THE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART II Calculators are NOT permitted Time allowed: 2 hours

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THE 06-07 KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART II Calculators are NOT permitted Time allowed: hours Let x, y, ad A all be positive itegers with x y a) Prove that there

of the flowers are blue, ad / of the flowers are yellow Mr Smith, beig a mathematicia, discovered that the probability that a radomly picked buch of three flowers cotais exactly red flower is greater

$are prime umbers Fid, with proof, all possible ordered pairs (p, q) such that the equatio P(x) = 0 has ratioal solutios Let x deote the fractioal part of the real umber x so that, for example, =, 5 5$

1 THE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART II Calculators are NOT permitted Time allowed: hours Let x, y, ad A all be positive itegers with x y a) Prove that there are ifiitely may ordered triples (x, y, A) for which b) If x, y, ad A are all less tha 0, fid oe ordered triple (x, y, A) for which + y A y A I Mr Smith s garde, exactly / of the flowers are red, / of the flowers are blue, ad / of the flowers are yellow Mr Smith, beig a mathematicia, discovered that the probability that a radomly picked buch of three flowers cotais exactly red flower is greater tha oe-half The same is true for the other two colors as well Compute, with proof, the largest umber of flowers that could be i Mr Smith s garde Cosider the polyomial P(x) = px qx + p, where p ad q are prime umbers Fid, with proof, all possible ordered pairs (p, q) such that the equatio P(x) = 0 has ratioal solutios Let x deote the fractioal part of the real umber x so that, for example, =, 5 5 = 0, ad π = π Fid, with proof, the smallest positive real umber x, larger tha, such that x + = x 5 Triagle ABC has side legths a, b, ad c, where a, b, ad c are cosecutive itegers with a < b < c A media draw to oe side of ΔABC divides ΔABC ito two triagles, at least oe of which is isosceles Compute, with proof, all possible ordered triples (a, b, c)

2 Solutios k a) Multiplyig 9 by ay umber of the form r, where r is a positive iteger will produce a k perfect square ( r ) Let k be a multiple of The k = m, where m is a positive iteger, ad 9( k k k k 6m 6m m m r ) = ( + )( r ) = r + r = r + r = ( r ) + ( r ) m m m Therefore, the ordered triple (x, y, A) = ( r, r, r ) satisfies the give equatio for all positive itegers m More specifically, if m =, the ordered triple ( r, r, r ) satisfies the give equatio Thus there are ifiitely may such ordered triples b) Factorig, we obtai ( A y = ( x y)( x + xy + y ) A Let x y = + xy + y ) Substitutig = x y ad simplifyig, we obtai The x x + = A x ( x ) + = A Thus we are lookig for two cosecutive positive itegers (if ay exist) whose product, whe multiplied by, is oe less tha a perfect square Sice x, y, ad A are all less tha 0, a little trial ad error shows that whe 8, A = (8)(7) + = 69 = Therefore, (8, 7, ) is a solutio (Note: the oly other solutio where x, y, ad A are all less tha 0 is (0, 6, 8)) (Note also that if we do t require x, y, ad A to be less tha 0, (05, 0, 8) also works Are there ifiitely may solutios?) Let R deote the umber of red flowers ad N the total umber of flowers i the garde The probability of pickig red flower whe we pick a buch of three flowers is R(N R)(N R) P N = = N(N )(N ) Sice / of all flowers are red we have that P N = R = N Substitutig ad simplifyig, N(N ) 9(N )(N ) Sice we are told that the probability of pickig exactly oe red flower i a buch of three is greater tha /, N(N ) 9(N )(N ) > Simplifyig, we obtai N 5N + 8 < Usig the quadratic formula we fid N is betwee ad Sice N is a iteger ad also a multiple of three, the largest possible N =

