Math Day at the Beach 2016 Solutions

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1 Math Day at the Beach 06 Solutions Benjamin Michelle Victor Michael Diao, Jack Sun and Jason Ye Individual Round Problem. (Problem ) What is isthe median of ofthe following five values: sin 89,,,,, sin 45, sin 65, cos 7,? (A) sin 89 (B) sin 45 (C) sin 65 (D) cos 7 (E) Solution. We can rewrite sin(89 )) as assin( 7 sin( 7 ), ), sin(45 )) as assin(5 ), ), sin(65 )) as assin(5 ), ), cos(7 )) as assin(8 ), ), and as as sin(0 ). sin(0 ). Since sin( ) sin(θ) is isincreasing on onthe interval ( 90 ( 90,, ), ), the five values arranged in inincreasing order is is sin( 7 sin( 7 ), ), sin(5 ), ), sin(8 ), ), sin(0 ), ), sin(5 ). ). The median is issin(8 ), ), giving answer choice D.. Solved by by7 7out outof of8 8individuals Problem. (Problem ) sin( fiπ? 6 + arcsin(x))? 6 x x (A) +x (B) Ô +x (C) Ô Ô x x + x (D) Ô x + Ô x (E) Ô Ô x x and Solution. Note that sin(arcsin(x)) x and cos(arcsin(x)) Ô x.. We We can cansee seethis thisby byletting θ arcsin(x), so sosin( ) sin(θ) x, x, and anddrawing a right triangle: Ô x x Then, Then, apply apply the the angle angle sum sum identity identity for for sin sin to to get get sin(fi/6 sin(π/6 + arcsin(x)) sin(fi/6) sin(π/6) cos(arcsin(x)) + cos(fi/6) cos(π/6) sin(arcsin(x)) Ô x Ô x + x

2 MDatB 06 Solutions giving the answer D. Solved by 7 out of 8 individuals Problem. (Problem ) Choose a string of 0 digits in such a way that each digit is chosen independently with each of {0,,..., 9} being equally likely. Let P be the probability that each digit appears in this string exactly once. (For instance, ) Let L log 0 P. Which of the following is true? (A) L 5 (B) 5 < L 4 (C) 4 < L (D) < L (E) < L 0 Solution. We begin by computing P. In total, there are 0 0 strings of digits from 0 to 9, since there are 0 choices for the value of each of the 0 digits in the string. The number of such strings which use each digit exactly once is equal to the number of permutations of 0,,,..., 9, which is just 0!. Thus P 0! 0 0. Now we use the properties of logarithms to estimate L. Notice that L log 0 P log 0 0! 0 0 log 0 0! log log 0 0! 0. It remains to approximate log 0 0!. Since 0! 68800, we know that 6 < log 0 0! < 7, meaning that 4 < log 0 0! 0, giving answer choice C. Solved by 78 out of 8 individuals Problem.4 (Problem 4) f is a quadratic polynomial of the form f(x) ax + bx + c such that f() 0, f() 0, and f( ) 4. If N f(5), find the remainder left when N is divided by 5. (A) 0 (B) (C) (D) (E) 4 Solution. Since f() f() 0, and are the roots of f. This means that f(x) k(x )(x ) for some constant k. Now since we are given that f( ) 4, we have that f( ) k( )( ) k( )( 4) 8k.

