2017 Canadian Team Mathematics Contest
Solutions

Individual Problems

1. If ${\displaystyle \frac{8}{x}}+6=8$, then ${\displaystyle \frac{8}{x}}=2$ and so $x=4$.

   Answer: 4

2. Since $AB$ and $CD$ are parallel, then $\mathrm{\angle }BAF$ and $\mathrm{\angle }AFD$ are co-interior angles.
   Therefore, $\mathrm{\angle }BAF+\mathrm{\angle }AFD={180}^{\circ }$, and so ${30}^{\circ }+3x^{\circ }={180}^{\circ }$.
   This gives $3x=150$ or $x=50$.

   Answer: 50

3. Since Ivan and Jackie are each 175 cm tall, then their average height is 175 cm.
   We are told that the average height of Ivan, Jackie and Ken together is 4% larger than 175 cm, which equals $1.04\times 175\text{ cm}=182\text{ cm}$.
   Since the average height of 3 people is 182 cm, then the sum of their heights is $3\times 182\text{ cm}$ or $546\text{ cm}$.
   Since Ivan and Jackie are each 175 cm tall, then Ken’s height is $546\text{ cm}-2\times 175\text{ cm}=196\text{ cm}$.

   Answer: 196 cm

4. There are $99-10+1=90$ positive integers between 10 and 99, inclusive.
   If a positive integer is between 10 and 99, the minimum possible sum of its digits is $1+0=1$ and the maximum possible sum of its digits is $9+9=18$.
   The multiples of 7 in the range 1 to 18 are 7 and 14.
   This means that we want to determine the number of positive integers between 10 and 99, inclusive, with sum of digits equal to 7 or 14.
   The integers with sum of digits equal to 7 are 16, 25, 34, 43, 52, 61, 70, of which there are 7.
   The integers with sum of digits equal to 14 are 59, 68, 77, 86, 95, of which there are 5.
   Thus, there are $7+5=12$ integers in the desired range with sum of digits divisible by 7, and so the probability of choosing one of these integers at random is $\frac{12}{90}=\frac{2}{15}$.

   Answer: $\frac{2}{15}$

5. The car takes 10 minutes to travel from the point at which the minivan passes it until it arrives in Betatown.
   Since the car drives at 40 km/h and since 10 minutes equals $\frac{1}{6}$ hour, then the car travels $40\text{ km/h}\cdot \frac{1}{6}\text{ h}=\frac{20}{3}\text{ km}$ in these 10 minutes.
   Thus, the distance between the point where the vehicles pass and Betatown is $\frac{20}{3}$ km.
   Since the minivan travels at 50 km/h, it covers this distance in ${\displaystyle \frac{203\text{ km}}{50\text{ km/h}}}=\frac{2}{15}\text{ h}$.
   Now $\frac{2}{15}\text{ h}=\frac{8}{60}\text{ h}$ which equals 8 minutes, and so the minivan arrives in Betatown $10-8=2$ minutes before the car.

   Answer: 2 minutes

6. Since Ruxandra wants to visit 5 countries, then there are $5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$ orders in which she can visit them with no restrictions.
   Of these ways, there are $1\cdot 4\cdot 3\cdot 2\cdot 1=24$ ways in which she visits Mongolia first. (Visiting Mongolia first is fixed and so there is 1 choice for the first country; there are still 4 choices for the second country, 3 for the second country, and so on.)
   Similarly, there are $24$ ways in which she visits Bhutan last.
   Both of these totals of 24 include the ways in which she visits Mongolia first and visits Bhutan last. There are 6 such ways, since there is 1 choice for the first country and 1 choice for the last country, and then 3 choices for the second country, 2 choices for the third country, and 1 for the fourth country.
   Therefore, the number of ways in which she either visits Mongolia first or Bhutan last (or both) is $24+24-6=42$.
   Therefore, the number of ways with the condition that she does not visit Mongolia first and she does not visit Bhutan last is $120-42=78$.

   Answer: 78

7. Since taps A, B and C can fill the bucket in 16 minutes, 12 minutes and 8 minutes, respectively, then in 1 minute taps A, B and C fill $\frac{1}{16}$, $\frac{1}{12}$ and $\frac{1}{8}$ of the bucket, respectively.
   Similarly, in 1 minute, $\frac{1}{6}$ of the bucket drains out through the hole.
   Therefore, in 1 minute with the three taps on and the hole open, the fraction of the bucket that fills is $\frac{1}{16}+\frac{1}{12}+\frac{1}{8}-\frac{1}{6}=\frac{3+4+6-8}{48}=\frac{5}{48}$.
   Thus, to fill the bucket under these conditions will take $\frac{48}{5}$ minutes.

   Answer: $\frac{48}{5}$ minutes

8. Let $A$ be the area of a regular octagon with side length l.
   Then the area of a regular octagon with side length 2 equals $2^{2}A=4A$.
   Thus, the area between the octagons is $4A-A=3A$.
   This means that it is sufficient to find the area of a regular octagon with side length 1.
   Label the octagon $PQRSTUVW$. Join $PU$, $QT$, $WR$, $VS$.
   

   

   We note that the sum of the angles in an octagon is $6\cdot {180}^{\circ }={1080}^{\circ }$ and so in a regular octagon, each interior angle equals $\frac{1}{8}({1080}^{\circ })={135}^{\circ }$.
   Therefore, $\mathrm{\angle }WPQ=\mathrm{\angle }PQR={135}^{\circ }$.
   By symmetry, $\mathrm{\angle }PWR=\mathrm{\angle }QRW$. Looking at quadrilateral $PQRW$, each of these angles equals $\frac{1}{2}({360}^{\circ }-2\cdot {135}^{\circ })={45}^{\circ }$.
   Since $\mathrm{\angle }PWR+\mathrm{\angle }WPQ={45}^{\circ }+{135}^{\circ }={180}^{\circ }$, then $PQ$ is parallel to $WR$. Using a similar argument, $PQ$, $WR$, $VS$, and $UT$ are all parallel, as are $WV$, $PU$, $QT$, and $RS$.
   Since $\mathrm{\angle }PWR={45}^{\circ }$ and $\mathrm{\angle }PWV={135}^{\circ }$, then $\mathrm{\angle }RWV={90}^{\circ }$, which means that $WR$ is perpendicular to $WV$.
   This means that $PQ$, $WR$, $VS$, and $UT$ are all perpendicular to $WV$, $PU$, $QT$, and $RS$.
   Therefore, the innermost quadrilateral is a square with side length 1. (Note that its sides are parallel and equal in length to $PQ$, $RS$, $UT$, $WV$, each of which has length 1.) Its area is 1.
   The four right-angled triangles with hypotenuses $QR$, $ST$, $UV$, and $WP$ are all isosceles right-angled triangles with hypotenuse of length 1.
   These can be pieced together to form a square with side length 1, as shown.

