10! = ?

AwesomeMath Team Contest 013 Solutions Problem 1. Define the value of a letter as its position in the alphabet. For example, C is the third letter, so its value is 3. The value of a word is the sum of the value of the letters in it. Find the value of MATH. 4 M is the 13th letter, A is the 1st letter, T is the 0th letter, H is the 8th letter. 13 + 1 + 0 + 8 = 4. Problem. 10 The sum of three consecutive integers is 15. Determine their product. Let the numbers be n 1, n and n + 1. Then their sum is 3n = 15, which implies that n = 5. The numbers are thus 4, 5, 6. Their product is 4 5 6 = 10. Problem 3. How many different positive integers divide 70 10! = 1 3 4 5 6 7 8 9 10? Note that 10! = 8 3 4 5 7. When constructing a divisor, we can choose to have 0 to 8 factors (9 choices) of. Similarly, we can choose to have 0 to 4 factors (5 choices) of 3. Continuing this argument, the number of choices for divisors for 10! is (8 + 1)(4 + 1)( + 1)(1 + 1) = 70. Problem 4. Four rectangular strips, each measuring 4 by 16 inches, are laid out with two vertical strips crossing two horizontal strips forming a single polygon which looks like a tic-tac-toe pattern. What is the perimeter of this polygon? 96 Each of the strips have a perimeter of (4 + 16) = 40, so the total perimeter of the four separate strips is 4 40 = 160. When placed in the configuration shown, for every crossing, a 4 4 square is eliminated from the perimeter. Since there are four 4 crossings, each of which have a perimeter of 16, the desired perimeter is 160 4 16 = 96. Problem 5. Evaluate the product ( 1 + ) ( 1 + ) ( 1 + ) ( 1 + ) 3 4 5 98 85 ( 1 + ) ( 1 + ) ( 1 + ) ( 1 + ) = 5 3 4 5 98 3 6 4 7 5 100 98 All the factors cancel except for the 99 and 100 in the numerator and the 3 and 4 in the denominator. The answer is thus 99 100 3 4 = 85. Problem 6. Let p 1 =, p = 3, p 3 = 5, be the sequence of prime numbers. Find the least positive even integer n such that p 1 + p + + p n is not a prime. 8 Let S n = p 1 + p + + p n. Calculating, we find S = 5, S 4 = 17, S 6 = 41, S 8 = 77, which is composite. Thus the answer is 8. 1

Problem 7. In ABC, the measure of ABC is 10. Point D is chosen in the triangle so that line DA bisects BAC and line DC bisects BCA. Find the degree measure of ADC. 150 From triangle ABC, we have x + y + 10 = 180. From triangle ADC, we have x + y + α = 180, implying x + y + α = 360. Subtracting the first equation, we have α 10 = 180, implying α = 150. Problem 8. polygon have? 13 A regular polygon has five times as many diagonals as it has sides. How many vertices does the If a regular polygon has n sides, it has n(n 3) diagonals. This can be seen by pairing each vertex with each other vertex not counting itself or its two neighbors. We are thus solving the equation n(n 3) = 5n, which implies n(n 3) = 10n, or n 3 = 10 implying n = 13. Problem 9. The faces of a cube contain the numbers 1,, 3, 4, 5, 6 such that the sum of the numbers on each pair of opposite faces is 7. For each of the cube s eight corners, we multiply the three numbers on the faces incident to that corner, and write down its value. (In the diagram, the value of the indicated corner is 1 3 = 6.) What is the sum of the eight values assigned to the cube s corners? 343 The left picture shows the corners and faces touching the side labeled 1, and the right picture shows the opposite side of the die, whose label is 6. Computing the eight numbers individually and summing them, we can find that the sum of numbers is 343. Alternatively, since no corner contains two numbers that sum to 7, the sum can be computed as (1+6)(+5)(3+4) = 7 3 = 343.

