. [5] Evaluate + 5 + 8 + + 0. 75 There are 0 HMMT November 0 Saturday 9 November 0 Guts Round = 4 terms with average +0, so their sum is 7 0 = 75.. [5] Two fair six-sided dice are rolled. What is the probability that their sum is at least 0? 6 There are,, outcomes with sum 0,,, so the probability is ++ 6 = 6.. [5] A square is inscribed in a circle of radius. Find the perimeter of the square. 4 OR 8 The square has diagonal length, so side length and perimeter 4. 4. [6] Find the minimum possible value of (x + 6x + ) over all real numbers x. 0 This is ((x + ) 7) 0, with equality at x + = ± 7. 5. [6] How many positive integers less than 00 are relatively prime to 00? (Two numbers are relatively prime if their greatest common factor is.) 40 n < 00 is relatively prime to 00 if and only if it s relatively prime to 00 (00, 00 have the same prime factors). Thus our answer is φ(00) = 00 4 5 = 40. 6. [6] A right triangle has area 5 and a hypotenuse of length 5. Find its perimeter. 5 + 5 If x, y denote the legs, then xy = 0 and x + y = 5, so x + y + x + y = (x + y ) + xy + 5 = 45 + 5 = 5 + 5. 7. [7] Marty and three other people took a math test. Everyone got a non-negative integer score. The average score was 0. Marty was told the average score and concluded that everyone else scored below average. What was the minimum possible score Marty could have gotten in order to definitively reach this conclusion? 6 Suppose for the sake of contradiction Marty obtained a score of 60 or lower. Since the mean is 0, the total score of the 4 test takers must be 80. Then there exists the possibility of students getting 0, and the last student getting a score of 0 or higher. If so, Marty could not have concluded with certainty that everyone else scored below average. With a score of 6, any of the other three students must have scored points lower or equal to 9 points. Thus Marty is able to conclude that everyone else scored below average. 8. [7] Evaluate the expression, where the digit appears 0 times. 0 04 Let f(n) denote the corresponding expression with the digit appearing exactly n times. Then f() = and for n >, f(n) = f(n ). By induction using the identity = N N+, f(n) = n n+ for all n, so f(0) = 0 04. N N
9. [7] Find the remainder when + + 5 + + 99 is divided by 000. 650 We have S = 49 i=0 (i + ) = 49 i=0 4i + 4i + = 4 49 50 99 6 + 4 49 50 + 50 700 + 900 + 50 (mod 000) 650 (mod 000). 0. [8] How many pairs of real numbers (x, y) satisfy the equation y 4 y = xy xy = x y xy = x 4 x = 0? 9 We can see that if they solve the first and fourth equations, they are automatically solutions to the second and third equations. Hence, the solutions are just the = 9 points where x, y can be any of, 0,.. [8] David has a unit triangular array of 0 points, 4 on each side. A looping path is a sequence A, A,..., A 0 containing each of the 0 points exactly once, such that A i and A i+ are adjacent (exactly unit apart) for i =,,..., 0. (Here A = A.) Find the number of looping paths in this array. 60 There are 0 times as many loop sequences as loops. To count the number of loops, first focus on the three corners of the array: their edges are uniquely determined. It s now easy to see there are loops (they form V -shapes ), so the answer is 0 = 60.. [8] Given that 6 + = 878, find positive integers (n, m) such that n + m = 964. (0, 9) OR (9, 0) If a + b = c, then ( a+b ) + ( a b ) = a +b 4 = a +b = c. Thus, n = 6+ = 9 and m = 6 = 0 works.. [9] Let S = {,,..., 0}. Find the number of ordered triples (A, B, C) of subsets of S such that A B and A B C = S. 5 0 OR 5 67 Let n = 0. Each of the n elements can be independently placed in 5 spots: there are choices with element x in at least one set, and we subtract the choices with element x in set A but not B. Specifying where the elements go uniquely determines A, B, C, so there are 5 n = 5 0 ordered triples. 4. [9] Find all triples of positive integers (x, y, z) such that x + y z = 00 and x + y z = 4. (,, 57) Cancel z to get 4 = (y x)(y + x ). Since x, y are positive, we have y + x + > 0, so 0 < y x < y + x. But y x and y + x have opposite parity, so (y x, y + x ) {(, 4), (, 8)} yields (y, x) {(, ), (6, )}. Finally, 0 < z = x + y 00 forces (x, y, z) = (,, 57). 5. [9] Find all real numbers x between 0 and 60 such that cos 0 = cos 40 + sin x. 70, 0 (need both, but order doesn t matter) Note that = cos 0, so sin x = cos 0 = x {70, 0}. 