Answer National Level Solution ....... 2 1. The region in which the two squares overlap with the rectangle is a right triangle, which we shall call T. Because the squares each have area 1, the area of T is also 1. However, upon closer inspection we see that the base and height of the rectangle are the same as the (slanted) base and height of triangle T. Since area(T)=bh/2=1, we find bh=2, so the area of the rectangle is 2. Alternate solution: can the reader figure out how to dissect the two squares into pieces that can be reassembled into the rectangle, thereby showing that they must have the same area? . 60 2. Since 30 must divide 72N, there must be factors of 2, 3, and 5 in 72N. The 2 and 3 are already present in 72, so N must be divisible by 5. In the same manner, since 30N is a multiple of 72, N must be a multiple of 12. The smallest positive integer divisible by both 5 and 12 is 60. As a check, we note that 60 divides evenly into (30)(72), so N=60 satisfies all the requirements. . 70 3. In the time it takes the older sister to reach the top, the up escalator has carried her forward 30 steps, because the total length of the escalator is 40 steps and she took 10 steps herself. Therefore in this same time the down escalator pushes the younger sister back 30 steps. To make up this set-back and cover the original 40 steps separating her from the top the younger sister must make a total of 70 steps. . 1/12 4. If I choose a shot that I will make with probability p (where p is between 0 and 1/3), then Michael Jordan will make the same shot with probability 3p. Hence the probability that I make a shot that Jordan subsequently misses is p(1-3p). The graph of this function is a parabola which equals zero when p=0 and p=1/3. By symmetry the vertex (maximum) is midway between the two, at p=1/6. Hence the best chance I have of winning the game is 1/12. Alternate solution: try applying the arithmetic-geometric mean inequality to the numbers 3p and (1-3p) to find the maximal value of p(1-3p). . 29pi 5. Notice that a right triangle with legs whose squares are 17 and 99 has the same hypotenuse as a right triangle with legs whose squares are 19 and 97, namely the square root of 116 in both cases. Now place these two right triangles next to one another (not overlapping) so that their hypotenuses coincide. The circle with the common hypotenuse as diameter will neatly circumscribe the resulting quadrilateral since any angle inscribed in a semicircle must be a right angle, and vice versa. To summarize: we have created a quadrilateral with the appropriate side lengths inscribed in a circle with diameter equal to the square root of 116, and hence with area 29pi. . 12 6. It turns out to be advantageous to analyze this problem from a polar coordinate perspective. For starters, line l1 is angled 60 degrees from the horizontal, while line l2 is inclined 30 degrees, because the slopes of those lines create 30-60-90 triangles. Also, reflecting a point through either line does not change its distance from the origin. (Why?) Finally, if a point P has polar angle x, then the reflection of P through line l1 has angle 120-x. This formula stems from the fact that the average of the polar angles before and after reflecting comes to exactly 60 degrees, the inclination of line l1. Similarly, reflecting P through line l2 results in an angle of 60-x. Therefore we choose to keep track of points in set S according to their polar angle. Let x be the angle out to the point (6,5), which is a little less than 45 degrees. We claim that exactly twelve different angles arise by reflecting the point (6,5) over the two lines repeatedly. They are x, 60+x, 120+x, ..., 300+x, -x, 60-x, ..., and 300-x. Clearly these angles are all distinct. Furthermore, reflecting any of these points over either line yields a polar angle already listed in our set. For example reflecting the point with angle 300+x in line l2 produces a point with angle 60-(300+x) = -240-x = 120-x. Finally, all of these points can be reached through successive reflections, since reflecting in line l2 followed by line l1 produces a net change of 120-(60-x) = 60+x. In other words, if a point with angle x is in set S, then so is the point with angle 60 degrees greater. Both x and 60-x are in set S, so the remaining ten angles must be also, for a total of 12 points. . -1/2 7. One effective technique for predicting the next value of a polynomial function is the method of finite differences. For ease of presentation let us suppose that the first three values of f(x) are f(0)=a, f(1)=b, and f(2)=0. Writing these values out as the first row in a table gives (a, b, 0). Subtracting each number from the one immediately following it yields (b–a, -b) in the second row. Subtracting these differences produces the third row containing just a-2b. One main result in the theory of finite differences states that for a quadratic function, this third row will contain all the same numbers. Therefore the next entry in the third row is also a-2b, so we may work backwards to find that the next entries in the second and first rows are both a-3b. In other words, f(3)=a–3b. In our case a=cos340 and b=(cos40)(sin240), so f(3)=cos340–3(cos40)(sin240). This is in fact the triple angle formula for cosine (check!), so f(3)=cos 120= -1/2.