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Solutions to 47(2) HEXA Challenge Problems

Challenge 1: Since the point is chosen randomly, the probabilities of being from the three different regions are proportional to the areas of the three regions. Once we have computed the areas of the three regions, we divide , the total area of circle , to get the probabilities. To compute these areas, we first need the length of the edge of the inscribed square . The line segments con necting two consecutive corners of the square and the center of the circle form a right isosceles triangle where the two equal sides are radii of length . The length of the third side is therefore . Since this is the length of one edge of the square, the area of the square is . This length is also a diameter of the small er circle , and so its radius is , giving an area of . Thus, the areas ranked by their probability are

Inside . Probability

Outside . Probability

Outside , inside . Probability

Challenge 2: The watch gains 1 second every 5 hours, or equivalently, it takes 300 hours to gain 1 minute. Since I reset the watch after increments of 4 minutes of extra time, 1200 hours or 50 days will pass before the watch gains 4 minutes. Fifty days after noon on January 1 is noon on February 20.

Challenge 3:

Since

, we may substitute

. Similarly, since

, we get

.

Thus, we need to solve the linear system

.

Subtracting the second equation from the first, we have y= 40, and then x= - 231

Challenge 4:

Denote the first and second numbers as x and y , respectively. Therefore x + y = 62. We also have y = 2 + 3x. Substituting this into the first equation, we have x + 2 + 3 x = 62 , or 4 x = 60, so that x =15 . Thus, y = 2+3(15) = 47

Virginia Mathematics Teacher vol. 48, no. 1

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