2025 Hypatia Contest
Solutions
(Grade 11)

Thursday, April 3, 2025
(in North America and South America)

Friday, April 4, 2025
(outside of North American and South America)

©2025 University of Waterloo

1. 1. After \(7\) minutes, Abi has a total of \(30+(7\times2)=30+14=44\) tokens.

   2. After \(12\) minutes, Desiree had received \(12\times2=24\) additional tokens. After \(12\) minutes, she had \(37\) tokens in total, and so Desiree started with \(37-24=13\) tokens.
      A total of \(80\) tokens were initially distributed, and so Carl started with \(80-13=67\) tokens.

   3. At the start, Essi received \(12\) tokens and Francis received \(100-12=88\) tokens.
      After \(t\) minutes, Essi has \(12+2t\) tokens and Francis has \(88+2t\) tokens.
      After \(t\) minutes, Francis has \(3\) times as many tokens as Essi, and so \(88+2t=3(12+2t)\). Solving for \(t\), we get \(88+2t=36+6t\) or \(52=4t\), and so \(t=\dfrac{52}{4}=13\).

2. 1. Solution 1:

      Since \(10\%\) of \(5 \text{ cm}\) is \(\dfrac{10}{100}\times5\text{ cm}=0.1\times5\text{ cm}=0.5\text{ cm}\), then the length of the resulting

      rectangle is \(5\text{ cm}+0.5\text{ cm}=5.5\text{ cm}\), and its area is \(5.5\text{ cm}\times4\text{ cm}=22\text{ cm}^2\).

      Solution 2:

      When \(5 \text{ cm}\) is increased by \(10\%\), the resulting length is \(\left(1+\dfrac{10}{100}\right)\times5\text{ cm}\) or \(1.1\times5\text{ cm}\) which is equal to \(5.5 \text{ cm}\).
      Thus, the area of the resulting rectangle is \(5.5\text{ cm}\times4\text{ cm}=22\text{ cm}^2\).

   2. Solution 1:

      A square with area \(100 \text{ cm}^2\) has both length and width equal to \(\sqrt{100\text{ cm}^2}=10\text{ cm}\).
      Since \(30\%\) of \(10\text{ cm}\) is \(\dfrac{30}{100}\times10\text{ cm}=0.3\times10\text{ cm}=3\text{ cm}\), then the length of the resulting rectangle is \(10\text{ cm}+3\text{ cm}=13\text{ cm}\), and its width is \(10\text{ cm}-3\text{ cm}=7\text{ cm}\).
      The area of the resulting rectangle is \(13\text{ cm}\times7\text{ cm}=91\text{ cm}^2\) which is \(\dfrac{91\text{ cm}^2}{100\text{ cm}^2}\times100\%=91\%\) of the area of the original square.
      Therefore, the area decreased by \(100\%-91\%=9\%\).

      Solution 2:

      A square with area \(100 \text{ cm}^2\) has both length and width equal to \(\sqrt{100\text{ cm}^2}=10\text{ cm}\).
      When \(10 \text{ cm}\) is increased by \(30\%\), the resulting length is \(\left(1+\dfrac{30}{100}\right)\times10\text{ cm}\) or \(1.3\times10\text{ cm}\) which is equal to \(13 \text{ cm}\).
      When \(10 \text{ cm}\) is decreased by \(30\%\), the resulting length is \(\left(1-\dfrac{30}{100}\right)\times10\text{ cm}\) or \(0.7\times10\text{ cm}\) which is equal to \(7 \text{ cm}\).
      Thus, the area of the resulting rectangle is \(13\text{ cm}\times7\text{ cm}=91\text{ cm}^2\) which is \(100\text{ cm}^2-91\text{ cm}^2=9\text{ cm}^2\) less than the area of the original square.
      Therefore, the area decreased by \(\dfrac{9\text{ cm}^2}{100\text{ cm}^2}\times100\%=9\%\).

