vmt-award-2024_47--2-_purple

25 / (169 + b 2 ) + (144 / b 2 ) = 1. Getting a common denominator and cross - multiplying yield 25 b 2 + 144 × 169 + 144 b 2 = 169 b 2 + b 4 . This can be solved to get Therefore, the ellipse is ( x 2 / 325) + ( y 2 / 156) = 1. The hyperbola is similar except that its standard form is ( x 2 / a 2 ) – ( y 2 / b 2 ) = 1, and the foci (± c , 0) have This hyperbola also has b 2 = 156 and has the equation ( x 2 / 13) – ( y 2 / 156) = 1.

For the point of intersection ( x , y ) = (12, 5), the solution method is the same and yields the ellipse

( x 2 / 234) + ( y 2 / 65) = 1 and the hyperbola ( x 2 / 104) - ( y 2 / 65) = 1.

Challenge 5: There are five different “ patterns ” that four - digit numbers may fit with respect to this matching based on the number of repeated digits. The more repeated digits a number has, the lower the probability of finding a match. A number that is one digit repeated four times, like 3333, matches only itself, but a number with a digit three repeated only three times, like 3334, matches four differ ent numbers: itself, 3343, 3433, and 4333. The following table summarizes the patterns, the number of matches for a number that fits that pattern, and the total number of numbers that fit that pattern.

Pattern

Example

Number of matches for a number

Number of numbers fitting this pattern

All the same

1111

1

10

Three the same

1112

4

Two pairs

1122

One pair

1123

All different

1234

The probability of the first number fitting each pattern is the last column divided by 10 4 . Given a first number in a known pattern, the probability of the second number matching the first is then the third column divided by 10 4 . Therefore, the total probability of getting a match is

Challenge 6: Since AD passes through the center of the circle, its length is 2 r , where r is the radius of the circle, and so AB has length r . If O is the center of the circle, then BO must also have length r , meaning that ABO is an equilateral triangle with edge length r and ∠ AOB = 60°. Since BC is parallel to AD , ∠ OBC must also be 60°. Thus, since CO is a radius, triangle BOC is also equilateral. This tells us that since ∠ AOB = ∠ BOC = 60°, ∠ COD must also be 60°, which means that COD is a third equilateral trian gle. Each of these triangles has base length r and height so that the area of the trapezoid is Since the area of the circle is 2020 π , r 2 = 2020, and the area of the trapezoid is

Virginia Mathematics Teacher vol. 47, no. 2

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