2018 Hypatia Contest

Transcription

1 The CENTRE f EDUCATION in MATHEMATICS and COMPUTING cemcuwaterlooca 01 Hypatia Contest Thursday, April 1, 01 (in Nth America and South America) Friday, April 13, 01 (outside of Nth America and South America) Solutions 01 University of Waterloo

2 01 Hypatia Contest Solutions Page (a) The average of Aneesh s first six test sces was = = 15 (b) After Jon s third test, his average sce was 1, and so the sum of the sces on his first three tests was 1 3 = The sum of his sces on his first two tests was = 9, and so the sce on his third test was 9 = (We may check that the average of 17, 1 and 13 is = 1) 3 (c) Dina wrote six tests followed by n me tests, f a total of n + 6 tests After Dina s first 6 tests, her average sce was 1, and so the sum of the sces on her first 6 tests was 1 6 = Dina sced 0 on each of her next n tests, and so the sum of the sces on her next n tests was 0n Therefe, the sum of the sces on these n + 6 tests was + 0n After Dina s n + 6 tests, her average sce was 1, and so the sum of the sces on her n + 6 tests was 1(n + 6) Thus, + 0n = 1(n + 6) + 0n = 1n + 10 n =, and so n = 1 (a) The distance from Botown to Aville is 10 km Jessica drove this distance at a speed of 90 km/h, and so it took Jessica = 3 hours 60 = 0 minutes 3 (b) The distance from Botown to Aville is 10 km The car predicted that Jessica would drive this distance at a speed of 0 km/h, and so it predicted that it would take Jessica 10 0 = 3 hours 3 60 = 90 minutes The ETA displayed by her car at 7:00 am was :30 am (c) Jessica drove from 7:00 am to 7:16 am (f 16 minutes) at a speed of 90 km/h, and so she travelled a distance of = km 60 At 7:16 am, Jessica had a distance of 10 km km = 96 km left to travel The car predicted that Jessica would drive this distance at a speed of 0 km/h, and so it predicted that it would take Jessica 96 0 = 6 5 hours 6 60 = 7 minutes to complete 5 the trip The ETA displayed by her car at 7:16 am was 7 minutes later : am (d) As in part (b), the car predicted that it would take Jessica 90 minutes 15 hours to travel from Botown to Aville Let the distance that Jessica travelled at 100 km/h be d km, and so the distance that Jessica travelled at 50 km/h was (10 d) km d The time that Jessica drove at 100 km/h was 100 hours The time that Jessica drove at 50 km/h was 10 d hours 50 Since the time predicted by her car is equal to the actual time that it took Jessica to travel from Botown to Aville, then d d = Solving f d, we get d + (10 d) = d + 0 = 150, and so d = 90 km Therefe, Jessica drove a distance of 90 km at a speed of 100 km/h

3 01 Hypatia Contest Solutions Page 3 3 (a) We are given that T 1 = 1, T = and T 3 = 3 Evaluating, we get T = 1 + T 1 T T 3 = 1 + (1)()(3) = 7, and T 5 = 1 + T 1 T T 3 T = 1 + (1)()(3)(7) = 3 (b) Solution 1 Each term after the second is equal to 1 me than the product of all previous terms in the sequence Thus, T n = 1 + T 1 T T 3 T n 1 F all integers n, we use the fact that T n = 1 + T 1 T T 3 T n 1 to get RS = T n T n + 1 = T n (T n 1) + 1 = T n (1 + T 1 T T 3 T n 1 1) + 1 = T n (T 1 T T 3 T n 1 ) + 1 = T 1 T T 3 T n 1 T n + 1 = T n+1 = LS Solution F all integers n, we use the fact that T n = 1 + T 1 T T 3 T n 1 to get LS = T n+1 = 1 + T 1 T T 3 T n 1 T n = 1 + (T 1 T T 3 T n 1 )T n = 1 + (T n 1)T n = T n T n + 1 = RS (c) Using the result from part (b), we get T n + T n+1 = T n + T n T n + 1 = T n + 1, f all integers n Similarly, T n T n+1 1 = T n (T n T n + 1) 1 = T 3 n T n + T n 1 = T n(t n 1) + T n 1 = (T n 1)(T n + 1) Since T n + T n+1 = T n + 1 and T n + 1 is a fact of T n T n+1 1, then T n + T n+1 is a fact of T n T n+1 1 f all integers n (d) Using the result from part (b), we get T 01 = T 017 T Since T 017 is a positive integer greater than 1, then T 017 T > T 017 T and T 017 T < T 017 That is, T 017 T < T 017 T < T 017, and so (T 017 1) < T 01 < T 017 Since T and T 017 are two consecutive positive integers, then (T 017 1) and T 017 are two consecutive perfect squares, and so T 01 lies between two consecutive perfect squares and thus is not a perfect square