Problem 10. Given three distinct digits a, b, and c, it is possible, by choosing two distinct digits at a time, to form six two-digit numbers. Determine all possible sets {a, b, c} for which the sum of the six two-digit numbers is 484. {6, 7, 9}, {5, 8, 9} (order of elements does not matter: accept all permutations) The six possible numbers are 10a + b, 10a + c, 10b + a, 10b + c, 10c + a, 10c + b. The sum of all six numbers is (a + b + c). Since the digits are distinct, the possible sets are {6, 7, 9} and {5, 8, 9}. Problem 11. In a certain football league, the only way to score is to kick a field goal for 3 points or score a touchdown for 7 points. Thus the scores 1, 4 and 8 are not possible. How many positive scores are not possible? 6 Checking directly, we obtain that 3, 6, 7, 9, 10, 1, 13, 14 points are achievable. Since we have 3 consecutive scores, 1, 13, 14 which are achievable, it follows that every score > 14 is achievable. Note that if n is an achievable score, then n + 3 is achievable. Hence 1 + 3 = 15, 13 + 3 = 16, 14 + 3 = 17 are all achievable scores. Continuing this argument, all positive integer scores greater than 14 are achievable. Consequently the only scores that are not achievable are 1,, 4, 5, 8, 11. The answer is 6. Problem 1. c? 166 The two roots of the quadratic equation x 85x + c = 0 are prime numbers. What is the value of By Vieta s relations, if a and b are the roots of the quadratic, we have a + b = 85 and ab = c. Since a+b is odd, one root of equation must be even. One root must be, which implies that the other is 83. c = 83 = 166. Problem 13. Simplify the product ( 5 + 7 + 11)( 5 + 7 11)( 5 7 + 11)( 5 + 7 + 11) 139 Rewrite the given product as ( 5 + 7 + 11)( 5 + 7 11)( 11 + 5 7)( 11 ( 5 7)) = [( 5 + 7) 11][11 ( 5 7) ] This can be computed directly, coming out to = 11( 5 + 7) (5 7) 11 + 11( 5 7) 139 Problem 14. What is the sum of all the different solutions to the following equation? (x + 1)(x 4 + 1)(x 6 + 1) x + 1 + (x 1) = 0 0 Mltiplying by x + 1, we find that the roots of the above equation are the same as the roots of (x + 1)(x 4 + 1)(x 6 + 1) + (x 1) = 0 Where we have used the fact that 1 is not a root of the above equation. Note that if r is a root, then so is r. Consequently the sum of the roots if 0. 3

Problem 15. How many triangles are there in the diagram below? 56 To count the triangles, notice that all triangles are right triangles. There are 10 triangles with a right angle at the intersection of two diagonal lines and a hypotenuse along the left vertical side of the rectangle. By symmetry, there are just as many triangles with a hypotenuse along the top, right, or bottom side of the rectange. This accounts for 40 triangles. There are 4 triangles with a right angle at the lower right corner of the rectangle. Again by symmetry, there are four times this many triangles or 16. Thus the diagram contains 40 + 16 = 56 triangles. Problem 16. How many sequences a 1, a, a 3, a 4, a 5 satisfying a 1 < a < a 3 < a 4 < a 5 can be formed if each a i must be chosen from the set {1,, 3, 4, 5, 6, 7, 8, 9}? 