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2022 Hypatia Contest
Solutions
(Grade 11)

Tuesday, April 12, 2022
(in North America and South America)

Wednesday, April 13, 2022
(outside of North American and South America)

©2022 University of Waterloo


    1. Regular hexagon ABCDEF has side length 2x, and so AB=2x.
      Since OAB is equilateral, then OA=OB=AB=2x.
      The radius of the circle is equal to OA and thus is 2x.

    2. Since M is the midpoint of AB and OA=OB, then OM is perpendicular to AB.
      Since M is the midpoint of AB, then AM=12AB=x.
      Using the Pythagorean Theorem in right-angled OAM, we get OA2=OM2+AM2 or (2x)2=OM2+x2, and so OM2=3x2 or OM=3x (since OM>0).
      Alternatively, notice that OAM is a 30-60-90 triangle, and so AM:OA:OM=1:2:3=x:2x:3x.

    3. The diagonals AD, BE and CF divide ABCDEF into six congruent equilateral triangles.
      Thus the area of ABCDEF is six times the area of OAB (having base AB and height OM), or 6×12×AB×OM=3×2x×3x=63x2

    4. The area of the shaded region is determined by subtracting the area of ABCDEF from the area of the circle with centre O and radius 2x.
      Thus, the area of the shaded region is π(2x)263x2=4πx263x2=(4π63)x2 The area of this shaded region is 123, and so (4π63)x2=123 or x2=1234π63.
      Since x>0, we get x=1234π63 and so x=7.5 when rounded to the nearest tenth.

    1. With 1 kg of muffin batter, exactly 24 mini muffins and 2 large muffins can be made. Thus with 2 kg of muffin batter, exactly 2×24=48 mini muffins and 2×2=4 large muffins can be made, and so n=4.

    2. Solution 1

      With 2 kg of muffin batter, exactly 36 mini muffins and 6 large muffins can be made. With 2 kg of muffin batter, exactly 48 mini muffins and 4 large muffins can be made. Adding these, we get that with 2 kg+2 kg=4 kg of muffin batter, exactly 36+48=84 mini muffins and 6+4=10 large muffins can be made, and so x=4.

      Solution 2

      Let m be the number of kilograms of muffin batter needed to make 1 mini muffin. Let be the number of kilograms of muffin batter needed to make 1 large muffin. Since 1 kg of muffin batter makes exactly 24 mini muffins and 2 large muffins, then 1=24m+2.
      Since 2 kg of muffin batter makes exactly 36 mini muffins and 6 large muffins, then 2=36m+6.
      Subtracting the second equation from 3 times the first equation, we get 3×12=3×(24m+2)(36m+6) or 1=36m, and so m=136.

      Substituting m=136 into the second equation, we get 2=36(136)+6 or 1=6, and so =16.

      Since 136 kg of muffin batter is needed to make 1 mini muffin, then 84136=73 kg of muffin batter is needed to make 84 mini muffins.
      Since 16 kg of muffin batter is needed to make 1 large muffin, then 1016=53 kg of muffin batter is needed to make 10 large muffins.
      Thus, exactly 84 mini muffins and 10 large muffins can be made with 73+53=123=4 kg of muffin batter, and so x=4.

    3. In part (b) Solution 2, it was determined that 136 kg of muffin batter is needed to make 1 mini muffin, and 16 kg of muffin batter is needed to make 1 large muffin.
      Therefore, the number of kilograms of muffin batter needed to make 1 large muffin is 6 times the number of kilograms of batter needed to make 1 mini muffin (6×136=16).
      In other words, the batter needed to make 1 large muffin can make 6 mini muffins.
      Then, using the amount of batter that makes 7 large muffins, 6×7=42 mini muffins can be made.

    1. If the first number in a sequence is 3 and the sequence is generated by the function x23x+1, then the second number in the sequence is 323(3)+1=1, and the third number in the sequence is 123(1)+1=1, and the fourth number in the sequence is (1)23(1)+1=5. The first four numbers in the sequence are 3,1,1,5.

    2. Let the first and second numbers in the sequence generated by the function x24x+7 be f and s, respectively.
      Then, the first three numbers in the sequence are f,s,7.
      Since the third number in the sequence is 7, then s24s+7=7.
      Solving this equation, we get s24s=0 or s(s4)=0, which has solutions s=0 and s=4, and so the first three numbers in the sequence could be f,0,7 or f,4,7.
      If the second number in the sequence is 0, then f24f+7=0.
      The discriminant of this equation is (4)24(1)(7)=12 (less than zero) and so there are no real solutions.
      Thus, there is no first number in this sequence for which the second number is 0.

      If the second number in the sequence is 4, then f24f+7=4, or f24f+3=0 and so (f1)(f3)=0, which has solutions f=1 and f=3.
      Therefore, if 7 is the third number in a sequence generated by the function x24x+7, then the first three numbers in the sequence could be 1,4,7 or 3,4,7, and so the possible first numbers in the sequence are 1 and 3.

