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2022 Canadian Team Mathematics Contest
Solutions

May 2022

© 2022 University of Waterloo

Individual Problems

  1. From the bar graph, 11 students chose chocolate ice cream, 5 students chose strawberry ice cream, and 8 students chose vanilla ice cream.
    The total number of students is 11+5+8=24, so the answer is 1124.

    Answer: 1124

  2. Since EB=BC=CE, BCE is equilateral so BEC=60.
    Since DF is parallel to EB, FDC=BEC=60.
    Since AB is parallel to DC and AD is perpendicular to AB, AD is also perpendicular to DC, so ADC=90.
    Therefore, FDA=ADCFDC=9060=30.

    Answer: 30

  3. Let x the real number satisfying BC=6x.
    From AB:BC=1:2 we get that AB=3x.
    From BC:CD=6:5 we get that CD=5x.
    This means AD=AB+BC+CD=3x+6x+5x=14x.
    We are given that AD=56, so 56=14x or x=4.
    Therefore, AB=3x=3(4)=12.

    Answer: 12

  4. The terms being added and subtracted are the integers 2(1), 2(2), 2(3), and so on up to 2(1011).
    This means there are 1011 terms in total, which is an odd number of terms.
    We will now insert parentheses to group the terms in pairs starting with the first two terms, then the second two terms, and so on.
    Since the total number of terms is odd, 2022 will not be grouped with another term: S=(24)+(68)+(1012)++(20142016)+(20182020)+2022 Each of the parenthetical expressions is equal to 2, and the number of parenthetical expressions is 101112=505.
    Therefore, S=505(2)+2022=1010+2022=1012

    Answer: 1012

  5. Suppose 12n=k2 for some integer k. Then k2 is even and so k must be even, which means k2 is an integer.
    Dividing both sides of 12n=k2 by 4 gives 3n=(k2)2, and since k2 is an integer, this means 3n is a perfect square.
    We are given that 200<n<250, which implies 600<3n<750.
    The perfect squares between 600 and 750 are 625, 676, and 729, among which 729 is the only multiple of 3, so 3n=729 or n=243.
    This value of n satisfies 200<n<250, and 12n=2916=542.
    We have shown that n=243 is the only n that satisfies the conditions, so it must be the largest.

    Answer: 243

  6. Substituting B=3C into D=2BC gives D=2(3C)C=5C.
    Substituting D=5C and B=3C into A=B+D gives A=3C+5C=8C.
    Since A=8C is a two-digit integer, we must have 108C99, so 108C998.
    Since C is an integer, this implies 2C12.
    We also know that C is a two-digit integer, so 10C, which means that C=10, C=11, or C=12.
    Using the equations from earlier, we have A+B+C+D=8C+3C+C+5C=17C, so the larger C is, the larger A+B+C+D is.
    With C=12, we get A=8(12)=96, B=3(12)=36, and D=5(12)=60, which are all two-digit integers.
    Therefore, the conditions are satisfied when C=12, its largest possible value, which means C=12 corresponds to the largest possible value of A+B+C+D.
    Therefore, the answer is A+B+C+D=17C=17(12)=204.

    Answer: 204

  7. We can rewrite the given expression as follows: 3×105002022×104972022=3000×104972022×104972022=(30002022)×104972022=978×104972022=(978×104971)2021 The integer 978×104971 is the 500-digit integer with leading digits 977 followed by 497 9’s.
    Thus, the digits of the given integer, from left to right, are 9, 7, and 7 followed by 4974=493 9’s, and the final four digits are 92=7, 90=9, 92=7, and 91=8.
    Therefore, the sum of the digits of the integer is 9+7+7+(493×9)+7+9+7+8=4491

    Answer: 4491

  8. By rearranging, we get 2n3+3n2+an+bn2+1=2n3+2n+3n2+3+(a2)n+(b3)n2+1=2n(n2+1)+3(n2+1)+(a2)n+(b3)n2+1=2n+3+(a2)n+(b3)n2+1 The quantity 2n+3 is an integer if n is an integer, so the given expression in n is an integer exactly when (a2)n+(b3)n2+1 is an integer.
    Suppose that a20 and consider the functions f(x)=x2+1, g(x)=(a2)x+b3, and h(x)=f(x)g(x)=x2(a2)x+4b.
    The graph of h(x) is a parabola with a positive coefficient of x2, which means that there are finitely many integers n with the property that h(n)0 (there could be no such integers).
    Since h(n)=f(n)g(n), we have shown that there are only finitely many integers n for which g(n)<f(n) is false.
    The graph of g(x) is a line that is neither vertical nor horizontal since a20, so there must be infinitely many integers for which g(n)>0.
    Since infinitely many integers satisfy 0<g(n) and only finitely many integers fail g(n)<f(n), there must exist an integer n with the property that 0<g(n)<f(n).
    This means there exists an integer n for which 0<g(n)f(n)<1 but this means g(n)f(n)=(a2)n+b3n2+1 is not an integer. Therefore, we have shown that if a20, then there is an integer n for which the expression in the problem is not an integer, so we conclude that a2=0 or a2.
    We now have that the expression in the question is equivalent to 2n+3+b3n2+1 which is an integer exactly when b3n2+1 is an integer.
    If b3 is positive, then n can be chosen so that n2+1 is larger than b3, which would mean that the positive expression b3n2+1 is not an integer.
    Similarly, if b3 is negative, then there are integers n for which b3n2+1 is not an integer.
    This shows that if b30, then there are integers n for which the expression in the question is not an integer.
    We can conclude that b3=0, or b=3.
    Therefore, we have that a=2 and b=3, so the expression given in the problem is equivalent to 2n+3+(22)n+(33)n2+1=2n+3 Thus, the answer to the question is 2(4)+3=11.

