2018 Canadian Team Mathematics Contest
Solutions

Individual Problems

1. Since $(a,0)$ is on the line with equation $y=x+8$, then $0=a+8$ or $a=-8$.

   Answer: $-8$

2. Simplifying, $$\begin{array}{rl}x& =(1-\frac{1}{12})(1-\frac{1}{11})(1-\frac{1}{10})(1-\frac{1}{9})(1-\frac{1}{8})(1-\frac{1}{7})(1-\frac{1}{6})(1-\frac{1}{5})(1-\frac{1}{4})(1-\frac{1}{3})(1-\frac{1}{2})\\ & =\left(\frac{11}{12}\right)\left(\frac{10}{11}\right)\left(\frac{9}{10}\right)\left(\frac{8}{9}\right)\left(\frac{7}{8}\right)\left(\frac{6}{7}\right)\left(\frac{5}{6}\right)\left(\frac{4}{5}\right)\left(\frac{3}{4}\right)\left(\frac{2}{3}\right)\left(\frac{1}{2}\right)\\ & =\frac{11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2}\\ & =\frac{1}{12}\text{(dividing out common factors)}\end{array}$$

   Answer: $\frac{1}{12}$

3. The following pairs of rectangles are not touching: $AC,AE,CD$.
   There are 3 such pairs.

   Answer: 3

4. Let the side length of the square be $s$.
   Since the diagonal of length 10 is the hypotenuse of a right-angled triangle with two sides of the square as legs, then $s^2+s^2={10}^2$ or $2s^2=100$, which gives $s^2=50$.
   Since the area of the square equals $s^2$, then the area is $50$.

   Answer: 50

5. To make $n$ as large as possible, we make each of the digits $a$, $b$, $c$ as large as possible, starting with $a$.
   Since $a$ is divisible by $2$, its largest possible value is $a=8$, so we try $a=8$.
   Consider the two digit integer $8b$.
   This integer is a multiple of $3$ exactly when $b=1,4,7$.
   We note that $84$ is divisible by $6$, but $81$ and $87$ are not.
   To make $n$ as large possible, we try $b=7$, which makes $n=87c$.
   For $n=87c$ to be divisible by 5, it must be the case that $c=0$ or $c=5$.
   But $875=7\cdot 125$ so $875$ is divisible by 7.
   Therefore, for $n$ to be divisible by 5 and not by 7, we choose $c=0$.
   Thus, the largest integer $n$ that satisfies the given conditions is $n=870$.

   Answer: 870

6. First, we note that $a^2\ge 0$ for every real number $a$.
   Therefore, $(4x^2-y^2{)}^2\ge 0$ and $(7x+3y-39{)}^2\ge 0$.
   Since $(4x^2-y^2{)}^2+(7x+3y-39{)}^2=0$, then it must be the case that $(4x^2-y^2{)}^2=0$ and $(7x+3y-39{)}^2=0$.
   This means that $4x^2-y^2=0$ and $7x+3y-39=0$.
   The equation $4x^2-y^2=0$ is equivalent to $4x^2=y^2$ and to $y=\pm 2x$.
   If $y=2x$, then the equation $7x+3y-39=0$ becomes $7x+6x=39$ or $x=3$.
   This gives $y=2x=6$.
   If $y=-2x$, then the equation $7x+3y-39=0$ becomes $7x-6x=39$ or $x=39$.
   This gives $y=-2x=-78$.
   Therefore, the solutions to the original equation are $(x,y)=(3,6),(39,-78)$.

   Answer: $(x,y)=(3,6),(39,-78)$

7. In an arithmetic sequence with common difference $d$, the difference between any two terms must be divisible by $d$. This is because to get from any term in the sequence to any term later in the sequence, we add the common difference $d$ some number of times.
   In the given sequence, this means that $468-3=465$ is a multiple of $d$ and $2018-468=1550$ is a multiple of $d$.
   Thus, we want to determine the possible positive common divisors of $465$ and $1550$.
   We note that $465=5\cdot 93=3\cdot 5\cdot 31$ and that $1550=50\cdot 31=2\cdot 5^2\cdot 31$.
   Therefore, the positive common divisors of $465$ and $1550$ are $1,5,31,155$. (These come from finding the common prime divisors.)
   Since $d>1$, the possible values of $d$ are $5,31,155$. The sum of these is $5+31+155=191$.

   Answer: 191

8. Let points $P$ and $Q$ be on $AB$ so that $GP$ and $EQ$ are perpendicular to $AB$.
   Let point $R$ be on $DC$ so that $FR$ is perpendicular to $DC$.

   

   Note that each of $\mathrm{△}BCE$, $\mathrm{△}EQB$, $\mathrm{△}EQF$, $\mathrm{△}FRE$, $\mathrm{△}FRG$, and $\mathrm{△}FPG$ is right-angled and has height equal to $BC$, which has length $10$ m.
   Since $\mathrm{\angle }BEC=\mathrm{\angle }FEG$, then the angles of $\mathrm{△}BCE$ and $\mathrm{△}FRE$ are equal. Since their heights are also equal, these triangles are congruent.
   Since $FREQ$ and $BCEQ$ are rectangles (each has four right angles) and each is split by its diagonal into two congruent triangles, then $\mathrm{△}EQF$ and $\mathrm{△}EQB$ are also congruent to $\mathrm{△}BCE$.
   Similarly, $\mathrm{△}FPG$ and $\mathrm{△}FRG$ are congruent to these triangles as well.
   Let $CE=x$ m. Then $ER=RG=EC=x$ m.
   Since $\mathrm{△}HDG$ is right-angled at $D$ and $\mathrm{\angle }HGD=\mathrm{\angle }FGE$, then $\mathrm{△}HGD$ is similar to $\mathrm{△}FRG$.
   Since $HD:FR=6:10$, then $DG:RG=6:10$ and so $DG=\frac{6}{10}RG=\frac{3}{5}x$ m.
   Since $AB=18$ m and $ABCD$ is a rectangle, then $DC=18$ m.
   But $DC=DG+RG+ER+EC=\frac{3}{5}x+3x=\frac{18}{5}x$ m.
   Thus, $\frac{18}{5}x=18$ and so $x=5$.
   Since $x=5$, then by the Pythagorean Theorem in $\mathrm{△}BCE$, $$BE=\sqrt{B{C}^2+E{C}^2}=\sqrt{(10\text{ m}{)}^2+(5\text{ m}{)}^2}=\sqrt{125{\text{ m}}^2}=5\sqrt{5}\text{ m}$$ Now $FG=EF=BE=5\sqrt{5}\text{ m}$ and $GH=\frac{6}{10}FG=\frac{3}{5}(5\sqrt{5}\text{ m})=3\sqrt{5}\text{ m}$.
   Therefore, the length of the path $BEFGH$ is $3(5\sqrt{5}\text{ m})+(3\sqrt{5}\text{ m})$ or $18\sqrt{5}\text{ m}$.

