge length $2$ and a height of $3-2=1$, since $P$ is $3$ units above $G$ and the base of the pyramid is $2$ units above $G$.
The other pyramid points downwards and has fifth vertex $G$. This pyramid has a square base with edge length $2$ and a height of $2$.
Thus, the volume of the region is $\frac{1}{3} \cdot 2^2 \cdot 1 + \frac{1}{3} \cdot 2^2 \cdot 2 = \frac{4}{3} + \frac{8}{3} = 4$.

Answer: $4$

Part B

Since $AB$ is parallel to $DC$ and $AD$ is perpendicular to both $AB$ and $DC$, then the area of trapezoid $ABCD$ is equal to $\frac{1}{2}\cdot AD \cdot (AB + DC)$ or $\frac{1}{2} \cdot 10 \cdot (7 + 17) = 120$.

Alternatively, we could separate trapezoid $ABCD$ into rectangle $ABFD$ and right-angled triangle $\triangle BFC$.
We note that $ABFD$ is a rectangle since it has three right angles.
Rectangle $ABFD$ is $7$ by $10$ and so has area $70$.
$\triangle BFC$ has $BF$ perpendicular to $FC$ and has $BF = AD = 10$.
Also, $FC = DC - DF = DC - AB = 17 - 7 = 10$.
Thus, the area of $\triangle BFC$ is $\frac{1}{2} \cdot FC \cdot BF = \frac{1}{2} \cdot 10 \cdot 10 = 50$.

This means that the area of trapezoid $ABCD$ is $70 + 50 = 120$.
Since $PQ$ is parallel to $DC$, then $\angle BQP = \angle BCF$.
We note that $ABFD$ is a rectangle since it has three right angles. This means that $BF = AD = 10$ and $DF = AB = 7$.
In $\triangle BCF$, we have $BF = 10$ and $FC = DC - DF = 17 - 7 = 10$.
Therefore, $\triangle BCF$ has $BF = FC$, which means that it is right-angled and isosceles.
Therefore, $\angle BCF = 45\degree$ and so $\angle BQP = 45\degree$.
Since $PQ$ is parallel to $AB$ and $AP$ and $BT$ are perpendicular to $AB$, then $ABTP$ is a rectangle.
Thus, $AP = BT$ and $PT = AB = 7$.
Since $PT = 7$, then $TQ = PQ - PT = x - 7$.
Since $\angle BQT = 45\degree$ and $\angle BTQ = 90\degree$, then $\triangle BTQ$ is right-angled and isosceles.
Therefore, $BT = TQ = x-7$.
Finally, $AP = BT = x-7$.
Suppose that $PQ = x$.
In this case, trapezoid $ABQP$ has parallel sides $AB=7$ and $PQ = x$, and height $AP = x - 7$.
The areas of trapezoid $ABQP$ and trapezoid $PQCD$ are equal exactly when the area of trapezoid $ABQP$ is equal to half of the area of trapezoid $ABCD$.
Thus, the areas of $ABQP$ and $PQCD$ are equal exactly when $\frac{1}{2}(x-7)(x+7) = \frac{1}{2} \cdot 120$, which gives $x^2 - 49 = 120$ or $x^2 = 169$.
Since $x > 0$, then $PQ = x = 13$.

Alternatively, we could note that trapezoid $PQCD$ has parallel sides $PQ = x$ and $DC = 17$, and height $PD = AD - AP = 10 - (x-7) = 17 - x$.
Thus, the area of trapezoid $ABQP$ and the area of trapezoid $PQCD$ are equal exactly when $\frac{1}{2}(x-7)(x+7) = \frac{1}{2}(17 - x)(x + 17)$, which gives $x^2 - 49 = 17^2 - x^2$ or $x^2 - 49 = 289 - x^2$ and so $2x^2 = 338$ or $x^2 = 169$.
Since $x > 0$, then $PQ = x = 13$.

