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2020 Fermat Contest
Solutions
(Grade 11)

Tuesday, February 25, 2020
(in North America and South America)

Wednesday, February 26, 2020
(outside of North American and South America)

©2020 University of Waterloo


  1. Since OPQR is a rectangle with two sides on the axes, then its sides are horizontal and vertical. Since PQ is horizontal, the y-coordinate of Q is the same as the y-coordinate of P, which is 3. Since QR is vertical, the x-coordinate of Q is the same as the x-coordinate of R, which is 5. Therefore, the coordinates of Q are (5,3).

    Answer: (B)

  2. Calculating, 3×2020+2×20204×2020=2020×(3+24)=2020×1=2020 Alternatively, 3×2020+2×20204×2020=6060+40408080=101008080=2020

    Answer: (E)

  3. Expanding and simplifying, (x+1)2x2=(x2+2x+1)x2=2x+1.

    Answer: (A)

  4. Ewan’s sequence starts with 3 and each following number is 11 larger than the previous number. Since every number in the sequence is some number of 11s more than 3, this means that each number in the sequence is 3 more than a multiple of 11. Furthermore, every such positive integer is in Ewan’s sequence. Since 110=11×10 is a multiple of 11, then 113=110+3 is 3 more than a multiple of 11, and so is in Ewan’s sequence. Alternatively, we could write Ewan’s sequence out until we get into the correct range: 3,14,25,36,47,58,69,80,91,102,113,124,

    Answer: (A)

  5. Calculating, (81+812=9+92=9=3.

    Answer: (A)

  6. Since 12 and 21 are multiples of 3 (12=4×3 and 21=7×3), the answer is not (A) or (D). 16 is a perfect square (16=4×4) so the answer is not (C). The sum of the digits of 26 is 8, which is not a prime number, so the answer is not (E). Since 14 is not a multiple of a three, 14 is not a perfect square, and the sum of the digits of 14 is 1+4=5 which is prime, then the answer is (B) 14.

    Answer: (B)

  7. Since WXY is a straight angle, then p+q+r+s+t=180 and so p+q+r+s+t=180. To calculate the average of p, q, r, s, and t, we add the five numbers and divide by 5. Therefore, the average of p, q, r, s, and t is p+q+r+s+t5=1805=36.

    Answer: (B)

  8. Since 8=2×2×2=23, then 820=(23)20=23×20=260. Thus, if 2n=820, then n=60.

    Answer: (B)

  9. The Pythagorean Theorem tells us that if a right-angled triangle has sides of length a, b and c, with c the hypotenuse, then a2+b2=c2. Since the area of a square of side length a is a2, the Pythagorean Theorem can be re-phrased to say that the sum of the areas of the squares that can be drawn on the two shorter sides equals the area of the square that can be drawn on the hypotenuse. (In the figure below, this says that x+y=z where x, y and z are the areas of the squares, as shown.)

    A right-angled triangle surrounded by three different-sized squares. The legs of the right-angled triangle have lengths a and b, respectively, and the hypotenuse has length c. Each side of the triangle also shares the side of a square. The square with area x shares the a side, the square with area y shares the b side, and the square with area z shares the c side.

    In the given diagram, this means that the area of the unmarked, unshaded square is 8+32=40.

    This means that the area of the shaded square is 40+5=45.

    Answer: (B)

  10. We are given that s and t are positive integers and that s(st)=29. Since s and t are positive, then st is less than s. Since s is positive and 29 is positive and s(st)=29, then st must also be positive. Since 29 is a prime number, the only way that it can be written as a product of two positive integers is 29=291. Since s(st)=29 and s>st, then we must have s=29 and st=1. Since s=29 and st=1, we obtain t=28. Therefore, s+t=29+28=57.

    Answer: (C)

  11. Each of the first and second columns has 4 X’s in it, which means that at least 2 X’s need to be moved. We will now show that this can be actually done by moving 2 X’s. Each of the first and second rows has 4 X’s in it, so we move the two X’s on the main diagonals, since this will remove X’s from the first and second columns and the first and second rows simultaneously. The fifth column starts with one X in it, so we move the two X’s to the fifth column into the rows that only contain 2 X’s. Doing this, we obtain:

    O X X X  
    X O X   X
    X X     X
    X X   X  
        X X X 

    (The cells from which X’s have been removed are marked with O’s; the cells to which X’s are moved are marked with X’s.) Therefore, the smallest number of X’s that must be moved is 2.

