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2020 Hypatia Contest
Solutions
(Grade 11)

Wednesday, April 15, 2020 (in North America and South America)

Thursday, April 16, 2020 (outside of North American and South America)

©2020 University of Waterloo


    1. The cost of 12 bags of avocados is $5.00×12=$60.00.

      Thus, the chef spent $135.00$60.00=$75.00 on mangoes. The cost of each box of mangoes is $12.50, and so the chef purchased $75.00$12.50=6 boxes of mangoes.

    2. Solution 1

      A bag of avocados sells for $5.00, and so a 10% discount represents a savings of 10100×$5.00 or 0.10×$5.00 which equals $0.50.

      Thus, the discounted price for a bag of avocados is $5.00$0.50=$4.50.

      A box of mangoes sells for $12.50, and so a 20% discount represents a savings of 20100×$12.50 or 0.20×$12.50 which equals $2.50.

      Thus, the discounted price for a box of mangoes is $12.50$2.50=$10.00. On Saturdays, the total cost for 8 bags of avocados and 4 boxes of mangoes is $4.50×8+$10.00×4=$36.00+$40.00=$76.00

      Solution 2

      A bag of avocados sells for $5.00, and so a 10% discount is equivalent to paying 100%10%=90% of the regular price.

      Thus, the discounted price for a bag of avocados is 90100×$5.00=0.90×$5.00=$4.50.

      A box of mangoes sells for $12.50, and so a 20% discount is equivalent to paying 100%20%=80% of the regular price.

      Thus, the discounted price for a box mangoes is 80100×$12.50=0.80×$12.50=$10.00. On Saturdays, the total cost for 8 bags of avocados and 4 boxes of mangoes is $4.50×8+$10.00×4=$36.00+$40.00=$76.00

    3. Avocados are sold in bags of 6 and the chef needs 100 avocados.

      Since 6×16=96 and 6×17=102, the chef will need to purchase 17 bags of avocados (16 bags is not enough).

      Mangoes are sold in boxes of 15 and the chef needs 70 mangoes.

      Since 15×4=60 and 15×5=75, the chef will need to purchase 5 boxes of mangoes. The total cost of the purchase was $5.00×17+$12.50×5=$85.00+$62.50=$147.50.

    4. Avocados are sold for $5.00 per bag, and so the cost to purchase any number of bags is a whole number.

      Mangoes are sold for $12.50 per box, and so the cost to purchase mangoes is a whole number only when an even number of boxes are bought (1 box costs $12.50, 2 boxes cost $25.00, 3 boxes cost $37.50, and so on).

      Since the chef spends exactly $75.00 (a whole number), then the chef must purchase an even number of boxes of mangoes.

      If the chef purchases 2 boxes of mangoes, the cost is $25.00, which leaves $75.00$25.00=$50.00 to be used to purchase $50.00÷$5.00=10 bags of avocados.

      In this case, the chef has 10×6=60 avocados and 2×15=30 mangoes.

      Each tart requires 1 avocado and 2 mangoes and so the chef can make 30÷2=15 tarts (he has more than 15 avocados but only 30 mangoes).

      If the chef purchases 4 boxes of mangoes, the cost is 4×$12.50=$50.00, which leaves $75.00$50.00=$25.00 to be used to purchase $25.00÷$5.00=5 bags of avocados.

      In this case, the chef has 5×6=30 avocados and 4×15=60 mangoes.

      Each tart requires 1 avocado and 2 mangoes and so the chef can make 30 tarts.

      If the chef purchases 6 boxes of mangoes, the cost is 6×$12.50=$75.00, which leaves no money to purchase avocados.

      Purchasing more than 6 boxes of mangoes will cost the chef more than $75.00. Therefore, if the chef purchases 30 avocados (5 bags) and 60 mangoes (4 boxes), she will have spent exactly $75.00, have twice as many mangoes as avocados, and be able to make the greatest number of tarts, 30.

    1. The parabola y=14x2 and the parabolic rectangle are each symmetrical about the y-axis, and thus a second vertex of the rectangle lies on the parabola and has coordinates (6,9).

      A third vertex of the parabolic rectangle lies on the x-axis vertically below (6,9), and thus has coordinates (6,0). Similarly, the fourth vertex also lies on the x-axis vertically below (6,9), and thus has coordinates (6,0).

    2. If one vertex of a parabolic rectangle is (3,0), then a second vertex has coordinates (3,0), and so the rectangle has length 6.

