2020 Canadian Team Mathematics Contest
Solutions

Individual Problems

1. Solution 1
   The numerator is equal to $24+12=36$. The denominator is equal to $4^2-4=16-4=12$, so $${\displaystyle \frac{24+12}{4^2-4}}={\displaystyle \frac{36}{12}}=3$$

   Solution 2
   By factoring $4$ out of both the numerator and denominator, we get $${\displaystyle \frac{24+12}{4^2-4}}={\displaystyle \frac{4(6+3)}{4(4-1)}}={\displaystyle \frac{6+3}{4-1}}={\displaystyle \frac{9}{3}}=3.$$

   Answer: 3

2. Solution 1
   Solving for $k$, we have $k={\displaystyle \frac{10}{3}}$, so $${\displaystyle \frac{6}{5}}k-2={\displaystyle \frac{6}{5}}\times {\displaystyle \frac{10}{3}}-2={\displaystyle \frac{60}{15}}-2=4-2=2.$$ Solution 2
   By factoring and substituting, we get $${\displaystyle \frac{6}{5}}k-2={\displaystyle \frac{2}{5}}(3k)-2={\displaystyle \frac{2}{5}}(10)-2=4-2=2.$$

   Answer: 2

3. Solution 1
   From the information given, $E$ is the reflection in the $y$-axis of $A$ and $D$ is the reflection in the $y$-axis of $B$. Since the coordinates of $A$ are $(-4,6)$, this means the coordinates of $E$ are $(4,6)$. Similarly, since the coordinates of $B$ are $(-2,0)$, the coordinates of $D$ are $(2,0)$.

   

   The slope of segment $DC$ is ${\displaystyle \frac{0-4}{2-0}}=-2$. The slope of $ED$ is ${\displaystyle \frac{6-0}{4-2}}=3$, so the sum of the slopes is $-2+3=1$.

   Solution 2
   The slope of segment $AB$ is ${\displaystyle \frac{0-6}{-2-(-4)}}=-3$. Since $ED$ is the reflection of $AB$ in the $y$-axis, its slope is the negative of the slope of $AB$. This means the slope of $ED$ is $3$. Similarly, the slope of segment $BC$ is ${\displaystyle \frac{4-0}{0-(-2)}}=2$, so the slope of $DC$ is $-2$. Therefore, the sum of the slopes is $-2+3=1$.

   Answer: $1$

4. The odd numbers on the spinner are $1$, $3$, and $5$. The wedges labelled by $3$ and $5$ each take up ${\displaystyle \frac{1}{4}}$ of the spinner and so each will be spun with a probability of ${\displaystyle \frac{1}{4}}$. If we let the probability of spinning $1$ be $x$, then we have that the probability of spinning $2$ is $2x$. The probability of spinning either $1$ or $2$ is ${\displaystyle \frac{1}{4}}$, which means $x+2x={\displaystyle \frac{1}{4}}$ or $3x={\displaystyle \frac{1}{4}}$ so $x={\displaystyle \frac{1}{12}}$.

   Therefore, the probability of spinning an odd number is ${\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{12}}={\displaystyle \frac{7}{12}}$.

   Answer: ${\displaystyle \frac{7}{12}}$

5. Suppose $y=mx+b$ and $y=nx+c$ are the equations two lines. If the two lines have the same slope, then the only way they can intersect is if they have they are the same line. To see this, suppose $m=n$ and there is some $x$ such that $mx+b=nx+c$. Since $m=n$, this equation is the same as $b=c$, which means the lines only intersect if $b=c$, but this would make the lines identical.

   On the other hand, if $m\ne n$, then the two lines will intersect in exactly one point. To see this, suppose $m\ne n$ and $mx+b=nx+c$. Then $(m-n)x=c-d$, and since $m-n\ne 0$, we have $x={\displaystyle \frac{c-d}{m-n}}$.

   The six lines that Maggie graphs can be broken into two sets of three distinct but parallel lines. Each of the three lines with slope $1$ intersects each of the three lines with slope $-2$ exactly once by the previous paragraph. This gives a total of $3\times 3=9$ intersections.

   We must be sure that none of the points have been counted more than once. For this to have happened, we would need to have at least three of the six lines intersecting in the same point. There are only two possible slopes, so among any three of the lines, at least two must have the same slope. Therefore, if three of the lines intersect in the same point, then two distinct lines of equal slope intersect. From the previous paragraph, this cannot happen. Therefore, the nine points are distinct, so $n=9$.

   Answer: 9

6. The prime factorization of $60$ is $2^2\times 3\times 5$, so the prime factorization of ${60}^5$ is $$(2^2\times 3\times 5{)}^5=2^{10}{3}^5{5}^5.$$ Therefore, the divisors of ${60}^5$ are numbers of the form $2^a{3}^b{5}^c$ where $a$, $b$, and $c$ are integers with $0\le a\le 10$, $0\le b\le 5$, and $0\le c\le 5$.

   In order for a divisor to be a perfect square, each of $a$, $b$, and $c$ must be even.

   Therefore, the divisors of ${60}^5$ which are perfect squares are the numbers of the form $2^a{3}^b{5}^c$ where $a$ is equal to $0$, $2$, $4$, $6$, $8$, or $10$, $b$ is equal to $0$, $2$, or $4$, and $c$ is equal to $0$, $2$, or $4$. This gives a total of $6\times 3\times 3=54$ ways of choosing $a$, $b$, and $c$ so that $2^a{3}^b{5}^c$ is a perfect square. This includes the divisor with $a=b=c=0$ corresponding to the divisor $1$ which is a perfect square, so the answer is $54-1=53$.

   Answer: 53

7. Since $\sqrt[3]{27}=3$, the dimensions of the larger cube must be $3\times 3\times 3$.

   Therefore, each side of the cube has area $3\times 3=9$, so the total surface of the cube is made up of $3^2\times 6=54$ $1$ by $1$ squares. Each of these unit squares is a face of one of the unit cubes. Since ${\displaystyle \frac{1}{3}}$ of the surface area is red, this means ${\displaystyle \frac{54}{3}}=18$ of these unit squares must be red.

   For any unit cube, there are either 0, 1, 2, or 3 of its faces showing on the surface of the larger cube. The unit cube at the centre of the larger cube has none of its faces showing, the 6 unit cubes in the centres of the outer faces have exactly one face showing, the 12 unit cubes on the edge but not at a corner have two faces showing, one of each of two adjacent sides, and the 8 unit cubes at the corners each have three faces showing.