6. [0] A bug is on one exterior vertex of solid S, a cube that has its center cube removed, and wishes to travel to the opposite exterior vertex. Let O denote the outer surface of S (formed by the surface of the cube). Let L(S) denote the length of the shortest path through S. (Note that such a path cannot pass through the missing center cube, which is empty space.) Let L(O) denote the length of the shortest path through O. What is the ratio L(S) L(O)? 9 5 OR 45 5 By (*), the shortest route in O has length.5 + = 5. By (**), the shortest route overall (in S) has length.5 + + = + + 4 = 9. Therefore the desired ratio is 9 = 45 5 5.
(*) Suppose we re trying to get from (0, 0, 0) to (,, ) through O. Then one minimal-length path through O is (0, 0, 0) (.5, 0, ) (,, ). (**) Suppose we re trying to get from (0, 0, 0) to (,, ) through S. Then the inner hole is [, ] [, ] [, ], and one minimal-length path is (0, 0, 0) (.5,, ) (,, ). To justify (*), note that we must at some point hit one of the three faces incident to (,, ), and therefore one of the edges of those faces. Without loss of generality, the first of these edges (which must lie on a face incident to (0, 0, 0)) is {(t, 0, ) : 0 t }. Then the shortest path goes directly from the origin to the edge, and then directly to (,, ); t =.5 minimizes the resulting distance. (One may either appeal to the classic geometric unfolding argument, or just direct algebraic minimization.) To justify (**), consider the portion of S visible from (0, 0, 0). It sees mutually adjacent faces of the center cube hole (when looking inside the solid) and its 6 edges. (,, ) can also see these 6 edges. The shortest path through S must be a straight line from the start vertex to some point P on the surface of the center cube (+), and it s easy to see, using the triangle inequality, that this point P must be on one of the 6 edges. Without loss of generality, it s on the edge {(t,, ) : t }. The remaining path is a straight line to (,, ). Then once again, t =.5 minimizes the distance. (And once again, one may either appeal to the classic geometric unfolding argument though this time it s a little trickier or just direct algebraic minimization.) Comment. (+) can be proven in many ways. The rough physical intuition is that a fully stretched rubber band with ends at (0, 0, 0) and (,, ) must not have any wiggle room (and so touches the inner cube). Perhaps more rigorously, we can try shortening a path that does not hit the center cube, for instance by projecting the path down towards the closest edge of the inner cube. 7. [0] Find the sum of n of n terminates. over all positive integers n with the property that the decimal representation 5 The decimal representation of n terminates if and only if n = i 5 j for some nonnegative integers i, j, so our desired sum is i 0 j 0 i 5 j = i 0 i j 0 5 j = ( ) ( 5 ) = 5 4 = 5. 8. [0] The rightmost nonzero digit in the decimal expansion of 0! is the same as the rightmost nonzero digit of n!, where n is an integer greater than 0. Find the smallest possible value of n. 0 0! has more factors of than 5, so its rightmost nonzero digit is one of, 4, 6, 8. Notice that if the rightmost nonzero digit of 0! is k ( k 4), then 0! has rightmost nonzero digit 0(k) 4k (mod 0), and 0! has rightmost nonzero digit 0(4k) k (mod 0). Hence n = 0. 9. [] Let p, q, r, s be distinct primes such that pq rs is divisible by 0. Find the minimum possible value of p + q + r + s. 54 The key is to realize none of the primes can be,, or 5, or else we would have to use one of them twice. Hence p, q, r, s must lie among 7,,, 7, 9,, 9,.... These options give remainders of (mod ) (obviously),,,,,,,,... modulo, and,,,, 4,, 4,... modulo 5. We automatically have pq rs, and we have pq rs if and only if pqrs (pq) (mod ), i.e. there are an even number of (mod ) s among p, q, r, s. If {p, q, r, s} = {7,,, 7}, then we cannot have 5 pq rs, or else pqrs (pq) (mod 5) is a quadratic residue. Our next smallest choice (in terms of p + q + r + s) is {7,, 7, 9}, which works: 7 7 9 4 0 (mod 5). This gives an answer of 7 + 7 + + 9 = 54. 0. [] There exist unique nonnegative integers A, B between 0 and 9, inclusive, such that Find 0 A + B. (00 A + 0 B) = 57, 08, 49.