16 Any 5 numbers chosen from the set {1,,, 9} determine exactly one increasing sequence of a i. Thus to find the answer we can just count the number of ways to pick 5 numbers from the 9. This is just ( 9 5) = 16. Problem 17. If x, y, z are real numbers, find the minimal value of x + 5y + 10z xy 4yz 6zx + 3 3 Note that the given expression is equal to (x y) + (y z) + (3z x) + 3 The minimum of the sum of squares is 0, achieved when x = y = z = 0. Hence the minimum of the expression is 3. Problem 18. Find the minimum value of the following function: 1 f(x) = (x 5) + (x 7) (x 4) (x 8) + (x 3) + (x 9) The function f(x) is quadratic. Its leading coefficient is 1 + 1 1 1 + 1 + 1 = > 0, so the graph of f is a U -shaped or upward pointing parabola. The minimum is attained at its vertex. Since the function is symmetric about x = 6, the vertex must be at x = 6. We have f(6) = 1 + 1 4 4 + 9 + 9 = (1 + 9 4) = 1. Problem 19. 18 Let x 1 and x be the roots of the equation x + 3x + 1 = 0. Evaluate ( ) ( ) x1 x + x + 1 x 1 + 1 We have x 1 + 3x 1 + 1 = 0 and x + 3x + 1 = 0, which implies ( ) ( ) x1 x x 1 + = x + 1 x 1 + 1 x + x + 1 + x x 1 + x 1 + 1 = 3x 1 1 x + 3x 1 x 1 = 3(x 1 + x ) + x 1 + x x 1 x = 3(x 1 + x ) 6x 1 x + (x 1 + x ) x 1 x By Vieta s formulas, we find the value of this expression to be 18. 4

Problem 0. Find the smallest positive integer n such that 11 n 1 is divisible by 105. 6 The last digit of 11 n 1 is 0, so 5 always divides 11 n 1. Checking the remainders when the powers of 11 are divided by 3 and 7, we find that 3 divides 11 n 1 precisely when n is even, and 7 divides 11 n 1 when 3 n. The smallest value of n that satisfies these conditions is n = 6. Problem 1. In how many ways can we arrange 7 (identical) white balls and 5 (identical) black balls in a row so that there is at least one white ball between any two black balls? 56 Denote the white balls as W and the black ones as B. We must have BWBWBWBWB, leaving us with 3 spare W balls which can be placed anywhere. There are 6 different places to place each white ball (at either end or next to one of the existing W balls). If we put each W in a different place, then there are 6 5 4/3! = 0 possibilities. If we put two W in one place and one in another, there are 6 5 = 30 possibilities. If we put all of the W in the same place, there are 6 possibilities. Adding them, we have 56 possibilities total. Alternatively, place all the 7 white balls first. Then there are 8 places to put the 5 black ones, which each must be in a different place. The number of ways to do this is 8 7 6 5 4 5! = 56. Problem. The diagram shows a rectangle ABCD with AB = 16 and BC = 1. Angle ACE is a right angle and CE = 15. The line segments AE and CD meet at F. What is the area of triangle ACF? 75 Using the Pythagorean theorem in ABC and then in ACE we find AC = 0 and AE = 5. It follows that ABC ACE, as the sides are in the same ratio. Therefore, we have BAC = CAE. Of course, by alternate interior angles, BAC = ACF. It follows that CAF = ACF, implying that ACF is isosceles. Let M be the midpoint of AC and join M to F. This creates two more right angled triangles AMF and CMF also similar to ABC. We have MF MA = BC 15 CA which implies MF =. The area of ACF is thus 1 15 0 = 75. Problem 3. The integer m has ninety-nine digits, all of them nines. What is the sum of the digits of m? 891. We have that m = 10 99 1. m = (10 99 1) = 10 198 10 99 + 1 = 999...9998000...001 where there are 98 nines and 98 zeros. The sum of the digits is 98 9 + 8 + 1 = 891. Problem 4. Let ABCD be a square with A = (0, 0), B = (5, 0), C = (5, 5), D = (0, 5). Let P be a point on the line y = x. Find the value of x such that the sum of the squares of the distances from P to the vertices of the square is a minimum. 3 Let P = (x, x). We have P A + P B + P C + P D = x + 4x + (x 5) + 4x + (x 5) + (x 5) + x + (x 5) = 0 = 0x 60x + 100 ( ( x 3 ) ) + 11 4 The quadratic function is minimized at the vertex, i.e. when x = 3. 5