   This means that at most three faces of any unit cube are on the surface of the larger cube, so we cannot possibly have 18 of the unit squares being coloured red if there are fewer than ${\displaystyle \frac{18}{3}}=6$ red unit cubes. There are 8 unit cubes on the corners, so if we colour exactly 6 unit cubes red and the other 21 black, then arrange the cubes into a $3\times 3\times 3$ cube so that the 6 red unit cubes are at the corners, there will be exactly 18 of the unit squares on the surface coloured red. Therefore, the answer is $6$.

   Answer: 6

8. Since Gina’s rate is 7 minutes per kilometre, she runs ${\displaystyle \frac{1}{7}}$ kilometres every minute. Since there are $60$ seconds in a minute, $15$ seconds corresponds to $0.25$ minutes. As long as $t>0.25$ minutes, the distance Gina has travelled at time $t$ is equal to $${\displaystyle \frac{1}{7}}(t-0.25)={\displaystyle \frac{t-0.25}{7}}$$ For $t\le 0.25$ minutes, the distance travelled is $0$.

   The rate in minutes per kilometre that the watch will display at time $t$ is equal to $${\displaystyle \frac{\text{total number of minutes}}{\text{total number of kilometres}}}={\displaystyle \frac{t}{{\displaystyle \frac{t-0.25}{7}}}}={\displaystyle \frac{7t}{t-0.25}}$$ Note that this formula only works for $t>0.25$. We will now use this to find the times at which the watch displayed 7 minutes 30 seconds per kilometre and 7 minutes 5 seconds per kilometre.

   First, 7 minutes 30 seconds per kilometre is the same as ${\displaystyle \frac{15}{2}}$ minutes per kilometre, so the value of $t$ satisfying $${\displaystyle \frac{15}{2}}={\displaystyle \frac{7t}{t-0.25}}$$ is the time at which this is the rate displayed by the watch. This equation can be rearranged to $$15(t-{\displaystyle \frac{1}{4}})=14t$$ or $t={\displaystyle \frac{15}{4}}$ minutes.

   Next, 7 minutes 5 seconds is $7{\displaystyle \frac{5}{60}}=7{\displaystyle \frac{1}{12}}={\displaystyle \frac{85}{12}}$ minutes. To find the corresponding time, we set $${\displaystyle \frac{85}{12}}={\displaystyle \frac{7t}{t-0.25}}$$ and rearrange to get $$85(t-{\displaystyle \frac{1}{4}})=84t$$ so $t={\displaystyle \frac{85}{4}}$.

   This means that Gina was running for ${\displaystyle \frac{85}{4}}-{\displaystyle \frac{15}{4}}={\displaystyle \frac{70}{4}}$ minutes from the time when the app said 7 minutes 30 seconds per kilometre to the time when it said 7 minutes 5 seconds per kilometre.

   While in motion, she runs at a rate of 7 minutes per kilometre. This means she runs $${\displaystyle \frac{{\displaystyle \frac{70}{4}}}{7}}={\displaystyle \frac{10}{4}}=2.5\text{kilometres}$$ in this time period.

   Answer: 2.5 km

9. Connect $O$ to each of $A$, $C$, $D$, and $E$ as shown.

   

   Each of $OA$, $OC$, $OD$, and $OE$ are radii of the circle and so are equal. As well, $AC=CD=DE=EA$ since they are the side-lengths of a square, which means $\mathrm{△}AOC$, $\mathrm{△}COD$, $\mathrm{△}DOE$, and $\mathrm{△}EOA$ are all congruent. This means $$\mathrm{\angle }AOC=\mathrm{\angle }COD=\mathrm{\angle }DOE=\mathrm{\angle }EOA$$ and since the sum of these angles must be ${360}^{\circ }$, $$\mathrm{\angle }AOC=\mathrm{\angle }COD=\mathrm{\angle }DOE=\mathrm{\angle }EOA={\displaystyle \frac{{360}^{\circ }}{4}}={90}^{\circ }$$ By the Pythagorean theorem, $O{E}^2+O{D}^2=E{D}^2$. Since the radius of the circle is $\sqrt{2}$, we have ${\sqrt{2}}^2+{\sqrt{2}}^2=E{D}^2$ or $2+2=4=E{D}^2$, so $ED=2$ since $ED>0$.

   Now draw a line from $B$ through $O$ to meet $ED$ at $H$ and $AC$ at $J$.

   

   We know that $\mathrm{\angle }EOD={90}^{\circ }$, which means that $\mathrm{\angle }EOB+\mathrm{\angle }DOB={360}^{\circ }-{90}^{\circ }={270}^{\circ }$. Furthermore, $OE$, $OD$, and $OB$ are all radii of the circle and hence are equal and $BE=BD$ by construction, so $\mathrm{△}EOB$ and $\mathrm{△}DOB$ are congruent. This means $\mathrm{\angle }EOB=\mathrm{\angle }DOB$ and since their sum is ${270}^{\circ }$, we must have $\mathrm{\angle }EOB={\displaystyle \frac{{270}^{\circ }}{2}}={135}^{\circ }$.

   Since $BOH$ is a straight line, this means $\mathrm{\angle }EOH={180}^{\circ }-{135}^{\circ }={45}^{\circ }$. We previously established that $\mathrm{△}EOD$ is right and isosceles, which means $\mathrm{\angle }OED={45}^{\circ }$ as well. This means $\mathrm{\angle }OHE={180}^{\circ }-{45}^{\circ }-{45}^{\circ }={90}^{\circ }$.

   We now have that $\mathrm{△}EBH$ and $\mathrm{△}DBH$ are right triangles having a common hypotenuse length and a shared side, which means they are congruent. Therefore, $EH=HD$. It follows that $EH=1$ because $ED=2$.

   Line segments $AC$ and $DE$ are parallel because they are opposite sides of a square. Therefore, $\mathrm{\angle }BFJ=\mathrm{\angle }BEH$ and $\mathrm{\angle }BJF=\mathrm{\angle }BHE$. It follows that $\mathrm{△}BFJ\sim \mathrm{△}BEH$.

   Since $\mathrm{\angle }BHE={90}^{\circ }$, we also have that $AE\parallel BH$ which means $\mathrm{\angle }AEF=\mathrm{\angle }JBF$. Angles $\mathrm{\angle }AFE$ and $\mathrm{\angle }JFB$ are alternating and hence equal, and both $\mathrm{\angle }EAF$ and $\mathrm{\angle }BJF$ are right, which means $\mathrm{△}EFA\sim \mathrm{△}BFJ$. By similar reasoning, $\mathrm{△}DGC$ is also similar to $\mathrm{△}BFJ$, $\mathrm{△}BEH$, and $\mathrm{△}EFA$.