   Answer: $18\sqrt{5}\text{ m}$

9. The box contains $R+B$ balls when the first ball is drawn and $R+B-1$ balls when the second ball is drawn.
   Therefore, there are $(R+B)(R+B-1)$ ways in which two balls can be drawn.
   If two red balls are drawn, there are $R$ balls that can be drawn first and $R-1$ balls that can be drawn second, and so there are $R(R-1)$ ways of doing this.
   Since the probability of drawing two red balls is ${\displaystyle \frac{2}{7}}$, then ${\displaystyle \frac{R(R-1)}{(R+B)(R+B-1)}}={\displaystyle \frac{2}{7}}$.
   If only one of the balls is red, then the balls drawn are either red then blue or blue then red.
   There are $RB$ ways in the first case and $BR$ ways in the second case, since there are $R$ red balls and $B$ blue balls in the box.
   Since the probability of drawing exactly one red ball is ${\displaystyle \frac{1}{2}}$, then ${\displaystyle \frac{2RB}{(R+B)(R+B-1)}}={\displaystyle \frac{1}{2}}$.
   Dividing the first equation by the second, we obtain successively $$\begin{array}{rl}{\displaystyle \frac{R(R-1)}{(R+B)(R+B-1)}}\cdot {\displaystyle \frac{(R+B)(R+B-1)}{2RB}}& ={\displaystyle \frac{2}{7}}\cdot {\displaystyle \frac{2}{1}}\\ {\displaystyle \frac{R-1}{2B}}& ={\displaystyle \frac{4}{7}}\\ R& ={\displaystyle \frac{8}{7}}B+1\end{array}$$ Substituting into the second equation, we obtain successively $$\begin{array}{rl}{\displaystyle \frac{2(\frac{8}{7}B+1)B}{(\frac{8}{7}B+1+B)(\frac{8}{7}B+1+B-1)}}& ={\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{2(\frac{8}{7}B+1)B}{(\frac{15}{7}B+1)\left(\frac{15}{7}B\right)}}& ={\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{2(8B+7)}{15(\frac{15}{7}B+1)}}& ={\displaystyle \frac{1}{2}}\text{(since }B\ne 0\text{)}\\ 32B+28& =\frac{225}{7}B+15\\ 13& =\frac{1}{7}B\\ B& =91\end{array}$$ Since $R=\frac{8}{7}B+1$, then $R=105$ and so $(R,B)=(105,91)$.
   (We can verify that the given probabilities are correct with these starting numbers of red and blue balls.)

   Answer: $(R,B)=(105,91)$

10. Consider the front face of the tank, which is a circle of radius 10 m.
    Suppose that when the water has depth 5 m, its surface is along horizontal line $AB$.
    Suppose that when the water has depth $(10+5\sqrt{2})$ m, its surface is along horizontal line $CD$.
    Let the area of the circle between the chords $AB$ and $CD$ be $x{\text{ m}}^2$.
    Since the tank is a cylinder which is lying on a flat surface, the volume of water added can be viewed as an irregular prism with base of area $x{\text{ m}}^2$ and length 30 m.
    Thus, the volume of water equals $30x{\text{ m}}^3$. Therefore, we need to calculate the value of $x$.
    

    Let $O$ be the centre of the circle, $N$ be the point where the circle touches the ground, $P$ the midpoint of $AB$, and $Q$ the midpoint of $CD$.
    Join $O$ to $A$, $B$, $C$, and $D$. Also, join $O$ to $N$ and to $Q$.
    Since $Q$ is the midpoint of chord $CD$ and $O$ is the centre of the circle, then $OQ$ is perpendicular to $CD$.
    Since $P$ is the midpoint of $AB$, then $ON$ passes through $P$ and is perpendicular to $AB$.
    Since $AB$ and $CD$ are parallel and $OP$ and $OQ$ are perpendicular to these chords, then $QOPN$ is a straight line segment.
    

    

    
    Since the radius of the circle is 10 m, then $OC=OD=OA=ON=OB=10\text{ m}$.
    Since $AB$ is 5 m above the ground, then $NP=5\text{ m}$.
    Since $ON=10\text{ m}$, then $OP=ON-NP=5\text{ m}$.
    Since $CD$ is $(10+5\sqrt{2})$ m above the ground and $ON=10\text{ m}$, then $QO=5\sqrt{2}\text{ m}$.
    Since $\mathrm{△}AOP$ is right-angled at $P$ and $OP:OA=1:2$, then $\mathrm{△}AOP$ is a ${30}^{\circ }$-${60}^{\circ }$-${90}^{\circ }$ triangle with $\mathrm{\angle }AOP={60}^{\circ }$. Also, $AP=\sqrt{3}OP=5\sqrt{3}\text{ m}$.
    Since $\mathrm{△}CQO$ is right-angled at $Q$ and $OC:OQ=10:5\sqrt{2}=2:\sqrt{2}=\sqrt{2}:1$, then $\mathrm{△}CQO$ is a ${45}^{\circ }$-${45}^{\circ }$-${90}^{\circ }$ triangle with $\mathrm{\angle }COQ={45}^{\circ }$. Also, $CQ=OQ=5\sqrt{2}\text{ m}$.
    We are now ready to calculate the value of $x$.
    The area between $AB$ and $CD$ is equal to the area of the circle minus the combined areas of the region under $AB$ and the region above $CD$.
    The area of the region under $AB$ equals the area of sector $AOB$ minus the area of $\mathrm{△}AOB$.
    The area of the region above $CD$ equals the area of sector $COD$ minus the area of $\mathrm{△}COD$.
    Since $\mathrm{\angle }AOP={60}^{\circ }$, then $\mathrm{\angle }AOB=2\mathrm{\angle }AOP={120}^{\circ }$.
    Since $\mathrm{\angle }COQ={45}^{\circ }$, then $\mathrm{\angle }COD=2\mathrm{\angle }COQ={90}^{\circ }$.
    Since the complete central angle of the circle is ${360}^{\circ }$, then sector $AOB$ is $\frac{{120}^{\circ }}{{360}^{\circ }}=\frac{1}{3}$ of the whole circle and sector $COD$ is $\frac{{90}^{\circ }}{{360}^{\circ }}=\frac{1}{4}$ of the whole circle.
    Since the area of the entire circle is $\pi (10\text{ m}{)}^2=100\pi {\text{ m}}^2$, then the area of sector $AOB$ is $\frac{100}{3}\pi {\text{ m}}^2$ and the area of sector $COD$ is $25\pi {\text{ m}}^2$.
    Since $P$ and $Q$ are the midpoints of $AB$ and $CD$, respectively, then $AB=2AP=10\sqrt{3}\text{ m}$ and $CD=2CQ=10\sqrt{2}\text{ m}$.
    Thus, the area of $\mathrm{△}AOB$ is $\frac{1}{2}\cdot AB\cdot OP=\frac{1}{2}(10\sqrt{3}\text{ m})(5\text{ m})=25\sqrt{3}{\text{ m}}^2$.
    Also, the area of $\mathrm{△}COD$ is $\frac{1}{2}\cdot CD\cdot OQ=\frac{1}{2}(10\sqrt{2}\text{ m})(5\sqrt{2}\text{ m})=50{\text{ m}}^2$.
    This means that the area of the region below $AB$ is $(\frac{100}{3}\pi -25\sqrt{3}){\text{ m}}^2$ and the area of the region above $CD$ is $(25\pi -50){\text{ m}}^2$.
    Finally, this means that the volume of water added, in ${\text{m}}^3$, is $$\begin{array}{rl}30x& =30(100\pi -(\frac{100}{3}\pi -25\sqrt{3})-(25\pi -50))\\ & =3000\pi -1000\pi +750\sqrt{3}-750\pi +1500\\ & =1250\pi +1500+750\sqrt{3}\end{array}$$ Therefore, $a\pi +b+c\sqrt{p}=1250\pi +1500+750\sqrt{3}$ and so $(a,b,c,p)=(1250,1500,750,3)$.

