OLYMON. Produced by the Canadian Mathematical Society and the Department of Mathematics of the University of Toronto. Issue 9:2.

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1 OLYMON Poduced by the Canadian Mathematical Society and the Depatment of Mathematics of the Univesity of Toonto Please send you solution to Pofesso EJ Babeau Depatment of Mathematics Univesity of Toonto 40 St Geoge Steet Toonto, ON M5S E4 Issue 9: Mach, 008 no late than Mach 3, 008 It is impotant that you complete mailing addess and you addess appea on the font page If you do not wite you family name last, please undeline it 535 Let the tiangle ABC be isosceles with AB = AC Suppose that its cicumcente is O, the D is the midpoint of side AB and that E is the centoid of tiangle ACD Pove that OE is pependicula to CD 536 Thee ae cities, and seveal ailines ae esponsible fo connections between them Each ailine seves five cities with flights both ways between all pais of them Two o moe ailines may seve a given pai of cities Evey pai of cities is seviced by at least one diect etun flight What is the minimum numbe of ailines that would meet these conditions? 537 Conside all squae aays each of whose enties is eithe 0 o A pai (A, B of such aays is compatible if thee exists a 3 3 squae aay in which both A and B appea as subaays Fo example, the two matices 0 and ae compatible, as both can be found in the aay Detemine all pais of aays that ae not compatible In the convex quadilateal ABCD, the diagonals AC and BD ae pependicula and the opposite sides AB and DC ae not paallel Suppose that the point P, whee the ight bisectos of AB and DC meet, is inside ABCD Pove that ABCD is a cyclic quadilateal if and only if the tiangles ABP and CDP have the same aea 539 Detemine the maximum value of the expession ove all quatuple of eal numbes not all zeo xy + yz + zw x + y + z + w

2 540 Suppose that, if all plana coss-sections of a bounded solid figue ae cicles, then the solid figue must be a sphee 54 Pove that the equation x x + xx + + xx = xx + + has no solution fo which x, x,, x, x + ae all distinct nonzeo integes Solutions 58 Let the sequence {x n : n = 0,,, } be defined by x 0 = a and x = b, whee a and b ae eal numbes, and by 7x n = 5x n + x n fo n Deive a fomula fo x n as a function of a, b and n Solution This can be done by the standad theoy of solving linea ecusions The auxiliay equation is 7t 5t = 0, with oots and /7 Tying a solution of the fom x n = A n + B( /7 n and plugging in the initial conditions leads to A + B = a and A (/7B = b and the solution x n = a + 7b 9 + 7(a b 9 ( 7 n 59 Let, n be positive integes Define p n, = fo all n and p n, = 0 fo n + Fo n, we define inductively p n, = (p n, + p n, Pove, by mathematical induction, that p n, = ( ( n Solution Let =0 =0 =0 q n, = ( ( n When n =, we have that q, = ( 0 = and, fo, q, = ( ( = [( ( ] + [( ( ] = 0 =0 Also q n, = 0 n = fo n When (n, = (,, we have that q, = 0 ( = and p, = ( + 0 = When n = and 3, then q, = ( ( =0 = ( [ ( + ( ] =0 = ( + ( =0 =0 = ( + ( ( + ( ( 3 = ( 3 [ + + ] = 0 ( + ( =0 (

3 Thus, we have that p n, = q n, fo all n and = as well as fo n =, and all The emainde of the agument can be done by induction Suppose that n and that and that it has been shown that p n, = q n, and p n, = q n, Then as desied p n+, = (p n, + p n, [ ( ] = ( ( n + ( ( n =0 =0 [ ( ] = n + ( ( n + ( ( n = = [ [ ( ] = n + ]( ( n = = n+ + ( ( n = = n+ + = =0 ( = ( ( ( n+ ( ( n+ = q n+,, 530 Let {x, x, x 3,, x n, } be a sequence is distinct positive eal numbes Pove that this sequence is a geometic pogession if and only if fo all n n x x n = x n x x x x + x x Solution Necessity Suppose that x = a fo some numbes a and Then n x x x n = (n x x + n ( = ( n n 3 = (n = (n = x n x x x Sufficiency Suppose that the equations of the poblem holds When n =, both sides of the equation ae equal to egadless of the sequence When n = 3, the equation is equivalent to x x 3 x x x (x 3 + x = (x 3 x (x 3 + x 3 x 3 x Since x 3 +x 0 [why?], we can divide out this facto and multiply up the denominatos to get the equivalent x 3 (x x = x (x 3 x x 3 x = x x x x 3 = x, 3