    Answer: 11

  9. We will use the fact that if two circles are externally tangent, then the line connecting their centres passes through the point of tangency.
    The side lengths of ABC are AB=(31)+(33)=2, AC=(31)+(1+3)=23, and BC=(33)+(1+3)=4.
    Notice that AB2+AC2=22+(23)2=4+12=16=42=BC2 and so the sides of ABC satisfy BC2=AB2+AC2. By the converse of the Pythagorean theorem, ABC has CAB=90.
    Therefore, the area of ABC is 12×AB×AC=12×2×23=23.
    Since CAB=90, the unshaded sector that lies inside the triangle has area 90360=14 of the area of the circle centred at A.
    The area of the circle centred at A is π(31)2=π(423).
    The area of the shaded region is therefore 23π(423)4=23π+12π3 which means a=2, b=1, and c=12, so a+b+c=32.

    Answer: 32

  10. We will count the integers with the desired property by considering four cases, one for each possible leading digit.

    The leading digit is 1: In this situation, we are looking for integers of the form 1000+n where n is an integer from 1 to 999 inclusive so that 1000+n is a multiple of n.
    For 1000+n to be a multiple of n, there must be some integer k so that kn=1000+n, which can be rearranged to get n(k1)=1000.
    Thus, 1000+n will have the desired property exactly when n is a positive factor of 1000 that is less than 1000.
    The positive factors of 1000 that are less than 1000 are 1,2,4,5,8,10,20,25,40,50,100,125,200,250,500 There are 15 integers in the list above, so we get 15 integers in this case.

    The leading digit is 2: Similar to case 1, we need to find all integers n for which 1n999 and 2000+n is a multiple of n.
    The latter condition means there is an integer k with the property that 2000+n=kn which can be rearranged to get 2000=n(k1).
    The positive factors of 2000 that are less than 1000 are 1,2,4,5,8,10,16,20,25,40,50,80,100,125,200,250,400,500 of which there are 18. Therefore, there are 18 integers in this case.

    The leading digit is 3: As in cases 1 and 2, we need to find all integers n for which 1n999 and 3000+n is a multiple of n.
    Again, this means we need to count the positive factors of 3000 that are less than 1000.
    If m is a factor of 3000, then it is either a multiple of 3 or it is not.
    If it is not a multiple of 3, then it is one of the 15 factors of 1000 from case 1.
    Otherwise, it is 3 times one of the factors of 1000 from case 1.
    Among the factors from case 1, all except 500 remain less than 1000 when multiplied by 3.
    Therefore, there are 15+14=29 integers in this case.

    The leading digit is 4: In this case, we need to count the positive factors of 4000 that are smaller than 1000.
    The prime factorization of 4000 is 2553.
    Each factor of 4000 is either a power of 2, 5 times a power of 2, 25 times a power of 2, or 125 times a power of 2.
    Each of 1=20, 2=21, 4=22, 8=23, 16=24, and 32=25 is less than 1000, so 6 of the factors are powers of 2.
    Multiplying each of the powers of 2 in the previous line by 5 gives 5, 10, 20, 40, 80, and 160, each of which is less than 1000.
    Multiplying 5 by each of the factors in the previous line gives 25, 50, 100, 200, 400, and 800, each of which is less than 1000.
    Multiplying 5 by each of the factors in the previous line gives 125, 250, 500, 1000, 2000, 4000, three of which are less than 1000.
    This gives a total of 6+6+6+3=21 factors of 4000 that are less than 1000, so we get 21 integers in this case.

    Thus, the number of integers with the desired property is 15+18+29+21=83.

    Answer: 83

Team Problems

  1. Substituting x=40 into x=2z gives 40=2z so z=20.
    Substituting z=20 into y=3z1 gives y=3(20)1=59.

    Answer: 59

  2. The smallest two positive two-digit multiples of 6 are 12 and 18.
    Since 12 is a multiple of 4 and 18 is not, the answer is 18.

    Answer: 18

  3. Since the first digit of a four-digit integer cannot be 0, there are three integers with the given property: 2022, 2202, and 2220.
    The largest is 2220 and the smallest is 2022, so the answer is 22202022=198.

