Calculators are NOT permitted.

Transcription

1 THE 0-0 KEESW STTE UIVERSITY HIGH SHOOL THETIS OETITIO RT II In addition to scoring student responses based on whether a solution is correct and complete, consideration will be given to elegance, simplicity, originality, and clarity o presentation. alculators are OT permitted.. The equation y = x + ax + a represents a parabola or all real values o a. (a) rove that all o these parabolas pass through a common point and determine the coordinates o this point. (b) The vertices o all the parabolas lie on a curve. Find, with proo, the equation o this curve.. sequence o unctions is deined by the ollowing rules (i) ( and (ii) n ( n ) or n =,,,. ompute, with proo, 0(0).. onsider the three numbers 6n, n, and n, where n is a positive integer. (a) Find, with proo, all values o n or which all three numbers are prime numbers. (b) rove that there are ininitely many values o n or which none o the three numbers is a prime number. 4. Six hotel guests wanted to check out, but the ront desk clerk was nowhere to be ound. The guests each put their room key on the counter and let. When the clerk returned, he did not know which o the 6 keys went to which room. I the clerk randomly gave the keys to each o the next six guests, compute, with proo, the probability that none o the new guests received the correct room key.. In the igure, is a square, is the midpoint o, and is the midpoint o. and intersect at point. ompute, with proo, the ratio o the area o quadrilateral to the area o square.

2 THE 0-0 KEESW STTE UIVERSITY HIGH SHOOL THETIS OETITIO RT II SOLUTIOS. a) We want to show that there is a point x, ) which passes through all parabolas ( y with equations o the orm y x ax a. Let m and n be two distinct real numbers. Then we want y x mx m and y x nx n. Subtracting the second equation rom the irst gives mx nx ( m n) 0 x ( m n) ( m n). Since m n, we must have x, so y. Thus the desired point is (, ). 4 4 b) To ind the vertex o each parabola y x ax a, we complete the square, giving y x ax a a a ( x a) ( a a ). Thereore, the vertex has coordinates ( a, a a ). Thus, the vertices lie on a parabola with equation y x x.. Let s see i there is a pattern to the sequence. x x, ( ( ). In a similar manner we ind, x x x x ( ). Similarly, x ( 4,, x x x 6 x, and ( 7. So we see that 6 n x or n =,,,. 0 0 Since 0 = (6)() +, we have 0( 0) (0). 0 0

3 . y inspection, I n = the three numbers are,, and, all prime. I n =, the three numbers are 9,, and, again all prime. However when n=, we get {9,, 0}, when n = 4 we get {0,, 7}, and when n = we get {,, 6}, so that each set consists o at least one non-prime. ore speciically, each o the sets o three numbers listed above contains a multiple o. (a) We will prove that one o the three numbers must be a multiple o or n >. ny integer must have one o the orms k, k, or k. I n = k, 6n 0k (0k ) which is clearly a multiple o. I n = k, then n (k ) 0k 0k (0k 4k ), again a multiple o. I n = k, then n (k ) k 0k (4k 4k ), again a multiple o. In all three cases, thereore, one o the three numbers 6n, n, and n is a multiple o. Since each number is larger than or n >, the only values o n or which all three numbers are prime are n = and n =. (b) I n is an odd multiple o, 6n is a multiple o, and n is even. I n is an odd multiple o, n is a multiple o, and n is even. Hence, i n is any odd multiple o, 6n is a multiple o (and thereore not prime), n is a multiple o (and thereore not prime), and n is even (and thereore not prime). Thereore, there are ininitely many values o n or which none o the three numbers is prime.

4 4. It is easier to compute the probability o the opposite event, namely the one where at least one o the new guests receives the correct room key. enote that probability by p(). I the desired probability is p(), than we compute it as p()= p(). Let probability that guest k received the correct room key, probability that guests k and l got the correct room keys and so on analogously. For example denotes the probability that guests, and received the correct room key. Using the principle o inclusion-exclusion we have otice that and that or all choices o k,l,m,n, and so on. Hence, ow we compute,. Thereore, It ollows.. ethod onstruct diagonal. Since and are medians o triangle, point is the centroid o triangle, so that = ⅓. onstruct a perpendicular rom to. and Q are similar, Q = ⅓. Since Q and are the altitudes o and, respectively, and they have the same base ( ), the area o is ⅓ the area o Q Thus, area o = ⅓( ½ ) = / 6 ( ). Similarly, the area o = / 6 ( ). Thereore, rea o quad = rea o (area o area o ) = [ / 6 ( ) + / 6 ( )] = ⅔ ( ). Thereore, the desired ratio is.

5 ethod onstruct diagonal. Let m = m = and m = m = θ. Using quadrilateral, θ = 70. Sinθ = sin(70 ) = sin70cos cos70sin = cos Without loss o generality, let =, so that = and =. Using right, sin =. Thereore, using the appropriate double angle ormula, cos = - sin = Thereore sinθ = =. θ Since and are medians o triangle, point is the centroid o the triangle, so that = = ⅔ = ⅔. Thereore, the area o = ½ ()()sinθ = ½ (⅔ )( ⅔ )( ) =. The area o = ½ () =, and the area o quad = + = 8. 8 Thereore, the desired ratio is 4. ethod lace the igure in the coordinate plane so the vertices o the square have the ollowing coordinates: = (0, 0), = (, 0), = (, ), = (0, ). Then, segment alls on the line y x and segment alls on the line y x E F Since x x x, the coordinates o are (/, /) raw a horizontal line through that intersects at E = (0, /) and at F = (, /) Since triangle E is congruent to triangle F, the area o quadrilateral equals the area o rectangle EF. EF has an area o / and the area o is. So, / is the desired ratio.