3 MDatB 06 Solutions This means that 8k 4, and so k. It remains to compute f(5) modulo 5. f(5) (5 )(5 ) ( ) ( ) 9 4 (mod 5). Our answer is E. Solved by 40 out of 8 individuals Problem.5 (Problem 5) There are four cities linked by roads as in the picture. Each segment of road linking adjacent cities has, independently, a probability of being shut down. What is the probability that one can travel from city A to city D? B A D C (A) 8 (B) (C) 5 8 (D) 4 (E) 7 8 Solution. Suppose that there is no road between B and C, which has a probability of. Then, at least one of the paths {(A, B), (B, D)} or {(A, C), (C, D)} must not be closed. Roads (A, B) and (B, D) are both open with a probability of ( ( ) ). If one of {(A, B), (B, D)} is closed, 4 then both of {(A, C), (C, D)} must be open with a probability of. This case has a probability ( 4 of + ( ( )) 4 4) 4 7 of succeeding. Next, suppose that there is a road between B and C. Then, at least one of {(A, B), (A, C)} and at least one of {(B, D), (C, D)} must be open. Since the probability that given roads are both closed is, the probability that at least one is open is. Hence, this case has a probability of (( ) ( )) of succeeding. In total, the probability that one can travel from A to D is 7 + 9, giving answer B Solution. We proceed using complementary probability. configurations for which A and D are not connected. Our goal is to find the number of If all roads are shut down, there is no way to get from A to D, contributing configuration. Similarly, there is no way to get from A to D if only one road is open, giving 5 configurations. Any choice of only two open roads will fail unless A is connected to B and B is connected to D, or if A is connected to C and C is connected to D, giving ( ) 5 8 configurations. There are only failing configurations if three roads are open, which occur when all roads but those connecting to A are open, or when all roads but those connecting to D are open. If four roads are open, A and D must be connected.

4 MDatB 06 Solutions Thus, in total, there are failing configurations. The probability that one can travel from A to D is then 6, so answer choice B is correct. 5 Solved by 48 out of 8 individuals Problem.6 (Problem 6) If AB BC and all of the lines shown meet at D, which of the following must be an arithmetic progression? D α β γ A B C (A) sin α, sin β, sin γ (B) cos α, cos β, cos γ (C) tan α, tan β, tan γ (D) cot α, cot β, cot γ (E) α, β, γ Solution. Drop a perpendicular from D to line AC at point E. For there to be an arithmetic progression, we need the three numbers in the progression to involve CE, BE, and AE, each over a constant denominator. This only occurs with the proportions AE, BE, CE. This is equivalent to DE DE DE cot α, cot β, cot γ, or D. Solved by 74 out of 8 individuals Problem.7 (Problem 7) Calculate the area in the plane of the set of all points (x, y) such that x + y. (A) 4 (B) (C) (D) 4 (E) The area is infinite. Solution. When graphed, the solution area is: y x The rhombus has height and width. Therefore, its area is, giving answer choice B. 4

5 MDatB 06 Solutions Solved by 85 out of 8 individuals Problem.8 (Problem 8) At a certain meeting, there are twice as many women present as men. of the men are 4 wearing some item of blue clothing. If half of all the people present are wearing blue, what is the percentage of the women wearing blue? (A) 0% (B) 5% (C) 7.5% (D) 4.7% (E) 50% Solution. Let m be the number of men, and m be the number of women present. Then m are 4 men wearing blue. We have a total of m m people at the meeting wearing blue, so there have to be m women wearing blue. The proportion of women wearing blue is thus m 4 4 m, or 8 7.5% of the women: C. Solved by 67 out of 8 individuals Problem.9 (Problem 9) How many integers n are there with n 00 such that the smallest prime factor of n is greater or equal to 7? (A) (B) (C) (D) 4 (E) 5 Solution. There are 5 primes less than 00, of which are greater than or equal to 7. Now we compute the number of composites less than 00 which only have prime factors greater than or equal to 7. These are , 7 77, and 7 9, giving + 5 solutions. Our answer is E. Solution. We can also solve this problem using complementary counting by looking at how many integers in the given range have their smallest prime factor less than 7. This is simply all numbers divisible by,, or 5. Using PIE, there are numbers in the range that are divisible by,, or 5. Because there are 99 total integers in the given range, our answer is Solved by 79 out of 8 individuals 5

6 MDatB 06 Solutions Problem.0 (Problem 0) Assume a and b are real numbers, not both zero. Suppose that the function ax + by + 5 has its maximum value on the region enclosed by the regular octagon shown below at the point B and nowhere else. Find the vertex at which 5 bx ay has its maximum value. E F D C G B H A (A) A (B) B (C) C (D) D (E) E Solution. Let f(x, y) ax + by + 5. Then, f(x, y) aî + bĵ. Since f attains its largest value at B, we know a > 0 and b < 0, with a > b. Let g(x, y) 5 bx ay. So g(x, y) bî aĵ. Since b > 0, a < 0, and b < a, the gradient points closest to A. Therefore, 5 bx ay attains its maximum at point A in the octagon. The correct answer is choice A. Solved by out of 8 individuals Individual Round Problem. (Problem ) If 05 b a, for positive integers b and a, what is the smallest possible value for b? Solution. Factoring the right hand side of the given equation, we find that 05 (b + a)(b a). Since a and b are both positive integers, we know that b + a and b a are corresponding divisors of 05. Let these divisors be m and n, respectively. Solving for a and b in terms of m, n, we see that b m + n a m n. Thus b attains a minimum when m + n is minimized. The pair of divisors (m, n) of 05 for which their sum is minimized is (65, ). We finish by substituting (65, ) as (m, n) back into our equations to get b