   Since $\mathrm{△}DGC\sim \mathrm{△}EFA$ and $AE=BC$, we have that $\mathrm{△}DGC\cong \mathrm{△}EFA$, which means $AF=GC$.

   Since $\mathrm{△}EFA\sim \mathrm{△}BEH$, we get that $${\displaystyle \frac{AF}{AE}}={\displaystyle \frac{HE}{HB}}={\displaystyle \frac{HE}{HO+OB}}={\displaystyle \frac{1}{1+\sqrt{2}}}$$ Since $AE=2$, this means $$AF={\displaystyle \frac{2}{1+\sqrt{2}}}={\displaystyle \frac{2}{1+\sqrt{2}}}\times {\displaystyle \frac{1-\sqrt{2}}{1-\sqrt{2}}}=2\sqrt{2}-2$$

   Since $AF=GC$, we have that $GC=2\sqrt{2}-2$ as well, and using that $AC=2$, we have $$FG=AC-AF-GC=2-(2\sqrt{2}-2)-(2\sqrt{2}-2)=6-4\sqrt{2}$$

   Since $AC\parallel ED$, we have that $\mathrm{△}BFG\sim \mathrm{△}BED$. This means their heights are in the same ratio as their sides. Therefore, $${\displaystyle \frac{BJ}{BH}}={\displaystyle \frac{FG}{ED}}$$ and since $BH=1+\sqrt{2}$, $FG=6-4\sqrt{2}$, and $ED=2$, this means $$BJ=BH{\displaystyle \frac{FG}{ED}}=(1+\sqrt{2}){\displaystyle \frac{6-4\sqrt{2}}{2}}=(1+\sqrt{2})(3-2\sqrt{2})=-1+\sqrt{2}$$ Therefore, the area of $\mathrm{△}BFG$ is $${\displaystyle \frac{1}{2}}(FG)(BJ)={\displaystyle \frac{1}{2}}(6-4\sqrt{2})(-1+\sqrt{2})=(3-2\sqrt{2})(-1+\sqrt{2})=-7+5\sqrt{2}.$$ The answer is $(-7,5)$.

   Answer: $(-7,5)$

10. By adding $-2$, $8$, and $-24$ to both sides of the first, second, and third equations, respectively, we get $$\begin{array}{rl}ab+2a-b-2& =56\\ bc+4b+2c+8& =308\\ cd-6c+4d-24& =77\end{array}$$ The left sides of these equations can now be factored. The integers on the right sides of the equations have been factored into products of primes as well. $$\begin{array}{rl}(a-1)(b+2)& =2^3\times 7\\ (b+2)(c+4)& =2^2\times 7\times 11\\ (c+4)(d-6)& =7\times 11\end{array}$$ We are given that $c$ and $d$ are positive integers which means $c+4$ is a positive integer and $d-6$ is an integer. This means $c+4$ is a positive divisor of $77$, so we must have that $c+4$ is one of $1$, $7$, $11$, and $77$. We cannot have $c+4=1$ since this would mean $c=-3$ which is negative.

    If $c+4=7$, then $$b+2={\displaystyle \frac{2^2\times 7\times 11}{7}}=2^2\times 11$$ but this would mean $$a-1={\displaystyle \frac{2^3\times 7}{2^2\times 11}}={\displaystyle \frac{2\times 7}{11}}$$ which is not an integer, but $a-1$ must be an integer because $a$ is an integer. Therefore, we cannot have $c+4=7$.

    If $c+4=11$, then $b+2={\displaystyle \frac{2^2\times 7\times 11}{11}}=2^2\times 7$, and so $a-1={\displaystyle \frac{2^3\times 7}{2^2\times 7}}=2$. Also, $d-6={\displaystyle \frac{7\times 11}{11}}=7$, so we have $$(a-1,b+2,c+4,d-6)=(2,28,11,7)$$ which gives $$(a,b,c,d)=(3,26,7,13)$$ If $c+4=77$, then $d-6=1$, $b+2={\displaystyle \frac{2^2\times 7\times 11}{77}}=4$, and $a-1={\displaystyle \frac{2^3\times 7}{4}}=14$. Therefore, $$(a-1,b+2,c+4,d-6)=(14,4,77,1)$$ which gives $$(15,2,73,7)$$

    Answer: $(3,26,7,13)$, $(15,2,73,7)$

Team Problems

1. Since $11<16<19$, we have $\sqrt{11}<\sqrt{16}<\sqrt{19}$ which is the same as $\sqrt{11}<4<\sqrt{19}$.

   Answer: 4

2. By factoring $3^3$ out of the numerator, we have $$\begin{array}{rl}{\displaystyle \frac{3^5-3^4}{3^3}}& ={\displaystyle \frac{3^3(3^2-3)}{3^3}}\\ & =3^2-3\\ & =9-3\\ & =6\end{array}$$

   Answer: 6

3. Let the measure of the smallest interior angle be $x^{\circ }$. Then the other three interior angles measure $2x^{\circ }$, $3x^{\circ }$, and $4x^{\circ }$.
   The interior angles of a quadrilateral sum to ${360}^{\circ }$, which means $${360}^{\circ }=x^{\circ }+2x^{\circ }+3x^{\circ }+4x^{\circ }=10x^{\circ }$$ Therefore, $360=10x$, so $x=36$.

   Answer: ${36}^{\circ }$

4. The numbers being summed are the multiples of $101$ from $10\times 101$ through $20\times 101$. Thus, if we set $S$ to be the sum in question, we can factor out $101$ to get $$S=101(10+11+12+13+14+15+16+17+18+19+20)$$ It can easily be checked by hand that $10+11+\cdots +19+20=165$. Alternatively, using the fact that for any positive integer $n$ we have $$1+2+3+4+\cdots +(n-1)+n={\displaystyle \frac{n(n+1)}{2}}$$ the sum can be computed as $$\begin{array}{rl}10+11+\cdots +19+20& =(1+2+3+\cdots +19+20)-(1+2+\cdots +8+9)\\ & ={\displaystyle \frac{20(21)}{2}}-{\displaystyle \frac{9(10)}{2}}\\ & =10\times 21-9\times 5\\ & =210-45\\ & =165\end{array}$$ Therefore, $S=101\times 165=(100+1)\times 165=16500+165=16665$.

   Answer: $16665$

5. The number $10004444$ is an eight-digit number having exactly four digits which are 4. We will show that this is the number we seek by showing that any other such number is larger than 10004444.
   Suppose we have an eight-digit number having exactly four digits which are 4. If a digit of 4 occurs as any of the first (leftmost) four digits, the number will be bigger than 10004444 (note that 1 is the smallest that the first digit can be since the number has eight digits).
   Otherwise, the four digits that are 4 occur as the last four digits. The smallest eight digit number ending in 4444 is 10004444.

