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However, when a =2, these merge with the vertex of the parabola intersecting the top of the circle at (0,2) for the only point of intersection. When a >2, the parabola lies entirely above the circle so that they do not inter sect.

The number of intersections points as a function of a is summarized by the table

Challenge 3: Can a number x be added to the set {9, 11, 14, 20, 31, 40} such that both the mean and the median of the new set are equal to x ? If so, find the value. If not, why not?

SOLUTION: No, there is no such number. In order for x to be both a value in the new set and its mean, we must have (9+11+14+20+31+40+ x )/7= x , or 125+ x =7 x . Thus, x =125/6. However, since this number is larger than 20, the median of the new set is 20.

Challenge 4: The dimensions of a rectangle are A and B . Find the dimensions of a rectangle that has exactly half the area and exactly half the perimeter of the first rectangle.

SOLUTION: Let a and b be the dimensions of the smaller rectangle. For the area and perimeter of this rec tangle to be half the values of the larger values, a and b must solve the nonlinear system ab =(1/2) AB 2( a+b )=(1/2)(2( A+B ))= A+B Solving the first equation for b=AB/a and substituting it into the second yields 2a+AB/a=A+B 2 a 2 −( A+B ) a+AB =0 This equation has two solutions from the quadratic formula

Given that b=AB/ 2 a , it can be shown that

so that the two solutions from the quadratic equation represent only one distinct pair of dimensions.

Virginia Mathematics Teacher vol. 47, no. 1

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