Consider the plane containing both the axis of the cone and two opposite vertices of the cube's bottom face. The cross section of the cone and the cub

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1 Solutions to the Twenty-Eighth University of Melbourne/Hewlett-Packard Mathematics Competition Senior Division, Two mathematicians, Walter and Hyam, are working at their respective homes on a problem on which they are collaborating. They both get a good idea, and decide to walk over to the home of the other to discuss their insights. They leave home simultaneously, and walk, at uniform velocity, along the same road to the home of the other. Being engrossed in the problem, they fail to notice that they actually pass one another! However, 3 minutes after passing one another, Hyam arrived at Walter's home, whereas Walter arrived at Hyam's home one minute after their meeting. How long had each of them walked? The distance apart doesn't matter it just scales the speed of the two walkers. However, let's call it d metres. Let Hyam walk with speed h m/min. and Walter with speed w m/min. Let them meet on the path after t min. of walking. Thus ht + wt = d: But we are told that ht +3h = d; and also that wt + t = d: Thus wt + t = ht +3h= ht + wt: Hence 3h = ht 2 : This gives h =0;a non-physical solution, or t = p 3: Hence Hyam walks for 3+ p 3 minutes, while Walter walks for 1+ p 3 minutes. Alternatively, let Walter and Hyam meet after t minutes of walking. Let d 1 and d 2 be the distances that Walter and Hyam each walk respectively from their homes to the point they pass each other. Then Walter covers distance d 1 in t minutes and d 2 in 1 minute, while Hyam covers distance d 2 in t minutes and d 1 in 3 minutes. As they are each walking at constant speeds, d 1 d2 = t 1 = 3 t, from which t = p 3. Hence Hyam walks for 3 + p 3 minutes, while Walter walks for 1 + p 3 minutes. 2 Three grasshoppers A, B and C are placed on a line. Grasshopper B sits at the midpoint between A and C. Every second, one of the grasshoppers jumps over one of the others to the symmetrical point on the other side (if X jumps over Y to point X', then XY=YX'). After several jumps it so happened that they returned to the three initial points (but maybe in different order). Prove that B returns to her initial position. Label the points corresponding to the initial positions of A; B; C as 1; 0; 1 respectively. Then by the rules of the game, A and B will continue to occupy points with odd numbered coordinates, while B will occupy points with even numbered coordinates. This is because every grasshopper moves an even number of steps on eachmove, leaving their parity unchanged. Thus if they return to the initial coordinates, B must occupy her original spot, being the only grasshopper whose initial coordinate was even. 3 A right circular cone has base of radius 1 and height 3. The largest possible cube is inscribed in the cone so that one face of the cube is contained in the base of the cone. What is the side-length of the cube? 1

2 Consider the plane containing both the axis of the cone and two opposite vertices of the cube's bottom face. The cross section of the cone and the cube in this plane consists of a rectangle of sides s and s p 2 inscribed in an isosceles triangle of base 2 and height 3, where s is the side-length of the cube. (The s p 2 side of the rectangle lies on the base of the triangle.) Similar triangles yield s=3 =(1 s p 2=2)=1, or s =(9 p 2 6)=7: (Many students visualised in cross section a square rather than a rectangle being inscribed in the triangle. This led to the incorrect answer of s =6=5.) 4 If S n denotes the sum of the decimal digits of the number 2 n ; either show that S n 6= S n+1 for all positive integers n; or find a counter example. Let 2 n = a 0 +10a 1 +:::+10 r a r = Pr a i=0 i10 i, where the a i 's are the digits of 2 n. Then S n = a 0 + a 1 + :::+a r = Pr a i=0 i. As 10 i 1 is divisible by 9 for any non-negative integer i, 9 is a factor of Pr a i=0 i(10 i 1) = Pr a i=0 i10 i Pr a i=0 i = 2 n S n. Similarly 9 divides 2 n+1 S n+1. Therefore 9 divides their difference which is (2 n+1 S n+1 ) (2 n S n ) = 2 n (as S n = S n+1 by assumption). But 9 cannot divide 2 n, which only has powers of two as factors. From this contradiction we conclude that there is no such n for which S n = S n+1. 5 Let s be any arc of the unit circle lying entirely in the first quadrant. Denote its endpoints by D and E. Let A be the area of the region lying below s; bounded by s; the x-axis and vertical lines through D and E. Let B be the area of the region lying to the left of s, and bounded by the y axis and horizontal lines through D and E. Prove that A + B depends only on the arc length, and not on the position, of s. First solution, without calculus: to fix notation, let A be the area of region DEFG, and B be the area of DEIH; further let C denote the area of sector ODE, which only depends on the arc length of s. If [XY Z] denotes the area of triangle [XY Z], then we have A = C +[OEG] [ODF] and B = C +[ODH] [OEI]. But clearly [OEG] =[OEI] and [ODF] =[ODH], and so A + B =2C. H I D E O F G Second solution, with calculus: We may parameterize a point in s by any of x, y, or = tan 1 (y=x). Then A and B are just the integrals of ydx and xdy over the appropriate intervals; thus A + B is the integral of xdy ydx (minus because the limits of integration are reversed). But d = xdy ydx, and so A + B = is precisely the radian measure of s. (Of course, one can perfectly well do this problem by computing the two integrals separately. But what's the fun in that?) 2

