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2018 Euclid Contest
Solutions

Wednesday, April 11, 2018
(in North America and South America)

Thursday, April 12, 2018
(outside of North American and South America)

©2018 University of Waterloo


    1. When x=11, x+(x+1)+(x+2)+(x+3)=4x+6=4(11)+6=50 Alternatively, x+(x+1)+(x+2)+(x+3)=11+12+13+14=50

    2. We multiply the equation a6+618=1 by 18 to obtain 3a+6=18.
      Solving, we get 3a=12 and so a=4.

    3. Solution 1:

      Since the cost of one chocolate bar is $1.00 more than that of a pack of gum, then if we replace a pack of gum with a chocolate bar, then the price increases by $1.00.
      Starting with one chocolate bar and two packs of gum, we replace the two packs of gum with two chocolate bars.
      This increases the price by $2.00 from $4.15 to $6.15.
      In other words, three chocolate bars cost $6.15, and so one chocolate bar costs 13($6.15) or $2.05.

      Solution 2:

      Let the cost of one chocolate bar be $x.
      Let the cost of one pack of gum be $y.
      Since the cost of one chocolate bar and two packs of gum is $4.15, then x+2y=4.15.
      Since one chocolate bar costs $1.00 more than one pack of gum, then x=y+1.
      Since x=y+1, then y=x1.
      Since x+2y=4.15, then x+2(x1)=4.15.
      Solving, we obtain x+2x2=4.15 or 3x=6.15 and so x=2.05.
      In other words, the cost of one chocolate bar is $2.05.

    1. Suppose that the five-digit integer has digits abcde.
      The digits a,b,c,d,e are 1,3,5,7,9 in some order.
      Since abcde is greater than 80 000, then a8, which means that a=9.
      Since 9bcde is less than 92000, then b<2, which means that b=1.
      Since 91cde has units (ones) digit 3, then e=3.
      So far, the integer is 91cd3, which means that c and d are 5 and 7 in some order.
      Since the two-digit integer cd is divisible by 5, then it must be 75.
      This means that the the five-digit integer is 91753.

    2. By the Pythagorean Theorem in ADB, AD2=AB2BD2=132122=169144=25 Since AD>0, then AD=25=5.
      By the Pythagorean Theorem in CDB, CD2=BC2BD2=(122)2122=122(2)122=122 Since CD>0, then CD=12.
      Therefore, AC=AD+DC=5+12=17.

    3. Solution 1:

      The area of the shaded region equals the area of square OABC minus the area of OCD.
      Since square OABC has side length 6, then its area is 62 or 36.
      Also, OC=6.
      Since the equation of the line is y=2x, then its slope is 2.
      Since the slope of the line is 2, then OCCD=2.
      Since OC=6, then CD=3.
      Thus, the area of OCD is 12(OC)(CD)=12(6)(3)=9.
      Finally, the area of shaded region must be 369=27.

      Solution 2:

      Since square OABC has side length 6, then OA=AB=CB=OC=6.
      Since the slope of the line is 2, then OCCD=2.
      Since OC=6, then CD=3.
      Since CB=6 and CD=3, then DB=CBCD=3.
      The shaded region is a trapezoid with parallel sides DB=3 and OA=6 and height AB=6.
      Therefore, the area of the shaded region is 12(DB+OA)(AB)=12(3+6)(6)=27.

    1. Calculating, (4+4)4=(4+2)4=(6)4=((6)2)2=62=36.

    2. Since y is an integer, then 8y2 is an integer.
      Therefore, 23x is an integer which means that 23x is a perfect square.
      Since x is a positive integer, then 23x<23 and so 23x must be a perfect square that is less than 23.
      We make a table listing the possible values of 23x and the resulting values of x, 23x=8y2, y2, and y:

      23x x 23x=8y2 y2 y
      16 7 4 4 ±2
      9 14 3 5 ±5
      4 19 2 6 ±6
      1 22 1 7 ±7
      0 23 0 8 ±8
      Since x and y are positive integers, then we must have (x,y)=(7,2).
      (We note that since we were told that there is only one such pair, we did not have to continue the table beyond the first row.)