      The vertex that lies vertically above (3,0) has x-coordinate 3.

      This vertex lies on the parabola y=14x2 and thus has y-coordinate equal to 14(3)2=94.

      The width of the rectangle is equal to this y-coordinate 94, and so the area of the parabolic rectangle having one vertex at (3,0) is 6×94=544=272.

    3. Let a vertex of the parabolic rectangle be the point (p,0), with p>0.

      A second vertex (also on the x-axis) is thus (p,0), and so the rectangle has length 2p.

      The width of this rectangle is given by the y-coordinate of the point that lies on the parabola vertically above (p,0), and so the width is 14p2.

      The area of a parabolic rectangle having length 2p and width 14p2 is 2p×14p2=12p3.

      If such a parabolic rectangle has length 36, then 2p=36, and so p=18.

      The area of this rectangle is 12(18)3=2916.

      If such a parabolic rectangle has width 36, then 14p2=36 or p2=144, and so p=12 (since p>0).

      The area of this rectangle is 12(12)3=864. The areas of the two parabolic rectangles that have side length 36 are 2916 and 864.

    4. Let a vertex of the parabolic rectangle be the point (m,0), with m>0.

      A second vertex (also on the x-axis) is thus (m,0), and so the rectangle has length 2m.

      The width of this rectangle is given by the y-coordinate of the point that lies on the parabola vertically above (m,0), and so the width is 14m2.

      The area of a parabolic rectangle having length 2m and width 14m2 is 2m×14m2=12m3.

      If the length and width of such a parabolic rectangle are equal, then 14m2=2mm2=8mm28m=0m(m8)=0 Thus m=8 (since m>0), and so the area of the parabolic rectangle whose length and width are equal is 12(8)3=256.

    1. For n3 and k0, the value of T(n,k) is constant for all possible locations of the k interior points and all possible triangulations. Thus, we may use the triangulation shown to determine that T(3,2)=5.

    2. A triangle with two interior points. One of the interior points is connected to all three vertices of the triangle. The second interior point is connected to two vertices of the triangle. The two interior points are also connected. The interior of the triangle is divided into five triangular regions.

    3. We begin by drawing triangulations to determine the values of T(4,k) for k=0,1,2,3.

      Descriptions of the triangulations follow.

      Although we would obtain these same four answers by positioning the interior points in different locations, or by completing the triangulations in different ways, the diagrams above were created to help visualize a pattern.

      From the answers shown, we see that T(4,k+1)=T(4,k)+2, for k=0,1,2.

      We must justify why this observation is true for all k0 so that we may use the result to determine the value of T(4,100).

      Notice that each triangulation (after the first) was created by placing a new interior point inside the previous triangulation.

      Further, each square is divided into triangles, and so each new interior point is placed inside a triangle of the previous triangulation (since no 3 points may lie on the same line).

      For example, in the following diagrams, we observe that P lies in triangle t of the previous triangulation.

      The triangulation of a square, with two interior points. The interior of the square is divided into six triangular regions, one of which is highlighted and labelled t. This triangulation is labelled T(4,2) equals 6. A second square showing the same triangulation only the point P has been put inside the triangle that was labelled t, and this region has been divided into three smaller triangular regions. The diagram is labelled T(4,3) equals 8.

      Also, each of the triangles outside of t is untouched by the addition of P, and thus they continue to contribute the same number of triangles (5) to the value of T(4,3) as they did to the value of T(4,2).

      Triangle t contributes 1 to the value of T(4,2).

      To triangulate the region defined by triangle t, P must be joined to each of the 3 vertices of triangle t (no other triangulation of this region is possible).

      Thus, the placement of P divides triangle t into 3 triangles for every possible location of P inside triangle t.

      That is, t contributes 1 to the value of T(4,2), but the region defined by t contributes 3 to the value of T(4,3) after the placement of P. To summarize, the value of T(4,k+1) is 2 more than the value of T(4,k) for all k0 since:

      • the (k+1)st interior point may be placed anywhere inside the triangulation for T(4,k) (provided it is not on an edge)

      • specifically, the (k+1)st interior point lies inside a triangle of the triangulation which gives T(4,k)

      • this triangle contributed 1 to the value of T(4,k)

      • after the (k+1)st interior point is placed inside this triangle and joined to each of the 3 vertices of the triangle, this area contributes 3 to the value of T(4,k+1)

      • this is a net increase of 2 triangles, and thus T(4,k+1)=T(4,k)+2, for all k0.