   3. Suppose the length of the original rectangle is \(\ell\) and its width is \(w\).
      Then the length of the resulting rectangle is \(\left(1+\dfrac{x}{100}\right)\times\ell\), and its width is \(\left(1-\dfrac{20}{100}\right)\times w\) or \(\dfrac{8}{10}w\).
      The area of the original rectangle, \(\ell w\), is equal to the area of the resulting rectangle \(\left(1+\dfrac{x}{100}\right)\times\ell \times \dfrac{8}{10}w\).
      Setting the areas equal and simplifying, we get the following equivalent equations: \[\begin{align*} \left(1+\dfrac{x}{100}\right)\times\ell \times \dfrac{8}{10}w &= \ell w \\ \left(1+\dfrac{x}{100}\right)\times\dfrac{8}{10}\times \ell w &= \ell w \\ \left(1+\dfrac{x}{100}\right)\times\dfrac{8}{10} &= 1 \ \ \text{ (since $\ell w>0$)}\\ 1+\dfrac{x}{100} &= \dfrac{10}{8} \\ \dfrac{x}{100} &= \dfrac{10}{8} -\dfrac{8}{8}\\ \dfrac{x}{100} &= \dfrac{2}{8}\\ x&= \dfrac{1}{4}\times100\end{align*}\] and so \(x=25\).

3. 1. We begin by determining where the parabola and the line intersect. Setting the right-hand sides of the equations equal and solving, we get \[\begin{align*} x^2+3x-12&=2x\\ x^2+x-12&=0\\ (x+4)(x-3)&=0\end{align*}\] and so \(x=-4\) and \(x=3\).
      Applying the given formula with \(p=3\) and \(q=-4\), the area enclosed by the parabola and the line is \(\dfrac{(3-(-4))^3}{6}=\dfrac{7^3}{6}=\dfrac{343}{6}\).

   2. Suppose the \(x\)-coordinates of \(V\) and \(W\) are the integers \(v\) and \(w\) respectively, with \(v>w\).
      Then \(v\) and \(w\) are the solutions of the equation \(mx-6=-x^2+7x-90\), which when simplified is \(x^2+(m-7)x+84=0\).
      Therefore, the equation \(x^2+(m-7)x+84=0\) is equivalent to \((x-v)(x-w)=0\) or \(x^2-(v+w)x+vw=0\). Thus, the product of the roots, \(vw\), is equal to 84.
      The area enclosed by the parabola and the line is equal to \(\dfrac{(v-w)^3}{6}\), which is as small as possible when \(v-w\) is as small as possible.
      To determine the smallest possible enclosed area, we must determine the smallest possible value of \(v-w\) for integers \(v\) and \(w\) with \(vw=84\) and \(v>w\).
      The positive factor pairs \((w,v)\) of 84 are \((1,84)\), \((2,42)\), \((3,28)\), \((4,21)\), \((6,14)\), and \((7,12)\).
      The smallest possible value of \(v-w\) is equal to 5, which occurs when \(v=12\) and \(w=7\).
      Each of the factors listed can also be negative, from which we similarly determine that the smallest possible value of \(v-w\) is equal to \(5\) when \(v=-7\) and \(w=-12\).
      Returning to the equivalent equations \(x^2+(m-7)x+84=0\) and \(x^2-(v+w)x+vw=0\), we have \(m-7=-(v+w)\), and so \(m=7-(v+w)\).
      When \(v=12\) and \(w=7\), we get \(m=7-(12+7)=-12\).
      When \(v=-7\) and \(w=-12\), we get \(m=7-(-7-12)=26\).
      The two possible values of \(m\) for which the area enclosed by the line and the parabola is as small as possible are \(-12\) and \(26\).

   3. Suppose the \(x\)-coordinates of \(T\) and \(U\) are the integers \(t\) and \(u\) respectively, with \(t>u\).
      Then \(t\) and \(u\) are the solutions of the equation \(x^2+(g+h)x+9=-x^2+gx+h\), which when simplified is \(2x^2+hx+9-h=0\).
      Therefore, the equation \(2x^2+hx+9-h=0\) is equivalent to \((x-t)(x-u)=0\), and so \(x^2+\dfrac{h}{2}x+\dfrac{9-h}{2}=0\) is equivalent to \(x^2-(t+u)x+tu=0\).
      Equating the constant terms, we get \(tu=\dfrac{9-h}{2}\) (equation \((1)\)), and equating the coefficients of the linear terms, we get \(-(t+u)=\dfrac{h}{2}\) or \(t+u=-\dfrac{h}{2}\) (equation \((2)\)).

      Consider the line passing through points \(T\) and \(U\), as shown.