    3. Since the line with equation y=mx+2 passes through (1,5), then 5=m+2 and so m=3.
      Since the parabola with equation y=ax2+5x2 passes through (1,5), then 5=a+52 and so a=2.
      To find the coordinates of Q, we determine the second point of intersection of y=3x+2 and y=2x2+5x2 by equating values of y: 2x2+5x2=3x+22x2+2x4=0x2+x2=0(x+2)(x1)=0 Therefore, x=1 or x=2.
      Since P has x-coordinate 1, then Q has x-coordinate 2.
      Since Q lies on the line with equation y=3x+2, we have y=3(2)+2=4.
      In summary, (i) m=3, (ii) a=2, and (iii) the coordinates of Q are (2,4).

    1. Since 80=245, its positive divisors are 1,2,4,5,8,10,16,20,40,80.
      For an integer n to share exactly two positive common divisors with 80, these divisors must be either 1 and 2 or 1 and 5. (1 is a common divisor of any two integers. The second common divisor must be a prime number since any composite divisor will cause there to be at least one more common divisor which is prime.)
      Since 1n30 and n is a multiple of 2 or of 5, then the possible values of n come from the list 2,4,5,6,8,10,12,14,15,16,18,20,22,24,25,26,28,30 We remove the multiples of 4 from this list (since they would share at least the divisors 1,2,4 with 80) and the multiples of 10 from this list (since they would share at least the divisors 1,2,5,10 with 80).
      This leaves the list 2,5,6,14,15,18,22,25,26 The common divisors of any number from this list and 80 are either 1 and 2 or 1 and 5.
      There are 9 such integers.

    2. We start with f(50) and apply the given rules for the function until we reach f(1): f(50)=f(25)(since 50 is even and 12(50)=25)=f(24)+1(since 25 is odd and 251=24)=f(12)+1(12(24)=12)=f(6)+1(12(12)=6)=f(3)+1(12(6)=3)=(f(2)+1)+1(31=2)=f(1)+1+1(12(2)=1)=1+1+1(f(1)=1)=3 Therefore, f(50)=3.

    1. Since the hexagon has perimeter 12 and has 6 sides, then each side has length 2.
      Since equilateral PQR has perimeter 12, then its side length is 4.
      Consider equilateral triangles with side length 2.
      Six of these triangles can be combined to form a regular hexagon with side length 2 and four of these can be combined to form an equilateral triangle with side length 4.

      To form the hexagon, six triangles are placed in a circle so they all share a vertex at the centre the hexagon. To form the triangle, three triangles are placed in a bottom row, oriented to point up then down then up, and one triangle is placed on top of the middle triangle in the row, pointing up.

      Note that the six equilateral triangles around the centre of the hexagon give a total central angle of 660°=360° (a complete circle) and the three equilateral triangles along each side of the large equilateral triangle make a straight angle of 180° (since 360°=180°).
      Also, the length of each side of the hexagon is 2 and the measure of each internal angle is 120°, which means that the hexagon is regular. Similarly, the triangle is equilateral.
      Since the triangle is made from four identical smaller triangles and the hexagon is made from six of these smaller triangles, the ratio of the area of the triangle to the hexagon is 4:6 which is equivalent to 2:3.

    2. Since sector AOB is 16 of a circle with radius 18, its area is 16(π182) or 54π.
      For the line AP to divide this sector into two pieces of equal area, each piece has area 12(54π) or 27π.
      We determine the length of OP so that the area of POA is 27π.
      Since sector AOB is 16 of a circle, then AOB=16(360°)=60°.
      Drop a perpendicular from A to T on OB.

      Recall that P lies on AB.

      The area of POA is 12(OP)(AT).
      AOT is a 30°-60°-90° triangle.
      Since AO=18, then AT=32(AO)=93.
      For the area of POA to equal 27π, we have 12(OP)(93)=27π which gives OP=54π93=6π3=23π.
      (Alternatively, we could have used the fact that the area of POA is 12(OA)(OP)sin(POA).)