    Answer: $(1250,1500,750,3)$

Team Problems

1. Evaluating, $$\sqrt{1+2+3+4+5+6+7+8}=\sqrt{36}=6$$

   Answer: 6

2. Since the bucket is $\frac{2}{3}$ full and contains 9 L of maple syrup, then if it were $\frac{1}{3}$ full, it would contain $\frac{1}{2}(9\text{ L})=4.5\text{ L}$.
   Therefore, the capacity of the full bucket is $3\cdot (4.5\text{ L})=13.5\text{ L}$.

   Answer: 13.5 L

3. Since the four integers are consecutive odd integers, then they differ by 2.
   Let the four integers be $x-6,x-4,x-2,x$.
   Since the sum of these integers is 200, then $(x-6)+(x-4)+(x-2)+x=200$.
   Simplifying and solving, we obtain $4x-12=200$ and $4x=212$ and $x=53$.
   Therefore, the largest of the four integers is 53.

   Answer: 53

4. Since $80=20\cdot 4$, then to make $80=20\cdot 4$ thingamabobs, it takes $20\cdot 11=220$ widgets.
   Since $220=44\cdot 5$, then to make $220=44\cdot 5$ widgets, it takes $44\cdot 18=792$ doodads.
   Therefore, to make 80 thingamabobs, it takes 792 doodads.

   Answer: 792

5. Since $B{P}_1=P_1{P}_2=P_2{P}_3=P_3{P}_4=P_4{P}_5=P_5{P}_6=P_6{P}_7=P_7{P}_8$, then each of $\mathrm{△}B{P}_1{P}_2$, $\mathrm{△}{P}_1{P}_2{P}_3$, $\mathrm{△}{P}_2{P}_3{P}_4$, $\mathrm{△}{P}_3{P}_4{P}_5$, $\mathrm{△}{P}_4{P}_5{P}_6$, $\mathrm{△}{P}_5{P}_6{P}_7$, and $\mathrm{△}{P}_6{P}_7{P}_8$ is isosceles.
   Since $\mathrm{\angle }ABC=5^{\circ }$, then $\mathrm{\angle }B{P}_2{P}_1=\mathrm{\angle }ABC=5^{\circ }$.
   Next, $\mathrm{\angle }{P}_2{P}_1{P}_3$ is an exterior angle for $\mathrm{△}B{P}_1{P}_2$.
   Thus, $\mathrm{\angle }{P}_2{P}_1{P}_3=\mathrm{\angle }{P}_{1}B{P}_2+\mathrm{\angle }{P}_1{P}_{2}B={10}^{\circ }$.
   (To see this in another way, $\mathrm{\angle }B{P}_1{P}_2={180}^{\circ }-\mathrm{\angle }{P}_{1}B{P}_2-\mathrm{\angle }{P}_1{P}_{2}B$ and $$\mathrm{\angle }{P}_2{P}_1{P}_3={180}^{\circ }-\mathrm{\angle }B{P}_1{P}_2={180}^{\circ }-({180}^{\circ }-\mathrm{\angle }{P}_{1}B{P}_2-\mathrm{\angle }{P}_1{P}_{2}B)=\mathrm{\angle }{P}_{1}B{P}_2+\mathrm{\angle }{P}_1{P}_{2}B$$ The first of these equations comes from the sum of the angles in the triangle and the second from supplementary angles.)
   Continuing in this way, $$\begin{array}{rl}\mathrm{\angle }{P}_2{P}_3{P}_1& =\mathrm{\angle }{P}_2{P}_1{P}_3={10}^{\circ }\\ \mathrm{\angle }{P}_3{P}_2{P}_4& =\mathrm{\angle }{P}_2{P}_3{P}_1+\mathrm{\angle }{P}_{2}B{P}_3={15}^{\circ }\\ \mathrm{\angle }{P}_3{P}_4{P}_2& =\mathrm{\angle }{P}_3{P}_2{P}_4={15}^{\circ }\\ \mathrm{\angle }{P}_4{P}_3{P}_5& =\mathrm{\angle }{P}_3{P}_4{P}_2+\mathrm{\angle }{P}_{3}B{P}_4={20}^{\circ }\\ \mathrm{\angle }{P}_4{P}_5{P}_3& =\mathrm{\angle }{P}_4{P}_3{P}_5={20}^{\circ }\\ \mathrm{\angle }{P}_5{P}_4{P}_6& =\mathrm{\angle }{P}_4{P}_5{P}_3+\mathrm{\angle }{P}_{4}B{P}_5={25}^{\circ }\\ \mathrm{\angle }{P}_5{P}_6{P}_4& =\mathrm{\angle }{P}_5{P}_4{P}_6={25}^{\circ }\\ \mathrm{\angle }{P}_6{P}_5{P}_7& =\mathrm{\angle }{P}_5{P}_6{P}_4+\mathrm{\angle }{P}_{5}B{P}_6={30}^{\circ }\\ \mathrm{\angle }{P}_6{P}_7{P}_5& =\mathrm{\angle }{P}_6{P}_5{P}_7={30}^{\circ }\\ \mathrm{\angle }{P}_7{P}_6{P}_8& =\mathrm{\angle }{P}_6{P}_7{P}_5+\mathrm{\angle }{P}_{6}B{P}_7={35}^{\circ }\\ \mathrm{\angle }{P}_7{P}_8{P}_6& =\mathrm{\angle }{P}_7{P}_6{P}_8={35}^{\circ }\\ \mathrm{\angle }A{P}_7{P}_8& =\mathrm{\angle }{P}_7{P}_8{P}_6+\mathrm{\angle }{P}_{7}B{P}_8={40}^{\circ }\end{array}$$