    Answer: (B)

  12. Since Harriet ran 720 m at 3 m/s, then this segment took her 720 m3 m/s=240 s. In total, Harriet ran 1000 m in 380 s, so the remaining part of the course was a distance of 1000 m720 m=280 m which she ran in 380 s240 s=140 s. Since she ran this section at a constant speed of v m/s, then 280 m140 s=v m/s which means that v=2.

    Answer: (A)

  13. Since the sum of any two adjacent numbers is constant, then 2+x=x+y. This means that y=2 and makes the list 2,x,2,5. This means that the sum of any two adjacent numbers is 2+5=7, and so x=5. Therefore, xy=52=3.

    Answer: (C)

  14. If 27 of the roses are to be yellow, then the remaining 57 of the roses are to be red. Since there are 30 red roses and these are to be 57 of the roses, then 17 of the total number of roses would be 30÷5=6, which means that there would be 6×7=42 roses in total. If there are 42 roses of which 30 are red and the rest are yellow, then there are 4230=12 yellow roses. Since there are 19 yellow roses to begin, then 1912=7 yellow roses are removed.

    Answer: (E)

  15. When N=3x+4y+5z with each of x, y and z equal to either 1 or 1, there are 8 possible combinations of values for x, y and z:

    x y z N
    1 1 1 12
    1 1 -1 2
    1 -1 1 4
    1 -1 -1 -6
    -1 1 1 6
    -1 1 -1 -4
    -1 -1 1 -2
    -1 -1 -1 -12
    From this information, N cannot equal 0, N is never odd, N can equal 4, and N is always even. Therefore, exactly one of the four given statements is true.

    Answer: (B)

  16. We note that x+yx=xx+yx=1+yx. The greatest possible value of x+yx=1+yx thus occurs when yx is as great as possible. Since x is always negative and y is always positive, then yx is negative. Therefore, for yx to be as great as possible, it is as least negative as possible (i.e. closest to 0 as possible). Since x is negative and y is positive, this happens when x is as negative as possible and y is as small as possible – that is, when x=4 and y=2. Therefore, the greatest possible value of x+yx is 1+24=12.

    Answer: (E)

  17. Since PQR is right-angled at Q, its area equals 12PQQR. Since its area is 30 and PQ=5, then 125QR=30 and so QR=3025=12. By the Pythagorean Theorem, we know that PR2=PQ2+QR2=52+122=25+144=169 Since PR>0, then PR=169=13. If we now consider PQR as having base PR and perpendicular height QS, we see that its area equals 12PRQS. Since its area is 30 and PR=13, then 1213QS=30 which gives QS=30213=6013.

    Answer: (A)

  18. Suppose that the four teams in the league are called W, X, Y, and Z. Then there is a total of 6 games played:

    W against X, W against Y, W against Z, X against Y, X against Z, Y against Z

    In each game that is played, either one team is awarded 3 points for a win and the other is awarded 0 points for a loss (for a total of 3 points between the two teams), or each team is awarded 1 point for a tie (for a total of 2 points between the two teams). Since 6 games are played, then the theoretical maximum number of points that could be awarded is 63=18 and the theoretical minimum number of points that can be awarded is 62=12. In particular, this means that it is not possible for the total number of points to be 11. We can show that each of the possibilities from 12 to 18 points, inclusive, is actually possible. Therefore, S cannot equal 11.

    Answer: (C)

  19. When (3+2x+x2)(1+mx+m2x2) is expanded, the terms that include an x2 will come from multiplying a constant with a term that includes x2 or multiplying two terms that includes x. In other words, the term that includes x2 will be 3m2x2+2xmx+x21=3m2x2+2mx2+x2=(3m2+2m+1)x2 From the condition that the coefficient of this term equals 1, we see that 3m2+2m+1=1 which gives 3m2+2m=0 or m(3m+2)=0, which means that m=0 or m=23. The sum of these possible values of m is 23.

    Answer: (B)

  20. When a dot is removed from a face with an even number of dots, that face then has an odd number of dots. When a dot is removed from a face with an odd number of dots, that face then has an even number of dots. Initially, there are 3 faces with an even number of dots and 3 faces with an odd number of dots. If a dot is removed from a face with an even number of dots, there are then 4 faces with an odd number of dots and 2 faces with an even number of dots. This means that the probability of rolling an odd number after a dot is removed is 46 in this case. If a dot is removed from a face with an odd number of dots, there are then 2 faces with an odd number of dots and 4 faces with an even number of dots. This means that the probability of rolling an odd number after a dot is removed is 26 in this case. Since there are 2+3+4+5+6+7=27 dots on the faces, then the probability that a dot is removed from the face with 2 dots is 227, from the face with 3 dots is 327, and so on. Thus, the probability that a dot is removed from the face with 2 dots and then an odd number is rolled is the product of the probabilities, which is 22723, since there are now 4 odd faces and 2 even faces. Similarly, the probability that a dot is removed from the face with 3 dots and then an odd number is rolled is 32713. Continuing in this way, the probability of rolling an odd number after a dot is removed is 22723+32713+42723+52713+62723+72713. This equals 23(227+427+627)+13(327+527+727)=231227+131527=827+527=1327.