    1. Let θ=10k°.
      The given inequalities become 0°<θ<180° and 5sinθ2sin2θ2.
      When 0°<θ<180°, sinθ0.
      This means that we can can multiply both sides by sin2θ>0 and obtain the equivalent inequalities: 5sinθ2sin2θ25sinθ22sin2θ02sin2θ5sinθ+20(2sinθ1)(sinθ2) Since sinθ1, then sinθ21<0 for all θ.
      Therefore, (2sinθ1)(sinθ2)0 exactly when 2sinθ10.
      Note that 2sinθ10 exactly when sinθ12.
      Therefore, the original inequality is true exactly when 12sinθ1.
      Note that sin30°=sin150°=12 and 0°<θ<180°.
      When θ=0°, sinθ=0.
      From θ=0° to θ=30°, sinθ increases from 0 to 12.
      From θ=30° to θ=150°, sinθ increases from 12 to 1 and then decreases to 12.
      From θ=150° to θ=180°, sinθ decreases from 12 to 0.
      Therefore, the original inequality is true exactly when 30°θ150° which is equivalent to 30°10k°150° and to 3k15.
      The integers k in this range are k=3,4,5,6,,12,13,14,15, of which there are 13.

    2. Suppose that Karuna and Jorge meet for the first time after t1 seconds and for the second time after t2 seconds.
      When they meet for the first time, Karuna has run partway from A to B and Jorge has run partway from B to A.

      At this instant, the sum of the distances that they have run equals the total distance from A to B.
      Since Karuna runs at 6 m/s for these t1 seconds, she has run 6t1 m.
      Since Jorge runs at 5 m/s for these t1 seconds, he has run 5t1 m.
      Therefore, 6t1+5t1=297 and so 11t1=297 or t1=27.
      When they meet for the second time, Karuna has run from A to B and is running back to A and Jorge has run from B to A and is running back to B. This is because Jorge gets to A halfway through his run before Karuna gets back to A at the end of her run.

      Since they each finish running after 99 seconds, then each has 99t2 seconds left to run.
      At this instant, the sum of the distances that they have left to run equals the total distance from A to B.
      Since Karuna runs at 6 m/s for these (99t2) seconds, she has to run 6(99t2) m.
      Since Jorge runs at 7.5 m/s for these (99t2) seconds, he has to run 7.5(99t2) m.
      Therefore, 6(99t2)+7.5(99t2)=297 and so 13.5(99t2)=297 or 99t2=22 and so t2=77.
      Alternatively, to calculate the value of t2, we note that when Karuna and Jorge meet for the second time, they have each run the distance from A to B one full time and are on their return trips.
      This means that they have each run the full distance from A to B once and the distances that they have run on their return trip add up to another full distance from A to B, for a total distance of 3297 m=891 m.
      Karuna has run at 6 m/s for t2 seconds, for a total distance of 6t2 m.
      Jorge ran the first 297 m at 5 m/s, which took 2975 s and ran the remaining (t22975) seconds at 7.5 m/s, for a total distance of (297+7.5(t22975)) m.
      Therefore, 6t2+297+7.5(t22975)=89113.5t2=891297+7.5297513.5t2=1039.5t2=77 Therefore, Karuna and Jorge meet after 27 seconds and after 77 seconds.

    1. Solution 1:

      Among a group of n people, there are n(n1)2 ways of choosing a pair of these people:

        There are n people that can be chosen first.
        For each of these n people, there are n1 people that can be chosen second.
        This gives n(n1) orderings of two people.
        Each pair is counted twice (given two people A and B, we have counted both the pair AB and the pair BA), so the total number of pairs is n(n1)2.