   

   Therefore, their combined area is 1.
   The four remaining rectangles are identical. Each has one side of length 1 (one of the sides of the octagon) and one side equal to the shorter side length of an isosceles right-angled triangle with hypotenuse 1, which equals $\frac{1}{\sqrt{2}}$.
   Therefore, the combined area of these rectangles is $4\cdot 1\cdot \frac{1}{\sqrt{2}}=2\sqrt{2}$.
   Putting these pieces together, we obtain $A=2+2\sqrt{2}$, and so the area between the octagons is $3A=6+6\sqrt{2}$.

   Answer: $6+6\sqrt{2}$

9. Let $M$ be the midpoint of $AB$.
   Let the coordinates of $A$, $B$ and $M$ be $(x_A,y_A)$, $(x_B,y_B)$ and $(x_M,y_M)$, respectively.
   Since $M$ is the midpoint of $AB$, then $x_M={\displaystyle \frac{x_A+x_B}{2}}$ and $y_M={\displaystyle \frac{y_A+y_B}{2}}$.
   We want $x_M+y_M=2017$, which is equivalent to ${\displaystyle \frac{x_A+x_B}{2}}+{\displaystyle \frac{y_A+y_B}{2}}=2017$.
   Thus, $x_A+x_B+y_A+y_B=4034$.
   Since $A$ and $B$ are the points of intersection of the parabolas with equations $y=x^2-2x-3$ and $y=-x^2+4x+c$, then $x_A$ and $x_B$ are the solutions of the equation $x^2-2x-3=-x^2+4x+c$ or $2x^2-6x-(3+c)=0$ and so $x^2-3x-\frac{1}{2}(3+c)=0$.
   The sum of the roots of this quadratic equation is $3$ and the product of the roots is $-\frac{1}{2}(3+c)$.
   In other words, $x_A+x_B=3$ and $x_A{x}_B=-\frac{1}{2}(3+c)$.
   Since points $A(x_A,y_A)$ and $B(x_B,y_B)$ are on the parabola with equation $y=x^2-2x-3$, then $y_A=x_A^2-2x_A-3$ and $y_B=x_B^2-2x_B-3$.
   Therefore, the following equations are equivalent: $$\begin{array}{rl}{x}_A+x_B+y_A+y_B& =4034\\ x_A+x_B+(x_A^2-2x_A-3)+(x_B^2-2x_B-3)& =4034\\ x_A^2+x_B^2-(x_A+x_B)-6& =4034\\ (x_A+x_B{)}^2-2x_A{x}_B-3-6& =4034\\ 3^2-2(-\frac{1}{2}(3+c))& =4043\\ 3+c& =4034\end{array}$$ and so $c=4031$.

   Answer: $4031$

10. Since the line with equation $b=ax-4y$ passes through the point $(r,0)$, then $b=ar-0$ and so $r={\displaystyle \frac{b}{a}}$.
    Since $0\le r\le 3$, then $0\le {\displaystyle \frac{b}{a}}\le 3$ and so $0\le b\le 3a$. (Since $a>0$, we can multiply the inequalities by $a$ without switching the direction of the inequalities.)
    Since the line with equation $b=ax-4y$ passes through the point $(s,4)$, then $b=as-4(4)$ and so $as=b+16$ or $s={\displaystyle \frac{b+16}{a}}$.
    Since $2\le s\le 4$, then $2\le {\displaystyle \frac{b+16}{a}}\le 4$ and so $2a\le b+16\le 4a$ (again, $a>0$).
    Rearranging, we get $2a-16\le b\le 4a-16$.
    We now proceed by determining, for each integer $a$ with $1\le a\le 100$, the number of values of $b$ that satisfy both $0\le b\le 3a$ and $2a-16\le b\le 4a-16$.
    To do this, we need to compare the possible lower and upper bounds in these inequalities.
    We note that, to satisfy both pairs of inequalities, we need both $0\le b$ and $2a-16\le b$ as well as both $b\le 3a$ and $b\le 4a-16$.
    Thus, we need to compare $0$ with $2a-16$ and $3a$ with $4a-16$.
    We note that $0\le 2a-16$ is equivalent to $16\le 2a$ or $8\le a$.
    Therefore, when $1\le a\le 7$, to satisfy both $0\le b$ and $2a-16\le b$, it is sufficient to look at $0\le b$, and when $8\le a\le 100$, it is sufficient to look at $2a-16\le b$.
    We note that $3a\le 4a-16$ is equivalent to $16\le a$.
    Therefore, when $1\le a\le 15$, it is sufficient to look at $b\le 4a-16$ and when $16\le a\le 100$, it is sufficient to look at $b\le 3a$.
    Putting this all together:

    • When $1\le a\le 7$, it is sufficient to count the values of $b$ that satisfy $0\le b\le 4a-16$.
      When $a=1,2,3$, $4a-16$ is negative, so there are no values of $b$ that work.
      When $a=4$, we obtain $0\le b\le 0$, and so there is 1 value of $b$.
      When $a=5$, we obtain $0\le b\le 4$, and so there are 5 values of $b$.
      When $a=6$, we obtain $0\le b\le 8$, and so there are 9 values of $b$.
      When $a=7$, we obtain $0\le b\le 12$, and so there are 13 values of $b$.
      Since $1+5+9+13=28$, then there are $28$ pairs $(a,b)$ in this case.