   Answer: 10004444

6. There were 200 visitors on Saturday, so there were 100 visitors the day before. Since tickets cost $\mathrm{\$}9$ on Fridays, the total money collected on Friday was $\mathrm{\$}900$.

   Therefore, the amount of money collected from ticket sales on the Saturday was $\frac{4}{3}(\mathrm{\$}900)=\mathrm{\$}1200$.

   Since there were 200 visitors on Saturday, the price of tickets on Saturday on that particular Saturday was $\frac{\mathrm{\$}1200}{200}=\mathrm{\$}6$. Therefore, the value of $k$ is 6.

   In fact, if all conditions are kept the same in this problem but the number of ticket sales on Saturday changes, the answer will still be $6$. You might like to try to work out why this is true.

   Answer: 6

7. An integer is divisible by both $5$ and $9$ exactly when it is divisible by $45$. Since we seek a number having only even digits, its last digit is one of $0,2,4,6,$ or $8$, so the number itself is even. Thus, we are looking for an even multiple of $45$. An even number is a multiple of $45$ exactly when it is a multiple of $90$.
   Therefore, we are looking for the smallest 4-digit multiple of 90 that has only even digits.
   The smallest 4-digit multiple of 90 is 1080, but the first digit of this number is 1, which is odd.
   Each of the next 10 multiples of 90 has a first digit equal to 1, so the number we seek must be larger than 2000. The multiples of 90 that are at least 2000 are $$2070,2160,2250,2340,2430,2520,2610,2700,2790,2880,2970$$ and the only number in this list that has all even digits is 2880. Therefore, 2880 is the smallest 4-digit number that is a multiple of 5, a multiple of 9, and has only even digits.

   Answer: 2880

8. The larger semicircle has diameter $64$ units. The length of a semicircular arc is $\pi {\displaystyle \frac{d}{2}}$ where $d$ is the diameter of the circle. This means the length of the large semicircular arc is $$\pi {\displaystyle \frac{d}{2}}=\pi {\displaystyle \frac{64}{2}}=32\pi$$ The diameter of each small semicircular arc is ${\displaystyle \frac{64}{4}}=16$ units, so each small semicircular arc has length $$\pi {\displaystyle \frac{16}{2}}=8\pi$$ There are four of these small arcs, so the perimeter of the figure is $$32\pi +4(8\pi )=64\pi$$

   Answer: $64\pi$

9. By factoring and using exponent rules, we have ${20}^{10}=(2\times 10{)}^{10}=2^{10}\times {10}^{10}$. Therefore, ${20}^{10}=1024\times {10}^{10}$, which is the integer 1024 followed by ten zeros. Thus, ${20}^{10}$ has eleven digits that are $0$.

   Answer: 11

10. Using exponent laws, we have $$\begin{array}{rl}{\displaystyle \frac{{16}^{x+1}}{8^{2y-1}}}& ={\displaystyle \frac{2^{4(x+1)}}{2^{3(2y-1)}}}\\ & =2^{(4x+4)-(6y-3)}\\ & =2^{-2(3y-2x)+7}\end{array}$$ Using the given condition that $3y-2x=4$, we get $$\begin{array}{rl}{\displaystyle \frac{{16}^{x+1}}{8^{2y-1}}}& =2^{-2(3y-2x)+7}\\ & =2^{-2(4)+7}\\ & =2^{-1}\\ & ={\displaystyle \frac{1}{2}}\end{array}$$

    Answer: ${\displaystyle \frac{1}{2}}$

11. The sum of the areas of the triangles is equal to the area of $\mathrm{△}A{Q}_1{Q}_7$. This is because the triangles share vertex $A$, the base of $\mathrm{△}A{Q}_2{Q}_3$ begins where the base of $\mathrm{△}A{Q}_1{Q}_2$ ends, the base of $\mathrm{△}A{Q}_3{Q}_4$ begins where the base of $\mathrm{△}A{Q}_2{Q}_3$ ends, and so on with all of $Q_1$, $Q_2$, $Q_3$, $Q_4$, $Q_5$, $Q_6$, and $Q_7$ lie on the $x$-axis.
    The height of $\mathrm{△}A{Q}_1{Q}_y$ is the $y$-coordinate of $A$, which is $2$. The first seven primes are $2,3,5,7,11,13,17$, so $p_1=2$ and $p_7=17$. The base of $\mathrm{△}A{Q}_1{Q}_7$ has length $p_7-p_1=17-2=15$. Therefore, the sum of the areas of the triangles is $${\displaystyle \frac{1}{2}}\times 15\times 2=15$$

    Answer: 15

12. Since $90$ is a multiple of $3$, there are ${\displaystyle \frac{90}{3}}=30$ multiples of $3$ in the list. Since $90$ is a multiple of $5$, there are ${\displaystyle \frac{90}{5}}=18$ multiples of $5$ in the list. A multiple of $5$ is a multiple of $3$ exactly when it is a multiple of $15$, which means the multiples of $15$ were already crossed out when the multiples of $3$ were crossed out. There are ${\displaystyle \frac{90}{15}}=6$ multiples of $15$ in the list, so this means there are $18-6=12$ multiples of $5$ remaining in the list once the multiples of $3$ have been crossed out. Therefore, there are $90-30-12=48$ numbers in the list once all of the multiples of $3$ and $5$ have been crossed out.

    Answer: 48

13. By completing the square, we have $y=-{\displaystyle \frac{1}{4}}(x-10{)}^2+4$, which means the vertex of the parabola is at $A(10,4)$.
    If we set $y=0$, we have $-{\displaystyle \frac{1}{4}}(x-10{)}^2+4=0$ or $-{\displaystyle \frac{1}{4}}(x-10{)}^2=-4$, so $(x-10{)}^2=16$. Taking square roots of both sides, we have $x-10=\pm 4$ or $x=10\pm 4$. Therefore, the parabola passes through $(6,0)$ and $(14,0)$. From the information given, this means the coordinates of $B$ are $(6,0)$ and the coordinates of $F$ are $(14,0)$.
    The second parabola has its vertex at $B(6,0)$, so its equation is $y=a(x-6{)}^2$ for some real number $a$. Using the fact that it passes through $A(10,4)$, we get $4=a(10-6{)}^2=16a$, or $a={\displaystyle \frac{1}{4}}$.
    Therefore, the equation of the second parabola is $y={\displaystyle \frac{1}{4}}(x-6{)}^2$. Substituting $x=0$, we get $y={\displaystyle \frac{1}{4}}(-6{)}^2={\displaystyle \frac{36}{4}}=9$. Therefore, the second parabola crosses the $y$-axis at $D(0,9)$.