The lattice points inside the region $A$ are precisely those lattice points whose coordinates $(r,s)$ satisfy $1 \leq r \leq 99$ and $1 \leq s \leq 99$.
Each point on the line with equation $y = 2x + 5$ is of the form $(a, 2a + 5)$ and so each lattice point on the line with equation $y = 2x + 5$ is of the form $(a, 2a+5)$ for some integer $a$.
For such a lattice point to lie in region $A$, we need $1 \leq a \leq 99$ and $1 \leq 2a + 5 \leq 99$.
The second pair of inequalities is equivalent to $-4 \leq 2a \leq 94$ and thus to $-2 \leq a \leq 47$.
Since we need both $1 \leq a \leq 99$ and $-2 \leq a \leq 47$ to be true, we have $1 \leq a \leq 47$.
Since there are $47$ integers $a$ in this range, then there are $47$ lattice points in the region $A$ and on the line with equation $y = 2x + 5$.
These are the points $(1,7), (2,9), (3,11), \ldots, (47,99)$.
Consider a lattice point $(r,s)$ that lies on the line with equation $y = \frac{5}{3}x + b$.
In this case, we must have $s = \frac{5}{3}r + b$ and so $\frac{5}{3}r = s - b$.
Since $s$ and $b$ are both integers, then $\frac{5}{3}r$ is an integer.
Since $r$ is an integer and $\frac{5}{3}r$ is an integer, then $r$ is a multiple of 3.
We write $r = 3t$ for some integer $t$ which means that $s = \frac{5}{3} \cdot 3t + b = 5t + b$.
Thus, the lattice point $(r,s)$ can be re-written as $(3t, 5t+b)$.
For $(3t, 5t+b)$ to lie within $A$, we need $1 \leq 3t \leq 99$.

Since $t$ is an integer, this means that $1 \leq t \leq 33$.
When $b=0$, these points are the points of the form $(3t,5t)$; these lie within $A$ when $1 \leq t \leq 19$. In other words, there are $19$ points in $A$ when $b=0$, which means that the greatest possible value of $b$ is at least $0$.
We note that $5t + b$ is increasing as $t$ increases.
When $b \geq 0$ and $t \geq 1$, we have $5t+b \geq 5$ and so if any points lie within $A$, then the point with $t=1$ must lie within $A$. This means that for at least $15$ of the points $(3t,5t+b)$ to lie within $A$, the points corresponding to $t = 1, 2, \ldots, 14, 15$ must all lie within $A$.
Since $5t+b$ is increasing, the largest value of $b$ should correspond to the largest value of $5(15)+b$ that does not exceed $99$.
When $b = 24$, we note that the points for $t=1, 2, \ldots, 14, 15$ are \[(r,s)=(3,29), (6,34), \ldots, (42,94), (45,99)\] which means that exactly $15$ points lie within $A$.
We note that if $b \geq 25$ and $t \geq 15$, then $5t + b \geq 100$ and so the point $(3t,5t+b)$ is not within $A$; in other words, if $b \geq 25$, there are fewer than $15$ points on the line that lie within $A$.
Therefore, $b = 24$ is indeed the largest possible value of $b$ that satisfies the given requirements.
Consider a line with equation $y = mx + 1$ for some value of $m$.
Regardless of the value of $m$, the point $(0,1)$ lies on this line. This point is not in the region $A$, but is right next to it.
Consider the line with equation $y = \frac{3}{7}x+1$ (that is, $m = \frac{3}{7}$).
The point $(7,4)$ is a lattice point in $A$ that lies on this line.
This means that $m = \frac{3}{7}$ cannot be in the final range of values, and so $n$ cannot be greater than $\frac{3}{7}$.
Consider the points on the line with equation $y = mx+1$ with $x$-coordinates from 1 to 99, inclusive. These are the points \[(1, m+1), (2, 2m+1), (3, 3m+1), \ldots, (98,98m+1), (99,99m+1)\] Since $m < n \leq \frac{3}{7}$, then $99m + 1 < 99 \cdot \frac{3}{7} + 1 < 99$ and so each of these $99$ points are in the region $A$.
This means that we need to ensure that none of $m+1, 2m+1, 3m+1, \ldots, 98m+1, 99m+1$ is an integer.
In other words, we want to determine the greatest possible real number $n$ for which none of $m+1, 2m+1, 3m+1, \ldots, 98m+1, 99m+1$ is an integer whenever $\frac{2}{7} < m < n$.
Since real numbers $s$ and $s+1$ are either both integers or both not integers, then we want to determine the greatest possible real number $n$ for which none of $m, 2m, 3m, \ldots, 98m, 99m$ is an integer whenever $\frac{2}{7} < m < n$.
The fact that none of $m, 2m, 3m, \ldots, 98m, 99m$ can be an integer is equivalent to saying that $m$ is not equal to a rational number of the form $\dfrac{c}{d}$ where $c$ is an integer and $d$ is equal to one of $1, 2, 3, \ldots, 98, 99$.
This means that the value of $n$ that we want is the largest real number $n$ with the property that there are no rational numbers $m = \dfrac{c}{d}$ with $c$ and $d$ integers and $1 \leq d \leq 99$ in the interval $\dfrac{2}{7} < m < n$.
Let $s$ be the smallest rational number of the form $\dfrac{c}{d}$ with $c$ and $d$ integers and $1 \leq d \leq 99$ that is greater than $\dfrac{2}{7}$.
Then it must be the case that $n = s$.
To see why this is true, we note that $s$ has the property that there are no rational numbers $m$ with the above restrictions between $\dfrac{2}{7}$ and $s$ by the definition of $s$, and also that any number larger than $s$ does not have this property because $s$ would be between it and $\frac{2}{7}$. Therefore, $n=s$.
This means that we need to determine the smallest rational number of the form $\dfrac{c}{d}$ with $c$ and $d$ integers and $1 \leq d \leq 99$ that is greater than $\dfrac{2}{7}$.
To do this, we minimize the value of $\dfrac{c}{d} - \dfrac{2}{7} = \dfrac{7c - 2d}{7d}$ subject to the conditions that $c$ and $d$ are positive integers with $1 \leq d \leq 99$ and that $\dfrac{c}{d} - \dfrac{2}{7} = \dfrac{7c - 2d}{7d} > 0$, which also means that $7c - 2d > 0$.