      

      The area enclosed by this line and the parabola with equation \(y=-x^2+gx+h\) is equal to \(\dfrac{(t-u)^3}{6}\).
      Similarly, the area enclosed by this line and the parabola with equation \(y=x^2+(g+h)x+9\) is also equal to \(\dfrac{(t-u)^3}{6}\).
      Thus, the area enclosed by the parabolas is equal to \(2\times\dfrac{(t-u)^3}{6}=\dfrac{(t-u)^3}{3}\).
      The area enclosed by the parabolas is \(\dfrac{3087}{8}\), and so \[\begin{align*} \dfrac{(t-u)^3}{3}&=\dfrac{3087}{8}\\ (t-u)^3&=\dfrac{9261}{8}\\ t-u&=\sqrt[3]{\dfrac{9261}{8}}\\ t&=u+\dfrac{21}{2} \text{ \ \ (equation $(3)$)}\end{align*}\] Substituting equation \((3)\) into \((2)\), we get \(u+\dfrac{21}{2}+u=-\dfrac{h}{2}\) or \(h=-4u-21\).
      Substituting \(h=-4u-21\) and equation \((3)\) into equation \((1)\) and solving, we get \[\begin{align*} tu&=\dfrac{9-h}{2}\\ \left(u+\dfrac{21}{2}\right)u&=\dfrac{9-(-4u-21)}{2}\\ (2u+21)u&=9+4u+21\\ 2u^2+17u-30&=0\\ (2u-3)(u+10)&=0\end{align*}\] and so the solutions are \(u=\dfrac32\) and \(u=-10\).
      When \(u=\dfrac32\), \(h=-4\cdot\dfrac{3}{2}-21=-27\) and when \(u=-10\), \(h=-4(-10)-21=19\).
      The possible values of \(h\) for which the area enclosed by the parabolas is \(\frac{3087}{8}\) are \(-27\) and \(19\).

4. 1. The third term of the sequence is \(45\cdot\frac35=27\) and the first term is \(45\div\frac35=45\cdot\frac53=75\).
      The sum of the first three terms of the sequence is \(75+45+27=147\).

   2. The common ratio, \(r\), of the geometric sequence \(x,12,y\) is \(r=\dfrac{12}{x}\), and also \(r=\dfrac{y}{12}\).
      Thus \(\dfrac{y}{12}=\dfrac{12}{x}\) or \(xy=144\).
      Substituting \(y=25-x\) into \(xy=144\) and solving, we get \[\begin{align*} x(25-x)&=144\\ 25x-x^2&=144\\ x^2-25x+144&=0\\ (x-9)(x-16)&=0\end{align*}\] and so \(x=9\) and \(x=16\). When \(x=9\), \(y=25-9=16\) and when \(x=16\), \(y=25-16=9\). The possible pairs of positive integers \((x,y)\) are \((9,16)\) and \((16,9)\).

   3. Suppose the geometric sequence has common ratio \(r\), so that \(b=ar\), \(c=ar^2\), and \(d=ar^3\).
      Since \(a+b+c+d=65\), then \[\begin{align*} a+ar+ar^2+ar^3&=65\\ a(1+r+r^2+r^3)&=65\\ a((1+r)+r^2(1+r))&=65\\ a(1+r)(1+r^2)&=65\end{align*}\] Since \(r=\dfrac{b}{a}\) and both \(a\) and \(b\) are non-zero integers, then \(r\) is a rational number.
      Suppose that \(r=\dfrac{m}{n}\) for some non-zero integers \(m\) and \(n\) having no common factors (that is, \(\gcd(m,n)=1\)).
      Then \(a(1+r)(1+r^2)=65\) becomes \[\begin{align*} a\left(1+\dfrac{m}{n}\right)\left(1+\left(\dfrac{m}{n}\right)^2\right)&=65\\ a\left(\dfrac{m+n}{n}\right)\left(\dfrac{m^2+n^2}{n^2}\right)&=65\\ \dfrac{a}{n^3}(m+n)(m^2+n^2)&=65\\\end{align*}\] Since \(d\) is an integer, and \(d=ar^3=a\cdot\dfrac{m^3}{n^3}\), then \(a\cdot\dfrac{m^3}{n^3}\) is an integer.

      Since \(\gcd(m,n)=1\), then \(m^3\) and \(n^3\) have no common divisors. We can conclude that\(n^3\) divides \(a\) which means that \(\dfrac{a}{n^3}\) is an integer.
      Suppose \(p=\dfrac{a}{n^3}\), then \(p(m+n)(m^2+n^2)=65\) for integers \(m,n,p\).
      Each of \(p\), \(m+n\) and \(m^2+n^2\) is an integer, and thus a divisor of 65.
      The divisors of 65 are \(\pm1,\pm5,\pm13\), and \(\pm65\).
      Since \(m\) and \(n\) are non-zero integers, then \(m^2+n^2\geq 2\) and \(m^2+n^2\geq m+n\).
      We also note that the factors \(p\) and \(m+n\) are either both positive or they are both negative since \(m^2+n^2\) is positive for all values of \(m\) and \(n\). This leads us to consider \(3\) possible cases: \(m^2+n^2=65\), \(m^2+n^2=13\), and \(m^2+n^2=5\).