    Answer: $(0,9)$

14. Since the total mass of the first three fish was $1.5$ kg, the average mass of the first three fish was $0.5$ kg. Let $M$ be the total mass of all of the fish. Since the average mass or the first three fish was the same as the average mass of of all of the fish,. This means ${\displaystyle \frac{M}{21}}=0.5$ kg or $M=10.5$ kg. We can find the largest possible mass of any individual fish by determining the smallest possible mass of any twenty of the fish, then subtracting this mass from $10.5$ kg.

    We first note that the answer is not $0.5$ kg. This is because it is possible, for example, that 19 of the fish (including the first three) weighed $0.5$ kg each and the other two weighed $0.2$ kg and $0.8$ kg. This distribution of weights satisfies all of the conditions.

    The largest fish Jeff could have caught would therefore be if 17 of the 18 fish caught after the first three weighed the minimum of $0.2$ kg. This would mean the first three in addition to these 17 fish weigh $1.5$ kg$+17(0.2)$ kg$=4.9$ kg. The remaining fish would weigh $10.5$ kg$-4.9$ kg$=5.6$ kg.

15. We will assume $a\le b\le c\le d$. That is, that $d$ is the largest of the integers. First, observe that if $c\ge 6$, then $d\ge c\ge 6$, which means $a^2+b^2+c^2+d^2>c^2+d^2\ge 6^2+6^2=72>70$. We are assume $a^2+b^2+c^2+d^2=70$, so this means we cannot have $c\ge 6$. It follows that $a,b,$ and $c$ are all at most $5$.
    We will work through the possible values of $d$ for which $a,b,c,$ and $d$ can be chosen with $a\le b\le c\le d$ and $a^2+b^2+c^2+d^2=70$.
    Since $9^2=81>70$, we have that $d\le 8$.
    If $d=8$, then $a^2+b^2+c^2=70-8^2=6$. The only perfect squares that do not exceed $6$ are $1^2=1$ and $2^2=4$. It can easily be checked that $1^2+1^2+2^2=6$ is the only way to express $6$ as a sum of three positive perfect squares. Thus, we could have $(a,b,c,d)=(1,1,2,8)$ and this is the only possibility with $d=8$.
    If $d=7$, we have $a^2+b^2+c^2=70-7^2=21$. The squares less than or equal to $21$ are $1^2=1$, $2^2=4$, $3^2=9$, and $4^2=16$. The only way to write $21$ as a sum of numbers in this list is $1^2+2^2+4^2$. Thus, $(a,b,c,d)=(1,2,4,7)$ is a the only possibility with $d=7$.
    If $d=6$, we have $a^2+b^2+c^2=70-6^2=34$. If $c=5$, then $a^2+b^2=34-5^2=9$, and it can be checked that $9$ cannot be expressed as a sum of positive perfect squares. Therefore, $c\ne 5$, and since $c\le 5$, this means $c\le 4$. If $c<4$, then $c\le 3$, and since $a\le b\le c$, we have $a^2+b^2+c^2\le 3^2+3^2+3^2=27<34$. This means we must have $c=4$, so $a^2+b^2=34-4^2=18$. The only way to write $18$ as a sum of positive squares is $18=3^2+3^2$, so this means $(a,b,c,d)=(3,3,4,6)$ is the only possibility with $d=6$.
    If $d=5$, then $a^2+b^2+c^2=70-5^2=45$. If $c=5$, then $a^2+b^2=45-5^2=20$. It can be checked that $2^2+4^2=20$ is the only way to write $20$ as a sum of two positive perfect squares. This means $(a,b,c,d)=(2,4,5,5)$ is a possibility, and it is the only possibility with $c=d=5$.
    If $c=4$, then $a^2+b^2=45-16=29$, and it can be checked that $29$ cannot be written as the sum of two positive perfect squares, so $c\ne 4$. Thus, $c\le 3$, so $a^2+b^2+c^2\le 3(3^2)=27<45$. Therefore, $(a,b,c,d)=(2,4,5,5)$ is the only possibility with $d=5$.
    Finally, if $d\le 4$, then $a,b,c,$ and $d$ are each less than or equal to $4$. This would mean $a^2+b^2+c^2+d^2\le 4(4^2)=64<70$, so we cannot have $d\le 4$.
    Therefore, $(a,b,c,d)$ must be one of $$(1,1,2,8),(1,2,4,7),(3,3,4,6),(2,4,5,5)$$ and the sums of the entries in these quadruples are $12$, $14$, $16$, and $16$, respectively. Therefore, the largest possible sum is $a+b+c+d=16$ (and it is achievable in two different ways).

    Answer: 16

16. Let $C$ be the common centre of the two circles, $R$ be the radius of the larger circle, and $r$ be the radius of the smaller circle. Since $AB$ is tangent to the smaller circle, we have that $AC\perp AB$. This means $\mathrm{△}ABC$ is right-angled at $A$. By the Pythagorean theorem and since $AB=5$, we then have $A{C}^2+A{B}^2=B{C}^2$ or $r^2+5^2=R^2$. The area of the shaded region is equal to the area of the larger circle minus the area of the smaller circle, or $\pi R^2-\pi r^2=\pi (R^2-r^2)$. Rearranging $r^2+25=R^2$, we have $R^2-r^2=25$, so the area of the shaded region is $\pi (R^2-r^2)=25\pi$.

    Answer: $25\pi$

17. A quadratic function of the form $f(x)=a{x}^2+bx+c$ with $a>0$ attains its minimum at $x=-{\displaystyle \frac{b}{2a}}$. If $a<0$, the function attains its maximum at $-{\displaystyle \frac{b}{2a}}$.
    For $f(x)=x^2-2mx+8m+4$, the minimum occurs at $x=-{\displaystyle \frac{-2m}{2}}=m$. The minimum value of $f(x)$ is therefore $$f(m)=m^2-2m(m)+8m+4=-m^2+8m+4.$$ This is a quadratic function in the variable $m$. The leading coefficient is negative, so the maximum occurs at $m=-{\displaystyle \frac{8}{2(-1)}}=4$. Therefore, the maximum of these minimum values is $-4^2+8(4)+4=20$.