    • When $8\le a\le 15$, it is sufficient to count the values of $b$ that satisfy $2a-16\le b\le 4a-16$.
      For each of these $a$, there are $(4a-16)-(2a-16)+1=2a+1$ values of $b$.
      Thus, as $a$ ranges from 8 to 15, there are $17+19+21+23+25+27+29+31$ values of $b$.
      Since $17+19+21+23+25+27+29+31=\frac{8}{2}(17+31)=192$, there are 192 pairs $(a,b)$ in this case.

    • When $16\le a\le 100$, it is sufficient to count the values of $b$ that satisfy $2a-16\le b\le 3a$.
      For each of these $a$, there are $3a-(2a-16)+1=a+17$ values of $b$.
      Thus, as $a$ ranges from 16 to 100, there are $33+34+35+\cdots +115+116+117$ values of $b$ and so this many pairs $(a,b)$.
      Since $33+34+35+\cdots +115+116+117=\frac{85}{2}(33+117)=85(75)=6375$, then there are $6375$ pairs $(a,b)$ in this case.

    Having considered all cases, we see that there are $28+192+6375=6595$ pairs in total that satisfy the desired conditions.

    Answer: 6595

Team Problems

1. Since $ABC$ is a straight angle, then $\mathrm{\angle }DBA={180}^{\circ }-\mathrm{\angle }DBC={180}^{\circ }-{130}^{\circ }={50}^{\circ }$.
   Since the angles in a triangle add to ${180}^{\circ }$, then $$\mathrm{\angle }DAB={180}^{\circ }-\mathrm{\angle }ADB-\mathrm{\angle }DBA={180}^{\circ }-{90}^{\circ }-{50}^{\circ }={40}^{\circ }$$ (Alternatively, we could note that $\mathrm{\angle }DBC$ is an exterior angle for the triangle.
   This means that $\mathrm{\angle }DBC=\mathrm{\angle }DAB+\mathrm{\angle }ADB$, which gives $\mathrm{\angle }DAB=\mathrm{\angle }DBC-\mathrm{\angle }ADB={130}^{\circ }-{90}^{\circ }={40}^{\circ }$.

   Answer: ${40}^{\circ }$

2. Evaluating, $${[{\left({\displaystyle \frac{2017+2017}{2017}}\right)}^{-1}+{\left({\displaystyle \frac{2018}{2018+2018}}\right)}^{-1}]}^{-1}={[2^{-1}+{\left({\displaystyle \frac{1}{2}}\right)}^{-1}]}^{-1}={[{\displaystyle \frac{1}{2}}+2]}^{-1}={\left[{\displaystyle \frac{5}{2}}\right]}^{-1}={\displaystyle \frac{2}{5}}$$

   Answer: ${\displaystyle \frac{2}{5}}$

3. The six expressions that Bethany creates are $$\begin{array}{rl}2+0-1\times 7& =2+0-7=-5\\ 2+0\times 1-7& =2+0-7=-5\\ 2-0+1\times 7& =2-0+7=9\\ 2-0\times 1+7& =2-0+7=9\\ 2\times 0+1-7& =0+1-7=-6\\ 2\times 0-1+7& =0-1+7=6\end{array}$$ Of these, the maximum value is $M=9$ and the minimum value is $m=-6$.
   Thus, $M-m=9-(-6)=15$.

   Answer: 15

4. Since $\sqrt{2018}\approx 44.92$, the largest perfect square less than 2018 is ${44}^2=1936$ and the smallest perfect square greater than 2018 is ${45}^2=2025$.
   Therefore, $m^2=2025$ and $n^2=1936$, which gives $m^2-n^2=2025-1936=89$.

   Answer: 89

5. Since $\sqrt{12}+\sqrt{108}=\sqrt{N}$, then $\sqrt{2^2\cdot 3}+\sqrt{6^2\cdot 3}=\sqrt{N}$ or $2\sqrt{3}+6\sqrt{3}=\sqrt{N}$.
   This means that $\sqrt{N}=8\sqrt{3}=\sqrt{8^2\cdot 3}=\sqrt{192}$ and so $N=192$.

   Answer: 192

6. Since the ratio of the width to the height is $3:2$, we let the width be $3x$ cm and the height be $2x$ cm, for some $x>0$.
   In terms of $x$, the area of the screen is $(3x\text{ cm})(2x\text{ cm})=6x^2{\text{ cm}}^2$.
   Since the diagonal has length 65 cm, then by the Pythagorean Theorem, we obtain$(3x{)}^2+(2x{)}^2={65}^2$.
   Simplifying, we obtain $9x^2+4x^2={65}^2$ or $13x^2=65\cdot 65$.
   Thus, $x^2=5\cdot 65$ and so $6x^2=6\cdot 5\cdot 65=30\cdot 65=1950$.
   Therefore, the area of the screen is $1950{\text{ cm}}^2$.

   Answer: $1950{\text{ cm}}^2$

7. Since the three wheels touch and rotate without slipping, then the arc lengths through which each rotates will all be equal.
   Since wheel $A$ has radius 35 cm and rotates through an angle of ${72}^{\circ }$, then the arc length through which it rotates is ${\displaystyle \frac{{72}^{\circ }}{{360}^{\circ }}}(2\pi (35\text{ cm}))=14\pi \text{ cm}$. (This equals the fraction of the entire circumference as determined by the fraction of the entire central angle.)
   If wheel $C$ rotates through an angle of $\theta$, then we must have ${\displaystyle \frac{\theta }{{360}^{\circ }}}(2\pi (8\text{ cm}))=14\pi \text{ cm}$.
   Simplifying, we obtain ${\displaystyle \frac{\theta }{{360}^{\circ }}}={\displaystyle \frac{14}{16}}$ and so $\theta ={360}^{\circ }\cdot {\displaystyle \frac{7}{8}}={315}^{\circ }$.
   Therefore, wheel $C$ rotates through an angle of ${315}^{\circ }$.

