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Challenge 5: For two distinct positive real numbers a and b , denote the two solutions of the equation (ln a x )(ln b x )=1 as x 1 and x 2 . What condition must a and b satisfy in order for |ln x 1 | = |ln x 2 |?

SOLUTION: Using the product rule for logarithms, we may expand the logs in the equation to get

(ln a + ln x )(ln b + ln x)=1

(ln x ) 2 +(ln a + ln b )ln x + ln a ln b − 1=0

Thus, the equation is quadratic in ln x . A quadratic equation has two solutions, and in order for them to be equal in absolute value, the coefficient of the first - order term must be zero. In this case, this requires ln a + ln b= 0, which implies that ln a = − ln b , or a = 1/ b .

The two solutions are then

Challenge 6: In Bill ’ s trigonometry class one morning, he finally understood polar coordinates when he correctly plotted the graph of the cardioid r =2(1− sin θ). Later that afternoon in his statistics class, he was fascinated by how his teacher approximated π. She selected N points randomly from the unit box by select ing x i and y i uniformly from the interval [0,1]. She let p N count the number of these points that were also in side the unit circle—that is, x i 2 + y i 2 ≤1. She says (and he observes) that 4p N /N is approximately π due to the ratio of the areas of the quarter circle and the unit box being π/ 4. Bill realizes that his cardioid is contained within the circle of radius 4 centered at the origin, and so he de cides to estimate its area by selecting r i and θ i uniformly from the intervals [0,4] and[0,2 π] and then letting p N count the number of points in polar coordinates that satisfy r i ≤2(1− sin θ i ). Using N =10000, his repeatedly finds that 16 π p N /N ≈25 and assumes this is a good approximation of the area of the cardioid. Unfortunately, the actual area is 6 π ≈18.85. Why is Bill so far off? SOLUTION: Bill correctly understands that if p N counts the number of random points taken from the circle that are also inside the cardioid, he can approximate the cardioid ’ s area by multiplying the area of the circle (16 π) by the ratio p N /N . However, his choice to pick random polar coordinate points in the circle by select ing r i from [0,4] means he does not sample the circle uniformly. Using his method, half of his points should be inside the circle of radius 2 and half would be in the annulus with inner radius 2 and outer radius 4. The area of this smaller circle is 4 π while the area of the annulus is 12 π. This means he oversamples points near the origin, skewing the area too large. A proper sample will account for this by biasing the sample to make the likelihood a point is selected increase the farther it is away from the origin. Specifically, if instead r i 2 is chosen uniformly from the interval [0,16] so that r i =(√ r i 2 ), the points will uniformly sample the circle and lead to a good approximation.

Virginia Mathematics Teacher vol. 47, no. 1

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