    Answer: 20

18. We will require the following fact. If $a$ and $r$ are real numbers with $-1<r<1$, then the sum of the geometric series $$a+ar+a{r}^2+a{r}^3+a{r}^4+\cdots$$ is ${\displaystyle \frac{a}{1-r}}$.
    We will compute this sum by finding the sum $t_1+t_3+t_5+\cdots$ and $t_2+t_4+t_6+\cdots$ and adding the results. For the first sum, we have $${\displaystyle \frac{1}{7}}+{\displaystyle \frac{1}{7^3}}+{\displaystyle \frac{1}{7^5}}+\cdots$$ which can be rewritten as $${\displaystyle \frac{1}{7}}+{\displaystyle \frac{1}{7}}{\displaystyle \frac{1}{7^2}}+{\displaystyle \frac{1}{7}}{\left({\displaystyle \frac{1}{7^2}}\right)}^2+{\displaystyle \frac{1}{7}}{\left({\displaystyle \frac{1}{7^2}}\right)}^3+\cdots .$$ This is a geometric series with $a={\displaystyle \frac{1}{7}}$ and $r={\displaystyle \frac{1}{49}}$. We also have that $-1<{\displaystyle \frac{1}{49}}<1$, so the sum of the series above is $${\displaystyle \frac{{\displaystyle \frac{1}{7}}}{1-{\displaystyle \frac{1}{49}}}}={\displaystyle \frac{{\displaystyle \frac{49}{7}}}{49-1}}={\displaystyle \frac{7}{48}}.$$ We now compute $t_2+t_4+t_6+\cdots$. Similar to before, we can write the sum as $${\displaystyle \frac{2}{7^2}}+{\displaystyle \frac{2}{7^2}}{\displaystyle \frac{1}{7^2}}+{\displaystyle \frac{2}{7^2}}{\left({\displaystyle \frac{1}{7^2}}\right)}^2+{\displaystyle \frac{2}{7^2}}\left({\displaystyle \frac{1}{7^2}}\right)+\cdots$$ which is a geometric series with $a={\displaystyle \frac{2}{7^2}}$ and $r={\displaystyle \frac{1}{7^2}}$. Thus, its sum is $${\displaystyle \frac{{\displaystyle \frac{2}{49}}}{1-{\displaystyle \frac{1}{49}}}}={\displaystyle \frac{2}{49-1}}={\displaystyle \frac{2}{48}}.$$ The sum $t_1+t_2+t_3+\cdots$ is therefore ${\displaystyle \frac{7}{48}}+{\displaystyle \frac{2}{48}}={\displaystyle \frac{9}{48}}={\displaystyle \frac{3}{16}}$.

    Answer: ${\displaystyle \frac{3}{16}}$

19. We need to choose 2 points out of 24, and there are ${\displaystyle (\genfrac{}{}{0ex}{}{24}{2})}={\displaystyle \frac{24(23)}{2}}=12\times 23=276$ ways to do this. Two points will form a triangle together with point $A$ unless all three points lie on a common line. Therefore, we can get the answer by counting the number of ways to choose two points so that they and $A$ all lie on the same line and subtract this number from 276.
    We now draw every line through $A$ and some other point in the diagram. Note that many of these lines will be the same. For example, drawing a line through $A$ and any of the other four points in the bottom row gives the same line regardless of which of the four points is chosen. There are 13 lines in total:

    

    Eight of these lines pass through only one of the points other than $A$, so there cannot be a pair of points different from $A$ on such a line that both lie on a line through $A$. Eliminating these lines from the diagram, we have

    

    There are five lines remaining. Two of them contain exactly two points other than $A$, so each of these identifies one pair of points that, together with $A$, do not form a triangle.
    Each of the other three remaining lines contains four points other than $A$, so each of these three lines identifies ${\displaystyle (\genfrac{}{}{0ex}{}{4}{2})}=6$ pairs of points that, together with $A$, do not form a triangle. Therefore, there are $276-2(1)-3(6)=256$ ways to choose two points from the array that, together with $A$, form a triangle.

    Answer: 256

20. We will imagine that the track has been broken into 12 equal units. With this convention, Isabelle will begin to run once Lyla has travelled 4 units, and Isabelle will always run 5 units in the amount of time it takes for Lyla to run 4 units.
    When Lyla has run a total of 4 units, Isabelle has run 0 units. At this time, Isabelle starts to run, so when Lyla has run $4+4=8$ units, Isabelle will have run 5 units.
    When Lyla has run $8+4=12$ units, Isabelle will have run $5+5=10$ units.
    When Lyla has run $12+4=16$ units, Isabelle will have run $10+5=15$ units, and when Lyla has run $16+4=20$ units, Isabelle will have run $15+5=20$ units. At this point, Isabelle passes Lyla for the first time.
    Since Isabelle runs faster than Lyla, Isabelle will always have run farther than Lyla from this point on. The points at which Isabelle passes Lyla will occur when Isabelle has travelled an integer number of laps, or a multiple of 12 units farther than Lyla.
    This means the 5th times Isabelle passes Lyla will occur when Isabelle has gained an additional $4\times 12=48$ units on Lyla.
    Isabelle gains one unit on Lyla for every 4 units Lyla runs, which means that Isabelle will gain 48 units on Lyla when Lyla runs $4\times 48=192$ units.
    Therefore, when Isabelle passes Lyla for the 5th time, Lyla has run a total of $192+20=212$ units. There are 12 units in a lap, so this means Lyla will have run ${\displaystyle \frac{212}{12}}=17{\displaystyle \frac{8}{12}}$ laps.
    When Isabelle passes Lyla for the 5th time, Lyla will have completed 17 laps.

    Answer: 17

21. Using a log rule, we have that $\log x^2=2\log x$, so $$f(x)=2{\sin }^2(\log (x))+\cos (2\log (x))-5.$$ Using the trigonometric identity $\cos 2x={\cos }^{2}x-{\sin }^{2}x$, we get $$f(x)=2{\sin }^2(\log (x))+{\cos }^2(\log (x))-{\sin }^2(\log (x))-5$$ which simplifies to $$f(x)={\sin }^2(\log (x))+{\cos }^2(\log (x))-5$$ and finally using the identity ${\sin }^{2}u+{\cos }^{2}u=1$ for any real number $u$, we have $$f(x)=1-5=-4$$ and this holds for all $x>0$. Therefore, $f(\pi )=-4$.