      We label the four canoes W, X, Y, and Z.
      First, we determine the total number of ways to put the 8 people in the 4 canoes.
      We choose 2 people to put in W. There are 872 pairs. This leaves 6 people for the remaining 3 canoes.
      Next, we choose 2 people to put in X. There are 652 pairs. This leaves 4 people for the remaining 2 canoes.
      Next, we choose 2 people to put in Y. There are 432 pairs. This leaves 2 people for the remaining canoe.
      There is now 1 way to put the remaining people in Z.
      Therefore, there are 872652432=87654323=76543 ways to put the 8 people in the 4 canoes.
      Now, we determine the number of ways in which no two of Barry, Carrie and Mary will be in the same canoe.
      There are 4 possible canoes in which Barry can go.
      There are then 3 possible canoes in which Carrie can go, because she cannot go in the same canoe as Barry.
      There are then 2 possible canoes in which Mary can go, because she cannot go in the same canoe as Barry or Carrie.
      This leaves 5 people left to put in the canoes.
      There are 5 choices of the person that can go with Barry, and then 4 choices of the person that can go with Carrie, and then 3 choices of the person that can go with Mary.
      The remaining 2 people are put in the remaining empty canoe.
      This means that there are 432543 ways in which the 8 people can be put in 4 canoes so that no two of Barry, Carrie and Mary are in the same canoe.
      Therefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is 43254376543=43276=2442=47.

      Solution 2:

      Let p be the probability that two of Barry, Carrie and Mary are in the same canoe.
      The answer to the original problem will be 1p.
      Let q be the probability that Barry and Carrie are in the same canoe.
      By symmetry, the probability that Barry and Mary are in the same canoe also equals q as does the probability that Carrie and Mary are in the same canoe.
      This means that p=3q.
      So we calculate q.
      To do this, we put Barry in a canoe. Since there are 7 possible people who can go in the canoe with him, then the probability that Carrie is in the canoe with him equals 17. The other 6 people can be put in the canoes in any way.
      This means that the probability that Barry and Carrie are in the same canoe is q=17.
      Therefore, the probability that no two of Barry, Carrie and Mary are in the same canoe is 1317 or 47.

    2. Solution 1:

      Consider square WXYZ in the first quadrant. Suppose that WY makes an angle of θ with the horizontal.

      Place W at the bottom left corner. WX is parallel to the y-axis. WZ is parallel to the x-axis. Rotate the square slightly counterclockwise such that diagonal WY has slope 2.

      Since the slope of WY is 2, then tanθ=2, since the tangent of an angle equals the slope of a line that makes this angle with the horizontal.
      Since tanθ=2>1=tan45°, then θ>45°.
      Now WY bisects ZWX, which is a right-angle.
      Therefore, ZWY=YWX=45°.
      Therefore, WX makes an angle of θ+45° with the horizontal and WZ makes an angle of θ45° with the horizontal. Since θ>45°, then θ45°>0 and θ+45°>90°.
      We note that since WZ and XY are parallel, then the slope of XY equals the slope of WZ.
      To calculate the slopes of WX and WZ, we can calculate tan(θ+45°) and tan(θ45°).
      Using the facts that tan(A+B)=tanA+tanB1tanAtanB and tan(AB)=tanAtanB1+tanAtanB, we obtain: tan(θ+45°)=tanθ+tan45°1tanθtan45°=2+11(2)(1)=3tan(θ45°)=tanθtan45°1tanθtan45°=211+(2)(1)=13 Therefore, the sum of the slopes of WX and XY is 3+13=83.

      Solution 2:

      Consider a square WXYZ whose diagonal WY has slope 2.
      Translate this square so that W is at the origin (0,0). Translating a shape in the plane does not affect the slopes of any line segments.
      Let the coordinates of Y be (2a,2b) for some non-zero numbers a and b.
      Since the slope of WY is 2, then 2b02a0=2 and so 2b=4a or b=2a.
      Thus, the coordinates of Y can be written as (2a,4a).
      Let C be the centre of square WXYZ.
      Then C is the midpoint of WY, so C has coordinates (a,2a).
      We find the slopes of WX and XY by finding the coordinates of X.
      Consider the segment XC.
      Since the diagonals of a square are perpendicular, then XC is perpendicular to WC.
      Since the slope of WC is 2, then the slopes of XC and ZC are 12.
      Since the diagonals of a square are equal in length and C is the midpoint of both diagonals, then XC=WC.
      Since WC and XC are perpendicular and equal in length, then the “rise/run triangle” above XC will be a 90° rotation of the “rise/run triangle” below WC.