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

6. Suppose that the base of the pyramid has $n$ sides.
   The base will also have $n$ vertices. Since the pyramid has one extra vertex (the apex), then it has $n+1$ vertices in total.
   The pyramid has $n+1$ faces: the base plus $n$ triangular faces formed by each edge of the base and the apex.
   The pyramid has $2n$ edges: the $n$ sides that form the base plus one edge joining each of the $n$ vertices of the base to the apex.
   From the given information, $2n+(n+1)=1915$.
   Thus, $3n=1914$ and so $n=638$.
   Since the pyramid has $n+1$ faces, then it has $639$ faces.

   Answer: 639

7. Since $2^{11}=2048$ and $2^5=32$, the eight values are $$\begin{array}{cccc}{2}^{11}+2^5+2=2082& 2^{11}+2^5-2=2078& 2^{11}-2^5+2=2018& 2^{11}-2^5-2=2014\\ -2^{11}+2^5+2=-2014& -2^{11}+2^5-2=-2018& -2^{11}-2^5+2=-2078& -2^{11}-2^5-2=-2082\end{array}$$ The third largest value is $2^{11}-2^5+2=2018$.

   Answer: $2^{11}-2^5+2=2018$

8. For every real number $a$, $(-a{)}^3=-a^3$ and so $(-a{)}^3+a^3=0$.
   Therefore, $$(-n{)}^3+(-n+1{)}^3+\cdots +(n-2{)}^3+(n-1{)}^3+n^3+(n+1{)}^3$$ which equals $$((-n{)}^3+n^3)+((-n+1{)}^3+(n-1{)}^3)+\cdots +((-1{)}^3+1^3)+0^3+(n+1{)}^3$$ is equal to $(n+1{)}^3$.
   Since ${14}^3=2744$ and ${15}^3=3375$ and $n$ is an integer, then $(n+1{)}^3<3129$ exactly when $n+1\le 14$.
   There are $13$ positive integers $n$ that satisfy this condition.

   Answer: 13

9. Using the given definition, the following equations are equivalent: $$\begin{array}{rl}(2\mathrm{\nabla }x)-8& =x\mathrm{\nabla }6\\ (2x-4x)-8& =6x-6x^2\\ 6x^2-8x-8& =0\\ 3x^2-4x-4& =0\end{array}$$ The sum of the values of $x$ that satisfy the original equation equals the sum of the roots of this quadratic equation.
   This sum equals $-\frac{-4}{3}$ or $\frac{4}{3}$.
   (We could calculate the roots and add these, or use the fact that the sum of the roots of the quadratic equation $a{x}^2+bx+c=0$ is $-{\displaystyle \frac{b}{a}}$.)

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

10. Suppose Birgit’s four numbers are $a,b,c,d$.
    This means that the totals $a+b+c$, $a+b+d$, $a+c+d$, and $b+c+d$ are equal to $415$, $442$, $396$, and $325$, in some order.
    If we add these totals together, we obtain $$\begin{array}{rl}(a+b+c)+(a+b+d)+(a+c+d)+(b+c+d)& =415+442+396+325\\ 3a+3b+3c+3d& =1578\\ a+b+c+d& =526\end{array}$$ since the order of addition does not matter.
    Therefore, the sum of Luciano’s numbers is $526$.

    Answer: 526

11. Let $v_1$ km/h be Krikor’s constant speed on Monday.
    Let $v_2$ km/h be Krikor’s constant speed on Tuesday.
    On Monday, Krikor drives for 30 minutes, which is $\frac{1}{2}$ hour.
    Therefore, on Monday, Krikor drives $\frac{1}{2}{v}_1$ km.
    On Tuesday, Krikor drives for 25 minutes, which is $\frac{5}{12}$ hour.
    Therefore, on Tuesday, Krikor drives $\frac{5}{12}{v}_2$ km.
    Since Krikor drives the same distance on both days, then $\frac{1}{2}{v}_1=\frac{5}{12}{v}_2$ and so $v_2=\frac{12}{5}\cdot \frac{1}{2}{v}_1=\frac{6}{5}{v}_1$.
    Since $v_2=\frac{6}{5}{v}_1=\frac{120}{100}{v}_1$, then $v_2$ is 20% larger than $v_1$.
    That is, Krikor drives 20% faster on Tuesday than on Monday.