    Answer: (C)

  21. If the product of three numbers x, 36 and y is 2592, then x36y=2592 and so xy=259236=72. If x and y are positive integers with xy=72, then we have the following possibilities:

    x y x+y
    72 1 73
    36 2 38
    24 3 27
    18 4 22
    12 6 18
    9 8 17
    We have assumed that x>y since we have not assigned an order to x, 36 and y. In the given problem, we want to put four pairs of numbers in the outer circles so that the 9 numbers are different and the sum of the 9 numbers is as large as possible. Putting this another way, we want to choose 4 of the 6 pairs in the table above (knowing that we cannot choose the pair 36 and 2 since 36 is already in the middle circle) to make the sum as large as possible. Since we know the sums of the pairs, we choose the pairs with the four largest sums. This means that the sum of the 9 numbers will be (72+1)+(24+3)+(18+4)+(12+6)+36 which equals 73+27+22+18+36 or 176.

    Answer: (B)

  22. Since x2+3xy+y2=909 and 3x2+xy+3y2=1287, then (x2+3xy+y2)+(3x2+xy+3y2)=909+12874x2+4xy+4y2=2196x2+xy+y2=549 Since x2+3xy+y2=909 and x2+xy+y2=549, then (x2+3xy+y2)(x2+xy+y2)=9095492xy=360xy=180 Since x2+3xy+y2=909 and xy=180, then (x2+3xy+y2)xy=909180x2+2xy+y2=729(x+y)2=272 Therefore, x+y=27 or x+y=27. This also shows that x+y cannot equal any of 39, 29, 92, and 41. (We can in fact solve the system of equations x+y=27 and xy=180 for x and y to show that there do exist real numbers x and y that are solutions to the original system of equations.) Therefore, a possible value for x+y is (A) 27.

    Answer: (A)

  23. Solution 1:

    Since f(x)=ax+b for all real numbers x, then f(t)=at+b for some real number t. When t=bx+a, we obtain f(bx+a)=a(bx+a)+b=abx+(a2+b). We also know that f(bx+a)=x for all real numbers x. This means that abx+(a2+b)=x for all real numbers x and so (ab1)x+(a2+b)=0 for all real numbers x. For this to be true, it must be the case that ab=1 and a2+b=0. From the second equation b=a2 which gives a(a2)=1 and so a3=1, which means that a=1. Since b=a2, then b=1 as well, which gives a+b=2.

    Solution 2:

    Since f(x)=ax+b for all x, then when x=a, we obtain f(a)=a2+b. Since f(bx+a)=x for all x, then when x=0, we obtain f(a)=0. Comparing values for f(a), we obtain a2+b=0 or b=a2. This gives f(x)=axa2 for all real numbers x and f(a2x+a)=x for all real numbers x. Since f(a2x+a)=x for all x, then when x=1, we obtain f(a2+a)=1. Since f(x)=axa2 for all x, then when x=a2+a, we obtain f(a2+a)=a(a2+a)a2. Comparing values for f(a2+a), we obtain a(a2+a)a2=1 or a3=1. Since a is a real number, then a=1. Since b=a2, then b=1, which gives a+b=2. Checking, we see that if f(x)=x1, then f(x1)=(x1)1=x, as required.

    Answer: (E)

  24. Suppose the centre of the largest circle is O. Suppose that the circle with centre X touches the largest circle at S and the two circles with centres Y and Z at T and U, respectively. Suppose that the circles with centres Y and Z touch each other at A, and the largest circle at B and C, respectively. Join X to Y, X to Z, and Y to Z.

