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Solutions to 47(1) HEXA Challenge Problems

Challenge 1: Denote the two numbers by x and y . The two ratios would then be written as

Cross multiplying, we get two linear equations: 6 x – 5 y = – 4, 2 x – y = 4. Solving the second for y gives y = 2 x – 4, which can be substituted into the first equation to get 6 x – 5(2 x – 4) = – 4, which has the solu tion x = 6. The other number would then be y = 6.

Challenge 2: Let the integers b , c , and g be the prices per can in cents of Broth Bros. Tomato Soup, Sou perman ’ s Clam Chowder, and Souperman ’ s Gumbo, respectively. The roommates know that c > b , g > b , and b ≥ 98, and that 3 b + 2 c = 612 and 3 b + g = 452. These combine to give us 612 = 3 b + 2 c > 5 b and 452 = 3 b + g > 4 b . Thus,

However, since 3 b is equal to 612 – 2 c , which must be an even number, we may conclude that b must also be even. Therefore, the maximum cost of a can of Broth Bros. is $1.12. To determine which can of Souperman ’ s is more expensive, we can write:

Since b ≥ 98, c – g ≥ (3/2) × 98 – 146 = 1, and so we may conclude that one can of Souperman ’ s Clam Chowder must cost at least 1 cent more than a can of Souperman ’ s Gumbo.

Challenge 3: The sum of the exterior angles of the n - gon is θ + 2θ + ... + n θ = ( n ( n+ 1) / 2) × θ , but this must equal 360°. Therefore, θ = (720 ° / ( n ( n + 1)). This takes an integer value when both n and n + 1 are divisors of 720. This happens for n = 4, 5, 8, 9, and 15, with values of θ = 36 °, 24°, 10°, 8°, and 3°.

Challenge 4: There are two ellipse - hyperbola pairs that satisfy these conditions. Let ( x , y ) define the ver tex of the rectangle in the first quadrant so that the other vertices are ( – x , y ), ( – x , – y ), and ( x , – y ). The area of the rectangle is 4 xy = 240 so that xy = 60, and the perimeter is 4 x + 4 y = 68, or x + y = 17. Thus, this vertex can be either ( x , y ) = (5, 12) or (12, 5). We will show the details for (5, 12). For an ellipse in standard form ( x 2 / a 2 ) + ( y 2 / b 2 ) = 1, if a > b , the foci are (± c , 0) where Here, c = 13, and so a 2 = 169 + b 2 . Substituting this and the point (5, 12) into the standard form,

Virginia Mathematics Teacher vol. 47, no. 2

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