    Answer: 20%

12. Using logarithm laws, $$\begin{array}{rl}& \pi {\log }_{2018}\sqrt{2}+\sqrt{2}{\log }_{2018}\pi +\pi {\log }_{2018}\left({\displaystyle \frac{1}{\sqrt{2}}}\right)+\sqrt{2}{\log }_{2018}\left({\displaystyle \frac{1}{\pi }}\right)\\ & ={\log }_{2018}\left({\sqrt{2}}^{\pi }\right)+{\log }_{2018}\left({\pi }^{\sqrt{2}}\right)+{\log }_{2018}\left({\displaystyle \frac{1}{{\sqrt{2}}^{\pi }}}\right)+{\log }_{2018}\left({\displaystyle \frac{1}{{\pi }^{\sqrt{2}}}}\right)\\ & ={\log }_{2018}\left({\displaystyle \frac{{\sqrt{2}}^{\pi }\cdot {\pi }^{\sqrt{2}}}{{\sqrt{2}}^{\pi }\cdot {\pi }^{\sqrt{2}}}}\right)\\ & ={\log }_{2018}(1)\\ & =0\end{array}$$

    Answer: 0

13. We make a table that lists, for each possible value of $k$, the digits, the possible three-digit integers made by these digits, and $k+3$:

    $\mathit{k}$	$\mathit{k}\mathbf{,}\mathit{k}\mathbf{+}\mathbf{1}\mathbf{,}\mathit{k}\mathbf{+}\mathbf{2}$	Possible Integers	$\mathit{k}\mathbf{+}\mathbf{3}$
    $0$	$0,1,2$	$102,120,201,210$	$3$
    $1$	$1,2,3$	$123,132,213,231,312,321$	$4$
    $2$	$2,3,4$	$234,243,324,342,423,432$	$5$
    $3$	$3,4,5$	$345,354,435,453,534,543$	$6$
    $4$	$4,5,6$	$456,465,546,564,645,654$	$7$
    $5$	$5,6,7$	$567,576,657,675,756,765$	$8$
    $6$	$6,7,8$	$678,687,768,786,867,876$	$9$
    $7$	$7,8,9$	$789,798,879,897,978,987$	$10$

    When $k=0$, the sum of the digits of each three-digit integer is 3, so each is divisible by 3.
    When $k=1$, only two of the three-digit integers are even: 132 and 312. Each is divisible by 4.
    When $k=2$, none of the three-digit integers end in 0 or 5 so none is divisible by 5.
    When $k=3$, only two of the three-digit integers are even: 354 and 534. Each is divisible by 6.
    When $k=4$, the integer 546 is divisible by 7. The rest are not. (One way to check this is by dividing each by 7.)
    When $k=5$, only two of the three-digit integers are even: 576 and 756. Only 576 is divisible by 8.
    When $k=6$, the sum of the digits of each of the three-digit integers is 21, which is not divisible by 9, so none of the integers is divisible by 9.
    When $k=7$, none of the three-digit integers end in 0 so none is divisible by 10.
    In total, there are $4+2+2+1+1=10$ three-digit integers that satisfy the required conditions.

    Answer: 10

14. Suppose that $d$ is the common difference in this arithmetic sequence.
    Since $t_{2018}=100$ and $t_{2021}$ is 3 terms further along in the sequence, then $t_{2021}=100+3d$.
    Similarly, $t_{2036}=100+18d$ since it is 18 terms further along.
    Since $t_{2018}=100$ and $t_{2015}$ is 3 terms back in the sequence, then $t_{2015}=100-3d$.
    Similarly, $t_{2000}=100-18d$ since it is 18 terms back.
    Therefore, $$\begin{array}{rl}{t}_{2000}+5t_{2015}+5t_{2021}+t_{2036}& =(100-18d)+5(100-3d)+5(100+3d)+(100+18d)\\ & =1200-18d-15d+15d+18d\\ & =1200\end{array}$$

    Answer: 1200

15. The area of the square wall with side length $n$ metres is $n^2$ square metres.
    The combined area of $n$ circles each with radius 1 metre is $n\cdot \pi \cdot 1^2$ square metres or $n\pi$ square metres.
    Given that Mathilde hits the wall at a random point, the probability that she hits a target is the ratio of the combined areas of the targets to the area of the wall, or ${\displaystyle \frac{n\pi {\text{ m}}^2}{n^2{\text{ m}}^2}}$, which equals ${\displaystyle \frac{\pi }{n}}$.
    For ${\displaystyle \frac{\pi }{n}}>{\displaystyle \frac{1}{2}}$, it must be the case that $n<2\pi \approx 6.28$.
    The largest value of $n$ for which this is true is $n=6$.

    Answer: $n=6$

16. First, we count the number of factors of 7 included in $200!$.
    Every multiple of 7 includes least 1 factor of 7.
    The product $200!$ includes 28 multiples of 7 (since $28\times 7=196$).
    Counting one factor of 7 from each of the multiples of 7 (these are $7,14,21,\dots ,182,189,196$), we see that $200!$ includes at least 28 factors of 7.
    However, each multiple of $7^2=49$ includes a second factor of 7 (since $49=7^2,98=7^2\times 2$, etc.) which was not counted in the previous 28 factors.
    The product $200!$ includes 4 multiples of 49, since $4\times 49=196$, and thus there are at least 4 additional factors of 7 in $200!$.
    Since $7^3>200$, then $200!$ does not include any multiples of $7^3$ and so we have counted all possible factors of 7.
    Thus, $200!$ includes exactly $28+4=32$ factors of 7, and so $200!=7^{32}\times t$ for some positive integer $t$ that is not divisible by 7.
    Counting in a similar way, the product $90!$ includes 12 multiples of 7 and 1 multiple of 49, and thus includes $13$ factors of 7.
    Therefore, $90!=7^{13}\times r$ for some positive integer $r$ that is not divisible by 7.
    Also, $30!$ includes $4$ factors of 7, and thus $30!=7^4\times s$ for some positive integer $s$ that is not divisible by 7.
    Therefore, ${\displaystyle \frac{200!}{90!30!}}={\displaystyle \frac{7^{32}\times t}{(7^{13}\times r)(7^4\times s)}}={\displaystyle \frac{7^{32}\times t}{(7^{17}\times rs)}}={\displaystyle \frac{7^{15}\times t}{rs}}$.
    Since we are given that ${\displaystyle \frac{200!}{90!30!}}$ is equal to a positive integer, then ${\displaystyle \frac{7^{15}\times t}{rs}}$ is a positive integer.
    Since $r$ and $s$ contain no factors of 7 and $7^{15}\times t$ is divisible by $rs$, then it must be the case that $t$ is divisible by $rs$.
    In other words, we can re-write ${\displaystyle \frac{200!}{90!30!}}={\displaystyle \frac{7^{15}\times t}{rs}}$ as ${\displaystyle \frac{200!}{90!30!}}=7^{15}\times {\displaystyle \frac{t}{rs}}$ where ${\displaystyle \frac{t}{rs}}$ is an integer.
    Since each of $r$, $s$ and $t$ does not include any factors of 7, then the integer ${\displaystyle \frac{t}{rs}}$ is not divisible by 7.
    Therefore, the largest power of 7 which divides ${\displaystyle \frac{200!}{90!30!}}$ is $7^{15}$, and so $n=15$.