      T(4,0)=2 and each additional interior point increases the number of triangles by 2.

      Thus, k additional interior points increases the number of triangles by 2k, and so T(4,k)=T(4,0)+2k=2+2k for all k0. Using this formula, we get T(4,100)=2+2(100)=202.

    4. In the triangulation of a regular n-gon with no interior points, we may choose any one of the n vertices and join this vertex to each of the remaining n3 non-adjacent vertices.

      All such triangulations of a regular n-gon with no interior points creates n2 triangles, and so T(n,0)=n2 for all n3 (since T(n,0) is constant).

      The reasoning used in part (b) extends to any regular polygon having n3 vertices.

      That is, each additional interior point that is added to the triangulation for n3 vertices and k0 interior points gives a net increase of 2 triangles.

      Thus, T(n,k+1)=T(n,k)+2 for all regular polygons having n3 vertices and k0 interior points.

      So then k additional interior points increases the number of triangles by 2k, and so T(n,k)=T(n,0)+2k=(n2)+2k for all k0. Using this formula T(n,k)=(n2)+2k, we get T(n,n)=(n2)+2n=3n2 and 3n2=2020 when n=20223=674.

    1. Solution 1

      If x0 is even, then x02 is even, and so x1=x02+1 is odd.

      If x1 is odd, then x12 is odd, and so x2=x12+1 is even.

      Thus if x0 is even, then x2 is even and so x2x0 is even.

      If x0 is odd, then x02 is odd, and so x1=x02+1 is even.

      If x1 is even, then x12 is even, and so x2=x12+1 is odd.

      Thus if x0 is odd, then x2 is odd and so x2x0 is even.

      Therefore, for all possible values of x0, x2x0 is even.

      Solution 2

      Using the definition twice and simplifying, we get x2=(x1)2+1x2=((x0)2+1)2+1x2=(x0)4+2(x0)2+2x2=(x0)4+2((x0)2+1)x2x0=(x0)4+2((x0)2+1)x0 To show that x2x0 is even, we may show that (x0)4+2((x0)2+1)x0 is even (since the expressions are equal).

      Since 2((x0)2+1) is the product of some integer and 2, this term is even for all possible values of x0.

      If x0 is even, then (x0)4 is even, and so (x0)4+2((x0)2+1)x0 is the sum and difference of three even terms and thus is even.

      If x0 is odd, then (x0)4 is odd, (x0)4x0 is even, and so (x0)4+2((x0)2+1)x0 is even.

      Thus, x2x0 is even for all possible values of x0.

      Solution 3

      Using the definition twice and simplifying, we get x2=(x1)2+1x2=((x0)2+1)2+1x2=(x0)4+2(x0)2+2x2x0=(x0)4+2(x0)2x0+2x2x0=(x0)4+(x0)2+(x0)2x0+2x2x0=(x0)2((x0)2+1)+x0(x01)+2 To show that x2x0 is even, we may show that (x0)2((x0)2+1)+x0(x01)+2 is even (since they are equal).

      Since x01 is one less than x0, then x0 and x01 are consecutive integers and so one of them is even.

      Thus, the product x0(x01) is even.

      Similarly, (x0)2 is one less than (x0)2+1, and thus these are consecutive integers and so one of them is even.

      Therefore, the product (x0)2((x0)2+1) is even.

      Since (x0)2((x0)2+1)+x0(x01)+2 is the sum of three even integers, x2x0 is even for all possible values of x0.

    2. An integer is divisible by 10 exactly when its units (ones) digit is 0.

      The difference x2026x2020 has units digit 0 exactly when x2026 and x2020 have equal units digits.

      Thus, we will show that for all possible values of x0, the units digit of x2026 is equal to the units digit of x2020, and so x2026x2020 is divisible by 10.

      When a non-negative integer is divided by 10, the remainder is one of the integers from 0 through 9, inclusive.

      Thus for every possible choice for x0, there exists some non-negative integer k, so that x0 can be expressed in exactly one of the following ways: 10k, 10k+1, 10k+2, , 10k+8, 10k+9.

      If for example x0 has units digit 4, then x0=10k+4 for some non-negative integer k, and so x1=(10k+4)2+1=100k2+80k+17=10(10k2+8k+1)+7, and thus has units digit 7.

      Since x1 is determined by x0 only (x1=(x0)2+1), the units digit of x1 is uniquely determined by the units digit of x0.