    3. The first two numbers in the sequence are c,c, and so c27c48=c.
      Solving this equation, we get c28c48=0 or (c+4)(c12)=0, which has solutions c=4 and c=12.

    4. The first number in the sequence is a and the second number is b, and so a212a+39=b. The second number in the sequence is b and the third number is a, and so b212b+39=a. Subtracting the second equation from the first and simplifying, we get (a212a+39)(b212b+39)=baa2b212a+12b=baa2b211a+11b=0(ab)(a+b)11(ab)=0(ab)(a+b11)=0 Since ab, then ab0 and so a+b11=0 or b=11a.
      Substituting into the first equation, we get a212a+39=11a or a211a+28=0. Factoring gives (a4)(a7)=0 and so the possible values of a are 4 and 7.
      (Note that the two possible sequences are 4,7,4, and 7,4,7,.)

    1. Written as a product of its prime factors, 240=243151 and so f(240)=(1+4)(1+1)(1+1)=(5)(2)(2)=20

    2. Suppose f(N)=6 and N is refactorable. Then 6 divides N.
      N is divisible by 6 exactly when its prime factors include at least one 2 and at least one 3 (since 6=2×3).
      Assume that N contains an additional prime factor p, distinct from 2 and 3.
      In this case, the divisors of N are 1,2,3,6,p,2p,3p, and 6p, which is too many divisors.
      Thus, it must be that N is a positive integer of the form N=2u3v, where u and v are positive integers, and so f(N)=(1+u)(1+v)=6.
      Since u and v are positive integers, then 1+u2 and 1+v2 and so there are exactly two possibilities: u=1 and v=2 or u=2 and v=1.
      When u=1 and v=2, N=2132=18 and when u=2 and v=1, N=2231=12.
      The refactorable numbers N with f(N)=6 are 12 and 18.

    3. Since N is refactorable, then f(N)=256=28 divides N.
      Thus for some integer w8, N is of the form N=2wp1a1p2a2pkak, where 2<p1<p2<<pk are prime numbers, and a1,a2,,ak are positive integers.
      In this case, f(N)=(1+w)(1+a1)(1+a2)(1+ak)=28, and so each of the factors 1+w,1+a1,1+a2,,1+ak is a power of 2 (since their product is 28).
      This means that each of the exponents w,a1,a2,,ak is one less than a power of two.
      Since w8, then 1+w9 and so 1+w24 (the smallest power of 2 greater than 8).
      Since f(N)=(1+w)(1+a1)(1+a2)(1+ak)=28, then 241+w28 and 1(1+a1)(1+a2)(1+ak)24, and so w must be equal to 15,31,63,127, or 255, and each of a1,a2,,ak must be equal to 1,3,7, or 15.
      For example, if 1+w=28, then w=2561=255 and N=2255.
      If 1+w=24, then w=15 and N is of the form N=215p1a1p2a2pkak.
      Thus, the smallest N having the greatest number of distinct prime factors is N=215315171111, and we may confirm that f(N)=(1+15)(1+1)(1+1)(1+1)(1+1)=242222=28, as required.
      Comparing these first two values of N, we recognize that 215315171111 is significantly less than 2255.
      Further, we recognize that

      • all remaining possible values of N have exactly 2, 3 or 4 distinct prime factors,

      • where exponents are equal, smaller prime factors give smaller values of N,

      • the greatest exponents must occur on the smallest prime factors, and

      • we recall that each exponent is one less than a power of 2.

      In the table below, we use the above information to determine the smallest possible values of N having 1, 2, 3, 4, and 5 distinct prime factors.
      Further, we compare the size of each of these values of N within each of the 5 groups.

      Number of distinct prime factors of N Values of N
      1 2255
      2 215315<23137<26333<212731
      3 2153353<2153751<2313351<2633151
      4 215335171<231315171
      5 215315171111

      Finally, we compare the smallest values of N from each of the 5 rows in the table above.
      Since 215335171<215315171111<2153353<215315<2255, then the smallest refactorable number N with f(N)=256, is N=215335171=30965760.

    4. Let m be a positive integer of the form m=p1a1p2a2pkak, where p1<p2<<pk are prime numbers, and a1,a2,,ak are positive integers.
      We begin by stating a value of N and then proceed to show that it works.

      For each m=p1a1p2a2pkak, choose N=p1(p1a11)p2(p2a21)pk(pkak1).
      On the contest paper, the useful fact states that 2nn+1 for all positive integers n.
      Since p2 for all prime numbers p, then pn2nn+1.
      Since pnn+1, then pn1n, and so piai1ai for all integers i where 1ik.
      Thus, p1a1 divides p1(p1a11) since each is a power with base p1 and p1a11a1.
      Similarly, p2a2 divides p2(p2a21), and so on.
      Therefore, m=p1a1p2a2pkak divides N=p1(p1a11)p2(p2a21)pk(pkak1).

      Further, f(N)=(1+p1a11)(1+p2a21)(1+pkak1)=p1a1p2a2pkak=m.
      Thus for every integer m>1, there exists a refactorable number N such that f(N)=m.