2018 Hypatia Contest
Solutions
(Grade 11)

Thursday, April 12, 2018 (in North America and South America)

Friday, April 13, 2018 (outside of North American and South America)

©2018 University of Waterloo

1. 1. The average of Aneesh’s first six test scores was ${\displaystyle \frac{17+13+20+12+18+10}{6}}={\displaystyle \frac{90}{6}}=15$.

   2. After Jon’s third test, his average score was 14, and so the sum of the scores on his first three tests was $14\times 3=42$.

      The sum of his scores on his first two tests was $17+12=29$, and so the score on his third test was $42-29=13$. (We may check that the average of $17,12$ and 13 is ${\displaystyle \frac{17+12+13}{3}}=14$.)

   3. Dina wrote six tests followed by $n$ more tests, for a total of $n+6$ tests.

      After Dina’s first $6$ tests, her average score was 14, and so the sum of the scores on her first $6$ tests was $14\times 6=84$.

      Dina scored 20 on each of her next $n$ tests, and so the sum of the scores on her next $n$ tests was $20n$.

      Therefore, the sum of the scores on these $n+6$ tests was $84+20n$.

      After Dina’s $n+6$ tests, her average score was 18, and so the sum of the scores on her $n+6$ tests was $18(n+6)$. Thus, $84+20n=18(n+6)$ or $84+20n=18n+108$ or $2n=24$, and so $n=12$.

2. 1. The distance from Botown to Aville is 120 km. Jessica drove this distance at a speed of 90 km/h, and so it took Jessica ${\displaystyle \frac{120}{90}}={\displaystyle \frac{4}{3}}$ hours or ${\displaystyle \frac{4}{3}}\times 60=80$ minutes.

   2. The distance from Botown to Aville is 120 km.

      The car predicted that Jessica would drive this distance at a speed of 80 km/h, and so it predicted that it would take Jessica ${\displaystyle \frac{120}{80}}={\displaystyle \frac{3}{2}}$ hours or ${\displaystyle \frac{3}{2}}\times 60=90$ minutes. The ETA displayed by her car at 7:00 a.m. was 8:30 a.m..

   3. Jessica drove from 7:00 a.m. to 7:16 a.m. (for 16 minutes) at a speed of 90 km/h, and so she travelled a distance of ${\displaystyle \frac{16}{60}}\times 90=24$ km.

      At 7:16 a.m., Jessica had a distance of $120\text{ km}-24\text{ km}=96\text{ km}$ left to travel.

      The car predicted that Jessica would drive this distance at a speed of 80 km/h, and so it predicted that it would take Jessica ${\displaystyle \frac{96}{80}}={\displaystyle \frac{6}{5}}$ hours or ${\displaystyle \frac{6}{5}}\times 60=72$ minutes to complete the trip. The ETA displayed by her car at 7:16 a.m. was 72 minutes later or 8:28 a.m..

   4. As in part (b), the car predicted that it would take Jessica 90 minutes or 1.5 hours to travel from Botown to Aville.

      Let the distance that Jessica travelled at 100 km/h be $d$ km, and so the distance that Jessica travelled at 50 km/h was $(120-d)$ km.

      The time that Jessica drove at 100 km/h was ${\displaystyle \frac{d}{100}}$ hours.

      The time that Jessica drove at 50 km/h was ${\displaystyle \frac{120-d}{50}}$ hours.

      Since the time predicted by her car is equal to the actual time that it took Jessica to travel from Botown to Aville, then ${\displaystyle \frac{d}{100}}+{\displaystyle \frac{120-d}{50}}=1.5$.

      Solving for $d$, we get $d+2(120-d)=1.5\times 100$ or $-d+240=150$, and so $d=90$ km. Therefore, Jessica drove a distance of $90$ km at a speed of 100 km/h.

3. 1. We are given that $T_1=1,T_2=2$ and $T_3=3$. Evaluating, we get $$T_4=1+T_1{T}_2{T}_3=1+(1)(2)(3)=7,\text{ and}$$ $$T_5=1+T_1{T}_2{T}_3{T}_4=1+(1)(2)(3)(7)=43.$$

   2. Solution 1

      Each term after the second is equal to 1 more than the product of all previous terms in the sequence. Thus, $T_n=1+T_1{T}_2{T}_3\cdots T_{n-1}$.

