SMT 2011 General Test and Solutions February 19, F (x) + F = 1 + x. 2 = 3. Now take x = 2 2 F ( 1) = F ( 1) = 3 2 F (2)

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1 SMT 0 General Test and Solutions February 9, 0 Let F () be a real-valued function defined for all real 0, such that ( ) F () + F = + Find F () Answer: Setting =, we find that F () + F ( ) = Now take =, to get that F ( ) + F ( ) = Finally, setting =, we get that F ( ) + F () = 0 Then we find that ( ) ( ) F () = F = F ( ) = + F ( ) = F () F () = Alternate Solution: We can eplicitly solve for F () and then plug in = Notice that for 0,, F () + F ( ) = + so Thus ( F ) ( ) + F = + and F ( ) + F () = + ( ) ( ) ( ) ( ) F () = F () + F F F + F + F () ( = + + ) + + It follows that F () = = ( ) and the result follows by taking = Given that a =, a =, a n = a n + a n, what is a 00 + a 99? Answer: 98 5 a n = a n + a n a n + a n = (a n + a n ) = n (a + a ) So a 00 + a 99 = 98 5 Let sequence A be { 7, 7 6, 7 9, } where the ( jth term is given by a j = 7 ) j Let B be a sequence where the j th term is defined by b j = a j + a j What is the sum of all the terms in B? Answer: Split B into two series C and D where the terms of C are c j = a j = ( 9 ) j 6 9 and the terms of D are ( d j = a j = 7 ) j Since both C and D are geometric series with ratios less than, the sum of their terms yields 9/6 /9 = 7/ 80 and / = Find all rational roots of = 0 Answer: =, 0, Therefore, the sum of the terms in B equals 80 + = There are four intervals to consider, each with their own restrictions Consider the case in which > Then the equation becomes ( )( ) = ( )( + ) = 0 Thus, = is

2 SMT 0 General Test and Solutions February 9, 0 the only rational root for > Consider the case in which < < Then the equation becomes ( )( ) = ( )( + ) = 0 Thus, = 0 and = are the rational roots for < < Consider the case in which < or the case in which < < In these cases, the equation becomes ( )( ) = + + By the rational root theorem, the rational roots of this polynomial can only be ±, ±, ± and a quick check shows that none of these are roots, so this polynomial has no rational roots 5 Let S = {,,,, 5, 6, 7, 8, 9, 0} In how many ways can two (not necessarily distinct) elements a, b be taken from S such that a b is in lowest terms, ie a and b share no common divisors other than? Answer: 6 This amounts to determining, for a given numerator, how many elements in S are relatively prime to the numerator If we let f(n) be the number of positive integers relatively prime to n and less than or equal to 0, it is obvious that f() = 0, f() = f() = f(8) = 5, f() = f(9) = 7, f(5) = 8, f(7) = 9, f(6) =, and f(0) = Therefore, the answer is = 6 6 Find all square numbers which can be represented in the form a + b, where a, b are nonnegative integers You can assume the fact that the equation y = has no integer solutions if Answer:,, 5 For b = 0 one has a + = c, a = (c + )(c ) Thus both c + and c should be powers of The only possibility is c =, which gives a solution + 0 = 9 = For b, a + b is not divisible by, so it should be (mod ) This requires a to be even Let a = d, then b = c d = (c + d )(c d ) Let c + d = p and c d = q Eliminating c, one has d+ = p q For q the right-hand side is divisible by, so q = 0 From what we know, there are only two solutions (d, p) = (0, ), (, ) These solutions give 0 + = = and + = 5 = 5 respectively 7 A frog is jumping on the number line, starting at zero and jumping to seven He can jump from to either + or + However, the frog is easily confused, and before arriving at the number seven, he will turn around and jump in the wrong direction, jumping from to This happens eactly once, and will happen in such a way that the frog will not land on a negative number How many ways can the frog get to the number seven? Answer: 6 Let f n be the number of ways to jump from zero to n, ignoring for the time being jumping backwards We have f 0 =, f =, and f n = f n + f n when n Therefore, we have that f =, f =, f = 5, f 5 = 8, f 6 =, and f 7 = Note that we can describe the frog s jumping as jumping forward n numbers, jumping backward number, and jumping forward 8 n numbers Therefore, the desired 6 answer is simply f i f 8 i = 6 i= 8 Call a nonnegative integer k sparse when all pairs of s in the binary representation of k are separated by at least two zeroes For eample, 9 = 00 is sparse, but 0 = 00 is not sparse How many sparse numbers are less than 7? Answer: 87 Let a n denote the number of sparse numbers with no more than n binary digits In particular, for numbers with less than n binary digits after removing leading zeroes, append leading zeroes so all numbers have n binary digits when including sufficiently many leading zeroes We have that a 0 =, a =, and a = since for these lengths, either zero digits are or one digit is We claim that the recurrence a = a n + a n holds for n We split this analysis into two cases; numbers where the nth binary digit is 0 or When the nth binary digit is zero, we can remove that zero to get a valid number with n binary digits When the nth binary digit is one, it is known that the (n )th and (n )th digits are both zero, so we can truncate those to get a valid number with n binary digits Therefore, the recurrence holds With the given initial conditions, a 7 = 87