Problem 5. A circle has two parallel chords of length x that are x units apart. If the part of the circle included between the chords has area + π, find x. Let C be the area of the circle, S be the area of the square two of whose edges are chords, and A be the area of the part of the circle included between the chords. The radius of the circle is x, implying that C = π x and S = x. The area A is the area of the square plus one half of the difference between the areas of the circle and square. It follows that x = A 1+ π =. A = C S + S = C + S = 1 + π/ x Problem 6. How many subsets {a, b, c} of {1,, 3,..., 15} satisfy b a 3 and c b 3? 165 Let x = a, y = b, and z = c 4. Then x, y, and z are elements that satisfy 1 x < y < z 11, so the number of possible triples (x, y, z) is ( ) 11 3 = 165. Each triple (x, y, z) leads back to a possible triple (a, b, c), so the number of possible triples (a, b, c) is also 165. Problem 7. Find all ordered triples (x, y, z) of positive reals such that x+y+z = 7 and x +y +z xy yz zx = 0. (x, y, z) = (9, 9, 9) We have x + y + z xy yz zx = (x y) +(y z) +(z x) = 0, which implies x = y = z = 7 3 = 9. Problem 8. Three lines are drawn parallel to each of the sides of triangle ABC so that they intersect in the interior of ABC. The resulting three smaller traingles have areas 1, 4, and 9. Find the area of triangle ABC. 36 Name the three intersection points along the horizontal in the center of the triangle D, E, F, and name the two intersections along the base of the triangle G, H as shown. Since the areas of the shaded triangles are in the ratio 1 to 4 to 9, the sides of these similar triangles are in the ratio 1 to to 3. Thus DE + EF = GH. Since EF is parallel to DB, BG = DE. Similarly, HC = EF. Therefore BC = DF and ADF has half the altitude of ABC. Thus ADF = EGH, and [ABC] = 4 [EGH] = 36. 6

Problem 9. In a certain country, there are 100 senators, each of whom has 4 aides. These senators and aides serve on various committees. A committee may consist either of 5 senators, of 4 senators and 4 aides, or of senators and 1 aides. Every senator serves on 5 committees, and every aide serves on 3 committees. How many committees are there altogether? 160 If each senator gets a point for every committee on which she serves, and every aide gets 1 4 point for every committee on which he serves, then the 100 senators get 500 points altogether, and the 400 aides get 300 points altogether, for a total of 800 points. But each committee contributes 5 points, so there must be 800 5 = 160 committees. Problem 30. Let ACE be a triangle with a point B on segment AC and a point D on segment CE such that BD is parallel to AE. A point Y is chosen on segment AE and segment CY is drawn. Let X be the intersection of CY and BD. If CX = 5, XY = 3, what is the ratio of the area of trapezoid ABDE to the area of triangle BCD? 39 5 Draw the altitude from C to AE intersecting line BD at K and line AE at L. Then CK is the altitude of triangle BCD, so CKX CLY. Since CY/CX = 8/5, CL/CK = 8/5. Also CKB CLA, so that CA/CB = 8/5 and BCD ACE, so that AE/BD = 8/5. The area of ACE is (1/)(AE)(CL), and the area of BCD is (1/)(BD)(CK) so the ratio of the area of ACE to the area of BCD is 64/5. It follows that the ratio of the area of ABDE to the area of BCD is 39/5. Problem 31. The function f(x), defined for 0 x 1, has the following properties: (i) f(0) = 0. (ii) If 0 x < y 1, then f(x) f(y). (iii) f(1 x) = 1 f(x) for all 0 x 1. (iv) f( x 3 ) = f(x) for all 0 x 1. Express f( 7 ) in lowest terms. 3 8 From (iii), f(1) = 1. Then from (iv), f(1/3) = f(1)/ = 1/, and from (iii), f(/3) = 1 f(1/3) = 1/. Then from (ii), f(x) = 1/ for all 1/3 x /3. In particular, f(3/7) = 1/. Then from (iv), f(1/7) = f(3/7)/ = 1/4. Then from (iii), f(6/7) = 1 f(1/7) = 3/4. Finally, from (iv), f(/7) = f(6/7)/ = 3/8. 7

