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2021 Hypatia Contest
Solutions
(Grade 11)

April 2021 (in North America and South America)

April 2021 (outside of North American and South America)

©2021 University of Waterloo


    1. The total cost to rent a car is $180.00. If 4 people rent a car, the cost per person is $180.004=$45.00.

    2. Since the members of the group equally share the total cost to rent the vehicle, the smaller the group, the greater the cost per person.

      To rent an SUV, the smallest group size required is 5 passengers and the total cost is $200.00. Thus, the maximum possible cost per person to rent an SUV is $200.005=$40.00.

    3. Let the total cost to rent a van be v.

      When renting a van, the maximum possible cost per person occurs when the number of

      passengers is 9 (the fewest number possible), and so this maximum cost is v9.

      The minimum possible cost per person occurs when the number of passengers is 12 (the

      greatest number possible), and so this minimum cost is v12.

      Then, v9v12=$6.00 or 4v3v36=$6.00, and so v=$6.00×36. Thus, the total cost to rent a van is $216.00.

    1. Trapezoid ABCD is drawn, as shown.

      The trapezoid in the first quadrant, with each vertex labelled.

      The slope of line segments AB and CD are each zero and thus they are parallel.

      The length of AB is the difference between the x-coordinates of A and B, or 12. The length of CD is the difference between the x-coordinates of C and D, or 112=9.

      The height of the trapezoid is equal to the vertical distance between AB and CD, which is 5. The area of trapezoid ABCD is 52(AB+CD) or 52(21)=1052.

    2. The line passing through B and D intersects the y-axis at E. Let the coordinates of E be (0,e), as shown.

      The slope of the line through B and D is 50212=12.

      Solution 1

      Since E,D and B lie on the same line, then the slope of ED is equal to the slope of BD.

      Equating slopes, we get e502=12 or e52=12, and so e5=1 or e=6.

      Thus, point E has coordinates (0,6).

      Solution 2

      The line passing through B(12,0) and D(2,5) has slope 12, and thus has equation y5=12(x2).

      Rearranging, we get y5=12x+1 or y=12x+6.

      Since this line has y-intercept 6, then point E has coordinates (0,6).

      Solution 3

      The line passing through B(12,0) and D(2,5) has slope 12, and thus has equation y5=12(x2).

      This line passes through E(0,e), and so e5=12(02) or e=1+5=6. Thus, point E has coordinates (0,6).

    3. Sides AD and BC are extended to intersect at F, as shown.

      Solution 1

      Let the coordinates of F be (j,k).

      Since A,D and F lie on the same line, then the slope of AD is equal to the slope of AF.

      Point F lies above the trapezoid such that the trapezoid ABCD is contained within the triangle ABF.

      Equating slopes, we get 52=kj or k=52j.

      Since B,C and F lie on the same line, then the slope of BC is equal to the slope of BF.

      Equating slopes, we get 501112=k0j12 or 5=kj12, and so k=5(j12).

      Substituting k=52j, we get 52j=5(j12) or j=2(j12), and so 3j=24 or j=8.

      When j=8, k=52(8)=20, and so F has coordinates (8,20).

      Solution 2

      The line passing through A(0,0) and D(2,5) has slope 52 and y-intercept 0, and thus has equation y=52x.

      The line passing through B(12,0) and C(11,5) has slope 5 and thus has equation y=5(x12).

      These two lines intersect at F, and so the coordinates of F can be determined by solving the equation 52x=5(x12). Solving, we get x=2(x12) or 3x=24, and so x=8. When x=8, y=52(8)=20, and so F has coordinates (8,20).

    4. Let P have coordinates (r,s).

      Assume AB is the base of PAB.

      In this case, if the height of PAB is h, then the area of PAB is 12(AB)h=6h.

      The area of PAB is 42, and so 6h=42 or h=7.

      That is, P(r,s) is located a vertical distance of 7 units from the line through A and B, or 7 units from the x-axis.

      There are two possibilities: P(r,s) is located 7 units above the x-axis, and thus lies on the horizontal line y=7, or P(r,s) is located 7 units below the x-axis, and thus lies on the horizontal line y=7.

      In the first case, P has coordinates (r,7) and in the second case, P has coordinates (r,7).

      Recall that P lies on the line passing through B and D.

      The line passing through B(12,0) and D(2,5) has slope 12, and thus has equation y5=12(x2).

      If P(r,7) lies on this line, then 75=12(r2) or 4=r2, and so r=2.

      Similarly, if P(r,7) lies on this line, then 75=12(r2) or 24=r2, and so in this case, r=26. The points P that lie on the line passing through B and D, so that the area of PAB is 42, are (2,7) and (26,7).

    1. Since an=2n for n1, then a5=25=32.

      Since b2=1,b3=3, and bn=bn1+2bn2 for n3, then b4=b3+2b2=3+2(1)=5 and b5=b4+2b3=5+2(3)=11. Therefore, a5=32 and b5=11.