   Answer: ${315}^{\circ }$

8. The volume of a cylinder with radius $r$ and height $h$ is $\pi r^{2}h$.
   The volume of a sphere with radius $r$ is $\frac{4}{3}\pi r^3$.
   The given cylinder has radius 10 cm and height 70 cm and so has volume $\pi (10\text{ cm}{)}^2(70\text{ cm})=7000\pi {\text{ cm}}^3$.
   Since the radius of the spheres and the radius of the cylinder are equal, then we can view each of the spheres as having a “height" of $2\cdot 10\text{ cm}=20\text{ cm}$. (In other words, the spheres can only stack perfectly on top of each other.)
   Since the height of the cylinder is 70 cm and the height of each sphere is 20 cm, then a maximum of 3 spheres will fit inside the closed cylinder.
   Therefore, the volume of the cylinder not taken up by the spheres is $7000\pi {\text{ cm}}^3-3\cdot \frac{4}{3}\pi (10\text{ cm}{)}^3$ or $7000\pi {\text{ cm}}^3-4000\pi {\text{ cm}}^3$ or $3000\pi {\text{ cm}}^3$.

   Answer: $3000\pi {\text{ cm}}^3$

9. Removing the initial 0, the remaining 99 terms can be written in groups of the form $(3k-2)+(3k-1)-3k$ for each $k$ from $1$ to $33$.
   The expression $(3k-2)+(3k-1)-3k$ simplifies to $3k-3$.
   Therefore, the given sum equals $$0+3+6+\cdots +93+96$$ Since $k$ ran from 1 to 33, then this sum includes 33 terms and so equals $$\frac{33}{2}(0+96)=33(48)=33(50)-33(2)=1650-66=1584$$

   Answer: 1584

10. When the 5th chord is added, it is possible that it creates only 1 new region, which means that $m=9+1=10$.
    Since there are already 4 chords, then the maximum possible number of chords that the 5th chord intersects is 4.
    If the 5th chords intersects 4 chords, then it passes through 5 regions (one before the first intersection and one after each intersection) and it splits each of the 5 regions into 2 regions, which creates 5 new regions.
    Since the 5th chord cannot intersect more than 4 chords, it cannot pass through more than 5 regions.
    Diagrams that show each of these cases are shown below:

    

    Therefore, $M=9+5=14$ and so $M^2+m^2={14}^2+{10}^2=296$.

    Answer: 296

11. The product of the roots of the quadratic equation $a{x}^2+bx+c=0$ is ${\displaystyle \frac{c}{a}}$ and the sum of the roots is $-{\displaystyle \frac{b}{a}}$.
    Since the product of the roots of $2x^2+px-p+4=0$ is 9, then ${\displaystyle \frac{-p+4}{2}}=9$ and so $-p+4=18$, which gives $p=-14$.
    Therefore, the quadratic equation is $2x^2-14x+18=0$ and the sum of its roots is $-{\displaystyle \frac{(-14)}{2}}=7$.

    Answer: 7

12. The six pairs are $a+b,a+c,a+d,b+c,b+d,c+d$.
    The sum of the two smallest numbers will be the smallest sum. Thus, $a+b=6$.
    The second smallest sum will be $a+c$ since $a<b<c<d$ gives $a+b<a+c$ and $a+c<a+d$ and $a+c<b+c$ and $a+c<b+d$ and $a+c<c+d$. Thus, $a+c=8$.
    Using a similar argument, the largest sum must be $c+d$ and the second largest sum must be $b+d$.
    Therefore, the third and fourth sums (as listed in increasing order) are $a+d=12$ and $b+c=21$ or $b+c=12$ and $a+d=21$.
    If $b+c=21$, then we would have $(a+b)+(a+c)+(b+c)=6+8+21$ and so $2a+2b+2c=35$. This cannot be the case since $a,b,c$ are integers, which makes the left side even and the right side odd.
    Therefore, $b+c=12$ and $a+d=21$.
    This gives $(a+b)+(a+c)+(b+c)=6+8+12$ or $2a+2b+2c=26$ or $a+b+c=13$.
    Since $b+c=12$, then $a=(a+b+c)-(b+c)=13-12=1$.
    Since $a+d=21$, then $d=21-a=20$.
    (We can check that $a=1$, $b=5$, $c=7$, $d=20$ fit with the given information.)

    Answer: 20

13. Starting with the given equation, we obtain the following equivalent equations: $$\begin{array}{rl}{16}^x-\frac{5}{2}(2^{2x+1})+4& =0\\ (4^2{)}^x-\frac{5}{2}(2\cdot (2^2{)}^x)+4& =0\\ (4^x{)}^2-5(4^x)+4& =0\\ (4^x-4)(4^x-1)& =0\end{array}$$ Therefore, $4^x=4$ (which gives $x=1$) or $4^x=1$ (which gives $x=0$).
    Thus, the solutions are $x=0,1$.

    Answer: $0,1$

14. Since $f(k)=4$ and $f(f(k))=7$, then $f(4)=7$.
    Since $f(f(k))=7$ and $f(f(f(k)))=19$, then $f(7)=19$.
    Suppose that $f(x)=ax+b$ for some real numbers $a$ and $b$.
    Since $f(4)=7$, then $7=4a+b$.
    Since $f(7)=19$, then $19=7a+b$.
    Subtracting, we obtain $(7a+b)-(4a+b)=19-7$ and so $3a=12$ or $a=4$, which gives $b=7-4a=7-16=-9$.
    Therefore, $f(x)=4x-9$ for all $x$.
    Since $f(k)=4$, then $4=4k-9$ and so $4k=13$ or $k=\frac{13}{4}$.

    Answer: $\frac{13}{4}$

15. Each of the 27 smaller triangular prisms has 3 faces that are squares with side length 1 and 2 faces that are equilateral triangles with side length 1.
    Combined, these prisms have $27\cdot 3=81$ square faces and $27\cdot 2=54$ triangular faces.
    Each of the 3 square faces of the larger triangular prism is made up of 9 of the smaller square faces, which means that $3\cdot 9=27$ of these smaller square faces are painted.
    Each of the 2 triangular faces of the larger triangular prism is made up of 9 of the smaller square faces, which means that $2\cdot 9=18$ of these smaller triangular faces are painted.
    Therefore, 27 of the 81 smaller square faces are painted and 18 of the 54 smaller triangular faces are painted. In other words, $\frac{1}{3}$ of each type of face is painted, so $\frac{1}{3}$ of the total surface area of the smaller prisms is painted.