3 SMT 0 General Test and Solutions February 9, 0 9 Two ants begin on opposite corners of a cube On each move, they can travel along an edge to an adjacent verte Find the probability they both return to their starting position after moves Answer: 9 79 Let the cube be oriented so that one ant starts at the origin and the other at (,, ) Let, y, z be moves away from the origin and, y, z be moves toward the origin in each the respective directions Any move away from the origin has to at some point be followed by a move back to the origin, and if the ant moves in all three directions, then it can t get back to its original corner in moves The number of ways to choose directions is ( ) = and for each pair of directions there are!!! = 6 ways to arrange four moves a, a, b, b such that a precedes a and b precedes b Hence there are 6 = 8 ways to move in two directions The ant can also move in a, a, a, a (in other words, make a move, return, repeat the move, return again) in three directions so this gives 8 + = moves There are = 8 possible moves, of which return the ant for a probability of 8 = 7 7 Since this must happen simultaneously to both ants, the probability is = An unfair coin has a / probability of landing on heads If the coin is flipped 50 times, what is the probability that the total number of heads is even? Answer: +(/)50 The coin can turn up heads 0,,,, or 50 times to satisfy the problem Hence the probability is P = ( 50 0 ) ( ) 0 ( ) 50 + ( 50 ) ( ) ( ) ( ) ( ) 50 ( ) 0 Note that this sum is the sum of the even-powered terms of the epansion (/ + /) 50 To isolate these terms, we note that the odd-powered terms of (/ /) 50 are negative So by adding (/ + /) 50 + (/ /) 50, we get rid of the odd-powered terms and we are left with two times the sum of the even terms Hence the probability is P = (/ + /)50 + (/ /) 50 = + (/)50 Find the unique polynomial P () with coefficients taken from the set {, 0, } and with least possible degree such that P (00) (mod ), P (0) 0 (mod ), and P (0) 0 (mod ) Answer: P () = First suppose P () is constant or linear Then we have P (00) + P (0) = P (0), which is a contradiction because the left side is congruent to (mod ) and the right is congruent to 0 (mod ) So P must be at least quadratic The space of quadratic polynomials in is spanned by the polynomials f() =, g() =, and h() = Applying each of these to 00, 0, and 0, we have the mod equivalences: f(00, 0, 0) (,, ) g(00, 0, 0) (0,, ) h(00, 0, 0) (0,, ) Subtracting the third row from the first, we have P () = f() h() =, giving P (00, 0, 0) (, 0, 0) (mod ), as desired Uniqueness follows from the observation that the three vectors above form a basis for (Z/Z) Let a, b C such that a + b = a + b = i Compute Re(a) Answer: From a + b = i we can let a = i + and b = i Then a + b = (( i) + ) = ( ) = i So = + i = eiπ/, = ± +i Since Re(a) = Re(), the answer is =