    Answer: 15

17. Let $BD=h$.
    Since $\mathrm{\angle }BCA={45}^{\circ }$ and $\mathrm{△}BDC$ is right-angled at $D$, then $\mathrm{\angle }CBD={180}^{\circ }-{90}^{\circ }-{45}^{\circ }={45}^{\circ }$.
    This means that $\mathrm{△}BDC$ is isosceles with $CD=BD=h$.
    Since $\mathrm{\angle }BAC={60}^{\circ }$ and $\mathrm{△}BAD$ is right-angled at $D$, then $\mathrm{△}BAD$ is a ${30}^{\circ }$-${60}^{\circ }$-${90}^{\circ }$ triangle.
    Therefore, $BD:DA=\sqrt{3}:1$.
    Since $BD=h$, then $DA={\displaystyle \frac{h}{\sqrt{3}}}$.
    Thus, $CA=CD+DA=h+{\displaystyle \frac{h}{\sqrt{3}}}=h(1+{\displaystyle \frac{1}{\sqrt{3}}})={\displaystyle \frac{h(\sqrt{3}+1)}{\sqrt{3}}}$.
    We are told that the area of $\mathrm{△}ABC$ is $72+72\sqrt{3}$.
    Since $BD$ is perpendicular to $CA$, then the area of $\mathrm{△}ABC$ equals ${\displaystyle \frac{1}{2}}\cdot CA\cdot BD$.
    Thus, $$\begin{array}{rl}{\displaystyle \frac{1}{2}}\cdot CA\cdot BD& =72+72\sqrt{3}\\ {\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{h(\sqrt{3}+1)}{\sqrt{3}}}\cdot h& =72(1+\sqrt{3})\\ {\displaystyle \frac{h^2}{2\sqrt{3}}}& =72\\ h^2& =144\sqrt{3}\end{array}$$ and so $BD=h=12\sqrt[4]{3}$ since $h>0$.

    Answer: $12\sqrt[4]{3}$

18. Since each word is to be 7 letters long and there are two choices for each letter, there are $2^7=128$ such words.
    We count the number of words that do contain three or more A’s in a row and subtract this total from 128.
    There is 1 word with exactly 7 A’s in a row: AAAAAAA.
    There are 2 words with exactly 6 A’s in a row: AAAAAAB and BAAAAAA.
    Consider the words with exactly 5 A’s in a row.
    If such a word begins with exactly 5 A’s, then the 6th letter is B and so the word has the form AAAAAB$x$ where $x$ is either A or B. There are 2 such words.
    If such a word has the string of exactly 5 A’s beginning in the second position, then it must be BAAAAAB since there cannot be an A either immediately before or immediately after the 5 A’s.
    If such a word ends with exactly 5 A’s, then the 2nd letter is B and so the word has the form $x$BAAAAA where $x$ is either A or B. There are 2 such words.
    There are $5$ words wth exactly 5 A’s in a row.
    Consider the words with exactly 4 A’s in a row.
    Using similar reasoning, such a word can be of one of the following forms: AAAAB$xx$, BAAAAB$x$, $x$BAAAAB, $xx$BAAAA.
    Since there are two choices for each $x$, then there are $4+2+2+4=12$ words.
    Consider the words with exactly 3 A’s in a row.
    Using similar reasoning, such a word can be of one of the following forms: AAAB$xxx$, BAAAB$xx$, $x$BAAAB$x$, $xx$BAAAB, $xxx$BAAA.
    Since there are two choices for each $x$, then there appear to be $8+4+4+4+8=28$ such words. However, the word AAABAAA is counted twice. (This the only word counted twice in any of these cases.) Therefore, there are $27$ such words.
    In total, there are $1+2+5+12+27=47$ words that contain three or more A’s in a row, and so there are $128-47=81$ words that do not contain three or more A’s in a row.

    Answer: 81

19. Let $t=x^{1/5}$. Thus, $x^{2/5}=t^2$ and $x^{3/5}=t^3$.
    Therefore, the following equations are equivalent: $$\begin{array}{rl}{x}^{3/5}-4& =32-x^{2/5}\\ t^3+t^2-36& =0\\ (t-3)(t^2+4t+12)& =0\end{array}$$ Thus, $t=3$ or $t^2+4t+12=0$.
    The discriminant of this quadratic equation is $4^2-4(1)(12)<0$, which means that there are no real values of $t$ that are solutions. This in turn means that there are no corresponding real values of $x$.
    This gives $x^{1/5}=t=3$ and so $x=3^5=243$ is the only solution.

    Answer: $3^5=243$

20. Let $AC=x$.
    Thus $BC=AC-1=x-1$.
    Since $AC=AB-1$, then $AB=AC+1=x+1$.
    The perimeter of $\mathrm{△}ABC$ is $BC+AC+AB=(x-1)+x+(x+1)=3x$.
    By the cosine law in $\mathrm{△}ABC$, $$\begin{array}{rl}B{C}^2& =A{B}^2+A{C}^2-2(AB)(AC)\cos (\mathrm{\angle }BAC)\\ (x-1{)}^2& =(x+1{)}^2+x^2-2(x+1)x\left(\frac{3}{5}\right)\\ x^2-2x+1& =x^2+2x+1+x^2-\frac{6}{5}(x^2+x)\\ 0& =x^2+4x-\frac{6}{5}(x^2+x)\\ 0& =5x^2+20x-6(x^2+x)\\ x^2& =14x\end{array}$$ Since $x>0$, then $x=14$.
    Therefore, the perimeter of $\mathrm{△}ABC$, which equals $3x$, is $42$.

    Answer: 42

21. Since $f(2x-3)-2f(3x-10)+f(x-3)=28-6x$ for all real numbers $x$, then when $x=2$, we obtain $f(2(2)-3)-2f(3(2)-10)+f(2-3)=28-6(2)$ and so $f(1)-2f(-4)+f(-1)=16$.
    Since $f$ is an odd function, then $f(-1)=-f(1)$ or $f(1)+f(-1)=0$.
    Combining with $f(1)-2f(-4)+f(-1)=16$, we obtain $-2f(-4)=16$ and so $-f(-4)=8$.
    Since $f$ is an odd function, then $f(4)=-f(-4)=8$.

    Answer: 8

22. Let the radius of the small sphere be $r$ and the radius of the large sphere be $2r$.
    Draw a vertical cross-section through the centre of the top face of the cone and its bottom vertex.
    By symmetry, this will pass through the centres of the spheres.
    In the cross-section, the cone becomes a triangle and the spheres become circles.