      Above referes to triangle XCY. Below refers to triangle WCZ.

      This is because these triangles are congruent (each is right-angled, their hypotenuses are of equal length, and their remaining angles are equal) and their hypotenuses are perpendicular.
      In this diagram, we have assumed that X is to the left of W and Z is to the right of W. Since the slopes of parallel sides are equal, it does not matter which vertex is labelled X and which is labelled Z. We would obtain the same two slopes, but in a different order.
      To get from W(0,0) to C(a,2a), we go up 2a and right a.
      Thus, to get from C(a,2a) to X, we go left 2a and up a.
      Therefore, the coordinates of X are (a2a,2a+a) or (a,3a).
      Thus, the slope of WX is 3a0a0=3.
      Since XY is perpendicular to WX, then its slope is the negative reciprocal of 3, which is 13.
      The sum of the slopes of WX and XY is 3+13=83.

    1. Since the base of a logarithm must be positive and cannot equal 1, then x>0 and x12 and x13.
      This tells us that log2x and log3x exist and do not equal 0, which we will need shortly when we apply the change of base formula.
      We note further that 48=243 and 162=342 and 33=31/3 and 23=21/3.
      Using logarithm rules, the following equations are equivalent: log2x(4833)=log3x(16223)log(24331/3)log2x=log(34221/3)log3x(change of base formula)log(2434/3)log2+logx=log(3424/3)log3+logx(logab=loga+logb)log(24)+log(34/3)log2+logx=log(34)+log(24/3)log3+logx(logab=loga+logb)4log2+43log3log2+logx=4log3+43log2log3+logx(log(ac)=cloga) Cross-multiplying, we obtain (4log2+43log3)(log3+logx)=(4log3+43log2)(log2+logx) Expanding the left side, we obtain 4log2log3+43(log3)2+(4log2+43log3)logx Expanding the right side, we obtain 4log3log2+43(log2)2+(4log3+43log2)logx Simplifying and factoring, we obtain the following equivalent equations: 43(log3)243(log2)2=logx(4log3+43log24log243log3)43(log3)243(log2)2=logx(83log383log2)(log3)2(log2)2=2logx(log3log2)logx=(log3)2(log2)22(log3log2)logx=(log3log2)(log3+log2)2(log3log2)logx=log3+log22logx=12log6logx=log(6) and so x=6.

    2. Rectangle PQRS lies inside rectangle ABCD.

      Recall that P lies on AB, Q lies on BC, R lies on CD, and S lies on DA.

      Let BC=x, PB=b, and BQ=a.
      Since BC=x, then AD=PS=QR=x.
      Since BC=x and BQ=a, then QC=xa.
      Since AB=718 and PB=b, then AP=718b.
      Note that PQ=SR=250.
      Let BQP=θ.
      Since PBQ is right-angled at B, then BPQ=90°θ.
      Since BQC is a straight angle and PQR=90°, then RQC=180°90°θ=90°θ.
      Since APB is a straight angle and SPQ=90°, then APS=180°90°(90°θ)=θ.
      Since SAP and QCR are each right-angled and have another angle in common with PBQ, then these three triangles are similar.

      Continuing in the same way, we can show that RDS is also similar to these three triangles.
      Since RS=PQ, then RDS is actually congruent to PBQ (angle-side-angle).
      Similarly, SAP is congruent to QCR.
      In particular, this means that AS=xa, SD=a, DR=b, and RC=718b.
      Since SAP and PBQ are similar, then SAPB=APBQ=SPPQ.
      Thus, xab=718ba=x250.
      Also, by the Pythagorean Theorem in PBQ, we obtain a2+b2=2502.
      By the Pythagorean Theorem in SAP, x2=(xa)2+(718b)2x2=x22ax+a2+(718b)2(*)0=2ax+a2+(718b)2 Since a2+b2=2502, then a2=2502b2.
      Since 718ba=x250, then ax=250(718b).
      Therefore, substituting into (), we obtain 0=2(250)(718b)+2502b2+(718b)2b2=25022(250)(718b)+(718b)2b2=((718b)250)2(since y22yz+z2=(yz)2)b2=(468b)2b=468b(since bb468)2b=468b=234 Therefore, a2=2502b2=25022342=(250+234)(250234)=48416=22242=882 and so a=88.
      Finally, x=250(718b)a=25048488=1375. Therefore, BC=1375.