    Answer: $\frac{1}{3}$

16. Let $\theta$ be the angle (in radians) between the line with equation $y=kx$ and the $x$-axis.
    Then $k=\tan \theta$, since the tangent function can be viewed as measuring slope.
    The shaded area is equal to the area of a sector with central angle $\theta$ of the larger circle with radius 2 minus the area of a sector with central angle $\theta$ of the smaller circle with radius 1.
    Therefore, ${\displaystyle \frac{\theta }{2\pi }}\pi (2^2)-{\displaystyle \frac{\theta }{2\pi }}\pi (1^2)=2$.
    Multiplying both sides by $2$ and dividing out the factor of $\pi$ in numerators and denominators, we obtain $4\theta -\theta =4$ and so $3\theta =4$ or $\theta =\frac{4}{3}$.
    Thus, $k=\tan \frac{4}{3}$.

    Answer: $\tan \frac{4}{3}$

17. We start by creating a Venn diagram:

    

    Hide/Reveal Description of Venn Diagram
    • There are three groups in the Venn diagram: Canoeing, Swimming, and Fishing.
    • In Canoeing, we have a value of "a". In Swimming, we have "c". And in Fishing, we have "f".
    • "b" is shared by Canoeing and Swimming. "d" is shared by Canoeing and Fishing. And "e" is shared by Swimming and Fishing.
    • 0 is shared by Canoeing, Swimming and Fishing.
    • 9 is not in any section (not in Canoeing, Swimming or Fishing).

    ---

    We want to determine the minimum possible value of $a+b+c+d+e+f+9$.
    From the given information $a+b+d=15$ and $b+c+e=22$ and $d+e+f=12$.
    Adding these equations, we obtain $a+2b+c+2e+f+2d=49$ or $a+b+c+d+e+f+(b+d+e)=49$.
    Thus, $a+b+c+d+e+f=49-(b+d+e)$ and so we want to minimize $49-(b+d+e)$, which means that we want to maximize $b+d+e$.
    Since $a,b,c,d,e,f$ are non-negative integers, then $a+b+d=15$ means $b+d\le 15$. Similarly, $b+e\le 22$ and $d+e\le 12$.
    Adding these inequalities, we obtain $2b+2d+2e\le 49$ and so $b+d+e\le \frac{49}{2}$.
    Since $b,d,e$ are integers, then in fact $b+d+e\le 24$.
    It is possible to make $b+d+e=24$ by setting $a=1$, $c=0$ and $f=0$. This gives $b+d=14$ and $b+e=22$ and $d+e=12$, and so $b=12$ and $d=2$ and $e=10$.
    We can see this in the completed Venn diagram here:

    

    Hide/Reveal Description of Venn Diagram
    • There are three groups in the Venn diagram: Canoeing, Swimming, and Fishing.
    • In Canoeing, we have 1. In Swimming, we have 0. And in Fishing, we have 0.
    • 12 is shared by Canoeing and Swimming. 2 is shared by Canoeing and Fishing. And 10 is shared by Swimming and Fishing.
    • 0 is shared by Canoeing, Swimming and Fishing.
    • 9 is not in any section (not in Canoeing, Swimming or Fishing).

    ---

    Therefore, the maximum possible value of $b+d+e$ is 24 and so the minimum possible number of campers is $58-24=34$.

    Answer: 34

18. Let $C$ be the centre of the circle. Note that $C$ is in the fourth quadrant.
    Let $A$ and $B$ be the points where the circle intersects the $x$-axis and let $M$ be the midpoint of $AB$.

    

    We want to find the area of the region inside the circle and above $AB$.
    We calculate this area by finding the area of sector $ACB$ and subtracting the area of $\mathrm{△}ACB$.
    Join $C$ to $A$, $B$ and $M$.
    Since the radius of the circle is 1, then $CA=CB=1$.
    Since $M$ is the midpoint of $AB$ and $C$ is the centre, then $CM$ is perpendicular to $AB$, which means that $CM$ is vertical. Since the coordinates of $C$ are $(1,-\frac{\sqrt{3}}{2})$, then $CM=\frac{\sqrt{3}}{2}$.
    Since $AC:CM=1:\frac{\sqrt{3}}{2}$, then $\mathrm{△}CAM$ is a ${30}^{\circ }$-${60}^{\circ }$-${90}^{\circ }$ triangle, as is $\mathrm{△}BCM$.
    Therefore, $\mathrm{\angle }ACB={60}^{\circ }$, and thus the area of sector $ACB$ equals $\frac{{60}^{\circ }}{{360}^{\circ }}\pi (1^2)=\frac{1}{6}\pi$.
    Also, since $\mathrm{△}CAM$ is ${30}^{\circ }$-${60}^{\circ }$-${90}^{\circ }$, then $AM=\frac{1}{2}$. Similarly, $BM=\frac{1}{2}$.
    Therefore, the area of $\mathrm{△}ACB$ is $\frac{1}{2}AB\cdot CM=\frac{1}{2}(1)\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}$.
    Finally, this means that the area inside the circle and inside the first quadrant is $\frac{1}{6}\pi -\frac{\sqrt{3}}{4}$.

    Answer: $\frac{1}{6}\pi -\frac{\sqrt{3}}{4}$

19. When $f(x)$ is divided by $x-4$, the remainder is 102.
    This means that $f(x)-102$ has a factor of $x-4$.
    Similarly, $f(x)-102$ is divisible by each of $x-3$, $x+3$ and $x+4$.
    Since $f(x)$ is a quartic polynomial, then $f(x)-102$ is a quartic polynomial and so can have at most four distinct linear factors.
    Since the leading coefficient of $f(x)$ is 1, then the leading coefficient of $f(x)-102$ is also 1.
    This means that $f(x)-102=(x-4)(x-3)(x+3)(x+4)$.
    Thus, $f(x)=(x-4)(x-3)(x+3)(x+4)+102$.
    Therefore, the following equations are equivalent: $$\begin{array}{rl}f(x)& =246\\ (x-4)(x-3)(x+3)(x+4)+102& =246\\ (x-4)(x+4)(x-3)(x+3)& =144\\ (x^2-16)(x^2-9)& =144\\ x^4-25x^2+144& =144\\ x^4-25x^2& =0\\ x^2(x^2-25)& =0\\ x^2(x-5)(x+5)& =0\end{array}$$ Thus, the values of $x$ for which $f(x)=246$ are $x=0,5,-5$.

