(n 1)n(n + 1)(n + 2) + 1 = (n 1)(n + 2)n(n + 1) + 1 = ( (n 2 + n 1) 1 )( (n 2 + n 1) + 1 ) + 1 = (n 2 + n 1) 2.

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1 Paabola Volume 5, Issue (017) Solutions Q151 Take any fou consecutive whole numbes, multiply them togethe and add 1. Make a conjectue and pove it! The esulting numbe can, fo instance, be expessed as a squae: (n 1)n(n + 1)(n + ) + 1 = (n 1)(n + )n(n + 1) + 1 = ( (n + n 1) 1 )( (n + n 1) + 1 ) + 1 = (n + n 1). Q15 Let ABC be a tiangle with longest side BC and let P be a point in the inteio of the tiangle. Show that AP < BP + P C. Note that BP + P C > BC > max{ac, AB} > AP. Q15 (a) Show that if p is odd, then x p 1 = (x 1)(x p 1 + x p + + x + 1). (b) Hence, show that if x is odd, then the highest powe of which divides x p 1 also divides x 1. (c) Find the highest powe of which divides (a) Expand the ight-hand side of the equation. (b) If x is odd, then (x p 1 + x p + + x + 1) is odd since it is the sum of an odd numbe of odd numbes. (c) = (( ) 15 1) so, by (b), the highest powe of that we seek is the highest powe of dividing This numbe is ( )( )( )( )(1999 1). The fist tem is divisible by but not 4; this is easy to see if calculating modulo 8. This is also tue fo each of the othe tems. So the powe of is 8. Q154 Let ABCD be a convex quadilateal and let P, Q, R, S be points on AB, BC, CD, DA espectively, such that AP = 1AB, QC = 1BC, CR = 1CD and SA = 1DA (a) Show that P QRS is a paallelogam. (b) Find the atio of the aea of P QRS to that of ABCD. 1

2 (a) This is easily shown by similaity. (b) P QRS = 8 ABCD. Q155 Find all positive intege solutions to x xy + y = 65. We can ewite the equation as (x y) + x = 65. The only two squae intege pais that add up to 65 ae 1 = 1 and 64 = 8, and 16 = 4 and 49 = 7, and these pais in evese ode. The solutions (x, y) ae thus (8, 9), (1, 9), (7, ), (4, 7). Q156 The base notation a b appeaing in this poblem is mostly ecently explained in the Paabola aticle hee. (a) Show that whateve base b is used, the numbe (1) b is neve equal to twice (1) b. (b) Find all the numbes and all bases b 1 fo which thee exists a two digit numbe (ac) b which is twice the numbe obtained by evesing its digits. (c) Find all bases b and all numbes n = (ac) b such that n = (ac) b. (a) Now, (1) b = b + 1 and (1) b = b + ; howeve, (b + ) is even and cannot equal b + 1. (b) By tial and eo, we find b = 5, (ac) b = (1) b, b = 8, (ac) b = (5) b, and b = 11, (ac) b = (7) b. (c) The condition is ab + c = (cb + a), o a(b ) = c(b 1). Now, if d (b ) and d (b 1), then we have d (b ) (b 1) and so d ; hence, d = 1 o d =. If d = 1, then a = b 1 which is geate than b. If d =, then we can wite b = k which implies that ka = (k + 1)c and, since gcd(k, k + 1) = 1, the only solution with a, c < b is c = k and a = k + 1. Hence, the equied conditions ae b = k +, c = k, a = k + 1. Q157 Denote the top of a cube by ABCD and the bottom by A 1, B 1, C 1, D 1, so that A is diectly above A 1 and so on. Take midpoints of the six edges AB, BB 1, B 1 C 1, C 1 D 1, D 1 D and DA. Show that a plane containing any thee of these points contains them all and deduce that these points fom the vetices of a egula hexagon. Let P be the midpoint of AA 1, S the midpoint of CC 1, U the midpoint of AB and T the midpoint of BC. Extend B 1 B to X such that BX = AP. The est of the solution is staight-fowad.

3 Q158 (a) Simplify (a + b + c)(a + b + c ab ac bc). (b) Show that (a + b + c ab ac bc) 0. (c) Pove that if x, y, z ae positive eal numbes, then x+y+z xyz. (a) a + b + c abc (b) a + b + c ab ac bc = 1 ( (a + b ab) + (a + c ac) + (b + c bc) ) = 1 ( (a b) + (a c) + (b c) ) 0. (c) By pats (a) and (b), x + y + z xyz 0. Since the function f(u) = u is convex fo positive u, we have that x + y + z (x ) + y + z = x + y + z xyz. Q159 If we expand ( + x) 18 as a polynomial, then we obtain ( + x) 18 = a 0 + a 1 x + a x + + a 18 x 18, whee a 0, a 1,..., a 18 ae integes. Without using the Binomial Theoem, find a 0, a 1, a 18 and a 0 + a a 18. Wite out the poduct as ( + x)( + x) ( + x). }{{} 18 times Then clealy, a 0 = 18, a 18 = 1 and the tem with one x can be obtained fom 18 tems each with a coefficient of 17, so a 1 = To get the sum of the coefficients, set x = 1; this gives 18.

4 Q1540 Cicumscibe a ight cicula cone about a sphee of adius such that the cone has minimum volume. Pove that the cone has exactly double the volume of the sphee. A D O C B Since the adius of the cone must be geate than the adius of the sphee, let the cone s adius be a, such that a > 1 and a is the volume-minimizing adius of the cone. In addition, let AD = x. A x D O C a B Using the fact that ADO ABC, we can wite the equation x = x + +. a 4

5 We solve fo x in tems of a and, fist getting (ax ) = x + and thus x = The height of the cone is a a 1. ( ) a 4a + a 1 + = + (a 1) + = (a + 1) (a 1) a 1 +, (1) so the volume of the cone, expessed as a function f(a) in a, is f(a) = 1 π(a) ( (a + 1) a 1 ) +. Because the cone must have minimum volume, we must find the value of a that minimizes ou function, so we calculate the deivative of ou volume function: f (a) = π 4a (a ) (a 1). Setting this deivative equal to 0 and solving fo a, we find that a = 0, ±. Since a > 1, the coect solution is a =. Thus, the adius of the cone is and, by (1), the height of the cone is 4. Theefoe, the volume of the cone is 1 π( ) 4 = 8 π. Since the volume of the sphee is 4 π, the volume of the cone is exactly double the volume of the sphee, as claimed. 5