    1. Here is a tiling of a 3×8 rectangle:

      Four identical 3 by 2 rectangles, covered by two triominos each, are placed side by side to tile the 3 by 8 rectangle. The 3 by 2 rectangles are formed by a triomino oriented so it forms a capital L shape in the bottom left corner together with a triomino rotated 180 degrees in the top right corner.

      There are many other tilings.

    2. First, we note that it is possible to tile each of a 3×2 and a 2×3 rectangle:

      Consider two separate images showing the two stacked triominos forming a 3 by 2 rectangle. The second image is of a 3 by 2 rectangle rotated by 90 degrees.

      Next, we note that it is not possible to tile a 6×1 rectangle because each of the triominos needs a width of at least 2 to be placed.
      Finally, we show that it is possible to tile a 6×W rectangle for every integer W2.
      To do this, we show that such a 6×W rectangle can be made up from 3×2 and 2×3 rectangles. Since each of these types of rectangles can be tiled with triominos, then the larger rectangle can be tiled with triominos by combining these tilings.

      Case 1: W is even.
      Suppose that W=2k for some positive integer k.
      We build a 6×2k rectangle by placing k 6×2 rectangles side by side.
      Each 6×2 rectangle is built by stacking two 3×2 rectangles on top of each other.

      Therefore, each such rectangle can be tiled.

      Case 2: W is odd, W3.
      Suppose that W=2k+1 for some positive integer k.
      We build a 6×(2k+1) rectangle by building a 6×3 rectangle and then putting k1 6×2 rectangles next to it. Note that k10 since k1 and that 2k+1=3+2(k1).
      The 6×3 rectangle is built by stacking three 2×3 rectangles on top of each other.
      Each 6×2 rectangle is built by stacking two 3×2 rectangles on top of each other.

      Therefore, each such rectangle can be tiled.
      Thus, a 6×W rectangle can be tiled with triominos exactly when W2.

    3. Suppose that (H,W) is a pair of integers with H4 and W4.
      Since the area of each triomino is 3, then the area of any rectangle that can be tiled must be a multiple of 3 since it is completely covered by triominos with area 3.
      Since the area of an H×W rectangle is HW, then we need HW to be a multiple of 3, which means that at least one of H and W is a multiple of 3.
      Since a rectangle that is a×b can be tiled if and only if a rectangle that is b×a can be tiled (we see this by rotating the tilings by 90° as we did with the 3×2 and 2×3 rectangles above), then we may assume without loss of generality that H is divisible by 3.
      We show that if H is divisible by 3, then every H×W rectangle with H4 and W4 can be tiled.

      Case 1: H is divisible by 3, H is even.
      Here, H is a multiple of 6, say H=6m for some positive integer m.
      Since W4, we know that a 6×W rectangle can be tiled.
      By stacking m 6×W rectangles on top of each other, we obtain a 6m×W rectangle.
      Since each 6×W rectangle can be tiled, then the 6m×W rectangle can be tiled.

      Case 2: H is divisible by 3, H is odd, W is even.
      Suppose that H=3q for some odd positive integer q and W=2r for some positive integer r.
      To tile a 3q×2r rectangle, we combine qr 3×2 rectangles in q rows and r columns:

      Therefore, every rectangle in this case can be tiled. (Note that in this case the fact that q was odd was not important.)

      Case 3: H is divisible by 3, H is odd, W is odd.
      Since H4 and W4, the rectangle with the smallest values of H and W is 9×5 which can be tiled as shown:

      A description of the tiling follows.