75 We only need to bound for AB00; in other words, AB 570 but (AB +) 570. A quick check gives AB = 75. (Lots of ways to get this...). [] Suppose A, B, C, and D are four circles of radius r > 0 centered about the points (0, r), (r, 0), (0, r), and ( r, 0) in the plane. Let O be a circle centered at (0, 0) with radius r. In terms of r, what is the area of the union of circles A, B, C, and D subtracted by the area of circle O that is not contained in the union of A, B, C, and D? (The union of two or more regions in the plane is the set of points lying in at least one of the regions.) 8r Solution. Let U denote the union of the four circles, so we seek U ([O] U) = U [O] = [(r) + 4 πr ] π(r) = 8r. (Here we decompose U into the square S with vertices at (±r, ±r) and the four semicircular regions of radius r bordering the four sides of U.) Solution. There are three different kinds of regions: let x be an area of a small circle that does not contain the two intersections with the other two small circles, y be an area of intersection of two small circles, and z be one of those four areas that is inside the big circle but outside all of the small circles. Then the key observation is y = z. Indeed, adopting the union U notation from the previous solution, we have 4z = [O] U = π(r) U, and by the inclusion-exclusion principle, 4y = [A] + [B] + [C] + [D] U = 4π U, so y = z. Now U = (4x + 4y) (4z) = 4x. But the area of each x is simply r by moving the curved outward parts to fit into the curved inward parts to get a r r rectangle. So the answer is 8r.. [] Let S be a subset of {,,,..., } such that it is impossible to partition S into k disjoint subsets, each of whose elements sum to the same value, for any integer k. Find the maximum possible sum of the elements of S. 77 We note that the maximum possible sum is 78 (the entire set). However, this could be partitioned into subsets with sum 9: {,,, 0,, } and {4, 5, 6, 7, 8, 9}. The next largest possible sum is 77 (the entire set except ). If k subsets each had equal sum, then they would have to be 7 subsets with sum each or subsets with sum 7 each. However, the subset containing will have sum greater than ; hence there is no way to partition the subset {,..., } into equal subsets.. [] The number 989 00 007 + 0 can be written as the product of three distinct primes p, q, r with p < q < r. Find (p, q, r). (99, 997, 009) Let f(x) = x(x )(x + 6) + 0 = x 6x 7x + 0, so that f(00) = 989 00 007 + 0. But f(4) = 4( 8)(0) + 0 = 0, so f(x) = (x 4)(x x 80) = (x 4)(x 0)(x + 8). Thus f(00) = 99 997 009, as desired. 4. [] Find the number of subsets S of {,,... 6} satisfying the following conditions: S is non-empty. No subset of S has the property that the sum of its elements is 0. 4 We do casework based on the largest element of S. Call a set n-free if none of its subsets have elements summing to n. Case : The largest element of S is 6. Then 4 / S. If 5 / S, then we wish to find all 4-free subsets of {,, } (note that + + = 6 < 0). We just cannot include both,, so we have ( ) = 6 choices here. If 5 S, then we want 4, 5-free subsets of {,, }. The only 4-but-not-5-free subset is {, }, so we have 6 choices here, for a case total of 6 + 5 =.