      For example, if the units digit of x0 is 4, then the units digit of x1 is equal to the units digit of (4)2+1, which equals 7.

      More generally, if the units digit of xi (for a non-negative integer i) is equal to u, then the units digit of xi+1 is equal to the units digit of u2+1.

      (Can you explain why this is true?)

      For example, if we know that x15=29, then the units digit of x16 is equal to the units digit of 92+1=82, which is 2.

      Given that we know all possible units digits of x0, this provides an efficient method for determining the units digits of x1, x2, x3, and so on. In the table below, we list the units digits for the terms x1 through x7 for each of the 10 possible units digits of x0, 0 through 9 inclusive.

      x0 0 1 2 3 4 5 6 7 8 9
      x1 1 2 5 0 7 6 7 0 5 2
      x2 2 5 6 1 0 7 0 1 6 5
      x3 5 6 7 2 1 0 1 2 7 6
      x4 6 7 0 5 2 1 2 5 0 7
      x5 7 0 1 6 5 2 5 6 1 0
      x6 0 1 2 7 6 5 6 7 2 1
      x7 1 2 5 0 7 6 7 0 5 2

      Looking at the table, we see that for each of the possible units digits for x0, the units digit of x1 is equal to the units digit of x7.

      Thus beginning at x1, each column in the table will repeat every 6 terms, and so independent of the starting value x0, xi+6 and xi have equal units digits for all integers i1. Since 20262020=6, then x2026 and x2020 have equal units digits, and so x2026x2020 has units digit 0, and thus is divisible by 10.

    3. Since x115110=(x1155)105, then x115110 is divisible by 105 exactly when x1155 is divisible by 105.

      Further, 105=3×5×7 and each of 3,5,7 is a prime number, and so x1155 is divisible by 105 exactly when it is divisible by 3, 5 and 7.

      Every xi is a multiple of 3, 1 more than a multiple of 3, or 2 more than a multiple of 3.

      Suppose that xi is a multiple of 3. Then xi=3k for some non-negative integer k, and so xi+1=(xi)2+1=(3k)2+1=3(3k2)+1 which is 1 more than a multiple of 3.

      If xi is 1 more than a multiple of 3, then xi=3k+1 for some non-negative integer k, and so xi+1=(xi)2+1=(3k+1)2+1=9k2+6k+2=3(3k2+2k)+2 which is 2 more than a multiple of 3.

      If xi is 2 more than a multiple of 3, then xi=3k+2 for some non-negative integer k, and so xi+1=(xi)2+1=(3k+2)2+1=9k2+12k+5=3(3k2+4k+1)+2 which is 2 more than a multiple of 3.

      Each possible choice of x0 is a multiple of 3, 1 more than a multiple of 3, or 2 more than a multiple of 3.

      If x0 is a multiple of 3, then x1 is 1 more than a multiple of 3 and x2,x3,x4, and so on are each 2 more than a multiple of 3.

      If x0 is 1 or 2 more than a multiple of 3, then x1,x2,x3,x4, and so on are each 2 more than a multiple of 3.

      Therefore, x2,x3,x4, and so on are all 2 more than a multiple of 3 (independent of x0), and so for all i2, xi is a number of the form 3k+2 for some non-negative integer k.

      Therefore, x1155=3k+25=3(k1) is divisible by 3 for all possible choices of x0=n.

      Thus, we need to only determine when x1155 is divisible by 5 and 7.

      For which of the possible values of x0 is x1155 divisible by 5?

      x1155 is divisible by 5 exactly when x115 is divisible by 5.

      Every xi is either a multiple of 5, 1 more than a multiple of 5, 2 more than a multiple of 5, 3 more than a multiple of 5, or 4 more than a multiple of 5. With respect to division by 5, the table below gives the remainders of xi+1 given each of the 5 possible remainders of xi, 0 through 4 inclusive.

      xi xi+1=(xi)2+1
      5k 25k2+1=5(5k2)+1
      5k+1 25k2+10k+2=5(5k2+2k)+2
      5k+2 25k2+20k+5=5(5k2+4k+1)
      5k+3 25k2+30k+10=5(5k2+6k+2)
      5k+4 25k2+40k+17=5(5k2+8k+3)+2

      From this table, we make the following observations:

      • if xi is a multiple of 5, then xi+1 is 1 more than a multiple of 5

      • if xi is 1 more than a multiple of 5, then xi+1 is 2 more than a multiple of 5

      • if xi is 2 more than a multiple of 5, then xi+1 is a multiple of 5

      • if xi is 3 more than a multiple of 5, then xi+1 is a multiple of 5

      • if xi is 4 more than a multiple of 5, then xi+1 is 2 more than a multiple of 5

      Using these observations, we summarize the remainders of x1,x2,x3,x4 when dividing by 5 for each of the possible remainders for x0.

      x0 0 1 2 3 4
      x1 1 2 0 0 2
      x2 2 0 1 1 0
      x3 0 1 2 2 1
      x4 1 2 0 0 2

      With respect to division by 5, we see in the table above that for each of the possible remainders for x0, the remainder for x1 is equal to that of x4.

      Thus beginning at x1, each column in the table repeats every 3 terms, and so independent of the starting value x0, xi+3 and xi have equal remainders after division by 5 for all integers i1.

      Since 115=3(37)+4, then x115 and x4 have the same remainders after division by 5, and so x115 is divisible by 5 for all choices of x0=n which are either 2 more than a multiple of 5 or 3 more than a multiple of 5.

      Finally, we want to determine for which of the possible values of x0 is x1155 divisible by 7.

      x1155 divisible by 7 exactly when x115 is 5 more than a multiple of 7.

      Every xi is exactly one of: a multiple of 7, 1 more than a multiple of 7, 2 more than a multiple of 7, and so on up to 6 more than a multiple of 7. With respect to division by 7, the table below gives the remainders of xi+1 given each of the 7 possible remainders of xi, 0 through 6 inclusive.

      xi xi+1=(xi)2+1
      7k 49k2+1=7(7k2)+1
      7k+1 49k2+14k+2=7(7k2+2k)+2
      7k+2 49k2+28k+5=7(7k2+4k)+5
      7k+3 49k2+42k+10=7(7k2+6k+1)+3
      7k+4 49k2+56k+17=7(7k2+8k+2)+3
      7k+5 49k2+70k+26=7(7k2+10k+3)+5
      7k+6 49k2+84k+37=7(7k2+12k+5)+2

      From this table, we make the following observations:

      • if xi is a multiple of 7, then xi+1 is 1 more than a multiple of 7

      • if xi is 1 more than a multiple of 7, then xi+1 is 2 more than a multiple of 7

      • if xi is 2 more than a multiple of 7, then xi+1 is 5 more than a multiple of 7

      • if xi is 3 more than a multiple of 7, then xi+1 is 3 more than a multiple of 7

      • if xi is 4 more than a multiple of 7, then xi+1 is 3 more than a multiple of 7

      • if xi is 5 more than a multiple of 7, then xi+1 is 5 more than a multiple of 7

      • if xi is 6 more than a multiple of 7, then xi+1 is 2 more than a multiple of 7

      Using these observations, we summarize the remainders of x1,x2,x3,x4 when dividing by 7 for each of the possible remainders for x0.

      x0 0 1 2 3 4 5 6
      x1 1 2 5 3 3 5 2
      x2 2 5 5 3 3 5 5
      x3 5 5 5 3 3 5 5
      x4 5 5 5 3 3 5 5

      Looking at the table, we see that if x0 is 0,1,2,5, or 6 more than a multiple of 7, then xi is 5 more than a multiple of 7 for all i3.

      Also, if x0 is 3 or 4 more than a multiple of 7, then xi is 3 more than a multiple of 7 for all i1 (and thus never 5 more than a multiple of 7).

      Therefore, x1155 is a multiple of 7 exactly when x0 is not 3 or 4 more than a multiple of 7.

      Summary: x115110 is divisible by 105 exactly when

      • x0 is 2 or 3 more than a multiple of 5, and

      • x0 is a multiple of 7 or 1,2,5, or 6 more than a multiple of 7.

      The values of x0 in the range 1x035 satisfying these properties are: 2,7,8,12,13,22,23,27,28,33 The values of x0 in the range 36x0100 satisfying these properties must each be a mutiple of 5×7=35 greater than one the numbers in the above list.

      Thus, there are 10 possible values for x0 in the original list, 10 more from 2+35=37 to 33+35=68, and 9 more from 2+2(35)=72 to 28+2(35)=98, so 29 in total (note that 33+2(35)>100). If Parsa chooses an integer n with 1n100 at random and sets x0=n, the probability that x115110 is divisible by 105 is 29100.