    Answer: 198

  4. Let Y denote the number of people who answered “yes” and let N denote the number of people who answered “no”.
    The number of people is n, so n=Y+N. As well, since 76% of respondents said “yes”, we have that 76100=Yn=YY+N.
    The equation 76100=YY+N implies 76Y+76N=100Y which can be rearranged to 76N=24Y.
    Dividing both sides by 4 gives 19N=6Y.
    Since Y and N are numbers of people, they are non-negative integers. Furthermore, since a non-zero percentage of people responded with each of “yes” and “no”, they must be positive integers.
    Therefore, the equation 19N=6Y implies that Y is a positive multiple of 19 since 6 and 19 do not have any positive factors in common other than 1.
    The smallest that Y can be is 19, and if Y=19, then N=(6)(19)19=6.
    If Y=19 and N=6, then n=25, which means Yn=1925=0.76 and Nn=625=0.24.

    Therefore, Y=19 and N=6 satisfy the conditions, and we have already argued that Y cannot be any smaller.
    If Y is greater than 19, then n is greater than 2(19)=38 since Y must be a positive multiple of 19. Therefore, the answer is 25.

    Answer: 25

  5. In the diagram below, there are 6 squares shaded in such a way that no two shaded squares share an edge, which shows that the answer is at least 6.

    In the bottom row, the first, third and fifth squares are shaded. In the middle row, the first and third squares are shaded. In the top row, the one square is shaded.

    We will now argue that it is impossible to shade more than 6 square in such a way that no two shaded squares share an edge, which will show that the answer to the question is 6.
    If four or five squares in the bottom row are shaded, then at least two shaded squares will share an edge.
    Therefore, at most 3 squares can be shaded in the bottom row.
    If all three squares in the middle row are shaded, then there will be shaded squares that share an edge.
    Therefore, at most 2 squares can be shaded in the middle row.
    There is only one square in the top row, so at most 1 square can be shaded in the top row.
    Thus, the number of squares that can be shaded so that no two shaded squares share an edge is at most 3+2+1=6.

    Answer: 6

  6. Multiplying both sides of the equation by 12x gives 12+6+4=x, so x=22.

    Answer: 22

  7. An isosceles triangle must have at least two sides of equal length.
    If an isosceles triangle has at least one side of length 10 and at least one side of length 22, then its three side lengths are either 10, 10, and 22 or 10, 22, and 22.
    In any triangle, the sum of the lengths of any two sides must be larger than the length of the other side.
    Since 10+10<22, a triangle cannot have sides of length 10, 10, and 22.
    Therefore, the triangle’s sides have lengths 10, 22, and 22, so the answer is 10+22+22=54.

    Answer: 54

  8. The area of interest is a triangle with vertices at the origin, the x-intercept of y=2x+28, and the point of intersection of the lines with equations y=mx and y=2x+28.

    Two lines are graphed in the xy-plane. The line with equation y equals m times x passes through the origin and goes up to the right. The line with equation y equals negative 2 times x plus 28 goes down to the right and intersects the positive x-axis. The origin, this x-intercept, and the intersection point of the two lines are the vertices of a triangle.

    Setting y=0 in y=2x+28 gives 0=2x+28 which implies x=14.
    Therefore, the base of the triangle has length 140=14.
    Setting mx=2x+28 leads to x=28m+2, and substituting this into y=mx gives y=28mm+2, which is the height of the triangle.
    Using that the area of the triangle is 98, we have that 98=12×14×28mm+2 or 98=196mm+2.
    Dividing both sides of this equation by 98 gives 1=2mm+2, which can be rearranged to get m+2=2m, which implies that m=2.

    Answer: 2

  9. The side length of a cube with volume a3cm3 is acm.
    Since 103=1000 and 43=64, the side length of the original cube is 10cm and the side length of the smaller cube is 4cm.
    The new figure has nine faces: three 10cm×10cm squares, three 4cm×4cm squares, and three faces obtained by removing a 4cm×4cm square from the corner of a 10cm×10cm square.
    Thus, the surface area of the new figure in cm2 is 3(102)+3(42)+3(10242)=600 [Note that this is exactly the surface area of the original cube. Can you see a simple reason why this is true?]
    The shaded region is a 4cm×4cm square, so the ratio we seek is 42:600 or 2:75.
    Therefore, x=75.

    Answer: 75

  10. The expression 8(n1)(n1)(n2) is undefined when n=1. For every integer n1, the expression is equal to 8n2.
    Thus, if n has the property that 8(n1)(n1)(n2) is also an integer, then n must be an integer different from 1 with the property that n2 is a factor of 8.
    The factors of 8 are 8,4,2,1,1,2,4,8 The following table has these values of n2 in the first column, the corresponding values of n in the second column, and the corresponding values of 8(n1)(n1)(n2) in the third column.

    n2 n 8(n1)(n1)(n2)
    8 6 1
    4 2 2
    2 0 4
    1 1 undefined
    1 3 8
    2 4 4
    4 6 2
    8 10 1

    Other than when n=1, we get an integer value of 8(n1)(n1)(n2).
    The sum of the integer values of 8(n1)(n1)(n2) is 124+8+4+2+1=8.