    (Note that the diagram has been re-drawn here so that the circle with centre X actually appears to pass through the centre of the largest circle.) Since the circles are tangent at points T and U, line segments XY and XZ pass through T and U, respectively. Further, XY=XT+TY=1+r, since the circles with centres X and Y have radii 1 and r, respectively. Similarly, XZ=1+r. Also, YA=ZA=YB=ZC=r, since these are radii of the two circles. When one circle is inside another circle, and the two circles touch at a point, then the radii of the two circles that pass through this point lie on top of each other. This is because the circles have a common tangent at the point where they touch and this common tangent will be perpendicular to each of the radii. Since the circle with centre X touches the largest circle at S, then X lies on OS. In the largest circle, consider the diameter that passes through X. Since the circle with centre X passes through O, then the radius of the largest circle is twice that of the circle with centre X, or 2. It is also the case that XO=1. Next, we join O to B. Since the circles with centres O and Y touch at B, then OB passes through Y. This means that OY=OBBY=2r. Similarly, OZ=2r. Further, by symmetry in the largest circle, the diameter through X also passes through A, the point at which the two smallest circles touch:

    To see this more formally, draw the common tangent through A to the circles with centres Y and Z. This line is perpendicular to YZ, since it is tangent to both circles. Since OYZ is isosceles with OY=OZ, the altitude through the midpoint A of its base passes through O. Similarly, XYZ is isosceles with XY=XZ and so its altitude through A passes through X. Since the line perpendicular to YZ at A passes through both O and X, it is the diameter that passes through X.

    Now, we consider XYA and OYA, each of which is right-angled at A. By the Pythagorean Theorem, OA=OY2YA2=(2r)2r2=44r+r2r2=44r Again, using the Pythagorean Theorem, XA2+YA2=XY2(XO+OA)2+r2=(1+r)2(1+44r)2=1+2r+r2r21+244r+(44r)=1+2r244r=6r444r=3r244r=(3r2)2(squaring both sides)44r=9r212r+48r=9r2 Since r0, then 9r=8 and so r=890.889. Of the given choices, this is closest to (E) 0.89.

    Answer: (E)

  25. Consider a 1×1×1 cube. We associate a triple (x,y,z) of real numbers with 0x1 and 0y1 and 0z1 with a point inside this cube by letting x be the perpendicular distance of a point from the left face, y the perpendicular distance of a point from the front face, and z the perpendicular distance from the bottom face. We call this point (x,y,z). Choosing x, y and z randomly and independently between 0 and 1 is equivalent to randomly and uniformly choosing a point (x,y,z) on or inside the cube.

    The conditions that 12<xy<12 and 12<xz<12 restrict the values of x, y and z that can be chosen, which translates into restricting the points inside the cube that satisfy these conditions. Hence, these restrictions determine a region inside this cube. The probability that a point randomly chosen inside this cube satisfies the given conditions will be equal to the volume of the region defined by the conditions divided by the volume of the entire cube. Since the volume of the cube is 1, then the probability will equal the volume of the region defined by those conditions. Consider now the region in the xy-plane defined by 12<xy<12. Re-arranging these inequalities, we obtain x12<y<x+12, which means that a point (x,y) that satisfies these conditions lies above the line with equation y=x12 and below the line with equation y=x+12. Restricting to 0x1 and 0y1, we obtain the region shown:

    The shaded region is shaped like an elongated tilted hexagon lying within the first quadrant of an xy-plane and its vertices are as follows: (0,0), (0,1/2), (1/2,1), (1,1), (1,1/2), (1/2,0).

    Since a point (x,y,z) in the region satisifes 12<xy<12, these conditions allow us to “slice” the cube from above keeping the portion that looks like the region above. The points that remain are exactly those that satisfy this condition. Similarly, the conditions 12<xz<12 give x12<z<x+12, which has the same shape in the xz-plane. Therefore, we can slice the cube from front to back to look like this shape. Now, we need to determine the volume of the remaining region. To determine the volume of the region, we split the 1×1×1 cube into eight cubes each measuring 12×12×12.

    When this cube is sliced by the restrictions corresponding to x12<y<x+12, the back left and front right cubes on the top and bottom layers are sliced in half.

    When this cube is sliced by the restrictions corresponding to x12<z<x+12, the top left and bottom right cubes in the front and back are sliced in half. The eight little cubes are sliced as follows:

    Little cube Sliced by x12<y<x+12 Sliced by x12<z<x+12
    Bottom front left No No
    Bottom front right Yes Yes
    Bottom back left Yes No
    Bottom back right No Yes
    Top front left No Yes
    Top front right Yes No
    Top back left Yes Yes
    Top back right No No

    This means that we can consider the little cubes as follows:

    Therefore, the volume of the solid is 218+4116+2124=14+14+112=712. Finally, this means that the required probability is 712.

    Answer: (B)