4 SMT 0 General Test and Solutions February 9, 0 Let T n denote the number of terms in ( + y + z) n when simplified, ie epanded and like terms collected, for non-negative integers n 0 Find Answer: ( ) k T k k=0 = T 0 T + T T T 00 First note that the epression ( + y + z) n is equal to n! a!b!c! a y b z c where the sum is taken over all non-negative integers a, b, and c with a + b + c = n The number of non-negative integer solutions to a + b + c = n is ( ) n+, so Tk = ( ) k+ for k 0 It is easy to see that T k = (k + ), so T k is the (k + )st triangular number If k = n is odd, then for all positive integers i, T i T i = i + and therefore k n ( ) j T j = T 0 + (T j T j ) j=0 = + = n j= n (j ) Therefore, since T 00 is the 0th triangular number and 0 = (006), we can conclude that the desired sum is 006 Let M = (, ) and N = (, ) be two points in the plane, and let P be a point moving along the -ais When MP N takes on its maimum value, what is the -coordinate of P? Answer: Let P = (a, 0) Note that MP N is inscribed in the circle defined by points M, P, and N, and that it intercepts MN Since MN is fied, it follows that maimizing the measure of MP N is equivalent to minimizing the size of the circle defined by M, P, and N Since P must be on the -ais, we therefore want this circle to be tangent to the -ais Since the center of this circle must lie on the perpendicular bisector of MN, which is the line y =, the center of the circle has to be of the form (a, a), so a has to satisfy (a + ) + ( a) = (a ) Solving this equation gives a = or a = 7 Clearly choosing a = gives a smaller circle, so our answer is 5 Consider the curves + y = and + y + y y = 0 These curves intersect at two points, one of which is (, 0) Find the other one Answer: ( 5, ) 5 From the first equation, we get that y = Plugging this into the second one, we are left with ± + = 0 ( ) = ( ) j= = assuming + = 5 = 0 The quadratic formula yields that = ±8 0 =, 5 (we said that above but we see that it is still valid) If =, the first equation forces y = 0 and we easily see that this solves the second equation If = 5, then clearly y must be positive or else the second equation will sum five positive terms Therefore y = 95 = 6 5 = 5 Hence the other point is ( 5, 5) For a quick visual proof of this fact, we refer the reader to

5 SMT 0 General Test and Solutions February 9, 0 6 If r, s, t, and u denote the roots of the polynomial f() = + + +, find r + s + t + u Answer: 9 First notice that the polynomial g() = ( ) = is a polynomial with roots r, s, t, u Therefore, it is sufficient to find the sum of the squares of the roots of g(), which we will denote as r through r Now, note that r + r + r + r = (r + r + r + r ) (r r + r r + r r + r r + r r + r r ) = ( a a ) a a by Vieta s Theorem, where a n denotes the coefficient of n in g() Plugging in values, we get that our answer is ( ) 0 = 9 7 An icosahedron is a regular polyhedron with vertices, 0 faces, and 0 edges How many rigid rotations G are there for an icosahedron in R? Answer: 60 There are vertices, each with 5 neighbors Any verte and any of its neighbors can be rotated to any other verte-neighbor pair in eactly one way There are 5 = 60 verte-neighbor pairs 8 Pentagon ABCDE is inscribed in a circle of radius If DEA = EAB = ABC, m CAD = 60, and BC = DE, compute the area of ABCDE Answer: 8 Looking at cyclic quadrilaterals ABCD and ACDF tells us that m ACD = m ADC, so ACD is equilateral and m DEA = 0 Now, if we let m EAD = θ, we see that m CAB = 60 θ = m ACB = θ = AED = CBA Now all we have to do is calculate side lengths After creating some triangles, it becomes evident that AC = Now let AB =, so BC = By applying the Law of Cosines to triangle ABC, we find that = 7 Hence, the desired area (ABCDE) = (ACD) + (ABC) = ( ) + ()()(sin 0 ) = 8 9 Five students at a meeting remove their name tags and put them in a hat; the five students then each randomly choose one of the name tags from the bag What is the probability that eactly one person gets their own name tag? Answer: 8 Assume without loss of generality that the first person gets a correct nametag Let s call the other people B, C, D, and E We can order the four people in nine ways such that none of the persons gets his own nametag; CBED, CDEB, CEBD, DBEC, DEBC, DECB, EBCD, EDBC, EDCB Therefore, the desired probability is 9! = 8 Alternative Solution: The selection of random nametags amounts to a selection of a random permutation of the five students from the symmetric group S 5 The condition will be met if and only if the selected permutation σ has eactly one cycle of length one (ie, eactly one fied point) The only distinct cycle types with eactly one fied point are (, ) and (,, ) There are 5! = 0 permutations of the first type and 5! = 5 permutations of the second Thus, the desired probability is = 5! 8 0 Find the 0th-smallest, with >, that satisfies the following relation: sin(ln ) + cos( ln ) sin( ln ) = 0

