2003 Solutions Euclid Contest - CEMC - University of Waterloo

2003 Euclid Solutions 21. (a) Solution 1Since the x-intercepts of the parabola are 2 and 4, then the axis of symmetry of theparabola is x = 3.Since the point 08 ,x = 3 is the point ( 68 , ). Thus, a = 6 .( ) is on the parabola, its image after a reflection across the vertical lineSolution 2Since the x-intercepts of the parabola are 2 and 4, then the equation of the parabola is ofthe form y = A( x−2) ( x−4 ).Since ( 08 , ) lies on the parabola, then 8= A ( −2) ( −4)or A = 1.2Therefore, the parabola has equation y = ( x−2) ( x−4)= x − 6x+8.But the point ( a,8) lies on the parabola, so28= a − 6a+820 = a −6a0 = aa ( −6)Since a ≠ 0, then a = 6 . Answer: a = 6(b) Solution 1Since the quadratic equation has two equal roots, then the expression on the left must be aperfect square. Since the leading coefficient is 1 and the coefficient of the x-term is 6,2 2then the expression must be x+3 x 6x9. By comparing expressions, k = 9.( ) = + +Solution 2Since the quadratic equation has two equal roots, the discriminant is 0, ie.6 2 − 4()( 1 k )= 0 or 4k = 36 or k = 9. Answer: k = 92( ) lies on the parabola, so 4 = 1 − 31 ()+ c or(c) From the given information, the point 14 ,c = 6.We now find the points of intersection of the parabola and the line by equating:22x+ 2 = x − 3x+620 = x − 5x+40 = ( x−1) ( x−4)Thus the points of intersection have x-coordinates x = 1 and x = 4.Substituting x = 4 into the line y = 2x+2, we get the point ( 410 , ).Therefore, the second point of intersection is ( 410 , ).