Problem 3. 144 Then Let Then Problem 33. Let g(x) = 3x 3 x + 7x 9. Find g(x + 7) g(x + 6) g(x + 5) + g(x + 4) g(x + 3) + g(x + ) + g(x + 1) g(x). Let g 1 (x) = g(x + 1) g(x) = [3(x + 1) 3 (x + 1) + 7(x + 1) 9] (3x 3 x + 7x 9) = 9x + 5x + 8. g(x + 7) g(x + 6) g(x + 5) + g(x + 4) g(x + 3) + g(x + ) + g(x + 1) g(x) = g 1 (x + 6) g 1 (x + 4) g 1 (x + ) + g 1 (x). g (x) = g 1 (x + ) g 1 (x) = [9(x + ) + 5(x + ) + 8] (9x + 5x + 8) = 36x + 46. g 1 (x + 6) g 1 (x + 4) g 1 (x + ) + g 1 (x) = g (x + 4) g (x) = 36(x + 4) + 46 36x 46 = 144. Solve in complex numbers the system x(yz 1) = 3 y(zx 1) = 4 z(xy 1) = 5 (x, y, z) = (3,, 1), (i, 1 + i, + i), ( i, 1 i, i) Subtracting the first equation from the second, we obtain x y = 1, while subtracting the second equation from the third yields y z = 1. We have y = z + 1 which implies from the first equation x = z +. The third equation becomes z(z + 3z + 1) = 5 Moving all the terms to the left, we find that z 3 + 3z + z 5 = 0. Noting this cubic equation has 1 as a root, this is equivalent to (z 1)(z + 4z + 5) = 0 whose solutions are z 1 = 1 and z,3 = ± 1. The solutions to the system are thus (x, y, z) = (3,, 1), (i, 1 + i, + i), ( i, 1 i, i) Problem 34. When expressed in decimal, the number 6684 contains 013 digits. How many digits does the number 5 6684 contain? 467 In general, the number of digits in n is log 10 n + 1. Hence, log 10 6684 + 1 = 013. Since log 10 6684 is not an integer, it must be strictly between 01 and 013. But log 10 6684 + log 10 5 6684 = log 10 10 6684 = 6684, so log 10 5 6684 must be strictly between 6684 013 = 4671 and 6684 01 = 467. Then the number of digits in 5 6684 is log 10 5 6684 + 1 = 4671 + 1 = 467. 8

Problem 35. A perfect power is a number of the form m n, where m and n are positive integers greater than 1. How many positive integers less than or equal to 1 can be expressed as a perfect power? 81 For n 1, it is not difficult to list the perfect n th powers that are less than or equal to 1. n n th perfect powers less than or equal to 1 #, 3,..., 64 63 3 3, 3 3,..., 16 3 15 4 4, 3 4,..., 8 4 7 5 5, 3 5, 4 5 3 6 6, 3 6, 4 6 3 7 7, 3 7 8 8 1 9 9 1 10 10 1 11 11 1 1 1 1 This gives us a count of 63 + 15 + 7 + 3 + 3 + + 1 + 1 + 1 + 1 + 1 = 98. However, in the table, the number 1 appears five times, the number 10 appears twice, the number 9 appears twice, the number 8 appears three times, the numbers 6 and 3 6 appear three times, and the numbers 4, 3 4, 5 4, 6 4, and 7 4 appear twice, for a total of 98 4 1 1 1 1 1 1 1 = 81 distinct perfect powers. Problem 36. Find the sum of the coefficients of x 17 and x 18 in the expansion of (1 + x 5 + x 7 ) 0. 340 To get a term of x k in the expansion of (1 + x 5 + x 7 ) 0, we can choose a, b, and c terms of the form 1, x 5, and x 7, respectively, among the 0 factors of 1 + x 5 + x 7. Then 5b + 7c = k and a + b + c = 0. When k = 17, the only solution is a = 17, b =, and c = 1. Hence, the coefficient of x 17 is 0! 