      (There are also other ways to tile this rectangle.)
      Since H is an odd multiple of 3 and H4, we can write H=9+6s for some integer s0.
      Since W is odd and W5, we can write W=5+2t for some integer t0.
      Thus, the H×W rectangle is (9+6s)×(5+2t).
      We break this rectangle into three rectangles – one that is 9×5, one that is 9×2t, and one that is 6s×W:

      A 9 by 5 rectangle is on the left side of a 9 by 2t rectangle with their sides of length 9 running vertically and touching. A 6s by W rectangle is placed along the bottom side of these two rectangles. The top horizontal length of this third rectangle is the same length as the combined horizontal lengths of the first two rectangles.

      (If s=0 or t=0, there will be fewer than three rectangles.)
      The 9×5 rectangle can be tiled as we showed earlier.
      If t>0, the 9×2t rectangle can be tiled as seen in Case 2.
      If s>0, the 6s×W rectangle can be tiled as seen in Case 1.
      Therefore, the H×W rectangle can be tiled.
      Through these three cases, we have shown that any H×W rectangle with H4 and W4 can be tiled when H is a multiple of 3.
      Since the roles of H and W can be interchanged and since at least one of H and W must be a multiple of 3, then an H×W rectangle with H4 and W4 can be tiled exactly when at least one of H and W is a multiple of 3.

    1. We draw part of the array using the information that A0=A1=A2=0 and A3=1: A0A1A2A3A4A5B0B1B2B3B4B50001A4A5B0B1B2B3B4B5 Since A1 is the average of A0, B1 and A2, then A1=A0+B1+A23 or 3A1=A0+B1+A2.
      Thus, 3(0)=0+B1+0 and so B1=0.
      Since A2 is the average of A1, B2 and A3, then 3A2=A1+B2+A3 and so 3(0)=0+B2+1 which gives B2=1.
      Since B2 is the average of B1, A2 and B3, then 3B2=B1+A2+B3 and so 3(1)=0+0+B3 which gives B3=3.
      So far, this gives 0001A4A5B0013B4B5 Since A3 is the average of A2, B3 and A4, then 3A3=A2+B3+A4 and so 3(1)=0+(3)+A4 which gives A4=6.

    2. We draw part of the array: Ak1AkAk+1Bk1BkBk+1 Then 3Sk=3Ak+3Bk=3(Ak1+Bk+Ak+13)+3(Bk1+Ak+Bk+13)=Ak1+Bk+Ak+1+Bk1+Ak+Bk+1=(Ak1+Bk1)+(Ak+Bk)+(Ak+1+Bk+1)=Sk1+Sk+Sk+1 Since 3Sk=Sk1+Sk+Sk+1, then Sk+1=2SkSk1.

    3. Proof of statement (P)
      Suppose that all of the entries in the array are positive integers.
      Assume that not all of the entries in the array are equal.
      Since all of the entries are positive integers, there must be a minimum entry. Let m be the minimum of all of the entries in the array.
      Choose an entry in the array equal to m, say Ar=m for some integer r. The same argument can be applied with Br=m if there are no entries equal to m in the top row.
      If not all of the entries Aj are equal to m, then by moving one direction or the other along the row we will get to some point where At=m for some integer t but one of its neighbours is not equal to m. (If this were not to happen, then all of the entries in both directions would be equal to m.)
      If all of the entries Aj are equal to m, then since not all of the entries in the array are equal to m, then there will be an entry Bt which is not equal to m.
      In other words, since not all of the entries in the array are equal, then there exists an integer t for which At=m and not all of At1, At+1, Bt are equal to m.
      But 3m=3At and 3At=At1+Bt+At+1 so 3m=At1+Bt+At+1.
      Since not all of At1, Bt and At+1 are equal to m and each is at least m, then one of these entries will be greater than m.
      This means that At1+Bt+At+1m+m+(m+1)=3m+1>3m, which is a contradiction.
      Therefore our assumption that not all of the entries are equal must be false, which means that all of the entries are equal, which proves statement (P).
      Proof of statement (Q)
      Suppose that all of the entries are positive real numbers.
      Assume that not all of the entries in the array are equal.
      As in (b), define Sk=Ak+Bk for each integer k.
      Also, define Dk=AkBk for each integer k.