Case : The largest element of S is 5. We seek 5, 0-free subsets of {,,, 4}. We just cannot have both, 4 or both, (note that getting 0 requires the whole set), so we have ( ) = 9 subsets in this case. Case : The largest element of S is at most 4. (So we want a 4-free subset of {,,, 4}.) The only way to sum to 0 with,,, 4 is by using all the terms, so we simply discount the empty set and {,,, 4}, for a total of 4 = 4 subsets. In conclusion, the total number of subsets is + 9 + 4 = 4. 5. [] Let a, b be positive reals with a > b > a. Place two squares of side lengths a, b next to each other, such that the larger square has lower left corner at (0, 0) and the smaller square has lower left corner at (a, 0). Draw the line passing through (0, a) and (a + b, 0). The region in the two squares lying above the line has area 0. If (a, b) is the unique pair maximizing a + b, compute a b. 5 Let t = a b (, ); we will rewrite the sum a + b as a function of t. The area condition easily translates to a ab+b = 0, or b (t 406 t + ) = 406 b = t t+. Thus a + b is a function f(t) = ( + t) of t, and our answer is simply the value of t maximizing f, 406 t t+ or equivalently g(t) = f 406 = (+t) t t+, over the interval (, ). (In general, such maximizers/maximums need not exist, but we shall prove there s a unique maximum here.) We claim that λ = 6 7 is the maximum of (+t) t t+. Indeed, λ g(t) = (λ )t (λ + )t + (λ ) t t + = 9t 0t + 5 7 t = (t 5) t + 7 (t ) + 7 0 4 for all reals t (not just t (, )), with equality at t = 5 (, ). Comment. To motivate the choice of λ, suppose λ were the maximum of f, attained at t 0 (, ); then h(t) = λ(t t + ) (t + ) is quadratic and nonnegative on (, ), but zero at t = t 0. If g is a nontrivial quadratic (nonzero leading coefficient), then t 0 must be a double root, so g has determinant 0. Of course, g could also be constant or linear over (, ), but we can easily rule out both of these possibilities. Alternatively, we can simply take a derivative of f to find critical points. 6. [] Trapezoid ABCD is inscribed in the parabola y = x such that A = (a, a ), B = (b, b ), C = ( b, b ), and D = ( a, a ) for some positive reals a, b with a > b. If AD + BC = AB + CD, and AB = 4, what is a? 7 40 Let t = 4, so a + b = (a b) + (a b ) = t gives a = b = t and t = (a b) [ + (a + b) ] = (a b) [ + t ]. Thus a = t+ t +t = 4 + 5 = 7 40. 7. [] Find all triples of real numbers (a, b, c) such that a + b bc = 6 and ab c = 6. (4, 4, 4), ( 4, 4, 4) (need both, but order doesn t matter) a + b bc and ab c are both homogeneous degree polynomials in a, b, c, so we focus on the homogeneous equation a + b bc = ab c, or (a b) + (b c) = 0. So a = b = c, and a = ab c = 6 gives the solutions (4, 4, 4) and ( 4, 4, 4). 8. [5] Triangle ABC has AB = 4, BC =, and a right angle at B. Circles ω and ω of equal radii are drawn such that ω is tangent to AB and AC, ω is tangent to BC and AC, and ω is tangent to ω. Find the radius of ω. 5 7 Solution. Denote by r the common radius of ω, ω, and let O, O be the centers of ω and ω respectively. Suppose ω i hits AC at B i for i =,, so that O O = B B = r.