      For all integers $n\ge 2$, we use the fact that $T_n=1+T_1{T}_2{T}_3\cdots T_{n-1}$ to get $$\begin{array}{rl}RS& =T_n^2-T_n+1\\ & =T_n(T_n-1)+1\\ & =T_n(1+T_1{T}_2{T}_3\cdots T_{n-1}-1)+1\\ & =T_n(T_1{T}_2{T}_3\cdots T_{n-1})+1\\ & =T_1{T}_2{T}_3\cdots T_{n-1}{T}_n+1\\ & =T_{n+1}\\ & =LS\end{array}$$

      Solution 2

      For all integers $n\ge 2$, we use the fact that $T_n=1+T_1{T}_2{T}_3\cdots T_{n-1}$ to get $$\begin{array}{rl}LS& =T_{n+1}\\ & =1+T_1{T}_2{T}_3\cdots T_{n-1}{T}_n\\ & =1+(T_1{T}_2{T}_3\cdots T_{n-1})T_n\\ & =1+(T_n-1)T_n\\ & =T_n^2-T_n+1\\ & =RS\end{array}$$

   3. Using the result from part (b), we get $T_n+T_{n+1}=T_n+T_n^2-T_n+1=T_n^2+1$, for all integers $n\ge 2$.

      Similarly, $$\begin{array}{rl}{T}_n{T}_{n+1}-1& =T_n(T_n^2-T_n+1)-1\\ & =T_n^3-T_n^2+T_n-1\\ & =T_n^2(T_n-1)+T_n-1\\ & =(T_n-1)(T_n^2+1)\end{array}$$ Since $T_n+T_{n+1}=T_n^2+1$ and $T_n^2+1$ is a factor of $T_n{T}_{n+1}-1$, then $T_n+T_{n+1}$ is a factor of $T_n{T}_{n+1}-1$ for all integers $n\ge 2$.

   4. Using the result from part (b), we get $T_{2018}=T_{2017}^2-T_{2017}+1$.

      Since $T_{2017}$ is a positive integer greater than 1, then $T_{2017}^2-T_{2017}+1>T_{2017}^2-2T_{2017}+1$ and $T_{2017}^2-T_{2017}+1<T_{2017}^2$.

      That is, $T_{2017}^2-2T_{2017}+1<T_{2017}^2-T_{2017}+1<T_{2017}^2$, and so $(T_{2017}-1{)}^2<T_{2018}<T_{2017}^2$. Since $T_{2017}-1$ and $T_{2017}$ are two consecutive positive integers, then $(T_{2017}-1{)}^2$ and $T_{2017}^2$ are two consecutive perfect squares, and so $T_{2018}$ lies between two consecutive perfect squares and thus is not a perfect square.

4. 1. 1. By completing the square, the equations defining the two parabolas become $$y=x^2-8x+17=x^2-8x+16+1=(x-4{)}^2+1,\text{ and}$$ $$y=-x^2+4x+7=-(x^2-4x+4)+11=-(x-2{)}^2+11.$$ Thus, the parabola defined by the equation $y=x^2-8x+17$ has vertex $V_1(4,1)$, and the parabola defined by the equation $y=-x^2+4x+7$ has vertex $V_2(2,11)$.

      2. First, we determine the coordinates of the points of intersection $P$ and $Q$.

         When the two parabolas intersect, $$\begin{array}{rl}{x}^2-8x+17& =-x^2+4x+7\\ 2x^2-12x+10& =0\\ x^2-6x+5& =0\\ (x-5)(x-1)& =0,\end{array}$$ and so the two parabolas intersect at $P(5,2)$ and $Q(1,10)$.

         Next, we want to show why quadrilateral $V_{1}P{V}_{2}Q$ is a parallelogram.

         To do this, we will use the property that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

         The midpoint of diagonal $V_1{V}_2$ is $({\displaystyle \frac{4+2}{2}},{\displaystyle \frac{1+11}{2}})$ or $(3,6)$, and the midpoint of diagonal $PQ$ is $({\displaystyle \frac{5+1}{2}},{\displaystyle \frac{2+10}{2}})$ or $(3,6)$.

         Since the midpoint of each diagonal is the same point, $(3,6)$, then the diagonals bisect each other and so quadrilateral $V_{1}P{V}_{2}Q$ is a parallelogram.