When $d = 99$, we are minimizing $\dfrac{7c - 198}{693}$ which is the smallest possible when $c = 29$, giving a difference of $\dfrac{5}{693}$.
When $d = 98$, we are minimizing $\dfrac{7c - 196}{686}$ which is the smallest possible when $c = 29$, giving a difference of $\dfrac{7}{686}$.
When $d = 97$, we are minimizing $\dfrac{7c - 194}{679}$ which is the smallest possible when $c = 28$, giving a difference of $\dfrac{2}{679}$.
When $d = 96$, we are minimizing $\dfrac{7c - 192}{672}$ which is the smallest possible when $c = 28$, giving a difference of $\dfrac{4}{672}$.
When $d = 95$, we are minimizing $\dfrac{7c - 190}{665}$ which is the smallest possible when $c = 28$, giving a difference of $\dfrac{6}{665}$.
When $d = 94$, we are minimizing $\dfrac{7c - 188}{658}$ which is the smallest possible when $c = 27$, giving a difference of $\dfrac{1}{658}$.
We can check that $\dfrac{1}{658}$ is smaller than any of $\dfrac{5}{693}, \dfrac{7}{686}, \dfrac{2}{679}, \dfrac{4}{672}, \dfrac{6}{665}$.
Furthermore, if $d < 94$, then since $\dfrac{7c - 2d}{7d} \geq \dfrac{1}{7d} > \dfrac{1}{658}$ (noting that $7c - 2d \geq 1$) and so every other difference will be greater than $\dfrac{1}{658}$.
This means that $\dfrac{27}{94}$ is the smallest of this set of rational numbers, which means that $n = \dfrac{27}{94}$.