    Answer: $-4$

22. An important observation for this problem is that every integer is either a multiple of $3$, one more than a multiple of $3$, or two more than a multiple of $3$. Suppose $x+y+z$ is a multiple of $3$. If $x$ is a multiple of $3$, then $y+z$ must also be a multiple of $3$. There are three ways this can happen: both $y$ and $z$ are multiples of $3$, $y$ is one more than a multiple of $3$ and $z$ is two more than a multiple of $3$, or $y$ is two more than a multiple of $3$ and $z$ is one more than a multiple of $3$. If $x$ is one more than a multiple of $3$, then $y+z$ must be two more than a multiple of $3$. Again, there are three ways that this can happen. We summarize the possible remainders after $x$, $y$, and $z$ are divided by $3$ (either 0, 1, or 2) that will lead to $x+y+z$ being a multiple of $3$:

    $\mathit{x}$	$\mathit{y}$	$\mathit{z}$
    0	0	0
    0	1	2
    0	2	1
    1	0	2
    1	1	1
    1	2	0
    2	0	1
    2	1	0
    2	2	2

    Of the numbers from $1$ to $10$ inclusive, $3$, $6$, and $9$ are multiple of $3$ (a total of three), $1$, $4$, $7$, and $10$ are one more than a multiple of $3$ (a total of 4), and $2$, $5$, and $8$ are two more than a multiple of $3$ (a total of 3). Thus, for example, there are $3\times 3\times 3=27$ triples $(x,y,z)$ so that $x+y+z$ is a multiple of $3$ and each of $x$, $y$, and $z$ is a multiple of $3$. Computing the total for each row in the table above, we have

    $\mathit{x}$	$\mathit{y}$	$\mathit{z}$
    0	0	0	$3\times 3\times 3$	$=27$
    0	1	2	$3\times 4\times 3$	$=36$
    0	2	1	$3\times 3\times 4$	$=36$
    1	0	2	$4\times 3\times 3$	$=36$
    1	1	1	$4\times 4\times 4$	$=64$
    1	2	0	$4\times 3\times 3$	$=36$
    2	0	1	$3\times 3\times 4$	$=36$
    2	1	0	$3\times 4\times 3$	$=36$
    2	2	2	$3\times 3\times 3$	$=27$

    which means the total is $2(27)+64+6(36)=54+64+216=334$

    Answer: $334$

23. We begin with a diagram. Suppose $C$ has coordinates $(a,b)$ and is the vertex of the rectangle to the right of $B$ and on the line $x+2y-18=0$. We let $E$ be the point with coordinates $(a,2)$ and $F$ be the point with $x$-coordinate $a$ and lying on the line through $B$ and $D$.

    

    The area of rectangle $ABCD$ will be twice that of $\mathrm{△}BCD$, and we can compute the area of $\mathrm{△}BCD$ by subtracting the areas of $\mathrm{△}BEC$ and $\mathrm{△}CFD$ from the area of $\mathrm{△}BEF$.
    To compute these areas, we need to find the values of $a$, $b$, and $c$.
    First, we find the equation of the line through $B$ and $D$. Since this line passes through $(4,2)$ and $(12,8)$, it has slope ${\displaystyle \frac{8-2}{12-4}}={\displaystyle \frac{6}{8}}={\displaystyle \frac{3}{4}}$. Thus, its equation is of the form $y={\displaystyle \frac{3}{4}}x+k$ for some $k$. Substituting $(x,y)=(4,2)$, we have $2={\displaystyle \frac{3}{4}}(4)+k$ so $k=-1$, which means the equation of the line is $y={\displaystyle \frac{3}{4}}x-1$.
    To find the coordinates of $C$, we note that the point of intersection of the diagonals of a rectangle is equidistant from all four of its vertices.
    The intersection of the diagonals will be the intersection of the lines $y={\displaystyle \frac{3}{4}}x-1$ and $x+2y-18=0$. Substituting, we have $x+2({\displaystyle \frac{3}{4}}x-1)=18$ or $x+{\displaystyle \frac{3}{2}}x-2=18$. Multiplying through by $2$, we have $2x+3x=40$, so $x=8$ from which we get $y={\displaystyle \frac{3}{4}}(8)-1=5$. The coordinates of the centre of the rectangle are $(8,5)$. The distance from this point to point $B$ is $$\sqrt{(8-4{)}^2+(5-2{)}^2}=\sqrt{16+9}=\sqrt{25}=5.$$ Therefore, $C$ is the point on the line $x+2y=18$ that is $5$ units away from $(8,5)$. The condition on the distance gives us $$\sqrt{(a-8{)}^2+(b-5{)}^2}=5$$ or $(a-8{)}^2+(b-5{)}^2=25$. Since $(a,b)$ is on $x+2y=18$, we also have $a+2b=18$ which rearranges to $a-8=10-2b=2(5-b)$. Substituting leads to $$[2(5-b){]}^2+(b-5{)}^2=25$$ which can be rearranged to get $$5(b-5{)}^2=25.$$ Thus, $(b-5{)}^2=5$, so $b=5\pm \sqrt{5}$. The point $C$ is to the right of the centre of the rectangle, and since the line connecting $A$ and $C$ has negative slope, this means $b$ should be less than $5$. Therefore, $b=5-\sqrt{5}$. Substituting into $a+2b=18$, we have $a=18-2(5-\sqrt{5})=8+2\sqrt{5}$.
    We now know the coordinates of $C$, and therefore, the coordinates of $E$ are $(8+2\sqrt{5},2)$. To find the $y$-coordinate of $F$, we need only plug the known value of $a$ in for $x$ in the equation $y={\displaystyle \frac{3}{4}}x-1$ to get $c={\displaystyle \frac{3}{4}}(8+2\sqrt{5})-1=5+{\displaystyle \frac{3}{2}}\sqrt{5}$.
    The area of $\mathrm{△}BEF$ is $$\begin{array}{rl}{\displaystyle \frac{1}{2}}BE\cdot EF& ={\displaystyle \frac{1}{2}}(a-4)(c-2)\\ & ={\displaystyle \frac{1}{2}}(8+2\sqrt{5}-4)(5+{\displaystyle \frac{3}{2}}\sqrt{5}-2)\\ & =(2+\sqrt{5})(3+{\displaystyle \frac{3}{2}}\sqrt{5})\\ & =6+3\sqrt{5}+3\sqrt{5}+{\displaystyle \frac{15}{2}}\\ & ={\displaystyle \frac{27}{2}}+6\sqrt{5}\end{array}$$

    The area of $\mathrm{△}BEC$ is $$\begin{array}{rl}{\displaystyle \frac{1}{2}}BE\cdot EC& ={\displaystyle \frac{1}{2}}(a-4)(b-2)\\ & ={\displaystyle \frac{1}{2}}(8+2\sqrt{5}-4)(5-\sqrt{5}-2)\\ & =(2+\sqrt{5})(3-\sqrt{5})\\ & =6-2\sqrt{5}+3\sqrt{5}-5\\ & =1+\sqrt{5}\end{array}$$

    Finally, since the height of $\mathrm{△}CFD$ is the horizontal distance from $D$ to $FC$ which is $a-12$, we have that the area of $\mathrm{△}CFD$ is $$\begin{array}{rl}{\displaystyle \frac{1}{2}}FC(a-12)& ={\displaystyle \frac{1}{2}}(c-b)(a-12)\\ & ={\displaystyle \frac{1}{2}}(5+{\displaystyle \frac{3}{2}}\sqrt{5}-(5-\sqrt{5}))(8+2\sqrt{5}-12)\\ & ={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{5}{2}}\sqrt{5}\right)(-4+2\sqrt{5})\\ & =-5\sqrt{5}+{\displaystyle \frac{25}{2}}\end{array}$$

    Using these three ares, we can compute the area of $\mathrm{△}BCD$ as $$({\displaystyle \frac{27}{2}}+6\sqrt{5})-(1+\sqrt{5})-(-5\sqrt{5}+{\displaystyle \frac{25}{2}})=10\sqrt{5}$$ Thus, the area of rectangle $ABCD$ is $2(10\sqrt{5})=20\sqrt{5}$.