      Case 1: \(m^2+n^2=65\).

      If \(m^2+n^2=65\), then \(m+n=1\) and \(p=1\) or \(m+n=-1\) and \(p=-1\).
      If \(m^2+n^2=65\), the integer solutions are \((m,n)=(\pm1,\pm8)\) and \((\pm4,\pm7)\) and \((\pm8,\pm1)\) and \((\pm7,\pm4)\). For each of these, \(m+n\neq \pm1\), so there are no solutions when \(m^2+n^2=65\).

      Case 2: \(m^2+n^2=13\).

      When \(m^2+n^2=13\), the integer solutions are \((m,n)=(\pm3,\pm2)\) and \((\pm2,\pm3)\).
      This gives the following \(8\) possibilities:

      1. \((m,n)=(3,2)\), so \(m+n=5\) and \(p=1=\dfrac{a}{n^3}\), and so \(a=1\cdot2^3=8\).
         In this case, \(r=\dfrac{m}{n}=\dfrac32\), and so the quadruple \((a,b,c,d)=(8,12,18,27)\) satisfies the given conditions.

      2. \((m,n)=(-3,-2)\), so \(m+n=-5\) and \(p=-1=\dfrac{a}{n^3}\), and so \(a=-1\cdot(-2)^3=8\).
         In this case, \(r=\dfrac{m}{n}=\dfrac32\), and so we get the same quadruple as in (i).

      3. \((m,n)=(3,-2)\), so \(m+n=1\) and \(p=5=\dfrac{a}{n^3}\), and so \(a=5\cdot(-2)^3=-40\).
         In this case, \(r=\dfrac{m}{n}=-\dfrac32\), and so we get \((a,b,c,d)=(-40,60,-90,135)\).

      4. \((m,n)=(-3,2)\), so \(m+n=-1\) and \(p=-5=\dfrac{a}{n^3}\), and so \(a=-5\cdot(2)^3=-40\).
         In this case, \(r=-\dfrac32\), and so we get the same quadruple as in (iii).

      5. \((m,n)=(2,3)\), so \(m+n=5\) and \(p=1\), and so \(a=1\cdot3^3=27\).
         In this case, \(r=\dfrac23\), and so we get \((a,b,c,d)=(27,18,12,8)\).

      6. \((m,n)=(-2,-3)\) gives the same quadruple as in (v).

      7. \((m,n)=(-2,3)\), so \(m+n=1\) and \(p=5\), and so \(a=5\cdot3^3=135\).
         In this case, \(r=\dfrac{m}{n}=-\dfrac23\), and so we get \((a,b,c,d)=(135,-90,60,-40)\).

      8. \((m,n)=(2,-3)\) gives the same quadruple as in (vii).

      Case 3: \(m^2+n^2=5\).

      When \(m^2+n^2=5\), the integer solutions are \((m,n)=(\pm2,\pm1)\) and \((\pm1,\pm2)\).
      Since \(m^2+n^2\geq m+n\), then \(m+n=1\) and \(p=13\), or \(m+n=-1\) and \(p=-13\).
      This gives the following \(4\) possibilities:

      1. \((m,n)=(2,-1)\), so \(m+n=1\) and \(p=13=\dfrac{a}{n^3}\), and so \(a=13\cdot(-1)^3=-13\). In this case, \(r=-2\), and so we get \((a,b,c,d)=(-13,26,-52,104)\).

      2. \((m,n)=(-2,1)\) gives the same quadruple as in (i).

      3. \((m,n)=(-1,2)\), so \(m+n=1\) and \(p=13\), and so \(a=13\cdot2^3=104\).
         In this case, \(r=-\dfrac12\), and so we get \((a,b,c,d)=(104,-52, 26, -13)\).

      4. \((m,n)=(1,-2)\) gives the same quadruple as in (iii).

      The quadruples of integers \((a,b,c,d)\) so that \(a,b,c,d\) is a geometric sequence and \(a+b+c+d=65\) are: \((8,12,18,27)\), \((27,18,12,8)\), \((-40,60,-90,135)\), \((135,-90,60,-40)\), \((-13,26,-52,104)\), and \((104,-52, 26, -13)\). We can verify that each of these quadruples is a geometric sequence with a sum of \(65\).