    2. Since b1=p(a1)+q(1)1 and a1=2, then b1=2pq.

      From the definition of sequence B, we know b1=1, and so 2pq=1.

      Since b2=p(a2)+q(1)2 and a2=22=4, then b2=4p+q.

      From the definition of sequence B, we know b2=1, and so 4p+q=1.

      This gives two equations in two unknowns, p and q.

      Adding these two equations, we get 6p=2, and so p=13.

      Substituting, we get 2(13)q=1 or q=231=13. Thus, the real numbers p and q for which bn=p(an)+q(1)n for all n1 are p=13 and q=13.

    3. Using algebraic manipulation, and the definitions an=2n and bn=13(an)13(1)n, each for n1, we obtain the following equivalent equations, Sn=b1+b2+b3++bn=(13(a1)13(1))+(13(a2)13(1)2)+(13(a3)13(1)3)++(13(an)13(1)n)=13(a1+a2+a3++an)13((1)+(1)2+(1)3++(1)n)=13(2+22+23++2n)13(1+11++(1)n) Next, we consider each of the two expressions within parentheses, separately.

      The expression 2+22+23++2n is the sum of n terms of a geometric sequence with first term a=2 and common ratio r=2.

      Thus, 2+22+23++2n=2(12n12)=2(12n).

      The expression 1+11++(1)n is an alternating sum of the terms 1 and 1.

      This simplifies to 0 if there are an even number of terms, that is, if n is even, and simplifies to 1 if n is odd.

      Summarizing, we have Sn={13(2(12n)) , if n is even13(2(12n))+13 , if n is odd and simplifying, we get Sn={23(2n1) , if n is even23(2n1)+13 , if n is odd We want the smallest positive integer n that satisfies Sn162021 and note that the value of Sn increases as n increases.

      Since 16=24, then 162021=(24)2021=28084 and so we want the smallest positive integer n that satisfies Sn28084.

      When n is even, we get 23(2n1)280842n13280832n328083+1

      When n is odd, we get 23(2n1)+132808423(2n1)28084132n1328083122n328083+12

      Since 2n is an even integer, 328083+1 is an odd integer and 328083+12 is between an even integer and an odd integer, and thus the inequalities 2n328083+1 and 2n328083+12 are both equivalent to saying 2n>328083.

      Since 328083>228083, then simplifying, we get 328083>28084.

      Thus, we want the smallest positive integer n that satisfies 2n>328083>28084.

      When n8084, this inequality is not true.

      When n=8085, we get 28085=2228083=428083 which is greater than 328083, as required. Thus, the smallest positive integer n that satisfies Sn162021 is n=8085.

    1. In XYZ, x=20, y=21, and XZY= 90°, as shown.

      Triangle XZY is right-angled at Z. Sides YZ and XZ are 20cm and 21cm respectively.

      Using the Pythagorean Theorem, we get z=202+212=29 (since z>0). The value of A is

      A=12(y)(x)=12(21)(20)=210

      The value of P is P=z+x+y=29+20+21=70

    2. When A=336, we get 12xy=336, and so xy=672.

      By the Pythagorean Theorem, x2+y2=502, which when manipulated algebraically gives the following equivalent equations x2+y2=2500(x+y)22xy=2500(x+y)2=2500+2xy(x+y)2=2500+2(672)(x+y)2=3844 Since x+y>0, then x+y=3844=62.

      Thus, we get P=x+y+z=62+50=112. (The triangle satisfying these conditions has side lengths 14 cm, 48 cm and 50 cm.)

    3. Since A=3P, we get 12xy=3(x+y+z), and so xy=6(x+y+z).

      When xy=6(x+y+z) is manipulated algebraically, we get the following equivalent equations xy=6(x+y+z)xy6x6y=6z(xy6x6y)2=(6z)2(xy)212x(xy)12y(xy)+72xy+36x2+36y2=36z2(xy)212x(xy)12y(xy)+72xy+36x2+36y2=36(x2+y2)   (since x2+y2=z2)xy(xy12x12y+72)=0xy12x12y+72=0   (since xy0)x(y12)12y=72x(y12)12y+144=72+144x(y12)12(y12)=72(x12)(y12)=72 Since x and y are positive integers, then x12 and y12 are a factor pair of 72.

      The product (x12)(y12) is positive (since 72>0), and thus x12<0 and y12<0 or x12>0 and y12>0.

      If x12<0 and y12<0, then x<12 and y<12.

      There are exactly two Pythagorean triples (x,y,z) in which x<12 and y<12.

      In the first case, (x,y,z)=(3,4,5), which gives A=12(3)(4)=6, P=3+4+5=12, and so A3P.

      In the second case, (x,y,z)=(6,8,10), which gives A=12(6)(8)=24, P=6+8+10=24, and so A3P.

      Therefore, x12>0 and y12>0, and so x and y are each greater than 12.