4 whence x, x, x 3 ae in geometic pogession Suppose, as an induction hypothesis, fo n 4 we now that x = a fo suitable a and and =,,, n Let x n = au n fo some numbe u n Then u ( n n 5 + n = u n u n [u n( n 6 + n 3 u n ]( = (u n ( n 4 ( n 4 u n + ( n 3 n 3 u n = ( n 4 u n n 4 0 = u n ( n n 3 u n n 4 = (u n n (u n + n 3 The case u n = n 3 is ejected because of the condition that the sequence consists of positive tems Hence u n = n, as desied The esult follows Comment In the absence of the positivity contiion, the second oot of the quadatic can be used Fo example, the finite sequences {,,,, } and {,,,, } both satisfies the equations fo n 5 It would be inteesting to investigate the situation futhe 53 Show that the emainde of the polynomial p(x = x x x x x x + 007x is the same upon division by x(x + as upon division by x(x + Solution We have that p(x = (x x x (x x x (x x 00 + x (x 3 + x + x + 004x = x(x + (x x x (004x + 008, fom which the esult follows with emainde 004x The angle bisectos BD and CE of tiangle ABC meet AC and AB at D and E espectively and meet at I If [ABD] = [ACE], pove that AI ED Is the convese tue? and that Solution Obseve that [ADB] : [CBD] = AD : DC = AB : BC [ACE] : [BCE] = AE : EB = AC : BC Now [ABD] = [ACE] [DBC] = [ABC] [ABD] = [ABC] [ACE] = [EBC] ED BC AE : EB = AD : DC AB : BC = AC : BC AB = BC AI BC Both the esult and the convese is tue If [ABD] = [ACE], the foegoing chain of implications can be ead in the fowad diection to deduce that AI ED Note that AI bisects angle A in tiangle AED Thus, if AI ED, then it follows that tiangle AED is isosceles with AE = AD Then AE : DC = AD : DC = AB : BC and AE : EB = AC : BC, whence DC AB = AE BC = EB AC Theefoe DC (AE + EB = EB (AD + CD, so that DC AE = EB AD and DC = EB Theefoe AB = AC and, following the foegoing implication in the bacwads diection, we find that [ABD] = [ACE] 4

5 533 Pove that the numbe + ( is divisible by 008 Solution Let a = and b = 5 7, so that a + b = 0 and ab = 8 Define x n = a n + b n Then x = 0, x = (a + b ab = 96 and x n+ = a n+ + b n+ = (a + b(a n+ + b n+ ab(a n + b n = 0x n+ 8x n, fo n 0 Note that x is divisible by and x by 4 Suppose, as an induction hypothesis, that x n = n u and x n+ = n+ v, fo some 0 and integes u and v Then Hence, fo all positive integes n, n divides x n x n+ = 5 n+ n+3 = 3 n+ Obseve that (5 7 n = b n < fo each positive intege n and that a n + b n is a positive intege Theefoe x n = a n + b n > a n > a n + b n = x n, whence x n = + a n and the esult follows 534 Let {x n : n =,, } be a sequence of distinct positive integes, with x = a Suppose that fo n Detemine n x n xi = (n + x n Solution When n =, ( x + x = 3 x, whence x = x and x = 4x = 4a When n = 3, ( x + x + x 3 = 4 x 3 = x 3 = ( x + x = 6 x = x 3 = 9x = 9a We conjectue that x = a fo each positive intege Let m and suppose that x = a fo m whence x m = m a and x m = m a (m x m = (m + x m x m = ( x + + x m Thus, x = a fo all Theefoe = x ( (m = m(m a, n x = ( n a = n(n + (n + a 6 5

(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.

(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. Paabola Volume 5, Issue (017) Solutions 151 1540 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