17!!1! = 340. When k = 18, there is no solution in a, b, and c. Hence, the sum of the coefficients is 340. Problem 37. Simplify (17 + 5) 3/ (17 5) 3/ 17 First, we find 17 + 5. Let a + b = 17 + 5. Then a + ab + b = 17 + 5. We set a + b = 17 and ab = 5. Then ab = 5, so a b = 5. Hence, by Vieta s formulas, a and b are roots of the quadratic x 17x + 5 = (x 4)(x 13) = 0. Thus, we can take a = and b = 13, so 17 + 5 = + 13. Similarly, 17 5 = 13. Then (17 + 5) 3/ (17 5) 3/ = ( + 13) 3 ( 13 ) 3 = (8 + 3 13 + 3 13 + 13 3 ) ( 13 3 3 13 + 3 13 8) = 17. Problem 38. Suppose x 3 ax + bx 48 is a polynomial with three positive roots p, q, and r such that p < q < r. What is the minimum possible value of 1/p + /q + 3/r? 3 By Vieta s relations, we know pqr = 48. By AM-GM, we have 1 p + q + 3 6 r 3 3 pqr = 3 by AM-GM, with equality when 1/p = /q = 3/r. Indeed, equality is achievable when p =, q = 4, r = 6, which implies that 3/ is the minimum possible value. 9

Problem 39. You have infinitely many boxes, and you randomly put 3 balls into them. The boxes are labeled 1,,. Each ball has probability 1/ n of being put into box n. The balls are placed independently of each other. What is the probability that some box will contain at least balls? 5/7 The desired probability is the sum of the probabilities that boxes 1,, respectively contain at least balls. For box n, the probability of having at least balls is 3[(1/ n ) (1 1/ n )] + (1/ n ) 3 = 3/ n / 3n = 3/4 n /8 n Summing to infinity using the geometric series formula, we find that the answer is 3/4 1 1/4 + /8 1 1/8 = 5/7. Problem 40. A great circle is a circle drawn on a sphere whose center is also the center of the sphere. Eight great circles are drawn on a sphere, so that no three great circles pass through the same point. Find the number of regions these eight great circles divide the surface of the sphere into. 58 Let a n be the number of regions that n great circles divide the surface of the sphere into. Suppose that we have n 1 great circles. When we draw the n th great circle, this circle intersects each of the previous great circles at two points, so the n th great circle can be divided into n arcs, where each arc runs between two consecutive intersection points. Furthermore, each of these n arcs divides a previous region into two regions, so a n = a n 1 + n. Using this recursion, we find a 1 =, a = 4, a 3 = 8, a 4 = 14, a 5 =, a 6 = 3, a 7 = 44, and a 8 = 58. In general, a n = n n +. Problem 41. 7 When the product 1 3 5 7 99 is written as a number, what is its tens digit? Let P n = (10n + 1)(10n + 3)(10n + 5)(10n + 7)(10n + 9). Then Hence, P n 1 3 5 7 9 + 10n 1 3 5 7 + 10n 1 3 5 9 + 10n 1 3 7 9 + 10n 1 5 7 9 + 10n 3 5 7 9 45 + 90n (mod 100). 1 3 5 99 = P 0 P 1 P 9 45 35 5 15 5 95 85 75 65 55 (5 95) (15 85) (5 75) (35 65) (45 55) 75 75 75 75 75 75 (mod 100). Problem 4. Solve the system of equations { (x ) + y + (x + ) + y = 6 9x + 5y = 45 (x, y) = ( 3 3 3 3 3 3 3 3 70, 70), ( 70, 70), ( 70, 70), ( 70, 70) 14 14 14 14 14 14 14 14 Also accept the same solution set with 3 14 70 replaced with 3 5 14. From the second equation, we consider the ellipse x 5 + y 9 = 1. By the definition of the ellipse, the sum of distances from any point on the ellipse to each of its foci are constant. We calculate its foci as F 1 (0, ) and F (0, ), and for any point P on the ellipse, we have P F 1 + P F = 6. 10