3 6 (a) A 6 6 board is tiled with 2 1 dominoes. Prove that we can always divide the board into two rectangles each of which is tiled separately (with no domino crossing the dividing line). (b) Is this true for an 8 8 board? Solution provided by Chaitanya Rao: (a) Assume that no such division is possible. We say a domino bridges two columns if half the domino is in each column. We show by induction that for 0 < n < 6 the number of dominoes bridging columns n and n +1 must be at least 2 and even. Consider first n =1:There cannot be 3 dominoes entirely in column 1, or it would be separately tiled. So there must be at least one domino bridging columns 1 and 2. The number must be even, because it must equal the number of squares in column 1 (even) less twice the number of dominoes (entirely) in column 1. Now suppose it is true for n < 5 and consider column n +1:There must be at least one domino bridging columns n +1 and n +2; or columns 1 through n +1 would be separately tiled. The number must be even, because it must equal the number of squares in column n + 1 (even) less the number bridging n and n + 1 (even) less twice the number entirely in the column. By induction, it follows that at least 2 dominoes bridge columns n and n + 1 for 0 <n<6: So in total there are at least 5 2 = 10 dominoes bridging columns. By the same argument there are at least another 10 bridging rows, but there are only 18 dominoes in total, a contradiction. Solution based on that presented by Andrew Cheeseman: Assume that no such division is possible. Then any separation of the board into two rectangles must have at least one domino being cut in half. Let the number of such dominoes for a particular separation be k, where k > 0. The total number of squares in one of the rectangles is then twice the number of whole dominoes within it, plus k. But this total has to be even as the rectangle has 6 rows or columns. Hence k is even and so the number of dominoes bridging" any row or column must be at least 2. Therefore in total there are at least 5 2 = 10 dominoes bridging columns and at least another 10 bridging rows. But there are only 18 dominoes in total, a contradiction. (b) No. Many examples can be constructed, here is one: 7 A polynomial P (x) is said to be boring if it takes rational values for rational x and irrational values for irrational x: Prove that the coefficients of a boring polynomial must be rational. Hence, or otherwise, prove that all boring polynomials are linear. Solutions provided by Frank Calegari: Solution 1 Here is a solution of a slightly easier problem. Let f(x) be a polynomial with real coefficients of degree 1 such that for infinitely many rational numbers R, f(r) is also rational. Prove that f(x) has rational coefficients. 3