7 MDatB 06 Solutions Solved by out of 9 individuals Problem. (Problem ) Let S be the set of all positive integers n such that n 0 n + 4 sum of the elements of S. is a positive integer. Find the Solution. We begin by noting that n 0 n + 4 n 96 n n + 4 n n + 4. Since we are given that n is a positive integer, n 4 is also an integer. Thus 76 must be integral n+4 as well, and so we consider positive integral values of n for which n Factorizing 76, we find that n + 4 must lie in the set {,, 4, 8,, 6,, 44, 88, 76}, meaning that n {,, 0,,, 8, 0, 74, 6}. Since n is a positive integer, we can eliminate the negative elements from the set, leaving {, 8, 0, 74, 6}. It remains to check whether n 0 is a positive integer for these values. It turns out that there is n+4 one exception: n does not work since 0 6, resulting in a negative value of n 0. All n+4 other values of n from the set {8, 0, 74, 6} are valid, so our answer is Solved by 8 out of 9 individuals Problem. (Problem ) Two square pyramids with the same altitude are placed in space so that they intersect. Their bases lie in parallel planes, corresponding edges of their bases are parallel, and each pyramid s vertex lies at the center of the base of the other pyramid. Each pyramid has altitude 8. One pyramid has a base of side 6 and the other has a base of side 0. Compute the volume of the intersection of the two pyramids. Solution. Figure 7

8 MDatB 06 Solutions Figure E h M 8 x x E 5 M We are told that the two square pyramids intersect such that the bases are parallel and each vertex lies in in the center of of the other pyramid s base. This indicates that the pyramids share the same height, and that the shape of of the intersection of of the two pyramids is is a square with each vertex on one of of the edges of of the pyramids. Let us position the pyramids such that the pyramid with base of ofside length 0 ABCD has its vertex upward and the pyramid with base of of side length 6 EFGH has its vertex pointing downward. They intersect at square PPQRS, as shown in in Figure.. Figure shows a vertical cross section of of the two pyramids, where M M is is the height of of the pyramids, and E M and E M are parallel to the edges of of the square bases. E is is the midpoint of of FFG, and E is is the midpoint of of BC. Then, we see that E M has length 5 and E M has length.. We wish to find h, h, which will allow us to find the area of of the intersection of of the two pyramids. Using similar triangles, we see that 8 x h h. 8 5, and x 8 h. Adding the two equations, h of PQRS is (h) 4 ) 5 + h 8 5 h. Solve to find h 5 8. Thus, the area of P is (h) 4 ( 5 8 ) 5 6. The three dimensional intersection of of the two pyramids is is the union of of PQRSM PQRSM of is square pyramid P and square pyramid P (8 x). Thus, the volume of the intersection is x+ 5 6 (8 x) Solved by 5 out of of 9 individuals 8