Working with $\boldsymbol{x}$ in degrees

We know that $\sin \theta = 1$ exactly when $\theta = 90\degree + 360\degree k$ for some integer $k$.
Therefore, $\sin\left(\dfrac{x}{5}\right) = 1$ exactly when $\dfrac{x}{5} = 90\degree + 360\degree k_1$ for some integer $k_1$ which gives $x = 450\degree + 1800\degree k_1$.
Also, $\sin\left(\dfrac{x}{9}\right) = 1$ exactly when $\dfrac{x}{9} = 90\degree + 360\degree k_2$ for some integer $k_2$ which gives $x = 810\degree + 3240\degree k_2$.
Equating expressions for $x$, we obtain \[\begin{align*} 450\degree + 1800\degree k_1 & = 810\degree + 3240\degree k_2 \\ 1800k_1 - 3240k_2 & = 360 \\ 5k_1 - 9k_2 & = 1\end{align*}\] One solution to this equation is $k_1 = 2$ and $k_2=1$.
These give $x = 4050\degree$. We note that $\dfrac{x}{5} = 810\degree$ and $\dfrac{x}{9} = 450\degree$; both of these angles have a sine of $1$.

Working with $\boldsymbol{x}$ in radians

We know that $\sin \theta = 1$ exactly when $\theta = \dfrac{\pi}{2} + 2\pi k$ for some integer $k$.
Therefore, $\sin\left(\dfrac{x}{5}\right) = 1$ exactly when $\dfrac{x}{5} = \dfrac{\pi}{2} + 2\pi k_1$ for some integer $k_1$ which gives $x = \dfrac{5\pi}{2} + 10\pi k_1$.
Also, $\sin\left(\dfrac{x}{9}\right) = 1$ exactly when $\dfrac{x}{9} = \dfrac{\pi}{2} + 2\pi k_2$ for some integer $k_2$ which gives $x = \dfrac{9\pi}{2} + 18\pi k_2$.
Equating expressions for $x$, we obtain \[\begin{align*} \dfrac{5\pi}{2} + 10\pi k_1 & = \dfrac{9\pi}{2} + 18\pi k_2 \\ 10\pi k_1 - 18\pi k_2 & = 2\pi \\ 5k_1 - 9k_2 & = 1\end{align*}\] One solution to this equation is $k_1 = 2$ and $k_2=1$.
These give $x = \dfrac{45\pi}{2}$. We note that $\dfrac{x}{5} = \dfrac{9\pi}{2}$ and $\dfrac{x}{9} = \dfrac{5\pi}{2}$; both of these angles have a sine of $1$.

Therefore, one solution is $x = 4050\degree$ (in degrees) or $x = \dfrac{45\pi}{2}$ (in radians).
Suppose that $M$ and $N$ are positive integers.
We work towards determining conditions on $M$ and $N$ for which there is or is not an angle $x$ with $\sin\left(\dfrac{x}{M}\right) + \sin\left(\dfrac{x}{N}\right) = 2$.
Since $-1 \leq \sin \theta \leq 1$ for all angles $\theta$, then the equation $\sin\left(\dfrac{x}{M}\right) + \sin\left(\dfrac{x}{N}\right) = 2$ is equivalent to the pair of equations $\sin\left(\dfrac{x}{M}\right) = \sin\left(\dfrac{x}{N}\right) = 1$. (Putting this another way, there must be an angle $x$ which makes both sines 1 simultaneously.)
As in (a), the equation $\sin\left(\dfrac{x}{M}\right) = 1$ is equivalent to the statement that $\dfrac{x}{M} = 90\degree + 360\degree r$ or $\dfrac{x}{M} = \dfrac{\pi}{2} + 2\pi r$ for some integer $r$. (We will carry equations in degrees and in radians simultaneously for a time.)
These equations are equivalent to saying $x = 90\degree M + 360\degree rM$ or $x = \dfrac{M\pi}{2} + 2\pi rM$ for some integer $r$.
Similarly, the equation $\sin\left(\dfrac{x}{N}\right) = 1$ is equivalent to saying $x = 90\degree N + 360\degree sN$ or $x = \dfrac{N\pi}{2} + 2\pi sN$ for some integer $s$.