    Answer: 8

  11. Let a be the amount in dollars initially invested in Account A, let b be the amount in dollars initially invested in Account B, and let c be the amount in dollars initially invested in Account C.
    The given information leads to the equations a+b+c=425, 0.05a=0.08b, and 0.08b=0.1c.
    The second and third equations can be solved for a and b, respectively, to get a=85b and b=54c.
    Substituting b=54c into the equation a=85b, we get a=2c.
    Substituting a=2c and b=54c into the equation a+b+c=425 gives 425=a+b+c=2c+54c+c=174c Therefore, the amount in dollars initially invested in Account C is c=417×425=100.

    Answer: $100

  12. Solution 1
    The original line contains the points (0,6) and (1,9).
    Translating these points 3 units up gives the points (0,9) and (1,12).
    Translating these points 4 units to the left gives (4,9) and (3,12).
    Reflecting these points in the line y=x gives (9,4) and (12,3).
    The slope of the line through these two points is 3(4)129=13.
    Substituting x=9 and y=4 into y=13x+b gives 4=3+b or b=7.
    Therefore, the equation of the resulting line is y=13x7, so the answer is 7.
    Solution 2
    The equation of the line obtained by translating up by 3 units is y=3x+6+3 or y=3x+9.
    The equation of the line obtained by translating to the left by 4 units is y=3(x+4)+9 or y=3x+21.
    To find the equation of the line obtained by reflecting the line in y=x, we interchange x and y to get x=3y+21, then solve for y to get y=13x7. Therefore, the answer is 7.

    Answer: 7

  13. The sum of the integers in the list is (m+1)+2(m+2)+3(m+3)+4(m+4)+5(m+5)=15m+1+4+9+16+25=15m+55 The number of integers in the list is (m+1)+(m+2)+(m+3)+(m+4)+(m+5)=5m+15 The average of the integers in the list is 196, so this means 196=15m+555m+15=3m+11m+3 The above equation is equivalent to 19(m+3)=6(3m+11), or 19m+57=18m+66, which can be solved for m to get m=9.

    Answer: 9

  14. Suppose Devi was given the bill for Bohan’s meal. From the second bullet point, this would mean that Bohan and Devi were given each other’s bills.
    This means that Ann’s, Che’s, and Eden’s bills were distributed among each-other in some way.
    Che was given Ann’s bill by the fourth bullet point, so if Ann was given Che’s bill, then Eden would have been given her own bill.
    Since nobody was given their own bill, Ann must have been given Eden’s bill, so the only bill remaining for Eden to have been given is Che’s.
    Thus, assuming that Devi was given Bohan’s bill, we have deduced that Ann was given Eden’s bill, Eden was given Che’s bill, and Che was given Ann’s bill.
    By the first bullet point, Che and Eden ordered meals with the same cost.
    Since Eden was given Che’s bill, the third bullet point is violated.
    This means that Devi was not given Bohan’s bill.
    Because Che was given Ann’s bill, Devi was not given Ann’s bill.
    As well, Devi could not have been given Che’s bill because this would imply, by the second bullet point, that Che was given Devi’s bill, which would violate the fourth bullet point.

    By eliminating all other possibilities, we have deduced that Devi’s and Eden’s bills were interchanged.
    This means that the remaining three bills, belonging to Ann, Bohan, and Che, were distributed among the same three people.
    Similar to the reasoning from before, since nobody was given their own bill and Che was given Ann’s bill, the only possibility is that Bohan was given Che’s bill and Ann was given Bohan’s bill.

    Answer: Ann

  15. Solution 1

    The line segment PQ has slope 39=13.
    Since PQR=90, the slope of QR is the negative reciprocal of 13, which is 3.
    The line segment QR has slope 3(3)x11=6x11, so 6x11=3.

    This implies 6=3x33, so 3x=39 or x=13.

    Solution 2

    In any right-angled triangle, the three vertices are equidistant from the midpoint of the hypotenuse.
    The right angle is at Q, so this means that the hypotenuse of PQR is PR.
    The coordinates of P are (2,0) and the coordinates of R are (x,3), so the coordinates of the midpoint of PR is at (2+x2,32).
    The length of PR is (x2)2+32=x24x+13 The distance from Q to the midpoint of PR is half of x24x+13, which leads to the equations 12x24x+13=(112+x2)2+(332)2x24x+13=2(20x2)2+(92)2x24x+13=22(20x2)2+22(92)2x24x+13=(20x)2+(9)2x24x+13=x240x+400+8136x=468x=13 It can be checked that if the coordinates of R are (13,3), that the side lengths of PQR are 40, 90, and 130, which are indeed the lengths of the sides of a right-angled triangle.

    Answer: 13

  16. Using that 3 is a solution to x27x+k=0 we have that 327(3)+k=0 or k=219=12.
    This means x27x+k is x27x+12=(x3)(x4), the roots of which are 3 and 4, which means a=4.
    The polynomial x28x+k+1 factors as (xb)(xc), so upon expanding the latter expression, we have bc=k+1=12+1=13.
    Therefore, a+bc=4+13=17.

    Answer: 17

  17. For positive x, g(f(x))=log3(9f(x))=log3(9×9x)=log3(9x+1)=log3((32)x+1)=log3(32x+2)=(2x+2)log3(3)=2x+2 A similar calculation shows that f(g(2))=9g(2)=9log3(18)=32log3(18)=(3log3(18))2=182=324 Therefore, 2x+2=324 so x=32422=161.