The first equation states that the sum of distances from a point P on the ellipse to the points F 3 (, 0) and F 4 (, 0) is also 6. By symmetry it implies that the point must also be on the ellipse x 9 + y 5 = 1. Solving the two ellipse equations, the system is now easily solved. To simplify calculations, we can use the fact that x = y or x = y by geometrical symmetry (since F 1 F 3 F F 4 is a square). Calculating, we find that (x, y) = ( 3 3 3 3 3 3 3 3 70, 70), ( 70, 70), ( 70, 70), ( 70, 70) 14 14 14 14 14 14 14 14 Problem 43. Let x = 3 + 3 4. There exist unique integers a, b, and c such that x 5 = ax + bx + c. Find a + b + c. 78 Hence, Let α = 3 and β = 3 4. Then x = α + β, and x 3 = (α + β) 3 = α 3 + 3α β + 3αβ + β 3 = + 3αβ(α + β) + 4 = 6 + 6x. x 5 = x 3 x = (6x + 6)x = 6x 3 + 6x = 6(6x + 6) + 6x = 6x + 36x + 36. Thus, a = 6, b = 36, and c = 36, so a + b + c = 6 + 36 + 36 = 78 Problem 44. Let ABC be a triangle and let D and E be points on BC such that DAB = EAC. If BD = 1, DE = 3, and EC = 4, then what is AC/AB? 7 By the sine law on triangles ABD and ACD, AB/ sin ADB = BD/ sin BAD and AC/ sin ADC = CD/ sin CAD. From these equations, AB AC = BD sin ADB/ sin BAD CD sin ADC/ sin CAD. Since ADB + ADC = 180, sin ADB = sin ADC. Hence, AB AC = BD sin CAD CD sin BAD. Likewise, by the sine law on triangles ABE and ACE, AB/ sin AEB = BE/ sin BAE and AC/ sin AEC = CE/ sin CAE. From these equations, AB AC = BE sin AEB/ sin BAE CE sin AEC/ sin CAE. Since AEB + AEC = 180, sin AEB = sin AEC. Hence, AB AC = BE sin CAE CE sin BAE. But BAD = CAE, which means BAE = CAD. Therefore, Then 7AB = AC, so AC/AB = 7. AB AC AB AC = BD sin CAD CD sin BAD BE sin CAE CE sin BAE = BD CD BE CE = 1 7 4 4 = 1 7. 11

Problem 45. Let ABC be a triangle with AB =, CA = 3, BC = 4. Let D be the point diametrically opposite A on the circumcircle of ABC, and let E lie on line AD such that D is the midpoint of AE. Line l passes through E perpendicular to AE, and F and G are the intersections of the extensions of AB and AC with l. Compute F G. 104 45 Using Heron s formula we arrive at [ABC] = 3 15 ( ) 4. Employing the relation [ABC] = abc 4R, where R is the circumradius of ABC, we have R 3 = [ABC] = 64 15. Since ABD = 90, BDEF is cyclic. Hence AB AF = AD AE = R 4R = 51 51 15. Similarly, AC AG = 15. Hence BCGF is cyclic, implying ABC AGF. We then have F G = BC AF AC = 4 51 15 3 = 104 45. Problem 46. Points A, C, and B lie on a line in that order such that AC = 4 and BC =. Circles ω 1, ω, and ω 3 have BC, AC, and AB as diameters. Circle Γ is externally tangent to ω 1 and ω at D and E respectively, and is internally tangent to ω 3. Compute the circumradius of triangle CDE. 3 Let the center of ω i be O i for i = 1,, 3 and let O denote the center of Γ. Then O, D, and O 1 are collinear, as are O, E, and O. If we let r denote the radius of Γ, we have OO 1 = r + 1, OO = r +, OO 3 = 3 r, O 1 O 3 =, O 3 O = 1, O 1 O = 3. By Stewart s Theorem, we have This gives rise to the equation OO 1 O O 3 + OO O 3 O 1 = O O 3 O 3 O 1 O 1 O + OO 3 O 1 O (r + 1) 1 + (r + ) = 1 3 + (3 r) 3 Solving, we find r = 6 7. Noting that the circumcircle of CDE is the incircle of OO 1O, we can easily compute the sides of OO 1 O to be 13 7, 0 7, 3. By Heron s, the area of OO 1O is 18 7 7, but the semipereter s = 7, so the desired radius is 3. Problem 47. 3 If f(x) = x xxx, find the last two digits of f(17) + f(18) + f(19) + f(0). Our goal is to find f(17) + f(18) + f(19) + f(0) modulo 100. By the Chinese Remainder Theorem, it suffices to find the sum modulo 4 and 5. The sum 0 (mod 4) because 18 and 0 are raised to high powers and 19 and 17 are both raised to odd powers. Computing, we have 18 4 1 (mod 5) = f(18) 1 (mod 5). Also, clearly f(0) 0 (mod 5). To find f(17) modulo 5, we compute X 17 1717 mod φ(5) = 0. X 1 (mod 4) since 17 1 (mod 4). X (mod 5), so X 17 (mod 0). Since 3 17 1 (mod 5), we have 17 17 3 3 (mod 5). Thus f(17) (mod 5) To compute f(19) mod 5, we need to compute Y 19 1919 mod 0. We have 19 1 (mod 0), which implies Y 1 (mod 0). Therefore f(19) = 19 1 4 (mod 5). Thus f(17) + f(18) + f(19) + f(0) + 1 + 4 + 0 7 (mod 5). We have that this sum is 0 (mod 4). Hence f(17) + f(18) + f(19) + f(0) 3 (mod 100). ( Problem 48. If p(x) is a polynomial with integer coefficients, let q(x) = p(x) x(1 x). If q(x) = q and p() = 7, p(3) = 11, find p(10). 