9 MDatB 06 Solutions Problem.4 (Problem 4) An 8 8 checkerboard can be folded into 4 8 halves. On each half of the checkerboard, three small Velcro squares are glued randomly onto three board squares. (That is, three squares on one half and three more, complementary, squares on the other half.) Fold the board. Let P be the probability that the board sticks shut. (That only happens if there is Velcro glued to corresponding squares on each side.) If P a, where a and b are integers and the fraction is b in lowest terms, compute a + b. Solution. In order for the board to stick shut, we need at least one pair of the Velcro squares to connect. We find the complement, in which none of the squares stick. WLOG, given any configuration of Velcro squares on the left half of the checkerboard, there are possible squares for the three Velcro squares on( the ) right half such that none will stick. The probability that the board will not stick shut is ( ) Thus, the probability 480 that the board sticks shut is The answer is Solved by 9 out of 9 individuals Individual Round Problem. (Problem 5) If the point (a, b) is reflected in the line x + y, find the coordinates of the reflected point. Solution. We begin by noting that the image of (a, b) after reflection across the line y x is ( b, a). We can rearrange the equation x + y to y x +, which is the translation of y x one unit up. This motivates us to translate the coordinate plane one unit up as well. The line y x + becomes y x relative to the translated grid, and the new coordinates of (a, b) are (a, b ). Thus, relative to our translated grid, reflection across the line y x sends the point (a, b ) to ( b, a). Relative to the original axes, this reflected point is ( b, a), and we are done. Solved by 78 out of 9 individuals Problem. (Problem 6) Let S be the set of complex numbers of the form z x + iy such that x and y are integers and x y. Let N be the product of the members of S that lie closest to the origin. Find the number of positive integer factors of N. Solution. We want to find the members of S with the least magnitude. We see that picking y 0, ±, ±, ±, ±4, ±5 gives us what we want, since the magnitude only increases on either side 9

10 MDatB 06 Solutions of this range. Now, note that ((a ) + ai)((a ) ai) (a ) + a a 4 a +. Let s find N now. 5 N ( + 0i) a 4 a N has 4 prime factors, so it has 4 6 positive integer factors. a Solved by 5 out of 9 individuals Problem. (Problem 7) A robot is standing at an integer point on a number line. The robot flips two fair coins. If either or both come up heads, it take one step to the right, to the next larger integer. If both come up tails, it takes one step to the left, to the next smaller integer. If it ever reaches the point 0, the robot stops and shuts down. If the robot starts at point, what is the probability that it ever stops? Solution. Let P n be the probability that the robot ever reaches 0 when he is at the n th point. We want to find P. Since the probability of flipping two tails is equivalent to, there is a 4 probability that the robot advances one point to the right and a probability that it regresses one 4 to the left after flipping the two coins. Then P P P 4 P + 4 P P 4 P + 4 P 4 P n 4 P n + 4 P n+. We sum all of these equations together to obtain P k k 4 + P k + 4 k 4 Rearranging and simplifying the equation, we get 4 P 4, P k. k meaning that P. Now from the first equation, we find that P P, and so meaning that P 9 4 P,, which is what we set out to find. 0

11 MDatB 06 Solutions Solved by 0 out of 9 individuals Problem.4 (Problem 8) Let f(n) 4n + 4n n + + n. Compute f() + f() + f() + + f(). Solution. Let a n +, b n. Note that a b. We can rewrite f(n) as f(n) a + b + ab a + b (a + b + ab)(a b) (a + b)(a b) a b (n + ) (n ) We would like to calculate f() + f() + + f(). n (n + ) (n ) ( + ) ( ) 5 6 Solved by 7 out of 9 individuals 4 Team Round Problem 4. (Team problem ) In how many ways can you write 05 as a sum of nonnegative integer powers of so that each power of appears at most twice? Solution. We can write 05 as 0 +. From here, we use the fact that a a + a to expand this expression. Evidently, in order to satisfy the condition that each power of appears at most twice, when we expand from a + a, we must leave one of the a to get a + a + a, or else we would need to have 4 a. This obviously does not satisfy the indicated conditions. We can list out the possible ways:

12 Diao, Diao, Sun, Sun, Ye Ye MDatB MDatB Solutions Solutions Thus, Thus, there there are are 0 0 possible possible ways ways to to write write satisfying satisfying the the conditions. conditions. Solved Solved by by 6 6 out out of of 8 8 teams teams Problem (Team problem ) The great circle distance between city A and city B is of the circumference of the earth. A 6 satellite in orbit can see" both cities at the same time. If h is the smallest possible altitude above the earth s surface for the satellite, compute h, where R is the radius of the earth. R (Assume the earth is a sphere.) Solution. Let S be the point of the satellite, and let A be city A and B be city B. First, we argue that S must be in the same plane as the great circle that contains A and B. In order to see A, S must be located above the plane that is tangent to the earth at A. Similarly, S must be above the plane tangent at B. The intersection of these two planes is a line, and it is clear by symmetry that the point on the line closest to the sphere is in the same plane as the great circle containing A and B. S A B O Since Since the the great great circle circle distance distance between between A and and B is is the circumference of the earth, \BOA 6 60 Then, \SOA 0 Since OA R, OS Ô R Then, the the circumference height of the of satellite the earth, is BOA Then, SOA 0. Since OA R, OS R. Then, the height of the satellite is A Ô B ( ) OS h OS R R Thus, Thus, Ô h R. Solved Solved by by 8 8 out out of of 8 8 teams teams