    Answer: 161

  18. If 2022x=1, then x=0. This means we have 2sin2θ3sinθ+1=0.
    The expression on the left can be factored to get (sinθ1)(2sinθ1)=0.
    Therefore, the given equation is true exactly when sinθ=1 or sinθ=12.
    Since 0<θ<360, the only value of θ with sinθ=1 is θ=90.
    The only values of θ with sinθ=12 are θ=30 and θ=150.
    Therefore, the answer is 90+30+150=270.

    Answer: 270

  19. The distance from point A to each of points B, D, and I is 1.
    Since ABCD is a square of side length 1, the Pythagorean theorem implies that the distance from A to C is 12+12=2.
    A similar calculation shows that the distance from A to K is 2 and the distance from A to O is 2.
    To compute the distance from A to M, observe that AOM is right-angled at O with OM=1, so the distance from A to M is AO2+OM2=2+1=3.
    The distance from A to J is 2 which is greater than 2.
    Notice that J is the closest point to A among the points J, L, N, P, E, F, G, and H, so none of these points can be a distance of 2 from A.
    Therefore, the points that are at a distance of 2 from A are C, K, and O, for a total of 3.
    By symmetry, there will be a total of 3 points at a distance of 2 from each of B, C, D, E, F, G, and H.
    Next consider point K. By reasoning similar to that which is above, each of E, F, G, and H has a distance of at least 2>2 from K.
    Applications of the Pythagorean theorem can be used to show that the points that are 2 away from K are A, C, O, N, and J, for a total of 5 points.
    By symmetry, there are a total of 5 points at a distance of 2 from each of I, J, L, M, N, O, and P.
    We have counted 3 points at a distance of 2 from each of the 8 points A through H, and 5 points at a distance of 2 from each of the 8 points from I through P.
    This is a total of 8×3+8×5=64.
    The lengths of the line segments were counted once for each end point, which means the total of 64 counts every distance of 2 twice.
    Therefore, the answer is 642=32.

    Answer: 32

  20. Label the centre of the circle by O, the point of tangency of AB with the circle by E, and the point of intersection of OE with CD by F:

    Let EF=x. Then AD=x, so AB=4x, which means CD=4x because ABCD is a rectangle.
    Using circle properties, OE is perpendicular to AB, and since AB is parallel to CD, we also have that OE, and hence, OF, is perpendicular to CD.
    We now have that OFD is right-angled at F, and OFC is also right-angled at F.
    Since OC and OD are radii of the same circle, they must be equal.
    Triangles OFD and OFC also share side OF, so they are congruent by hypotenuse-side similarity.
    Therefore, F is the midpoint of CD, so DF=2x.
    It is given that the radius of the circle is 5, which implies that OF=5x and OD=5.
    By the Pythagorean theorem, OF2+DF2=OD2, and after substituting the lengths computed earlier, this leads to the following equivalent equations: (5x)2+(2x)2=52525x+x2+4x2=55x225x=0x(5x25)=0 this implies that either x=0 or 5x=25.
    We cannot have x=0 since x is the side-length of a rectangle. Therefore, 5x=25 or x=25.
    The area of ABCD is x(4x)=25×85=165.

    Answer: 165

  21. Solution 1

    Suppose ABCD has side length x and let G be on AC so that EG is perpendicular to AC. Set GE=y.

    When the paper is folded, B lands on AC and E does not move.
    This means that BAE=GAE. As well, AGE=ABE=90 by construction.
    Since ABE and AGE have two angles in common, they must have three angles in common.
    They also share side AE, so they must be congruent.
    The area of AECF is the sum of the areas of AEC and AFC.
    By symmetry, AFC is congruent to AEC, which means the area of AECF is 2(12×AC×GE)=(AC)(GE) The length of AC is 2x since it is a diagonal of a square with side length x.
    Therefore, the area of AECF is 2xy.
    The area of the shaded region is twice the area of AGE, and since AGE is congruent to ABE, we have AG=AB=x, so the shaded region has area 2(12×AG×GE)=(AG)(GE)=xy Therefore, the fraction we seek is xy2xy=12.

    Solution 2

    Let the side length of the square be 1, set BE=x, and define G as in Solution 1.
    Using the argument from Solution 1, we get that GE=x as well.
    Since AC is the diagonal of a square, ABC is isosceles with ABC=90, which means ACE=45.
    Since CGE=90 by construction, CGE is an isosceles right triangle, so CE=2GE.
    We also have that CE=BCBE=1x and GE=x, so this means 1x=2x.
    Solving for x, we get x=12+1=12+1×2121=21.