51 1 1 x ) for every x 0 Insert the equation for q(x) into itself, and obtain q((x 1)/x) = q(x). Write q(x) and q((x 1)/x) in terms of x and p(x). Multiplying and clearing denominators, we have p(x) = x 3 p((x 1/x)) = x 3 p(1 1/x). Since both sides of the equation must be polynomials, it follows that p(x) is a cubic. Let p(x) = ax 3 + bx + cx + d. Substituting, we can obtain two equations: b + c = 3a = 3d. The values of p() and p(3) give two more equations: 8a + 4b + c + d = 7 and 7a + 9b + 3c + d = 11. Solving the system of four equations, we obtain a = 1, b = 5, c =, d = 1. It follows that p(10) = 1000 500 + 0 + 1 = 51. 1

Problem 49. A set of six edges of a regular octahedron is called a Hamiltonian cycle if the edges in some order constitute a continuous loop that visits each vertex exactly once. How many ways are there to partition the twelve edges into two Hamiltonian cycles? 6 Call the octahedron ABCDEF where A, B, and C are opposite D, E, and F, respectively. Note that each Hamiltonian cycle can be described in terms of the order it visits vertices in exactly 1 different ways. Conversely, listing the six vertices in some order determines a Hamiltonian cycle precisely when no pair of opposite vertices are listed consecutively or first-and-last. Suppose we begin with AB. If D is listed third, then the final three letters are CEF or F EC. Otherwise C or F is listed next, and each gives three possibilities for the final three. (For example, ABC is to be followed by DEF, DF E, or EDF.) Thus, there are 6 4 ( + 3 + 3) = 19 listings. These correspond to 19/1 = 16 Hamiltonian cycles. Finally, the complement of all but four Hamiltonian cycles is a Hamiltonian cycle. Each vertex has degree four, so it is an endpoint of two edges in the complement of a Hamiltonian cycle, so it is also a Hamiltonian cycle unless it describes two opposite faces. It follows that there are six pairs of disjoint Hamiltonian cycles. Problem 50. and Find the number of quadruples (a, b, c, d) of non-zero integers satisfying both 1 a + 1 b + 1 c + 1 d = 1 (a + b + c + d) = ab + cd + (a + b)(c + d) + 1 48 Multiply the first equation by abcd and then rewrite it as Rewrite the second equation to get abcd (bcd + acd + abd + abc) = 0 Now note that (a + b + c + d) (ab + cd + ac + ad + bc + bd) = 1 (a )(b )(c )(d ) = abcd (bcd + acd + abd + abc) +4(ab + cd + ac + ad + bc + cd) 8(a + b + c + d) + 16 = [abcd (bcd + acd + abd + abc)] 4[(a + b + c + d) (ab + cd + ac + ad + bc + bd)] + 16 = [0] 4[1] + 16 = 1 There are four ways to factor 1 as the product of four positive integers: 1 1 1 1, 6 1 1, 4 3 1 1, and 3 1. If one takes one of these products and assigns the first factor to be a, the second factor to be b, the third to be c, and the fourth to be d, the assignment does not result in an assignment of values for a, b, c, d which satisfy the original two equations. On the other hand, if two or four of the four factors in some of the products are made negative, then some solutions can be found. There are three families of such solutions resulting from the products ( 3)()()( 1), ( 4)( 3)( 1)( 1), and (4)( 3)( 1)(1). There are 4! + 4!! + 4!! = 48 ways to permute these solutions, which gives rise to 48 quadruples. 13