    

    We label the vertices of the triangle $A$, $B$, $C$.
    We label the centres of the large circle and small circle $L$ and $S$, respectively.
    We label the point where the circles touch $U$.
    We label the midpoint of $AB$ (which represents the centre of the top face of the cone) as $M$.
    Join $L$ and $S$ to the points of tangency of the circles to $AC$. We call these points $P$ and $Q$.
    Since $LP$ and $SQ$ are radii, then they are perpendicular to the tangent line $AC$ at $P$ and $Q$, respectively.
    Draw a perpendicular from $S$ to $T$ on $LP$.
    The volume of the cone equals $\frac{1}{3}\pi \cdot A{M}^2\cdot MC$. We determine the lengths of $AM$ and $MC$ in terms of $r$.
    Since the radii of the small circle is $r$, then $QS=US=r$.
    Since $TPQS$ has three right angles (at $T$, $P$ and $Q$), then it has four right angles, and so is a rectangle.
    Therefore, $PT=QS=r$.
    Since the radius of the large circle is $2r$, then $PL=UL=ML=2r$.
    Therefore, $TL=PL-PT=2r-r=r$.
    Since $MC$ passes through $L$ and $S$, it also passes through $U$, the point of tangency of the two circles.
    Therefore, $LS=UL+US=2r+r=3r$.
    By the Pythagorean Theorem in $\mathrm{△}LTS$, $$TS=\sqrt{L{S}^2-T{L}^2}=\sqrt{(3r{)}^2-r^2}=\sqrt{8r^2}=2\sqrt{2}r$$ since $TS>0$ and $r>0$.
    Consider $\mathrm{△}LTS$ and $\mathrm{△}SQC$.
    Each is right-angled, $\mathrm{\angle }TLS=\mathrm{\angle }QSC$ (because $LP$ and $QS$ are parallel), and $TL=QS$.
    Therefore, $\mathrm{△}LTS$ is congruent to $\mathrm{△}SQC$.
    Thus, $SC=LS=3r$ and $QC=TS=2\sqrt{2}r$.
    This tells us that $MC=ML+LS+SC=2r+3r+3r=8r$.
    Also, $\mathrm{△}AMC$ is similar to $\mathrm{△}SQC$, since each is right-angled and they have a common angle at $C$.
    Therefore, ${\displaystyle \frac{AM}{MC}}={\displaystyle \frac{QS}{QC}}$ and so $AM={\displaystyle \frac{8r\cdot r}{2\sqrt{2}r}}=2\sqrt{2}r$.
    This means that the volume of the original cone is $\frac{1}{3}\pi \cdot A{M}^2\cdot MC=\frac{1}{3}\pi (2\sqrt{2}r{)}^2(8r)=\frac{64}{3}\pi r^3$.
    The volume of the large sphere is $\frac{4}{3}\pi (2r{)}^3=\frac{32}{3}\pi r^3$.
    The volume of the small sphere is $\frac{4}{3}\pi r^3$.
    The volume of the cone not occupied by the spheres is $\frac{64}{3}\pi r^3-\frac{32}{3}\pi r^3-\frac{4}{3}\pi r^3=\frac{28}{3}\pi r^3$.
    The fraction of the volume of the cone that this represents is ${\displaystyle \frac{\frac{28}{3}\pi r^3}{\frac{64}{3}\pi r^3}}=\frac{28}{64}=\frac{7}{16}$.

    Answer: $\frac{7}{16}$

23. Let $f(x)=-x^2+2ax+a$.
    Since $(x-a{)}^2=x^2-2ax+a^2$, then $$-x^2+2ax+a=-(x^2-2ax+a^2)+a^2+a=-(x-a{)}^2+a^2+a$$ Therefore, $M(t)$ is the maximum value of $-(x-a{)}^2+a^2+a$ over all real numbers $x$ with $x\le t$.
    Now the graph of $y=f(x)=-(x-a{)}^2+a^2+a$ is a parabola opening downwards with vertex at $(a,a^2+a)$.
    Since the parabola opens downwards, then the parabola reaches its maximum at the vertex $(a,a^2+a)$.
    Therefore, $f(x)$ is increasing when $x<a$ and decreasing when $x>a$.
    This means that, when $t<a$, the maximum of the values of $f(x)$ with $x\le t$ is $f(t)$ (because $f(x)$ increases until $f(t)$) and when $t\ge a$, the maximum of the values of $f(x)$ with $x\le t$ is $f(a)$ (because the maximum value of $f(x)$ is to the left of $t$).

       

    In other words, $$M(t)=\{\begin{array}{lll}f(t)& & t<a\\ f(a)& & t\ge a\end{array}$$ Since $a-1<a$, then $M(a-1)=f(a-1)$.
    Since $a+2>a$, then $M(a+2)=f(a)$.
    Therefore, $$\begin{array}{rl}M(a-1)+M(a+2)& =f(a-1)+f(a)\\ & =(-((a-1)-a{)}^2+a^2+a)+(-(a-a{)}^2+a^2+a)\\ & =-1+a^2+a-0+a^2+a\\ & =2a^2+2a-1\end{array}$$