    Using the argument from Solution 1, ABE, AGE, AGF, and ADF are all congruent.
    This means the area of the shaded region is equal to twice the area of ABE, which is 2(12×AB×BE)=x=21 The area of AEFC is the area of the square minus the combined area of ABE and ADF.
    This means the area of AEFC is 1 minus twice the area of ABE, or 1(21)=22.
    Therefore, the ratio is 2122=212(21)=12

    Answer: 12

  22. Factoring x2+xzxyyz, we have x(x+z)y(x+z) or (xy)(x+z).
    Therefore, the integers x, y, z, and p satisfy (xy)(x+z)=p or (yx)(x+z)=p.
    Because p is a prime number and yx and x+z are integers, it must be that (yx,x+z) is a factor pair of p.
    Observe that y+z=(yx)+(x+z), so y+z must be equal to the sum of the integers in a factor pair of p.
    Since p is a prime number, there are only two ways to express p as the product of two integers, and they are p=(1)×(p) and p=1×p.
    The sum of the factors in these products are p1=(p+1) and p+1, respectively.
    Therefore, y+z=±(p+1), so |y+z|=p+1.

    Answer: p+1

  23. We can categorize the 64 small cubes into four groups: 8 cubes that are completely in the interior of the larger cube (and are completely invisible), 4×6=24 “face cubes” that have exactly one of their faces showing, 2×12=24 “edge cubes” that have exactly two faces showing, and 8 “corner cubes” that have 3 faces showing.
    Notice that 8+24+24+8=64, so this indeed accounts for all 64 small cubes.
    The diagram below has all visible faces of the smaller cubes labelled so that faces of edge cubes are labelled with an E, faces of face cubes are labelled with an F, and faces of corner cubes are labelled with a C:

    Each visible face of the cube is divided into 4 rows of 4 squares. The squares in the first row and the fourth row, in order, are labelled C, E, E, and C. The squares in the second row and the third row are labelled E, F, F, and E.

    Each of the 8 corner cubes has 3 of its faces exposed.
    Because of the way the faces are labelled on the small cubes, each of the 8 corner cubes will always contribute a total of 1+1+2=4 to the sum of the numbers on the outside.
    Regardless of how the small cubes are arranged, the corner cubes contribute a total of 8×4=32 to the total of all numbers on the outside of the larger cube.
    The edge cubes can each show either a 1 and a 1 or a 1 and a 2, for a total of either 2 or 3.
    Thus, each edge cube contributes a total of either 2 or 3 to the total on the outside of the larger cube.
    This means the smallest possible total that the edge cubes contribute is 24×2=48, and the largest possible total that they contribute is 24×3=72.
    Notice that it is possible to arrange the edge cubes to show any total from 48 to 72 inclusive.
    This is because if we want the total to be 48+k where k is any integer from 0 to 24 inclusive, we simply need to arrange exactly k of the edge cubes to show a total of 3.
    Each of the face cubes shows a total of 1 or a total of 2. There are 24 of them, so the total showing on the face cubes is at least 24 and at most 24×2=48.
    Similar to the reasoning for the edge cubes, every integer from 24 to 48 inclusive is a possibility for the total showing on the face cubes.
    Therefore, the smallest possible total is 32+48+24=104, and the largest possible total is 32+72+48=152.
    By earlier reasoning, every integer between 104 and 152 inclusive is a possibility, so the number of possibilities is 152103=49.

    Answer: 49

  24. The sum of the positive integers from 1 to 9 inclusive is 45. Since each row sum must be the same, this means each row must be one third of 45 which is 15.
    In the second row, the first two cells contain the integers 1 and 8, so the third must contain 1581=6.
    The integers that still need to be placed are 2, 3, 4, 5, 7, and 9.
    Notice that 159=6 and that 2 and 4 are the only two integers in this list that have a sum of 6.
    Therefore, the integers 9, 4, and 2 must be in one of the remaining rows and the integers 3, 5, and 7 must be in the other.
    In fact, if the integers are placed in such a way that all conditions are satisfied and the top and bottom rows are exchanged, all properties will still be satisfied and the column products will not have changed.
    Therefore, we can assume that 2, 4, and 9 are in the top row.
    This gives six possible ways to fill in the top two rows. They are shown below.

    2 4 9
    1 8 6
    2 9 4
    1 8 6
    4 2 9
    1 8 6
    4 9 2
    1 8 6
    9 2 4
    1 8 6
    9 4 2
    1 8 6

    The digits that remain to be placed are 3, 5, and 7.
    We will first focus on the first partially filled in grid in which the first row contains 2, 4, and 9 from left to right.
    The third column already contains 9 and 6. The integer to be placed in the third column in the bottom row is at least 3, so this means the third column product is at least 9×6×3=162.
    By similar reasoning, the first column product is at most 2×1×7=14.
    Therefore, no matter how the remaining integers are placed, the difference between the largest and smallest column product will be at least 16214=148, which is greater than 40.
    Therefore, the 2, 4, and 9 cannot be placed in that order.
    In the second grid, the second column product is at least 9×8×3=216 and the first column product is at most 2×1×7=14, so the difference between the largest and smallest column product is at least 21614=202 which is greater than 40, so the 2, 4, and 9 cannot be placed in the order 2, 9, 4, either.
    By similar reasoning, the third and fourth possibilities above can be eliminated.
    Therefore, the top row must contain, from left to right, either 9, 2, and 4, or 9, 4, and 2.
    If the 5 is placed in the third column of the grid below,

    9 2 4
    1 8 6

    then the third column product will be 4×6×5=120.
    This would mean that the largest the first column product can be is 9×1×7=63.
    This would lead to the largest and smallest column products having a difference greater than 40, so 5 cannot be placed in the third column.
    If 7 is placed in the third column, the difference between the largest and smallest column products would be even larger, so the 3 must be placed in the third column.
    By similar reasoning, the 3 must be placed in the second column of the last grid.
    We have now reduced to four possible ways to place the integers, and they are shown below with the column products given below each column.