Extend angle bisector AO to hit BC at P. By the angle bisector theorem and triangle similarity AB O ABP, we deduce = BP AB = 4+5. Similarly, r CB = 4 +5, so or r = 5 7. r AB 5 = AC = AB + B B + B C = r + r + r = 7r, Solution. Use the same notation as in the previous solution, and let α = A. By constructing right triangles with hypotenuses AO, O O, and O C and legs parallel to AB and BC, we obtain But cot α = or r = 5 7. +cos α sin α = + 4 5 5 4 = AB = r cot α + r cos A + r. = and cos A = 4 5, so the above equation simplifies to 4 = r( + 8 5 + ) = 8r 5, 9. [5] Let XY Z be a right triangle with XY Z = 90. Suppose there exists an infinite sequence of equilateral triangles X 0 Y 0 T 0, X Y T,... such that X 0 = X, Y 0 = Y, X i lies on the segment XZ for all i 0, Y i lies on the segment Y Z for all i 0, X i Y i is perpendicular to Y Z for all i 0, T i and Y are separated by line XZ for all i 0, and X i lies on segment Y i T i for i. Let P denote the union of the equilateral triangles. If the area of P is equal to the area of XY Z, find XY Y Z. For any region R, let [R] denote its area. Let a = XY, b = Y Z, ra = X Y. Then [P] = [XY T 0 ]( + r + r 4 + ), [XY Z] = [XY Y X ]( + r + r 4 + ), Y Y = ra, and b = ra ( + r + r + ) (although we can also get this by similar triangles). Hence a 4 = (ra + a)(ra ), or r(r + ) = = r =. Thus XY Y Z = a b = r r =. 0. [5] Find the number of ordered triples of integers (a, b, c) with a, b, c 00 and a b + b c + c a = ab + bc + ca. 9800 This factors as (a b)(b c)(c a) = 0. By the inclusion-exclusion principle, we get 00 00 + 00 = 9800.. [7] Chords AB and CD of circle ω intersect at E such that AE = 8, BE =, CD = 0, and AEC = 90. Let R be a rectangle inside ω with sides parallel to AB and CD, such that no point in the interior of R lies on AB, CD, or the boundary of ω. What is the maximum possible area of R? 6 + 6 7 By power of a point, (CE)(ED) = (AE)(EB) = 6, and CE + ED = CD = 0. Thus CE, ED are, 8. Without loss of generality, assume CE = 8 and DE =. Assume our circle is centered at the origin, with points A = (, 5), B = (, 5), C = (5, ), D = ( 5, ), and the equation of the circle is x + y = 4. Clearly the largest possible rectangle must lie in the first quadrant, and if we let (x, y) be the upper-right corner of the rectangle, then the area of the rectangle is (x + )(y + ) = 9 + 6(x + y) + xy 9 + equality holds if and only if x = y = 7. x +y + x +y = 6 + 6 7, where. [7] Suppose that x and y are chosen randomly and uniformly from (0, ). What is the probability that x y is even? Hint: n= n = π 6. π 4 OR 4 π 4 Note that for every positive integer n, the probability that x y ( = n is just the area of the triangle formed between (0, 0), (, n ), (, (n+) ), which is just n (n+) ).