         (Note that we could have also shown that each pair of opposite sides of $V_{1}P{V}_{2}Q$ is parallel.)

   2. 1. By completing the square, the equation defining the parabola $y=-x^2+bx+c$ becomes $$\begin{array}{rl}y& =-x^2+bx+c\\ & =-(x^2-bx)+c\\ & =-(x^2-bx+{\displaystyle \frac{b^2}{4}}-{\displaystyle \frac{b^2}{4}})+c\\ & =-(x^2-bx+{\displaystyle \frac{b^2}{4}})+{\displaystyle \frac{b^2}{4}}+c\\ & =-(x-{\displaystyle \frac{b}{2}}{)}^2+{\displaystyle \frac{b^2}{4}}+c.\end{array}$$ The vertex of this parabola is $V_3({\displaystyle \frac{b}{2}},{\displaystyle \frac{b^2}{4}}+c)$ and the vertex of the parabola defined by the equation $y=x^2$ is $V_4(0,0)$.

         First, we determine the conditions on $b$ and $c$ so that the points of intersection $R$ and $S$ exist and are distinct from one another.

         When the two parabolas intersect, $-x^2+bx+c=x^2$ or $2x^2-bx-c=0$.

         This equation has two distinct real roots when its discriminant is greater than 0, or when $b^2-4(2)(-c)>0$. The points of intersection, $R$ and $S$, exist and are distinct from one another when $c>{\displaystyle \frac{-b^2}{8}}$.

         Next, we determine conditions on $b$ and $c$ so that each of $R$ and $S$ are distinct from both $V_3$ and $V_4$.

         The roots of the equation $2x^2-bx-c=0$ are given by the quadratic formula, and so$x={\displaystyle \frac{b\pm \sqrt{b^2+8c}}{4}}$. We let the $x$-coordinate of $R$ be $x_1={\displaystyle \frac{b+\sqrt{b^2+8c}}{4}}$ and the $x$-coordinate of $S$ be$x_2={\displaystyle \frac{b-\sqrt{b^2+8c}}{4}}$.

         Each of the points $R$ and $S$ is not distinct from $V_4$ when ${\displaystyle \frac{b\pm \sqrt{b^2+8c}}{4}}=0$ or $b=\mp \sqrt{b^2+8c}$ or $b^2=b^2+8c$, and so $c=0$.

         Thus, we require that $c\ne 0$.

         Similarly, each of the points $R$ and $S$ is not distinct from $V_3$ when ${\displaystyle \frac{b\pm \sqrt{b^2+8c}}{4}}={\displaystyle \frac{b}{2}}$ or $b\pm \sqrt{b^2+8c}=2b$ or $\pm \sqrt{b^2+8c}=b$ or $b^2+8c=b^2$, and so $c=0$.

         As before, we require that $c\ne 0$. (Note that since $R$ and $V_4$ lie on the same parabola, then if their $x$-coordinates are not equal, then they are distinct points – that is, we need not consider their $y$-coordinates. The same is true for points $S$ and $V_4$, $R$ and $V_3$, and $S$ and $V_3$.)

         Finally, we require that the vertices of the parabolas, $V_3({\displaystyle \frac{b}{2}},{\displaystyle \frac{b^2}{4}}+c)$ and $V_4(0,0)$, be distinct from one another.

         Vertices $V_3$ and $V_4$ are distinct provided that if their $x$-coordinates are equal, then their $y$-coordinates are not equal ($V_3$ and $V_4$ lie on different parabolas and so we must consider both $x$- and $y$-coordinates). If ${\displaystyle \frac{b}{2}}=0$ or $b=0$, then ${\displaystyle \frac{b^2}{4}}+c={\displaystyle \frac{0^2}{4}}+c=c$, and since we have the requirement (from earlier) that $c\ne 0$, then vertices $V_3$ and $V_4$ are certainly distinct when $c\ne 0$.

         If the two conditions $c>{\displaystyle \frac{-b^2}{8}}$ and $c\ne 0$ are satisfied, then for all pairs $(b,c)$, the points $R$ and $S$ exist, and the points $V_3,V_4,R,S$ are distinct.

      2. We begin by assuming that the conditions on $b$ and $c$ from part (b)(i) above are satisfied.

         Thus, the points $R$ and $S$ exist, and the points $V_3,V_4,R,S$ are distinct.