    Answer: $0,5,-5$

20. Since $-1\le \cos \theta \le 1$ for all angles $\theta$, then $0\le {\cos }^6\theta \le 1$ for all angles $\theta$.
    Also, ${\sin }^6\alpha \ge 0$ for all angles $\alpha$.
    Therefore, to have ${\cos }^6\theta -{\sin }^6\alpha =1$ for some angles $\theta$ and $\alpha$, then we must have ${\cos }^6\theta =1$ and ${\sin }^6\alpha =0$.
    From the given equation, this means that ${\cos }^6(1000x)=1$ and ${\sin }^6(1000y)=0$.
    Therefore, $\cos (1000x)=\pm 1$ and $\sin (1000y)=0$.
    Since $0\le x\le \frac{\pi }{8}$, then $0\le 1000x\le 125\pi$.
    Now $\cos \theta =\pm 1$ exactly when $\theta$ is a multiple of $\pi$.
    Therefore, the possible values of $1000x$ are $0,\pi ,2\pi ,\dots ,124\pi ,125\pi$.
    There are 126 such values.
    Since $0\le y\le \frac{\pi }{8}$, then $0\le 1000y\le 125\pi$.
    Now $\sin \alpha =0$ exactly when $\alpha$ is a multiple of $\pi$.
    Therefore, the possible values of $1000y$ are $0,\pi ,2\pi ,\dots ,124\pi ,125\pi$.
    There are 126 such values.
    Since there are 126 values of $x$ and 126 independent values of $y$, then there are ${126}^2=15876$ pairs $(x,y)$ that satisfy the original equation and restrictions.

    Answer: ${126}^2$

21. From the given information, Serge writes down the times that are $1+2+3+\cdots +(n-1)+n$ minutes after midnight for each successive value of $n$ from 1 up until the total is large enough to exceed 24 hours in minutes, which equals $24\cdot 60=1440$ minutes.
    Now $1+2+3+\cdots +(n-1)+n=\frac{1}{2}n(n+1)$ for each positive integer $n$.
    To find times that are “on the hour", we find the values of $n$ for which $\frac{1}{2}n(n+1)$ is a multiple of 60 (that is, the time is a multiple of 60 minutes after midnight).
    Since we want $\frac{1}{2}n(n+1)\le 1440$, then $n(n+1)\le 2880$.
    Since $52\cdot 53=2756$ and $53\cdot 54=2862$ and $n(n+1)$ increases as $n$ increases, then we only need to check values of $n$ up to and including $n=52$.
    For $\frac{1}{2}n(n+1)$ to be a multiple of 60, we need $\frac{1}{2}n(n+1)=60k$ for some integer $k$, or equivalently $n(n+1)=120k$ for some integer $k$.
    Note that $n$ and $n+1$ are either even and odd, or odd and even, respectively.
    Note also that $120k$ is divisible by 8.
    Thus, one of $n$ or $n+1$ is divisible by 8.
    Since the maximum possible value of $n$ is 52, we can check the possibilities by hand:

    $n$	$n(n+1)$	$\text{Divisible by }120?$	$\frac{1}{2}n(n+1)$
    $7$	$56$	No
    $8$	$72$	No
    $15$	$240$	Yes	$120$
    $16$	$272$	No
    $23$	$552$	No
    $24$	$600$	Yes	$300$
    $31$	$992$	No
    $32$	$1056$	No
    $39$	$1560$	Yes	$780$
    $40$	$1640$	No
    $47$	$2256$	No
    $48$	$2352$	No

    (Note that we could have ruled out the cases $n=7,8,16,23,31,32,47,48$ by noting that in these cases neither $n$ nor $n+1$ is divisible by 5 and so $n(n+1)$ cannot be a multiple of 120.)
    Therefore, the times that Serge writes down are those that are 120, 300 and 780 minutes after midnight, which are 2, 5 and 13 hours, respectively, after midnight.
    That is, Serge writes down 2:00 a.m., 5:00 a.m., and 1:00 p.m.

    Answer: 2:00 a.m., 5:00 a.m., 1:00 p.m.

22. Figure 0 consists of 1 square with side length 18. Thus, $A_0={18}^2$.
    Figure 1 consists of Figure 0 plus the addition of 2 squares with side length $\frac{2}{3}\cdot 18$.
    Thus, $A_1={18}^2+2{(\frac{2}{3}\cdot 18)}^2$.
    Figure 2 consists of Figure 1 plus the addition of 4 squares with side length $\frac{2}{3}\cdot \frac{2}{3}\cdot 18$.
    Thus, $A_2={18}^2+2{(\frac{2}{3}\cdot 18)}^2+4{(\frac{2}{3}\cdot \frac{2}{3}\cdot 18)}^2$.
    In general, since twice as many squares are added at each step as at the step before, then $2^n$ squares are added to Figure $n-1$ to make Figure $n$.
    Also, since the side length of the squares added at each step is $\frac{2}{3}$ of the side length of the squares added at the previous step, then the squares added for Figure $n$ have side length $18{\left(\frac{2}{3}\right)}^n$.
    Therefore, $$\begin{array}{rl}{A}_n& ={18}^2+2{(\frac{2}{3}\cdot 18)}^2+4{(\frac{2}{3}\cdot \frac{2}{3}\cdot 18)}^2+\cdots +2^n{\left(18{\left(\frac{2}{3}\right)}^n\right)}^2\\ & ={18}^2+2^1\cdot {18}^2{\left(\frac{2}{3}\right)}^2+2^2\cdot {18}^2{\left(\frac{2}{3}\right)}^4+\cdots +2^n\cdot {18}^2{\left(\frac{2}{3}\right)}^{2n}\end{array}$$ The right side is a geometric series with $n+1$ terms, first term $a={18}^2$, and common ratio $r=2{\left(\frac{2}{3}\right)}^2=\frac{8}{9}$.
    Therefore, $$A_n={\displaystyle \frac{{18}^2(1-{\left(\frac{8}{9}\right)}^{n+1})}{1-\frac{8}{9}}}=9\cdot {18}^2(1-{\left(\frac{8}{9}\right)}^{n+1})=2916(1-{\left(\frac{8}{9}\right)}^{n+1})$$ Since ${\left(\frac{8}{9}\right)}^{n+1}>0$ for all integers $n\ge 0$, then $1-{\left(\frac{8}{9}\right)}^{n+1}<1$ for all integers $n\ge 0$.
    Therefore, $A_n<2916\cdot 1=2916$ for all integers $n\ge 0$.
    But, as $n$ approaches infinity, ${\left(\frac{8}{9}\right)}^{n+1}$ approaches 0, which means that $1-{\left(\frac{8}{9}\right)}^{n+1}$ approaches 1, which means that $A_n$ approaches 2916.
    In other words, while $A_n$ can never equal 2916, it will eventually be infinitesimally close to 2916. We can check that if $n\ge 68$, then $A_n>2915$.
    Therefore, the smallest positive integers $M$ with the property that $A_n<M$ for all integers $n\ge 0$ is $M=2916$.