      In the table below, we use the positive factor pairs of 72 to determine all possible integer values of x, y and z. Further, we initially make the assumption that xy, recognizing that by the symmetry of the equation, the values of x and y may be interchanged with one another and doing so gives the same value for z and the same triangle.

      Factor pair x12 y12 x y z=x2+y2 (x,y,z)
      1 and 72 1 72 13 84 85 (13,84,85) or (84,13,85)
      2 and 36 2 36 14 48 50 (14,48,50) or (48,14,50)
      3 and 24 3 24 15 36 39 (15,36,39) or (36,15,39)
      4 and 18 4 18 16 30 34 (16,30,34) or (30,16,34)
      6 and 12 6 12 18 24 30 (18,24,30) or (24,18,30)
      8 and 9 8 9 20 21 29 (20,21,29) or (21,20,29)

      It is worth noting that instead of factoring the equation xy12x12y+72=0 as we did, we could have chosen to rewrite it as x(y12)=12y72x=12y72y12x=12(y12)+14472y12x=12+72y12 and considered that since x is a positive integer, then y12 is a divisor of 72.

    4. Since A=kP, we get 12xy=k(x+y+z), and so xy=2k(x+y+z).

      When xy=2k(x+y+z) is manipulated algebraically, we get the following equivalent equations xy=2k(x+y+z)xy2kx2ky=2kz(xy2kx2ky)2=(2kz)2(xy)24kx(xy)4ky(xy)+8k2xy+4k2x2+4k2y2=4k2z2(xy)24kx(xy)4ky(xy)+8k2xy+4k2x2+4k2y2=4k2(x2+y2)  (x2+y2=z2)xy(xy4kx4ky+8k2)=0xy4kx4ky+8k2=0   (since xy0)x(y4k)4ky=8k2x(y4k)4ky+16k2=8k2+16k2x(y4k)4k(y4k)=8k2(x4k)(y4k)=8k2 Since x, y and k are positive integers, then x4k and y4k are a factor pair of 8k2.

      We begin by assuming that k=2.

      Substituting, we get (x8)(y8)=32.

      The product (x8)(y8) is positive (since 32>0), and thus x8<0 and y8<0 or x8>0 and y8>0.

      If x8<0 and y8<0, then x<8 and y<8 which is not possible since P=510, and so x8>0 and y8>0.

      If (x8)(y8)=32 and xy, then x8 is equal to 1,2 or 4, which gives x=9,10,12 and y8 is equal to 32,16, 8, and so y=40,24,16, respectively.

      Using the Pythagorean Theorem, we get z=41,26,20, respectively.

      For each of the three possibilities, P=x+y+z510 and so we conclude k2.

      It can similarly be shown that k cannot equal 3, and thus k5 (since k is a prime number).

      Since k is a prime number and k5, the positive factor pairs of 8k2 are (1,8k2),(2,4k2),(4,2k2),(8,k2),(k,8k),(2k,4k) and the negative factor pairs of 8k2 are (1,8k2),(2,4k2),(4,2k2),(8,k2),(k,8k),(2k,4k) If for example x4k=1 and y4k=8k2, then y=4k8k2 which is less than zero for all k5.

      This is not possible since y>0.

      Assuming xy, it can similarly be shown that when x4k and y4k are equal to a negative factor pair of 8k2, then y0 for all values of k5.

      Thus, x4k and y4k must equal a positive factor pair of 8k2.

      Beginning with the fact that the perimeter of the triangle is 510 cm, we get the following equivalent equations x+y+z=510x+y=510zx2+2xy+y2=51021020z+z2   (squaring both sides)2xy=51021020z   (since x2+y2=z2)4A=51021020z   (A=12xy and so 4A=2xy)4(510k)=51021020z   (since A=kP and P=510)2k=255z2k=255(510xy)x+y2k=255 From the first factor pair, we get x4k=1 and y4k=8k2, and so (x,y)=(1+4k,8k2+4k) (assuming xy).

      Substituting x=1+4k and y=8k2+4k into x+y2k=255 and simplifying, we get 8k2+6k=254 or k(4k+3)=127, which has no solutions since 127 is a prime number.

      We continue our analysis of the remaining 5 factor pairs in the table below.

      As before, we make the assumption that xy, recognizing that the values of x and y may be interchanged with one another and doing so gives the same value(s) for k.

      Factor pair x y x+y2k=255Simplified Value(s) of k
      2,4k2 2+4k 4k2+4k 4k2+6k=253 No k (LS is even and the RS is odd)
      4,2k2 4+4k 2k2+4k 2k2+6k=251 No k (LS is even and the RS is odd)
      8,k2 8+4k k2+4k k(k+6)=247 k=13
      k,8k 5k 12k 15k=255 k=17
      2k,4k 6k 8k 12k=255 No k (LS is even and the RS is odd)

      Therefore, the values of k which satisfy the given conditions are k=13 and k=17.