13 MDatB 06 Solutions Problem 4. (Team problem ) There are 06 slips of paper with numbers written on them, numbered from through 06, laid out in order around the edge of a large round table. Sasha starts walking around the table picking up every other slip of paper she sees. She picks up #, skips over #, picks up #, skips over #4, picks up #5, and so on. She keeps walking around and around the table until she has picked up every slip of paper. What is the number on the very last slip of paper that she picks up? Solution. On her first trip around the table, Sasha picks up every number that is not divisible by. On her second trip, she picks up every number not divisible by 4. This pattern continues until Sasha begins picking up every number not divisible by 64. Since 06 6, for simplicity, we can divide the numbers on the table at that point in time by, so the numbers on the table are,,,..., 6. Then,,, 5,..., 6 are all picked up and the pattern breaks. Fortunately, the numbers are pretty small by this point so we can simulate the rest of Sasha s trips by hand. We end up with 6 being the last number remaining, so multiply by to get 984. Solved by 8 out of 8 teams Problem 4.4 (Team problem 4) In the figure to the right, area( BDX), area( DCX), and area( CEX). Compute area( ABC). A F X E B C D Solution. We shall use mass points. If we give point B a mass of, then point C will have a mass of. Since area( BCX) then BX : XE : and therefore E has a mass of. From here, area( ECX) we can see that point A has a mass of. So area( ABC) area( EBC) AC EC area( ABC) area( EBC). Solved by out of 8 teams

14 MDatB 06 Solutions Problem 4.5 (Team problem 5) Find the product of the slopes of the two tangent lines from the point (0, ) to the ellipse x 9 + y. Solution. Let the equations of the tangent lines be y ax + and y ax +. We find the intersection point of the tangent line y ax + and the ellipse by substituting. Then x 9 + (ax + ) x + 9(ax + ) 9 x + 9a x + 8 ax (9a + )x + 8 ax Since the line is tangent to the ellipse, there must be only one solution to this quadratic. Then we have to have (8 a) 4 9 (9a + ) 0. Solve to find that 9a, so a. We want to find the product of the slopes of the two tangent 9 lines, which is equivalent to a. Thus, our answer is 9. Solved by 4 out of 8 teams Problem 4.6 (Team problem 6) Suppose x, y, z are real numbers such that x + y + z and xy + xz + yz 99. Find the maximum possible value of xyz. Solution. Let x, y, z be the roots of the cubic polynomial p(x) x x + 99x k, where k xyz by Vieta s Formulas. We want to maximize k while keeping all three of x, y, z real. Increasing k will shift the graph vertically downward; it can be seen that p(x) is tangent to the x-axis when k reaches a maximum. This motivates us to take the derivative of p(x): the double root will be a zero of p (x). Now since we have that p (x) x 4x + 99, x 4x + (x )(x ) 0, meaning that and are possible double roots of p(x). Setting p(x) 0 and solving for k, we find that either k + 99 or k Clearly 5 is the larger possibility, so the largest possible value of xyz is 5. 4

15 Diao, Diao, Sun, Sun, Ye Ye MDatB MDatB Solutions Solutions Solved Solved by by out out of of 8 8 teams teams Problem 4.7 Problem 4.7 (Team problem 7) Let x 55 Let x 55, and after that, x n+ x n + x, and after that, x n+ x Œÿ n + x n. Compute. n. Computen +x n. n + x n Solution. The given recursion can be written as x n+ x Solution. The given recursion can be written as x n+ x n( + x n ) n( x n ) which can be rearranged to which can be rearranged to give give x n+ n ( n ) n+ x n ( + x n ) x n +x n + x n n A Œÿ ( Œÿ ) B n +x n + x n n n x n n x n n+ x n since x n grows without bound. Solved by 4 out of 8 teams Problem 4.8 (Team problem 8) Inside a square of ofarea,, an isosceles triangle is isinscribed with its unique vertex at a corner of of the square, and then a smaller square is isinscribed in inthe triangle, as shown. Find the largest possible area for the B E X C W Y F Z A D 5 5