    Answer: 2916

23. Brad’s answer to each question is either right or wrong, and so there are $2^8$ possible combinations of right and wrong answers for the remaining 8 questions.
    Since Brad has already answered exactly 1 question correctly, then ${\displaystyle (\genfrac{}{}{0ex}{}{8}{4})}$ of these combinations will include exactly 5 correct answers. (We choose 4 of the remaining 8 questions to be answered correctly.)
    We show that each of these ${\displaystyle (\genfrac{}{}{0ex}{}{8}{4})}$ combinations has an equal probability, $p$.
    This will mean that the probability that Brad has exactly 5 correct answers is ${\displaystyle (\genfrac{}{}{0ex}{}{8}{4})p}$.
    Consider a specific combination that includes exactly 5 correct answers.
    For each question from the 3rd to the 10th, the probability that his answer is correct equals the ratio of the number of problems that he has already answered correctly to the total number of problems that he has already answered.
    Since the probability that his answer is wrong equals 1 minus the probability that his answer is correct, then the probability that his answer is wrong will equal the number of problems that he has already answered incorrectly to the total number of problems that he has already answered. (This is because the total number of problems answered equals the number correct plus the number incorrect.)
    Therefore, for each problem from the 3rd to 10th, the associated probability is a fraction with denominator from 2 to 9, respectively (the number of problems already answered).
    Consider the positions in this combination in which he gives his 2nd, 3rd, 4th, and 5th correct answers. In these cases, he has previously answered 1, 2, 3, and 4 problems correctly, and so the numerators of the corresponding probability fractions will be 1, 2, 3, and 4.
    Similarly, in the positions where his answer is wrong, the numerators will be 1, 2, 3, and 4.
    Therefore, $p={\displaystyle \frac{(1\cdot 2\cdot 3\cdot 4)(1\cdot 2\cdot 3\cdot 4)}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9}}$ and so the probability that Brad answers exactly 5 out of the 10 problems correctly is $$(\genfrac{}{}{0ex}{}{8}{4}){\displaystyle \frac{(1\cdot 2\cdot 3\cdot 4)(1\cdot 2\cdot 3\cdot 4)}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9}}={\displaystyle \frac{8!}{4!4!}}\cdot {\displaystyle \frac{(1\cdot 2\cdot 3\cdot 4)(1\cdot 2\cdot 3\cdot 4)}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8\cdot 9}}={\displaystyle \frac{8!}{4!4!}}\cdot {\displaystyle \frac{4!4!}{9!}}={\displaystyle \frac{8!}{9!}}={\displaystyle \frac{1}{9}}$$