Since $x$ is common, then we can equate values of $x$ to say that if such an $x$ exists, then $90\degree M + 360\degree rM = 90\degree N + 360\degree sN$ or $\dfrac{M\pi}{2} + 2\pi rM = \dfrac{N\pi}{2} + 2\pi sN$.
It is also true that if these equations are true, then the existence of an angle $x$ that satisfies, say, $x = 90\degree M + 360\degree rM$ then guarantees the fact that the same angle $x$ satisfies $x = 90\degree N + 360\degree sN$.
In other words, the existence of an angle $x$ is equivalent to the existence of integers $r$ and $s$ for which $90\degree M + 360\degree rM = 90\degree N + 360\degree sN$ or $\dfrac{M\pi}{2} + 2\pi rM = \dfrac{N\pi}{2} + 2\pi sN$.
Dividing the first equation throughout by $90\degree$ and the second equation throughout by $\dfrac{\pi}{2}$ gives us the same resulting equation, namely $M + 4rM = N + 4sN$. Thus, we can not concern ourselves with using degrees or radians for the rest of this part.
At this stage, we know that there is an angle $x$ with the desired property precisely when there are integers $r$ and $s$ for which $M + 4rM = N + 4sN$.

Suppose that $M = 2^a c$ and $N = 2^b d$ for some integers $a$, $b$, $c$, $d$ with $a \geq 0$, $b \geq 0$, $c$ odd, and $d$ odd. Here, we are writing $M$ and $N$ as the product of a power of 2 and their "odd part".
Suppose that $a \neq b$; without loss of generality, assume that $a > b$.
Then, the following equations are equivalent: \[\begin{align*} M + 4rM & = N + 4sN \\ 2^a c + 4r \cdot 2^a c & = 2^b d + 4s \cdot 2^b d \\ 2^{a-b} c + 2^{2+a-b}rc & = d + 4sd \\ 2^{a-b} c + 2^{2+a-b}rc - 4sd & = d \end{align*}\] Since the right side of this equation is an odd integer and the left side is an even integer regardless of the choice of $r$ and $s$, there are no integers $r$ and $s$ for which this is true.
Thus, if $M$ and $N$ do not contain the same number of factors of 2, there is no angle $x$ that satisfies the initial equation.

To see this in another way, we return to the equation $M + 4rM = N + 4sN$, factor both sides to obtain $M(1 + 4r) = N(1+4s)$ which gives the equivalent equation $\dfrac{M}{N} = \dfrac{1+4s}{1+4r}$.
If integers $r$ and $s$ exist that satisfy this equation, then $\dfrac{M}{N}$ can be written as a ratio of odd integers and so $M$ and $N$ must contain the same number of factors of 2.
Putting this another way, if $M$ and $N$ do not contain the same number of factors of 2, then integers $r$ and $s$ do not exist and so the initial equation has no solutions.

To complete (b), we need to demonstrate the existence of a sequence $n_1, n_2, \ldots, n_{100}$ of positive integers for which $\sin\left(\dfrac{x}{n_i}\right) + \sin\left(\dfrac{x}{n_j}\right) \neq 2$ for all angles $x$ and for all pairs $1 \leq i < j \leq 100$.
Suppose that $n_i = 2^i$ for $1 \leq i \leq 100$.
In other words, the sequence $n_1, n_2, \ldots, n_{100}$ is the sequence $2^1, 2^2, \ldots, 2^{100}$.
No pair of numbers from the sequence $n_1, n_2, \ldots, n_{100}$ contains the same number of factors of 2, and so there is no angle $x$ that makes $\sin\left(\dfrac{x}{n_i}\right) + \sin\left(\dfrac{x}{n_j}\right) = 2$ for any $i$ and $j$ with $1 \leq i < j \leq 100$.
Therefore, the sequence $n_i = 2^i$ for $1 \leq i \leq 100$ has the desired property.
Suppose that $M$ and $N$ are positive integers for which there is an angle $x$ that satisfies the equation $\sin\left(\dfrac{x}{M}\right) + \sin\left(\dfrac{x}{N}\right) = 2$.
From (b), we know that $M$ and $N$ must contain the same number of factors of 2.
Again, suppose that $M = 2^a c$ and $N = 2^a d$ for some integers $a$, $c$, $d$ with $a \geq 0$, $c$ odd, and $d$ odd.
Then, continuing from earlier work, the following equations are equivalent: \[\begin{align*} M + 4rM & = N + 4sN \\ 2^a c + 4r \cdot 2^a c & = 2^a d + 4s \cdot 2^a d \\ c + 4rc & = d + 4sd \\ c - d & = -4rc + 4sd \end{align*}\] Since the right side is a multiple of $4$, then the left side must also be a multiple of $4$ and so $c$ and $d$ have the same remainder when divided by $4$.
(Using a more advanced result from number theory, it turns out that if $c - d$ is divisible by $4$, then this equation will always have a solution for the integers $r$ and $s$, but we do not need this precise fact.)