    Answer: $2a^2+2a-1$

24. We find the points of intersection of $y=2\cos 3x+1$ and $y=-\cos 2x$ by equating values of $y$ and obtaining the following equivalent equations: $$\begin{array}{rl}2\cos 3x+1& =-\cos 2x\\ 2\cos (2x+x)+\cos 2x+1& =0\\ 2(\cos 2x\cos x-\sin 2x\sin x)+\cos 2x+1& =0\\ 2((2{\cos }^{2}x-1)\cos x-(2\sin x\cos x)\sin x)+(2{\cos }^{2}x-1)+1& =0\\ 2(2{\cos }^{3}x-\cos x-2{\sin }^{2}x\cos x)+2{\cos }^{2}x& =0\\ 4{\cos }^{3}x-2\cos x-4(1-{\cos }^{2}x)\cos x+2{\cos }^{2}x& =0\\ 4{\cos }^{3}x-2\cos x-4\cos x+4{\cos }^{3}x+2{\cos }^{2}x& =0\\ 8{\cos }^{3}x+2{\cos }^{2}x-6\cos x& =0\\ 4{\cos }^{3}x+{\cos }^{2}x-3\cos x& =0\\ \cos x(4{\cos }^{2}x+\cos x-3)& =0\\ \cos x(\cos x+1)(4\cos x-3)& =0\end{array}$$ Therefore, $\cos x=0$ or $\cos x=-1$ or $\cos x=\frac{3}{4}$.
    Two of these points of intersection, $P$ and $Q$, have $x$-coordinates between $\frac{17\pi }{4}=4\pi +\frac{\pi }{4}$ and $\frac{21\pi }{4}=5\pi +\frac{\pi }{4}$.
    Since $\cos 4\pi =1$ and $\cos (4\pi +\frac{\pi }{4})=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}=\frac{2\sqrt{2}}{4}<\frac{3}{4}$, then there is an angle $\theta$ between $4\pi$ and $\frac{17\pi }{4}$ with $\cos \theta =\frac{3}{4}$ and so there is not such an angle between $\frac{17\pi }{4}$ and $\frac{21\pi }{4}$.
    Therefore, we look for angles $x$ between $\frac{17\pi }{4}$ and $\frac{21\pi }{4}$ with $\cos x=0$ and $\cos x=-1$.
    Note that $\cos 5\pi =-1$ and that $\cos \frac{18\pi }{4}=\cos \frac{9\pi }{2}=\cos \frac{\pi }{2}=0$.
    Thus, the $x$-coordinates of $P$ and $Q$ are $5\pi$ and $\frac{9\pi }{2}$.
    Suppose that $P$ has $x$-coordinate $5\pi$.
    Since $P$ lies on $y=-\cos 2x$, then its $y$-coordinate is $y=-\cos 10\pi =-1$.
    Suppose that $Q$ has $x$-coordinate $\frac{9\pi }{2}$.
    Since $Q$ lies on $y=-\cos 2x$, then its $y$-coordinate is $y=-\cos 9\pi =1$.
    Therefore, the coordinates of $P$ are $(5\pi ,-1)$ and the coordinates of $Q$ are $(\frac{9\pi }{2},1)$.
    The slope of the line through $P$ and $Q$ is ${\displaystyle \frac{1-(-1)}{\frac{9\pi }{2}-5\pi }}={\displaystyle \frac{2}{-\frac{\pi }{2}}}=-\frac{4}{\pi }$.
    The line passes through the point with coordinates $(5\pi ,-1)$.
    Thus, its equation is $y-(-1)=-\frac{4}{\pi }(x-5\pi )$ or $y+1=-\frac{4}{\pi }x+20$ or $y=-\frac{4}{\pi }x+19$.
    This line intersects the $y$-axis at $A(0,19)$. Thus, $AO=19$.
    This line intersects the $x$-axis at point $B$ with $y$-coordinate $0$ and hence the $x$-coordinate of $B$ is $x=19\cdot \frac{\pi }{4}=\frac{19\pi }{4}$. Thus, $BO=\frac{19\pi }{4}$.
    Since $\mathrm{△}AOB$ is right-angled at the origin, then its area equals $\frac{1}{2}(AO)(BO)=\frac{1}{2}(19)(\frac{19\pi }{4})=\frac{361\pi }{8}$.
    Answer: $\frac{361\pi }{8}$

25. Let $f(2n)$ be the number of different ways to draw $n$ non-intersecting line segments connecting pairs of points so that each of the $2n$ points is connected to exactly one other point.
    Then $f(2)=1$ (since there are only 2 points) and $f(6)=5$ (from the given example).
    Also, $f(4)=2$. (Can you see why?)
    We show that $$f(2n)=f(2n-2)+f(2)f(2n-4)+f(4)f(2n-6)+f(6)f(2n-8)+\cdots +f(2n-4)f(2)+f(2n-2)$$ Here is a justification for this equation:

    Pick one of the $2n$ points and call it $P$.
    $P$ could be connected to the 1st point counter-clockwise from $P$. This leaves $2n-2$ points on the circle. By definition, these can be connected in $f(2n-2)$ ways.
    $P$ cannot be connected to the 2nd point counter-clockwise, because this would leave an odd number of points on one side of this line segment. An odd number of points cannot be connected in pairs as required.
    Similarly, $P$ cannot be connected to any of the 4th, 6th, 8th, $(2n-2)$th points.
    $P$ can be connected to the 3rd point counter-clockwise, leaving $2$ points on one side and $2n-4$ points on the other side. There are $f(2)$ ways to connect the 2 points and $f(2n-4)$ ways to connect the $2n-4$ points. Therefore, in this case there are $f(2)f(2n-4)$ ways to connect the points. We cannot connect a point on one side of the line to a point on the other side of the line because the line segments would cross, which is not allowed.
    $P$ can be connected to the 5th point counter-clockwise, leaving $4$ points on one side and $2n-6$ points on the other side. There are $f(4)$ ways to connect the 4 points and $f(2n-6)$ ways to connect the $2n-6$ points. Therefore, in this case there are $f(4)f(2n-6)$ ways to connect the points.
    Continuing in this way, $P$ can be connected to every other point until we reach the last ($(2n-1)$th) point which will leave $2n-2$ points on one side and none on the other. There are $f(2n-2)$ possibilities in this case.
    Adding up all of the cases, we see that there are $$f(2n)=f(2n-2)+f(2)f(2n-4)+f(4)f(2n-6)+f(6)f(2n-8)+\cdots +f(2n-4)f(2)+f(2n-2)$$ ways of connecting the points.
    The figures below show the case of $2n=10$.

    

    We can use this formula to successively calculate $f(8),f(10),f(12),f(14),f(16)$: $$\begin{array}{rl}f(8)& =f(6)+f(2)f(4)+f(4)f(2)+f(6)\\ & =5+1(2)+2(1)+5\\ & =14\\ +f(2)f(6)+f(4)f(4)+f(6)f(2)+f(8)\\ & =14+1(5)+2(2)+5(1)+14\\ & =42\\ f(4)f(6)+f(6)f(4)+f(8)f(2)+f(10)\\ & =42+1(14)+2(5)+5(2)+14(1)+42\\ & =132\\ f(2)f(10)+f(4)f(8)+f(6)f(6)+f(8)f(4)+f(10)f(2)+f(12)\\ & =132+1(42)+2(14)+5(5)+14(2)+42(1)+132\\ & =429\\ +f(8)f(6)+f(10)f(4)+f(12)f(2)+f(14)\\ & =429+1(132)+2(42)+5(14)+14(5)+42(2)+132(1)+429\\ & =1430\end{array}$$ Therefore, there are 1430 ways to join the 16 points.
    (The sequence $1,2,5,12,42,\dots$ is a famous sequence called the Catalan numbers.)

    Answer: 1430