    The first column has the integers 9, 1, 5 with a product of 45. The second has 2, 8, 7 with product 112 and the third has 4, 6, 3 with product 72. The first column has the integers 9, 1, 7 with a product of 63. The second has 2, 8, 5 with product 80 and the third has 4, 6, 3 with product 72. The first column has the integers 9, 1, 5 with a product of 45. The second has 4, 8, 3 with product 96 and the third has 2, 6, 7 with product 84. The first column has the integers 9, 1, 7 with a product of 63. The second has 4, 8, 3 with product 96 and the third has 2, 6, 5 with product 60.

    The column products in the first grid are 45, 112, and 72. Since 45 and 112 differ by more than 40, this grid does not satisfy the conditions.
    The column products in the second grid are 63, 80, and 72.
    The difference between the largest and smallest column product is 8063=17 which is less than 40, so this grid satisfies all of the conditions.
    Its largest column product is 80, so 80 is one of the possibilities.
    The column products in the third grid are 45, 96, and 84. Since 96 and 45 differ by more than 40, this grid does not satisfy the conditions.
    The column products in the fourth grid are 63, 96, and 60.
    The difference between the largest and smallest column product is 9660=36 which is less than 40, so this grid satisfies all of the conditions.
    Its largest column product is 96, so 96 is another possibility.
    We have shown that there are exactly two ways to arrange the integers so that the conditions are satisfied, and the largest column products in these two arrangements are 80 and 96.

    Answer: 80, 96

  25. The solution will use two facts. The first is a general fact about triangles.

    Fact 1: Suppose QRS has X on QR and Y on SR so that XR=2QX and YR=2SY. If Z is the point of intersection of QY and SX and W is the point where the extension of RZ intersects QS, then WR=5WZ.

    The second fact that we will use is specific to the question.

    Fact 2: The point M is on the line segment connecting D to E so that MD=2EM.
    (Note: this justifies the fact that EH and CM intersect as implied by the problem statement.)

    We will answer the question before proving the two facts.
    To start, we will show that the altitude of tetrahedron ABCD is 5 times that of EBCP.
    Let K be the point where the line through DP intersects ABC and let L be the point in ABC so that DL is perpendicular to ABC, and let V be the point on ABC so that PV is perpendicular to ABC.
    By construction, DL is the altitude of ABCD from A to ABC and PV is the altitude of EBCP from P to EBC.

    By Fact 2, M is on DE so that MD=2EM and H is on DC so that HD=2CH by assumption.
    By Fact 1, KD=5KP.
    Since KPV and KDL share an angle at K and both have a right angle, the triangles are similar.
    This means PVDL=KPKD=15.
    Thus, the length of the altitude of ABCD is five times that of EBCP.
    Since E is the midpoint of AB, the area of EBC is half that of ABC.
    Thus, the base area of tetrahedron EBCP is half that of tetrahedron ABCD, while the heights are in ratio 1:5, which means that the volumes are in the ratio 1:10.
    Now for the proofs of the facts.

    Proof of Fact 1

    We will first show that the height of QZS from Z is one fifth the height of QRS from R.
    Suppose QZS has height h1 from Z and QRS has height h2 from R. The first goal is to show that h1h2=15.
    The area of QZS is 12(QS)h1 and the area of QRS is 12(QS)h2.
    Thus, if we take the ratio of their areas we get 12(QS)h112(QS)h2=h1h2 and so the ratio of their heights is the same as the ratio of their areas.
    This means we can show that h1h2=15 by showing that the area of QRS is 5 times the area of QZS.
    To do this, draw a line from R to Z and label the areas of QXZ, RXZ, RYZ, SYZ, and QZS by A1, A2, A3, A4 and A5, respectively.

    We will now make several observations about the areas A1 through A5.
    Since QXZ and RXZ have the same altitude from Z to QR and their bases satisfy XR=2QX, we get that (1)A2=2A1 By similar reasoning, we also get the equations (2)A3=2A4(3)A2+A3+A4=2(A1+A5)=2A1+2A5(4)A1+A2+A3=2(A4+A5)=2A4+2A5 Substituting Equation (1) into Equation (3) gives A2+A3+A4=A2+2A5 which simplifies to A3+A4=2A5.
    Substituting Equation (2) into Equation (4) gives A1+A2+A3=A3+2A5 which simplifies to A1+A2=2A5.
    Adding A3+A4=2A5 and A1+A2=2A5 gives A1+A2+A3+A4=4A5.
    Adding A5 to both sides gives A1+A2+A3+A4+A5=5A5, which exactly says that the area of QRS is five times that of QZS.
    We will now let T be the point on QS so that RT is perpendicular to QS and let U be on WT so that ZU is perpendicular to RW.