Thus the probability that x y is odd is k= ( (k ) ) (k) Thus our answer is just π 4. = = ( k= n= (k ) + (k) n 4 k= = π π 4 = π 4. k ) k= (k). [7] On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edge such that the interior of the triangle intersects the interior of the rectangle. What is the total area of all regions that are contained in exactly of the 4 equilateral triangles? 96 54 OR 88 54 OR 96 54 OR 96 54 Let the rectangle be ABCD with AB = 8 and BC = 6. Let the four equilateral triangles be ABP, BCP, CDP, and DAP 4 (for convenience, call them the P -, P -, P -, P 4 - triangles). Let W = AP DP, X = AP DP 4, and Y = DP 4 CP. Reflect X, Y over the line P P 4 (the line halfway between AB and DC) to points X, Y. First we analyze the basic configuration of the diagram. Since AB = 8 < 6, the P -, P 4 - triangles intersect. Furthermore, AP BP, so if T = BP AP, then BP = 6 < 4 = BT. Therefore P lies inside triangle P BA, and by symmetry, also triangle P DC. It follows that the area we wish to compute is the union of two (congruent) concave hexagons, one of which is W XY P Y X. (The other is its reflection over Y Y, the mid-line of AD and BC.) So we seek [W XY P Y X ] = ([W XP 4 X ] [P Y P 4 Y ]). It s easy to see that [W XP 4 X ] = [ADP 4] = 6 4 =, since W XP 4 X and its reflections over lines DW X and AW X partition ADP 4. It remains to consider P Y P 4 Y, a rhombus with (perpendicular) diagonals P P 4 and Y Y. If O denotes the intersection of these two diagonals (also the center of ABCD), then OP is P B AB = 4, the difference between the lengths of the P -altitude in CBP and the distance between the parallel lines Y Y, CB. Easy angle chasing gives OY = OP, so [P Y P 4 Y ] = 4 OP OY = OP = ( 4) = 86 48, and our desired area is [W XP 4 X ] [P Y P 4 Y ] = 6 7 96 = 96 54, or 88 54. 4. [0] Find the number of positive integers less than 000000 that are divisible by some perfect cube greater than. Your score will be max { 0, 0 00 k S }, where k is your answer and S is the actual answer. 68089 Using the following code, we get the answer (denoted by the variable ans): ans = 0 for n in xrange(,000000):
divisible_by_cube = True for i in xrange(,0): if n%(i*i*i)==0: divisible_by_cube = False break if divisible_by_cube: ans = ans + print ans This gives the output 68089 Alternatively, let N = 000000 and denote by P the set of primes. Then by PIE, the number of n (0, N) divisible by a nontrivial cube, or equivalently, by p for some p P, is which deviates from p P by at most the sum of p P N p N p p<q P N p q p<q P N p q ±, ± = (N )( p P ( p )) N / sup t R t t = N /, for terms N p p r with p p r (N ) /, and (N ) k>(n ) / k < (N )[(N ) + (N ) / x dx] = + (N ) (N ) / = O(N / ), for the remaining terms. So we are really interested in 0 6 0 6 p P ( p ) (which, for completeness, is 6809.67...). There are a few simple ways to approximate this: We can use a partial product of p P ( p ). Using just = 0.875 gives an answer of 5000 (this is also just the number of x N divisible by = 8), ( )( ) 0.84 gives 57000 (around the number of x divisible by or ), etc. This will give a lower bound, of course, so we can guess a bit higher. For instance, while 57000 gives a score of around 7, rounding up to 60000 gives 0. We can note that p P ( p ) = ζ() is the inverse of + + +. This is a bit less efficient, but successive partial sums (starting with + ) give around 000, 9000, 50000, 57000, etc. Again, this gives a lower bound, so we can guess a little higher. We can optimize the previous approach with integral approximation after the rth term: ζ() is the sum of + + +r plus something between r+ x dx = (r +) and x dx = r r. Then starting with r =, we get intervals of around (000, 4000), (5000, 00000), (6000, 79000), (65000, 7000), etc. Then we can take something like the average of the two endpoints as our guess; such a strategy gets a score of around 0 for r = already, and 7 for r =. 5. [0] Consider the following 4 by 4 grid with one corner square removed: You may start at any square in this grid and at each move, you may either stop or travel to an adjacent square (sharing a side, not just a corner) that you have not already visited (the square you start at is automatically marked as visited). Determine the distinct number of paths you can take. Your score will be max { 0, 0 00 k S }, where k is your answer and S is the actual answer. 4007
6. [0] Pick a subset of at least four of the following seven numbers, order them from least to greatest, and write down their labels (corresponding letters from A through G) in that order: (A) π; (B) + ; (C) 0; (D) 55 ; (E) 6 tan 5 4 tan 40 ; (F) ln(); and (G) e. If the ordering of the numbers you picked is correct and you picked at least 4 numbers, then your score for this problem will be (N )(N ), where N is the size of your subset; otherwise, your score is 0. F, G, A, D, E, B, C OR F < G < A < D < E < B < C OR C > B > E > D > A > G > F We have ln() < e < π < 55 < 6 tan 5 4 tan 40 < + < 0.