      Step 1: Prove that the numbers Sk form an arithmetic sequence

      From (b), Sk+1=2SkSk1.
      Re-arranging, we see Sk+1Sk=SkSk1 for each integer k, which means that the differences between consecutive pairs of terms are equal.
      Since this is true for all integers k, then the difference between each pair of consecutive terms through the whole sequence is constant, which means that the sequence is an arithmetic sequence.

      Step 2: Prove that Sk is constant

      Suppose that S0=c. Since A0>0 and B0>0, then S0=c>0.
      Since the terms Sk form an arithmetic sequence, then the sequence is either constant, increasing or decreasing.
      If the sequence of terms Sk is increasing, then the common difference d=S1S0 is positive.
      Note that S1=cd, S2=c2d, and so on.
      Since c and d are constant, then if we move far enough back along the sequence, eventually St will be negative for some integer t. This is a contradiction since At>0 and Bt>0 and St=At+Bt.
      Thus, the sequence cannot be increasing.
      If the sequence of terms Sk is decreasing, then the common difference d=S1S0 is negative.
      Note that S1=c+d, S2=c+2d, and so on.
      Since c and d are constant, then if we move far along the sequence, eventually St will be negative for some integer t. This is also a contradiction since At>0 and Bt>0 and St=At+Bt.
      Thus, the sequence cannot be decreasing.
      Therefore, since all of the entries are positive and the sequence Sk is arithmetic, then Sk is constant, say Sk=c>0 for all integers k.

      Step 3: Determine range of possible values for Dk

      We note that Sk=Ak+Bk=c for all integers k and Ak>0 and Bk>0.
      Since Ak>0, then Bk=SkAk=cAk<c.
      Similarly, Ak<c.
      Therefore, 0<Ak<c and 0<Bk<c.
      Since Dk=AkBk, then Dk<c0=c and Dk>0c=c.
      In other words, c<Dk<c.

      Step 4: Dk+1=4DkDk1

      Using a similar approach to our solution to (b), 3Dk=3Ak3Bk3Dk=(Ak1+Bk+Ak+1)(Bk1+Ak+Bk+1)3Dk=(Ak+1Bk+1)+(Ak1Bk1)(AkBk)3Dk=Dk+1+Dk1Dk4DkDk1=Dk+1 as required.

      Step 5: Final contradiction

      We want to show that Dk=0 for all integers k.
      This will show that Ak=Bk for all integers k.
      Since Sk=Ak+Bk=c for all integers k, then this would show that Ak=Bk=12c for all integers k, meaning that all entries in the array are equal.
      Suppose that Dk0 for some integer k.
      We may assume that D00. (If D0=0, then because the array is infinite in both directions, we can shift the numbering of the array so that a column where Dk0 is labelled column 0.)
      Thus, D0>0 or D0<0.
      We may assume that D0>0. (If D0<0, then we can switch the bottom and top rows of the array so that D0 becomes positive.)
      Suppose that D1D0>0.
      Then D2=4D1D04D1D1=3D1. Since D1>0, this also means that D2>D1>0.
      Similarly, D3=4D2D14D2D2=3D2>D2>0. Since D23D1, then D39D1.
      Continuing in this way, we see that D427D1 and D581D1 and so on, with Dk3k1D1 for each positive integer k2. Since the value of D1 is a fixed positive real number and Dk<c for all integers k, this is a contradiction, because the sequence of values 3k1 grows without bound.
      The other possibility is that D1<D0.
      Here, we re-arrange Dk+1=4DkDk1 to obtain Dk1=4DkDk+1.
      Thus, D1=4D0D1>4D0D0=3D0>D0>0.
      Extending this using a similar method, we see that Dj>3jD0 for all positive integers j which will lead to the same contradiction as above.
      Therefore, a contradiction is obtained in all cases and so it cannot be the case that Dk0 for some integer k.
      Since Dk=0 and Sk=c for all integers k, then Ak=Bk=12c for all integers k, which means that all entries in the array are equal.