6 SMT 0 General Test and Solutions February 9, 0 Answer: = e 0π/5 Set y = ln, and observe that cos(y) sin(y) = sin(y + y) sin(y y) = sin(5y) sin(y), so that the equation in question is simply sin(5y) = 0 The solutions are therefore ln = y = nπ 5 = = e nπ/5 for all n N An ant is leashed up to the corner of a solid square brick with side length unit The length of the ant s leash is 6 units, and it can only travel on the ground and not through or on the brick In terms of = arctan ( ), what is the area of region accessible to the ant? Answer: 79π + 5 Label the top left corner of the square as the origin O By keeping the leash straight, the ant can travel through of a circle of radius 6 (A = 6π = 7π) The ant can also bend the leash around the two nearest corners of the square to where it is leashed (A = 5 5π = π) However, this double counts the area enclosed by BCDE, which is equal to two times the area of BCD To calculate the latter, notice that ACD is the sector of the circle centered at A with radius 5 We can calculate the coordinates of D usign the two equations y = (from the symmetry) and + (y + ) = 5 which yields D = (, ) Since A = (0, ), the angle of sector ACD is arctan ( ) = The area of triangle ABD equals (base times height) so BCD has area 5 and BCDE has area 0 Hence, the total area is A + A ( 5 ) = 79π + 5 Compute the sum of all n for which the equation + y = n has eactly 0 nonnegative (, y 0) integer solutions Answer: 78 Observe that if the equation a + by = n has m solutions, the equation a + by = n + ab has m + solutions Also note that a + by = a 0 + by 0 for 0 0 < b, 0 y 0 < a has no other solution than (, y) = ( 0, y 0 ) (It is easy to prove both if you consider the fact that the general solution has form ( + bk, y ak)) So there are ab such n and their sum is 0 <b 0 y<a (a + by + 00ab) = 00a b + ab(ab a b) Let ABC be any triangle, and D, E, F be points on BC, CA, AB such that CD = BD, AE = CE and BF = AF AD and BE intersect at X, BE and CF intersect at Y, and CF and AD intersect at Z Find Area( ABC) Area( XY Z)

7 SMT 0 General Test and Solutions February 9, 0 Answer: 7 Using Menelaus s Theorem on ABD with collinear points F, X, C and the provided ratios gives DX/XA = / Using Menelaus s Theorem on ADC with collinear points B, Y, E gives AY/Y D = 6 We conclude that AX, XY, Y D are in length ratio : : By symmetry, this also applies to the segments CZ, ZX, XF and BY, Y Z, ZE Repeatedly using the fact that the area ratio of two triangles of equal height is the ratio of their bases, we find [ABC] = (/)[ADC] = (/)(7/)[XY C] = (/)(7/)()[XY Z] = 7[XY Z], or [ABC]/[XY Z] = 7 Alternate Solution Stretching the triangle will preserve ratios between lengths and ratios between areas, so we may assume that ABC is equilateral with side length We now use mass points to find the length of XY Assign a mass of to A In order to have X be the fulcrum of ABC, C have mass and B must have mass Hence, BX : XE = : and AX : XD = 6 :, the latter of which also equals BY : Y E by symmetry Hence, XY = 7BE To find BE, we apply the Law of Cosines to CBE to get that BE = + cos 60 = 7 = XY = 7 7 Since XY Z must be equilateral by symmetry, the desired ratio equals ( AB XY ) = 7 Let P () be a polynomial of degree 0 such that P () = 0, P () =, P () =,, and P ( 0 ) = 0 Compute the coefficient of the term in P () Answer: 00 We analyze Q() = P () P () One can observe that Q() has the powers of starting from,,,, up to 00 as roots Since Q has degree 0, Q() = A( )( ) ( 00 ) for some A Meanwhile Q(0) = P (0) P (0) = 0, so Q(0) = = A( )( ) ( 00 ) = (00 0)/ A Therefore A = (005 0) Finally, note that the coefficient of is same for P and Q, so it equals A( 0 )( ) ( 00 )(( 0 ) + ( ) + + ( 00 )) = A ( 0 ) = Find the maimum of for reals a, b, c, and d not all zero Answer: 5+ ab + bc + cd a + b + c + d One has ab t a + t b, bc b + c, and cd t c + t d by AM-GM If we can set t such that t = t +, it can be proved that ab+bc+cd a +b +c +d = t, and this is maimal because we can set a, b, c, d so that the equality holds in every inequality we used Solving this equation, we get t = + 5, so the maimum is t = 5+ a +b +c +d t (a +b +c +d )