3 If x < 0, the the left side of the equatio px qx + p= 0 is positive Therefore, If the equatio has real solutios, the x > 0 Let m where m ad are positive itegers ad m is i lowest terms (ie m ad have o commo factors other tha ) Substitutig m ito the equatio, ad multiplyig by to clear fractios, we obtai pm qm + p = 0 from which p = qm pm = m( q pm) p But sice m ad have o commo factors, m must divide p Therefore, m divides m Similarly must divide p as well Sice p is prime, =, p, or p If, the q = p, which is impossible sice q is prime If or p, the the quadratic p equatio yields q = p + If p is odd, the q is a eve umber 0, agai a cotradictio sice q is prime Therefore, p must be eve, so p = ad q = 5, ad the oly possible ordered pair (p, q) = (, 5) Thus, P(x) is x 5x +, ad the roots of P(x) = 0 are ad Let x > be a real umber satisfyig x + = If deotes the greatest iteger x i x, the + x, ad obviously Sice 0 < <, we have = x x x Thus x + = + x + = + Therefore, x is a solutio to the quadratic equatio x x x ( + ) x + = 0 Usig the quadratic formula, we fid Sice x >, we ca elimiate the mius sig, so that ( + ) ± ( + ) ( + ) + ( + ) If =, the, which is a cotradictio Thus, ad the potetially smallest x occurs + 5 whe = ad To see that this umber actually satisfies the give coditio, ote that x 6, so x = Also = = = x x Therefore, x + = + = Thus the smallest value of x is x

4 5 There are oly three such ordered triples: (,, ), (,, 5), ad (7, 8, 9) If ΔABC has sides of legth,, ad, the clearly the media to the side of legth creates a isosceles triagle with legs of legth If ΔABC has sides of legth,, ad 5, it is a right triagle The media to the hypoteuse of a right triagle is always half the legth of the hypoteuse Therefore, the media to the hypoteuse divides the triagle ito two isosceles triagles with legs of legth ½ Now represet the side legths of ΔABC by,, ad + Case : The media is draw to the shortest side Sice AM < AB, the oly way either ΔAMC or ΔAMB could be isosceles is if AM = or AM = Usig the law of cosies o ΔABC, ( + ) = + ( ) ( ) cosc from which cosc = Usig the law of cosies o ΔAMC, ( ) AM = + ( ) cosc Suppose AM = The ( ) = + ( )( ) cosc from which cosc = Therefore, = = 0 which has o iteger solutios Suppose AM = Agai usig the law of cosies o ΔAMC, ( ) = + ( ) cosc from which cosc = Therefore, =, whose oly solutio is = Thus the media caot be to the shortest side Case : The media is draw to the side of legth Sice BM < AB, the oly way either ΔAMC or ΔAMB could be isosceles is if BM = or BM = A A M C M + B C + B

5 From case, we kow that cosc = Usig the law of cosies o ΔBMC, Substitutig BM BM cosc =, we obtai = + ( ) ( ) = + ( ) ( )cosc, from which BM = + + If BM =, the =, which has o real solutios + If BM =, the ( ) = from which 8 = 0 ad = 8 yieldig a triagle with sides of legth 7, 8, ad 9 C Case : The media is draw to the logest side + If AM = MB = CB, the = from which = This gives a,, triagle, which we have already cosidered If CM = MB = AM, the triagle ABC is a right triagle ad the oly right triagle with side legths that are cosecutive itegers is the,, 5 triagle already cosidered A M + B The oly other possibilities for ΔAMC or ΔAMB to be isosceles is if CM = or Usig the law of cosies o ΔABC, + ( ) = + ( + ) ( + ) cos A from which cos A = + Usig the law of cosies o ΔAMC, CM = + cos A = + = ( ) + ( + ) If CM =, the = + This becomes + 6 = 0, which has o iteger solutios + ( + ) If CM =, the ( ) = + This becomes + = 0, which has o real solutios Therefore, the oly triagles with the give properties have sides of legths (,, ), (,, 5), ad

6 (7, 8, 9) The three triagles are show below ½ ½ ½ 7 7 9

= 4 and 4 is the principal cube root of 64.

= 4 and 4 is the principal cube root of 64. Chapter Real Numbers ad Radicals Day 1: Roots ad Radicals A2TH SWBAT evaluate radicals of ay idex Do Now: Simplify the followig: a) 2 2 b) (-2) 2 c) -2 2 d) 8 2 e) (-8) 2 f) 5 g)(-5) Based o parts a ad