16 MDatB 06 Solutions Solution. In the figure above, let x be the length of EC and CF. We want to find z, the length of a side of the inscribed square. We see that the height of the isosceles triangle is x, and the base of the isosceles triangle is x. We let h x for ease of computation. Using similar triangles, we have that z h z x h. Simplify to find that z( z + ). Solve for z, substituting for h, to find h z x + h x h x + h s x ( x) x + x( x). x + x We want to maximize the area of the smaller square, so we have to maximize the value of z. Thus, we set the derivative of z to 0, using the Chain Rule multiple times: ( x)(x + ) + x d dx [( x)(x + ) ] 0 ( x)(x + ) + 4 x (x + ) 0 (4 x ) 4 x (x + ) (x + ) (4 x ) 4 x x + 4x 4 0 x + ( + )( + ) Substitute to find that z the inscribed square is (6 8) Then the maximum area of Solution. Let the larger square be called ABCD, and let the triangle be called AEF, such that E is on BC and F is on CD. Then, let the smaller square be called W XY Z, with W on AE, X and Y on EF, and Z on AF. Then, let θ EW X, and let s be the side length of W XY Z. Draw the diagonal AC. Because AEF is isosceles, AC perpendicularly bisects ZW, EF, and XY. So, we have AC W X. Therefore, CAW θ as well, and EAB 45 θ. 6

17 MDatB 06 Solutions By trig, AE cos(45 θ), AW s, and W E s. Therefore, we have (with much trig) sin θ cos θ cos(45 θ) s sin θ + s cos θ cos θ + sin θ s cos 45 cos θ + sin 45 sin θ sin θ cos θ cos θ + sin θ s cos θ + sin θ sin θ s sin θ (cos θ + sin θ) (cos θ + sin θ) sin θ cos θ + sin θ cos θ + sin θ sin θ cos θ + sin θ s sin θ 4 cos θ + sin θ To maximize the area of W XY Z, we want to maximize s, so we want to maximize this value. We can take the derivative, and ignore its denominator, as we only want to find when the derivative is 0 (the denominator can t be 0, as θ ranges from 0 to 45). We can plug this back in to get: cos θ ( cos θ + sin θ) sin θ ( sin θ + 6 cos θ) 0 6 cos θ cos θ + 6 sin θ cos θ sin θ 6 sin θ cos θ 0 6 cos θ 0 cos θ. sin θ. s 4 sin θ cos θ + sin θ s A s Solution. Let the vertices of the triangle be (0, 0), (r, ), (, r). Note that 0 r <. The equations for the edges of the triangle are y rx, x ry, and x + y + r. Let (b, rb) be a point on the lower edge; we want to check whether it is a vertex of the smaller square. That square is set at 7

18 MDatB 06 Solutions a 45 degree angle to the axes (it s a diamond ) so the two adjacent vertices are (r t, rb + t) and (r + t, rb + t) for some positive t. But (r t, rb + t) has to be (rb, r) which gives us t ( r)b. Now we check that the other point we mentioned lies on x + y + r. b + t + rb + t + r ( + r)b + t + r ( + r)b + ( r)b + r ( r)b + r b + r r t r r So we seek to maximize t. Taking the derivative with respect to r and setting that derivative equal to zero gives us the equation r 6r + 0. That has two positive roots, one small, one large. Since we must have r between 0 and, we choose the small root, namely r. Then r 7 and t r r We do need to know that this is a maximum. We check that if r, then t 0 and if t 0, then r. We need to compare to 6 4.? < 6 4? < 8? < 7 88 < 89 So we have found a maximum value of t in the required interval. Now (again, using the diamond idea) the area of the small square is t ( 6 4 ) Solved by out of 8 teams 8