4 01 Hypatia Contest Solutions Page (a)(i) By completing the square, the equations defining the two parabolas become y = x x + 17 = x x = (x ) + 1, and y = x + x + 7 = (x x + ) + 11 = (x ) + 11 Thus, the parabola defined by the equation y = x x + 17 has vertex V 1 (, 1), and the parabola defined by the equation y = x + x + 7 has vertex V (, 11) (a)(ii) First, we determine the codinates of the points of intersection P and Q When the two parabolas intersect, x x + 17 = x + x + 7 x 1x + 10 = 0 x 6x + 5 = 0 (x 5)(x 1) = 0, and so the two parabolas intersect at P (5, ) and Q(1, 10) Next, we want to show why quadrilateral V 1 P V Q is a parallelogram To do this, we will use the property that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram ( + The midpoint of diagonal V 1 V is, ) (3, 6), and the midpoint of diagonal ( P Q is, + 10 ) (3, 6) Since the midpoint of each diagonal is the same point, (3, 6), then the diagonals bisect each other and so quadrilateral V 1 P V Q is a parallelogram (Note that we could have also shown that each pair of opposite sides of V 1 P V Q is parallel) (b)(i) By completing the square, the equation defining the parabola y = x + bx + c becomes y = x + bx + c = (x bx) + c = (x bx + b b ) + c = (x bx + b ) + b + c = (x b ) + b + c ( ) b The vertex of this parabola is V 3, b + c and the vertex of the parabola defined by the equation y = x is V (0, 0) First, we determine the conditions on b and c so that the points of intersection R and S exist and are distinct from one another When the two parabolas intersect, x + bx + c = x x bx c = 0 This equation has two distinct real roots when its discriminant is greater than 0, when b ()( c) > 0 The points of intersection, R and S, exist and are distinct from one another when c > b Next, we determine conditions on b and c so that each of R and S are distinct from both

5 01 Hypatia Contest Solutions Page 5 V 3 and V The roots of the equation x bx c = 0 are given by the quadratic fmula, and so x = b ± b + c We let the x-codinate of R be x 1 x = b b + c = b + b + c and the x-codinate of S be Each of the points R and S is not distinct from V when b ± b + c = 0 b = b + c b = b + c, and so c = 0 Thus, we require that c 0 Similarly, each of the points R and S is not distinct from V 3 when b ± b + c = b b ± b + c = b ± b + c = b b + c = b, and so c = 0 As befe, we require that c 0 (Note that since R and V lie on the same parabola, then if their x-codinates are not equal, then they are distinct points that is, we need not consider their y-codinates The same is true f points S and V, R and V 3, and S and V 3 ) ( ) b Finally, we require that the vertices of the parabolas, V 3, b + c and V (0, 0), be distinct from one another Vertices V 3 and V are distinct provided that if their x-codinates are equal, then their y-codinates are not equal (V 3 and V lie on different parabolas and so we must consider both x- and y-codinates) If b b = 0 b = 0, then + c = 0 + c = c, and since we have the requirement (from earlier) that c 0, then vertices V 3 and V are certainly distinct when c 0 If the two conditions c > b and c 0 are satisfied, then f all pairs (b, c), the points R and S exist, and the points V 3, V, R, S are distinct (b)(ii) We begin by assuming that the conditions on b and c from part (b)(i) above are satisfied Thus, the points R and S exist, and the points V 3, V, R, S are distinct F quadrilateral V 3 RV S to be a rectangle, it is sufficient to require that it be a parallelogram that has at least one pair of adjacent sides that are perpendicular to each other From (b)(i), the parabolas intersect at R(x 1, x 1) and S(x, x ) (R and S each lie on the parabola y = x, and thus the y-codinates are x 1 and x, respectively) Recall that x 1 and x are the distinct real roots of the quadratic equation x bx c = 0 The sum of the roots of the general quadratic equation Ax + Bx + C = 0 is equal to B A, and so x 1 + x = b The product of the roots of the general quadratic equation Ax + Bx + C = 0 is equal to C A, and so x 1x = c First, we will show that quadrilateral V 3 RV S is a parallelogram since its diagonals bisect each other

6 01 Hypatia Contest Solutions Page 6 ( b The midpoint of diagonal V 3 V is + 0 b, + c + 0 ) ( b, b + c ) ( ) b, b + c ( ) x1 + x The midpoint of diagonal RS is, x 1 + x However, x 1 + x = b ( ) ( ) b c and x 1 + x = (x 1 + x ) x 1 x =, and so the ( b ( b ) ) ( midpoint of RS is, + c b, b + c ) ( ) b, b + c Since the midpoint of diagonal V 3 V is equal to the midpoint of diagonal RS, then the diagonals bisect each other, and so V 3 RV S is a parallelogram Next, we require that any one pair of adjacent sides of quadrilateral V 3 RV S be perpendicular to each other (This will mean that all pairs of adjacent sides are perpendicular) The slope of V S is x 0 x 0 = x since x 0 (S(x, x ) and V (0, 0) are distinct points) Similarly, the slope of V R is x 1 0 x 1 0 = x 1 since x 1 0 (R(x 1, x 1) and V (0, 0) are distinct points) Sides V S and V R are perpendicular to each other if the product of their slopes, x 1 x, is equal to 1 Since x 1 x = c c, then = 1, and so c = In addition to the condition that c =, the two conditions from part (b)(i), c > b and c 0, must also be satisfied Clearly if c =, then c 0 Further, when c =, c > b becomes > b b > 16 which is true f all real values of b The points R and S exist, the points V 3, V, R, S are distinct, and quadrilateral V 3 RV S is a rectangle f all pairs (b, c) where c = and b is any real number