4 We can assume f(x) 2 Q[x]; by Lagrange's formula. We can assume f(x) 2 Z[x]; since if the property holds for f(x) it also holds for f(ax+b);a;b2 Q achange of variables then means we can assume that f(x) is monic also. Finally, iff(x)=x n +::: + c; then f n (x) = f(x) c n; has a rational solution, which (Gauss' lemma) says that f(z) contains c; c +1;c+2; :::; which is obvious. 2 An alternative proof is given below. Solution 2 The solution consists of two parts. (1) Show that p(x) has rational coefficients, (which follows immediately from the solution of the simpler problem above). (2) Show that if p(x) has rational coefficients, then p(x) is linear. (1) Consists of the following Lemma: Lemma 1 If p(x) is rational for all but finitely many rational x; then p(x) has rational coefficients.2 Proof by induction (on degree). If deg(p(x)) = 0; then the result is trivial. Otherwise, let R be a rational number such that p(r) is rational. Let p(r) =C: Then p(x) =C+(x R)q(x) and q(x) is rational for all x for which p(x) is rational, except possibly at R; and now we are done. (2). Multiplying p(x) by an integer does not affect its properties. Thus we may let p(x) =a n x n +::::: + a 0 ;a i 2 Z: 4

5 W.L.O.G, let a n be positive. Hence the image of p(x) consists (at least) of all positive real numbers greater or equal to a 0 (by Rolle's theorem). The idea is now that the polynomials p(x) a 0 ;p(x) a 0 1;p(x) a 0 2; :::p(x) a 0 k have tohave rational solutions for all x (since they have real solutions, and if they had irrational solutions then p(x) would be rational), and this will be impossible. Note that for p(x) ="integer"; one must have a n Λ x an integer. Hence the possible integer values of p(x) are contained within the set p(n=a n ); for n 2 Z: If p(x) is of degree > 1; then its pretty clear that this expression can't equal every positive integer greater than a 0 : The reason is that it is growing too fast, and we have forced the range to be discrete. 2 Based on a solution provided by Garth Gaudry: Let p(x) =a 0 +a 1 x+ +a n x n be a boring" polynomial. Then Lemma 2 The coefficients of p(x) are rational. Proof. There are various possibilities. For instance, use the fact that a polynomial of degree n is determined by its values at n +1 points, and write the interpolating polynomial down. (Digression. The interpolation polynomial is the unique polynomial which solves the following problem: Find, for a given function f(x) a polynomial of n th degree P n (x), which attains for n +1different values x k ;k = 0;1;::: ;n; of the independent variable x, the same values as the given function f(x): That is P n (x k )=f(x k ) for k =0;1;::: ;n: The proof of uniqueness and the construction of this polynomial is discussed in most textbooks on calculus or numerical analysis.) Alternatively, let x =0toget a 0 rational. We may suppose without loss of generality from now on that a 0 = 0. Then substituting x = 1; 2;:::n 1;m where m is a sufficiently large integer, we get rational values for p(x) for each such x, and hence a system of n a system of simultaneous linear equations in n unknowns with rational right-hand sides. The coefficient matrix is nonsingular if m is large enough, (more simply, wehavenindependent equations if m is large enough), and so we can solve for a 1 ;:::;a n. They are rational. 2 Theorem 1 A boring polynomial is linear. Proof. Recall that we are assuming that a 0 = 0; we may also assume that the polynomial has integer coefficients. Suppose that the degree of the polynomial is greater than 1. The polynomial takes all sufficiently large real values (all real values if n is odd), so the equation a n x n + a n 1x n 1 + +a 1 x= u v (1) 5

6 has a solution for all sufficiently large u=v. Take v to be a prime that does not divide a n, and let u be a large integer not divisible by v. Since (1) has a solution (assuming u is large enough), that root must be rational, say p=q, where (p; q) =1. This yields the equation and hence p n a n q + a n n 1 q + +a p n 1 1 q = u v p n 1 v(a n p n + a n 1p n 1 q + +a 1 pq n 1 )=uq n : Since v is prime, coprime to u, and divides the left-hand side, it must also divide the right-hand side, and hence q. There are at least two factors q on the right-hand side; so after division by v, the right- hand side is still divisible by v. This implies that vja n p n. But (p; q) =1,sovja n. This is a contradiction. 2 (Nobody completely solved the second part of this question.) 6