    Note that ZU is the height of QZS from Z and RT is the height of QRS from R, so we know that RT=5ZU.
    Since WUZ=WTR=90 and ZWU=RWT, ZWU is similar to RWT.
    Since ZURT=15, we get that WZWR=15, which proves the fact.

    Proof of Fact 2
    Consider ADB and FDG. It is given that FDAD=23=GDBD, and since the triangles have a common angle at D, they are similar by side-angle-side similarity.
    Let N be the point at which DE intersects FG.

    Since ADB is similar to FDG, DFN=DAE.
    We also have that FDN=ADE since they are the same angle, which implies that FDN is similar to ADE since if two triangles have two angles in common, they must have three angles in common.
    Since FDAD=23, similarity implies FNAE=23.
    By similar reasoning, GNBE=23.
    Rearranging these two equations leads to FN=23AE and GN=23BE, but E is the midpoint of AB, so AE=BE, and hence, FN=GN.
    This means N is the midpoint of FG, so M=N, which shows that M is on DE since N is on DE.
    Finally, since FDM and ADE are similar, that MDED=FDAD=23, which means that M is on DE so that MD=2EM.
    Answer:

    110

Relay Problems

(Note: Where possible, the solutions to parts (b) and (c) of each Relay are written as if the value of t is not initially known, and then t is substituted at the end.)

    1. Evaluating, 2+5×53=2+253=273=9.

    2. The area of a triangle with base 2t and height 2t6 is 12(2t)(2t6) or t(2t6).
      The answer to (a) is 9, so t=9 which means t(2t6)=9(12)=108.

    3. Since ABC is isosceles with AB=BC, it is also true that BCA=BAC.
      The angles in a triangle add to 180°, so 180°=ABC+BAC+BCA=ABC+2BAC=t°+2BAC The answer to (b) is 108, so t=108. Therefore, BAC=12(180°t°)=12(180°108°)=12(72°)=36°.

    Answer: 9, 108, 36°

    1. Since 33 is positive, 11<23 implies 33(11)<33(23) which simplifies to 3×<22.
      The largest multiple of 3 that is less than 22 is 3×7, so this means the number in the box cannot be larger than 7.

      Indeed, 711=2133 is less than 23=2233, so the answer is 7.

    2. Rearranging the equation, we have 2x=t9 or x=t92, so x+4=t92+4=t9+82=t12 Substituting t=7 into this equation gives x+4=712=4.

    3. The vertices of the triangle are the origin and the points at which the line intersects the x- and y-axes.
      When y=0, we get 0=tx+6 so x=6t.
      When x=0, we get y=t(0)+6 so y=6.
      The triangle is right-angled at the origin, its height is 6, and its base the distance from 6t to the origin.
      If t is negative, then the base is 6t, and if t is positive, then the base is 6t.
      Therefore, if t is negative, then the area of the triangle is 12×6×6t=18t and if t is positive, then the area of the triangle is 12×6×6t=18t Since t=4 which is negative, the area of the triangle is 184=92.

    Answer: (7,4,92)

    1. The prime numbers between 10 and 30 are 11, 13, 17, 19, 23, and 29, which means x=6.

      Factoring the numerator in x24x+2 gives (x2)(x+2)x+2 which is equal to x2 as long as x2.
      Since x=6, the answer is 62=4.

    2. Let A be the number of beans that Alida has, B be the number of beans that Bono has, and C be the number of beans that Cate has.
      The given information translates to A+B=6t+3A+C=4t+5B+C=6t Adding these three equations together gives 2(A+B+C)=16t+8, which implies A+B+C=8t+4.
      Subtracting A+C=4t+5 from A+B+C=8t+4 gives B=(A+B+C)(A+C)=(8t+4)(4t+5)=4t1 Substituting t=4 into B=4t1 gives B=4(4)1=15.

    3. Suppose x is the real number such that x2tx+36=0 and x28x+t=0.
      Therefore, we get that x2tx+36=x28x+t which can be rearranged to get 36t=tx8x=(t8)x.
      Dividing by t8 gives x=36tt8.
      Substituting t=15 gives x=3615158=217=3.

    Answer: (4,15,3)

    1. A cube with edge length x has six square faces with area x2, so its surface area is 6x2.
      This means 6x2=1014 so x2=169.
      Since x is an edge length, it is positive, so x=169=13.

    2. Multiplying through by 3(t+x) gives 3(5+x)=2(t+x).
      Expanding gives 15+3x=2t+2x which can be rearranged to x=2t15.
      Substituting t=13 gives x=2(13)15=11.

    3. Let E be on AD so that BE is perpendicular to AE.

      Quadrilateral BCDE has three right angles, which means that it must have four right angles.
      This means that quadrilateral BCDE is a rectangle, so BE=CD=t+13.
      We also have that ED=BC=4, so AE=ADDE=t4.
      By the Pythagorean theorem applied to AEB, we get AB2=(t4)2+(t+13)2=2t2+18t+185 Since AB is positive, AB=2t2+18t+185.
      Since t=11, AB=2(121)+18(11)+185=242+198+185=625=25

    Answer: (13,11,25)