2004 AMC 12B Solutions

Solutions 2004 55 th AMC 12 B 413. (A) Since f(f −1 (x)) = x, it follows that a(bx + a) + b = x, so ab = 1 anda 2 + b = 0. Hence a = b = −1, so a + b = −2.14. (D) Because △ABC, △NBK, and △AMJ are similar right triangles whosehypotenuses are in the ratio 13 : 8 : 1, their areas are in the ratio 169 : 64 : 1.The area of △ABC is 1 2(12)(5) = 30, so the areas of △NBK and△AMJ are641169(30) and169(30), respectively.Thus the area of pentagon CMJKN is (1 − 64169 − 1169)(30) = 240/13.15. (B) Let Jack’s age be 10x + y and Bill’s age be 10y + x. In five years Jack willbe twice as old as Bill. Therefore10x + y + 5 = 2(10y + x + 5),so 8x = 19y + 5. The expression 19y + 5 = 16y + 8 + 3(y − 1) is a multiple of 8if and only if y − 1 is a multiple of 8. Since both x and y are 9 or less, the onlysolution is y = 1 and x = 3. Thus Jack is 31 and Bill is 13, so the differencebetween their ages is 18.16. (C) From the definition of f,f(x + iy) = i(x − iy) = y + ixfor all real numbers x and y, so the numbers that satisfy f(z) = z are thenumbers of the form x + ix. The set of all such numbers is a line through theorigin in the complex plane. The set of all numbers that satisfy |z| = 5 is acircle centered at the origin of the complex plane. The numbers satisfying bothequations correspond to the points of intersection of the line and circle, of whichthere are two.17. (A) Let r 1 , r 2 , and r 3 be the roots. Thenso r 1 r 2 r 3 = 2 5 = 32. Since5 = log 2 r 1 + log 2 r 2 + log 2 r 3 = log 2 r 1 r 2 r 3 ,8x 3 + 4ax 2 + 2bx + a = 8(x − r 1 )(x − r 2 )(x − r 3 ),it follows that a = −8r 1 r 2 r 3 = −256.