         For quadrilateral $V_{3}R{V}_{4}S$ to be a rectangle, it is sufficient to require that it be a parallelogram that has at least one pair of adjacent sides that are perpendicular to each other.

         From (b)(i), the parabolas intersect at $R(x_1,x_1^2)$ and $S(x_2,x_2^2)$ ($R$ and $S$ each lie on the parabola $y=x^2$, and thus the $y$-coordinates are $x_1^2$ and $x_2^2$, respectively).

         Recall that $x_1$ and $x_2$ are the distinct real roots of the quadratic equation $2x^2-bx-c=0$.

         The sum of the roots of the general quadratic equation $A{x}^2+Bx+C=0$ is equal to ${\displaystyle \frac{-B}{A}}$, and so $x_1+x_2={\displaystyle \frac{b}{2}}$.

         The product of the roots of the general quadratic equation $A{x}^2+Bx+C=0$ is equal to ${\displaystyle \frac{C}{A}}$, and so $x_1{x}_2={\displaystyle \frac{-c}{2}}$.

         First, we will show that quadrilateral $V_{3}R{V}_{4}S$ is a parallelogram since its diagonals bisect each other.

         The midpoint of diagonal $V_3{V}_4$ is $({\displaystyle \frac{\frac{b}{2}+0}{2}},{\displaystyle \frac{\frac{b^2}{4}+c+0}{2}})$ or $({\displaystyle \frac{b}{4}},{\displaystyle \frac{b^2}{8}}+{\displaystyle \frac{c}{2}})$ or $({\displaystyle \frac{b}{4}},{\displaystyle \frac{b^2+4c}{8}})$.

         The midpoint of diagonal $RS$ is $({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{x_1^2+x_2^2}{2}})$.

         However, $x_1+x_2={\displaystyle \frac{b}{2}}$ and $x_1^2+x_2^2=(x_1+x_2{)}^2-2x_1{x}_2={\left({\displaystyle \frac{b}{2}}\right)}^2-2\left({\displaystyle \frac{-c}{2}}\right)$, and so the midpoint of $RS$ is $({\displaystyle \frac{\frac{b}{2}}{2}},{\displaystyle \frac{{\left(\frac{b}{2}\right)}^2+c}{2}})$ or $({\displaystyle \frac{b}{4}},{\displaystyle \frac{b^2}{8}}+{\displaystyle \frac{c}{2}})$ or $({\displaystyle \frac{b}{4}},{\displaystyle \frac{b^2+4c}{8}})$. Since the midpoint of diagonal $V_3{V}_4$ is equal to the midpoint of diagonal $RS$, then the diagonals bisect each other, and so $V_{3}R{V}_{4}S$ is a parallelogram.

         Next, we require that any one pair of adjacent sides of quadrilateral $V_{3}R{V}_{4}S$ be perpendicular to each other. (This will mean that all pairs of adjacent sides are perpendicular.)

         The slope of $V_{4}S$ is ${\displaystyle \frac{x_2^2-0}{x_2-0}}=x_2$ since $x_2\ne 0$ ($S(x_2,x_2^2)$ and $V_4(0,0)$ are distinct points).

         Similarly, the slope of $V_{4}R$ is ${\displaystyle \frac{x_1^2-0}{x_1-0}}=x_1$ since $x_1\ne 0$ ($R(x_1,x_1^2)$ and $V_4(0,0)$ are distinct points).

         Sides $V_{4}S$ and $V_{4}R$ are perpendicular to each other if the product of their slopes, $x_1{x}_2$, is equal to $-1$. Since $x_1{x}_2={\displaystyle \frac{-c}{2}}$, then ${\displaystyle \frac{-c}{2}}=-1$, and so $c=2$.

         In addition to the condition that $c=2$, the two conditions from part (b)(i), $c>{\displaystyle \frac{-b^2}{8}}$ and $c\ne 0$, must also be satisfied.

         Clearly if $c=2$, then $c\ne 0$.

         Further, when $c=2$, $c>{\displaystyle \frac{-b^2}{8}}$ becomes $2>{\displaystyle \frac{-b^2}{8}}$ or $b^2>-16$ which is true for all real values of $b$. The points $R$ and $S$ exist, the points $V_3,V_4,R,S$ are distinct, and quadrilateral $V_{3}R{V}_{4}S$ is a rectangle for all pairs $(b,c)$ where $c=2$ and $b$ is any real number.