Suppose that $m_1, m_2, \ldots, m_{100}$ is a list of $100$ distinct positive integers with the property that, for each integer $i = 1, 2, \ldots, 99$, there is an angle $x_i$ that satisfies the equation $\sin\!\left(\dfrac{x_i}{m_i}\right)\! + \sin\! \left(\dfrac{x_i}{m_{i+1}}\right) = 2$.
Suppose further that $m_1=6$.
Since $m_1 = 2^1 \cdot 3$ and there is an angle $x_1$ with $\sin\!\left(\dfrac{x_1}{m_1}\right)\! + \sin\! \left(\dfrac{x_1}{m_{2}}\right) = 2$, then from above, $m_2 = 2^1 \cdot c_2$ for some positive integer $c_2$ that is $3$ more than a multiple of $4$ (that is, $c_2$ has the same remainder upon division by $4$ as $3$ does).
Similarly, each integer in the list $m_1$, $m_2$, $\ldots$, $m_{100}$ can be written as $m_i = 2c_i$ where $c_i$ is a positive integer that is $3$ more than a multiple of $4$.
Define $t = \dfrac{3\pi}{2^{100}} \cdot m_1 m_2 \cdots m_{100}$.
Then $\dfrac{t}{m_i} = \dfrac{3\pi}{2 \cdot 2^{99}(2c_i)}(2c_1)(2c_2)\cdots(2c_{100}) = \dfrac{\pi}{2}\cdot \dfrac{3c_1c_2\cdots c_{100}}{c_i}$.
In other words, $\dfrac{t}{m_i}$ is equal to $\dfrac{\pi}{2}$ times the product of 100 integers each of which is $3$ more than a multiple of $4$. (Note that the numerator of the last fraction includes $101$ such integers and the denominator includes $1$.)
The product of two integers each of which is $3$ more than a multiple of $4$ is equal to an integer that is $1$ more than a multiple of $4$. This is because if $y$ and $z$ are integers, then \[(4y+3)(4z+3) = 16yz + 12y + 12z + 9 = 4(4yz+3y+3z+2) + 1\] Also, the product of two integers each of which is $1$ more than a multiple of $4$ is equal to an integer that is $1$ more than a multiple of $4$. This is because if $y$ and $z$ are integers, then \[(4y+1)(4z+1) = 16yz + 4y + 4z + 1 = 4(4yz+y+z) + 1\] Thus, the product of $100$ integers each of which is $3$ more than a multiple of $4$ is equal to the product of $50$ integers each of which is $1$ more than a multiple of $4$, which is equal to an integer that is one more than a multiple of $4$.
Therefore, $\dfrac{t}{m_i}$ is equal to $\dfrac{\pi}{2}$ times an integer that is $1$ more than a multiple of $4$, and so $\sin\left(\dfrac{t}{m_i}\right) = 1$, and so \[\sin\!\left(\dfrac{t}{m_1}\right)\! + \sin\!\left(\dfrac{t}{m_2}\right)\! + \cdots + \sin\!\left(\dfrac{t}{m_{100}}\right) = 100\] as required.
Therefore, for every such sequence $m_1, m_2, \ldots, m_{100}$, there does exist an angle $t$ with the required property.

2023 Canadian Senior Mathematics Contest Solutions

Part A

Part B

2023 Canadian Senior
Mathematics Contest
Solutions