    Answer: ${\displaystyle \frac{1}{9}}$

24. We proceed by (very carefully!) calculating the coordinates of points $A$, $B$ and $C$, and then equating $AB$ and $AC$ to solve for $t$.
    Point $A$ is the point of intersection of the lines with equations $x+y=3$ and $2x-y=0$.
    Since $x+y=3$ and $y=2x$, then $x+2x=3$ and so $x=1$ which gives $y=2x=2$.
    Therefore, $A$ has coordinates $(1,2)$.
    Point $B$ is the point of intersection of the lines with equations $x+y=3$ and $3x-ty=4$.
    Adding $t$ times the first equation to the second, we obtain $tx+3x=3t+4$ and so $x={\displaystyle \frac{3t+4}{t+3}}$, with the restriction that $t\ne -3$.
    Since $x+y=3$, then $y=3-{\displaystyle \frac{3t+4}{t+3}}={\displaystyle \frac{3t+9}{t+3}}-{\displaystyle \frac{3t+4}{t+3}}={\displaystyle \frac{5}{t+3}}$.
    Therefore, $B$ has coordinates $({\displaystyle \frac{3t+4}{t+3}},{\displaystyle \frac{5}{t+3}})$.
    Point $C$ is the point of intersection of the lines with equations $2x-y=0$ and $3x-ty=4$.
    Subtracting $t$ times the first equation from the second, we obtain $3x-2tx=4-0$ and so $x={\displaystyle \frac{4}{3-2t}}$ with the restriction that $t\ne \frac{3}{2}$.
    Since $y=2x$, then $y={\displaystyle \frac{8}{3-2t}}$.
    Therefore, $C$ has coordinates $({\displaystyle \frac{4}{3-2t}},{\displaystyle \frac{8}{3-2t}})$.
    We assume that $t\ne -3,\frac{3}{2}$.
    Since $AB$ and $AC$ are non-negative lengths, then the following equations are equivalent: $$\begin{array}{rl}AB& =AC\\ A{B}^2& =A{C}^2\\ {({\displaystyle \frac{3t+4}{t+3}}-1)}^2+{({\displaystyle \frac{5}{t+3}}-2)}^2& ={({\displaystyle \frac{4}{3-2t}}-1)}^2+{({\displaystyle \frac{8}{3-2t}}-2)}^2\\ {\left({\displaystyle \frac{(3t+4)-(t+3)}{t+3}}\right)}^2+{\left({\displaystyle \frac{5-(2t+6)}{t+3}}\right)}^2& ={\left({\displaystyle \frac{4-(3-2t)}{3-2t}}\right)}^2+{\left({\displaystyle \frac{8-(6-4t)}{3-2t}}\right)}^2\\ {\left({\displaystyle \frac{2t+1}{t+3}}\right)}^2+{\left({\displaystyle \frac{-2t-1}{t+3}}\right)}^2& ={\left({\displaystyle \frac{2t+1}{3-2t}}\right)}^2+{\left({\displaystyle \frac{4t+2}{3-2t}}\right)}^2\\ (2t+1{)}^2[{\left({\displaystyle \frac{1}{t+3}}\right)}^2+{\left({\displaystyle \frac{-1}{t+3}}\right)}^2]& =(2t+1{)}^2[{\left({\displaystyle \frac{1}{3-2t}}\right)}^2+{\left({\displaystyle \frac{2}{3-2t}}\right)}^2]\end{array}$$ $$\begin{array}{rl}(2t+1{)}^2[{\displaystyle \frac{2}{(t+3{)}^2}}-{\displaystyle \frac{5}{(3-2t{)}^2}}]& =0\\ (2t+1{)}^2\left[{\displaystyle \frac{2(3-2t{)}^2-5(t+3{)}^2}{(t+3{)}^2(3-2t{)}^2}}\right]& =0\\ (2t+1{)}^2\left[{\displaystyle \frac{2(4t^2-12t+9)-5(t^2+6t+9)}{(t+3{)}^2(3-2t{)}^2}}\right]& =0\\ (2t+1{)}^2\left[{\displaystyle \frac{3t^2-54t-27}{(t+3{)}^2(3-2t{)}^2}}\right]& =0\\ 3(2t+1{)}^2\left[{\displaystyle \frac{t^2-18t-9}{(t+3{)}^2(3-2t{)}^2}}\right]& =0\end{array}$$ Therefore, the values of $t$ for which $AB=AC$ are $t=-\frac{1}{2}$ and the two values of $t$ for which $t^2-18t-9=0$, which are $t={\displaystyle \frac{18\pm \sqrt{{18}^2-4(1)(-9)}}{2}}={\displaystyle \frac{18\pm \sqrt{360}}{2}}=9\pm 3\sqrt{10}$.

    Answer: $-\frac{1}{2},9\pm 3\sqrt{10}$

25. Since $\mathrm{△}ACB$ is isosceles with $AC=BC$ and $CD$ is a median, then $CD$ is perpendicular to $AB$.
    Since $CD$ is perpendicular to $AB$, then when $\mathrm{△}ABC$ is folded along $CD$, $CD$ will be perpendicular to the plane of $\mathrm{△}ADB$.
    Therefore, the volume of the tent can be calculated as $\frac{1}{3}$ times the area of $\mathrm{△}ADB$ times $CD$.
    Since $CD$ is fixed in length, then the volume is maximized when the area of $\mathrm{△}ADB$ is maximized.
    Since $AD=DB$ are fixed in length, the area of $\mathrm{△}ADB$ is maximized when $\mathrm{\angle }ADB={90}^{\circ }$. (One way to see this is to note that we can calculate the area of $\mathrm{△}ADB$ as $\frac{1}{2}(AD)(DB)\sin (\mathrm{\angle }ADB)$, which is maximized when $\sin (\mathrm{\angle }ADB)=1$.)
    Since $AB=6$ m, then $AD=DB=3$ m.
    Since $BD=3$ m and $BC=5$ m, then $CD=4$ m. ($\mathrm{△}BDC$ is a 3-4-5 triangle.)
    Thus, the maximum volume of the tent is $$\frac{1}{3}(\frac{1}{2}(AD)(DB))(DC)=\frac{1}{6}(AD)(DB)(DC)=\frac{1}{6}(3\text{ m})(3\text{ m})(4\text{ m})=6{\text{ m}}^3$$ To find the height, $h$, of $D$ above the base $\mathrm{△}ABC$, we note that the volume equals $\frac{1}{3}$ times the area of $\mathrm{△}ABC$ times the height, $h$.
    Since we know that the volume is $6{\text{ m}}^3$, then if we calculate the area of $\mathrm{△}ABC$, we can solve for $h$.
    Since $AD=DB=3$ m and $\mathrm{\angle }ADB={90}^{\circ }$, then $AB=3\sqrt{2}$ m in the maximum volume configuration.
    Therefore, $\mathrm{△}ABC$ has $AB=3\sqrt{2}$ m and $AC=BC=5$ m.
    Let $N$ be the midpoint of $AB$.

    

    Since $AC=BC$, then $CN$ is perpendicular to $AB$.
    Since $BC=5$ m and $NB=\frac{3\sqrt{2}}{2}$ m, then by the Pythagorean Theorem, $$C{N}^2=B{C}^2-N{B}^2=(5\text{ m}{)}^2-{\left(\frac{3\sqrt{2}}{2}\text{ m}\right)}^2=25{\text{ m}}^2-\frac{18}{4}{\text{ m}}^2=25{\text{ m}}^2-\frac{9}{2}{\text{ m}}^2=\frac{41}{2}{\text{ m}}^2$$ and so $CN=\sqrt{\frac{41}{2}}$ m.
    Thus, the area of $\mathrm{△}ABC$ is $\frac{1}{2}(AB)(CN)=\frac{1}{2}\left(3\sqrt{2}\text{ m}\right)\left(\frac{\sqrt{41}}{\sqrt{2}}\text{ m}\right)=\frac{3\sqrt{41}}{2}{\text{ m}}^2$.
    Finally, we have $\frac{1}{3}\left(\frac{3\sqrt{41}}{2}{\text{ m}}^2\right)h=6{\text{ m}}^3$ and so $h=\frac{12}{\sqrt{41}}$ m.

    Answer: $\frac{12}{\sqrt{41}}$ m