    Answer: $20\sqrt{5}$

24. By clearing denominators, the equation becomes $$2x(x^2+x+3)+3x(x^2+5x+3)=(x^2+5x+3)(x^2+x+3)$$ which can be expanded to get $$2x^3+2x^2+6x+3x^3+15x^2+9x=x^4+6x^3+11x^2+18x+9$$ and then rearranged to $$x^4+x^3-6x^2+3x+9=0$$ Recognizing that $x=-3$ is a solution to the equation above, $x+3$ must be a factor of the polynomial on the left, so we can factor to get $$(x+3)(x^3-2x^2+3)=0$$ Next, $x=-1$ is a root of $x^3-2x^2+3$, so we can further factor $x+1$ from the cubic on the left side of the above equation to get $$(x+3)(x+1)(x^2-3x+3)=0$$ Using the quadratic formula, the roots of the polynomial $x^2-3x+3$ are $$x={\displaystyle \frac{3+\sqrt{(-3{)}^2-4(3)}}{2}}={\displaystyle \frac{3+\sqrt{-3}}{2}}\text{and}{\displaystyle \frac{3-\sqrt{-3}}{2}}$$ neither of which are real. Therefore, the only real values of $x$ that can possibly satisfy the equation are $-3$ and $-1$. It is easily check that both $x=-1$ and $x=3$ satisfy the original equation, so the answer is $(-1)+(-3)=-4$.

    Answer: $-4$

25. We will suppose $O$ is at the origin, which gives rise to coordinates $A(-30,30)$, $B(30,30)$, $C(30,-30)$, and $D(-30,-30)$. Additionally, we will have $E(-30,0)$.

    

    The line passing through $A$ and $C$ also passes through the origin and has slope $m={\displaystyle \frac{-30-30}{30-(-30)}}={\displaystyle \frac{-60}{60}}=-1$, and hence has equation $y=-x$. The line through $B$ and $E$ has slope $m={\displaystyle \frac{30-0}{30-(-30)}}={\displaystyle \frac{30}{60}}={\displaystyle \frac{1}{2}}$ and passes through $(30,30)$, so if its equation is $y={\displaystyle \frac{1}{2}}x+b$ for some real number $b$. Substituting $(x,y)=(30,30)$, we have $30={\displaystyle \frac{1}{2}}(30)+b$ or $b=15$. Thus, the line passing through both $B$ and $E$ has equation $y={\displaystyle \frac{1}{2}}x+15$.

    We can now compute the coordinates of points $G$ and $H$. Since $H$ is the intersection of line segments $AC$ and $BE$, the $x$-coordinate of $H$ satisfies $-x={\displaystyle \frac{1}{2}}x+15$ or $-15={\displaystyle \frac{3}{2}}x$, which can be solved to get $x=-10$. Since $H$ lies on the line $y=-x$, this means $H$ has coordinates $(-10,10)$.

    The circle has diameter equal to the side length of the square, which is given to be $60$. Therefore, its radius is ${\displaystyle \frac{60}{2}}=30$, which means the equation of the circle is $x^2+y^2={30}^2$. Points $G$ and $E$ are the points of intersection with the circle and the line $y={\displaystyle \frac{1}{2}}x+15$, which means their $x$-coordinates can be found by substituting ${\displaystyle \frac{1}{2}}x+15$ in the equation of the circle for $y$ and solving for $x$. That is, the $x$-coordinates of $E$ and $G$ are the solutions to the equation $$x^2+{({\displaystyle \frac{1}{2}}x+15)}^2=900$$ which can be expanded to get $$x^2+{\displaystyle \frac{1}{4}}{x}^2+15x+225=900.$$ Multiplying through by $4$ and rearranging, we get $$5x^2+60x-2700=0$$ and after dividing through by $5$ we have $$x^2+12x-540=0.$$ We know that the $x$-coordinate of $E$ is a root, which means $x+30$ must be a factor of the polynomial on the right, which leads to $$(x+30)(x-18)=0.$$ Thus, the other solution is $x=18$, so the $x$-coordinate of $G$ must be $18$. Substituting into $y={\displaystyle \frac{1}{2}}x+15$, we get that $y={\displaystyle \frac{18}{2}}+15=24$, so $G$ has coordinates $(18,24)$.

    We next observe that the shaded region bund by the circle and the line segments $AF$ and $AB$ is congruent to the region bound by the circle, $AF$, and $AE$. Thus, the shaded region is equal to the sum of the areas of $\mathrm{△}HGO$ and $\mathrm{△}AHE$.

    If we take $AE$ as the base of $\mathrm{△}AHE$, then its height will be the horizontal distance from $H$ to line segment $AE$. This is equal to the difference between the $x$ coordinates of $A$ and $H$, or $-10-(-30)=20$. Thus, the area of $\mathrm{△}AHE$ is $${\displaystyle \frac{1}{2}}(AE)(20)=10AE=10(30)=300.$$

    To compute the area of $\mathrm{△}HGO$, we will subtract the area of $\mathrm{△}EHO$ from the area of $\mathrm{△}EGO$. Taking $EO$ as a common base, the heights of $\mathrm{△}EHO$ and $\mathrm{△}EGO$ are the $y$-coordinates of $H$ and $G$, respectively. These values are $10$ and $24$, so the area of $\mathrm{△}HGO$ is

    $$\begin{array}{rlr}{\displaystyle \frac{1}{2}}(EO)(24)-{\displaystyle \frac{1}{2}}(EO)(10)& =& {\displaystyle \frac{1}{2}}(EO)(24-10)\\ & =& {\displaystyle \frac{1}{2}}(30)(14)\\ & =& 210\end{array}$$

    Therefore, the total area of the shaded regions is $300+210=510$.

    Answer: 510