mosp 2004 solution Solutions to the Mathematics Olympiad Test 1

mosp 2004 solution Solutions to the Mathematics Olympiad Test 1

17 MOP 2004 - 2005 Solutions to the Mathematics Olympiad Test 1 1.1. Let n be a positive integer. There are n wood blocks, numbered 1, 2, . . . , n placed on a circle in clockwise order. Zachary is playing the following game. At each step, he is allowed to pick four consecutive blocks and reverse their order. For n = 20, determine if it is possible for Zachary to obtain blocks 6, 1, 2, 3, 4, 5, 7, 8, . . . , 20 in clockwise order in a finite number of steps. (Query: What if n = 19?) Solution: The answer is yes. We use letters a, b, . . . , t instead of numbers 1, 2, . . . , 20. Applying the sequence of moves bcde F → bFedc → deFbc → dcbFe → F bcde repeatedly, one can obtain a bcde F ghi jklm nopq rst → a F bcde ghi jklm nopq rst → a bcde ghi jklm nopq F rst → . . . → abcd eghi F jklm nopq rst → . . . → F abcd eghi jklm nopq rst. Starting with 1, 2, . . . , 20, one ends up exactly with the desired sequence 6, 1, 2, 3, 4, 5, 7, 8, . . . , 20. Note: If you think the bcdeF → Fbcde thing is tricky, you’re wrong. It’s entirely obvious. Given five blocks on a line, only two moves are possible: flip the first four, or flip the last four. Making the same choice twice in a row amounts to doing nothing. Therefore, the only sensible thing to do is to alternate between the two. In the case n = 19, one cannot obtain 6, 1, 2, 3, 4, 5, 7, 8, . . . , 19. The operation abcd → dcba is an even permutation, and so is the operation of cyclically permuting 19 blocks. Because those are the only two types of operations allowed, no odd permutation, such as the one desired, can be obtained. 18 1.2. Let ABCD be a convex quadrilateral with AB not parallel to CD, and let X be a point inside ABCD such that ∠ADX = ∠BCX < 90 ◦ and ∠DAX = ∠CBX < 90 ◦ . If the perpendicular bisectors of segments AB and CD intersect at Y , prove that ∠AY B = 2∠ADX. First Solution: A B C D X W Z Figure 1.1. Construct point Z inside ABCD such that AZ = ZB and ∠AZB = 2∠ADX. It suffices to show that DZ = CZ. Construct W within the non-reflex angle AZB so that triangle AZW is similar to triangle ADX or, equivalently, triangle BZW is similar to triangle BCX. Then triangle ADZ is similar to triangle AXW and triangle BCZ is similar to triangle BXW , implying that DZ XW = AD AX = BC BX = CZ W X , and DZ = CZ, as desired. Second Solution: (By Tony Zhang) Construct point M on the side of line AB opposite X such that triangle MAB is similar to triangle Y DC. Construct point N on the side of line CD opposite X such that triangle NCD is similar to triangle Y BA. Construct point X 1 on the same side of line AD as B such that triangle X 1 AD is similar to triangle CY N, and construct point X 2 on the same side of line BC as D such that triangle X 2 BC is similar to triangle DY N. It suffices to show X 1 = X 2 , because X is the unique point with the properties described in the problem. Note that the triangles AY M, BY M, DY N, CY N, AX 1 D and 19 BX 2 C are all similar. Now use complex numbers; each lowercase letter will denote the complex number coordinate of the point named by the corresponding uppercase letter. Set y = 0 and r = d/n. Then d = rn c = rn a = m −rm b = m −rm and x 1 = a + r(d −a) = m −rm + r(rn − m + rm) = m − rm + r(rn − m + rm) = b + r(c −b) = x 2 , as desired. 1.3. Let n be a positive integer, and let p 1 , p 2 , . . . , p n be distinct primes greater than 3. Prove that 2 p 1 p 2 ···p n + 1 has at least 4 n divisors. Note: We start with the following lemma. Lemma 1. For any odd positive integers a, b, gcd(2 a + 1, 2 b + 1) = 2 gcd(a,b) + 1. Proof: Suppose this were false; choose a counterexample where a + b is minimum. Obviously we must have a = b; assume a > b. If a −2b ≥ 0, 2 a + 1 ≡ 2 a + 1 −(2 b + 1)(2 b − 1)(2 a−2b + 1) ≡ 2 a + 1 −(2 2b − 1)(2 a−2b + 1) ≡ 2 a−2b + 1 (mod 2 b + 1), and if a −2b ≤ 0, 2 2b−a (2 a + 1) ≡ 2 2b−a (2 a + 1) −(2 b + 1)(2 b − 1) ≡ 2 2b−a + 1 (mod 2 b + 1). In either case, gcd(2 a + 1, 2 b + 1) = gcd(2 |a−2b| + 1, 2 b + 1), contradicting the assumption of the minimality of a + b. Please also compare this lemma to lemma 2 in the next session. In this problem, the full generality of lemma 1 is not necessary. 20 The following proofs rely only on the weaker statement that, for any relatively prime odd positive integers a, b not divisible by 3, gcd(2 a + 1, 2 b + 1) = 3. The proof of this is a bit simpler. Suppose that p is prime and 2 a ≡ 2 b ≡ −1 (mod p). Then if c is the minimum positive integer such that 2 c ≡ 1 (mod p), then c must divide both 2a and 2b, so c is either 1 or 2, which implies p = 3. It remains to note that 9 divides neither 2 a + 1 nor 2 b + 1. First Solution: Call an integer “tenebrous” if it is odd, square- free, not divisible by 3, and at least 5. For any integer m, let ψ(m) denote the number of distinct prime factors of m, and let d(m) denote the number of factors of m. We wish to prove that d(2 a + 1) ≥ 4 ψ(a) for all tenebrous integers a. Induct on ψ(a). For the base case ψ(a) = 1, 2 a + 1 is divisible by 3 exactly once and is greater than 3, so ψ(2 a + 1) ≥ 2 and d(2 a + 1) ≥ 4. Now let a, b be relatively prime tenebrous integers such that the claim holds for both a and b. Clearly 2 ab + 1 is divisible by both 2 a + 1 and 2 b + 1, so we can write 2 ab + 1 = C ·lcm[2 a + 1, 2 b + 1]. Because ab −2a −2b −4 = (a −2)(b −2) − 8 > 0, 2 ab + 1 > 2 2a+2b+4 > (2 a + 1) 2 (2 b + 1) 2 > lcm[2 a + 1, 2 b + 1] 2 , so C ≥ lcm[2 a + 1, 2 b + 1]. From the comment, we have gcd(2 a + 1, 2 b + 1) = 3, so as 3 divides each of 2 a + 1 and 2 b + 1 exactly once, d(lcm[2 a + 1, 2 b + 1]) = d(2 a + 1)d(2 b + 1) 2 ≥ 2 2ψ(a)+2ψ(b)−1 . For every divisor m of lcm[2 a + 1, 2 b + 1], both m and Cm are divisors of 2 ab + 1. As C > lcm[2 a + 1, 2 b + 1], d(2 ab + 1) ≥ 2 ·d(lcm[2 a + 1, 2 b + 1]) ≥ 4 ψ(a)+ψ(b) , completing the induction. Second Solution: Following the notation of the first solution, a stronger claim is that, for any tenebrous integer a, ψ(2 a + 1) ≥ 2ψ(a). 21 We proceed by induction on ψ(a). The base case is the same as in the first solution. Now let a, b be coprime tenebrous integers. We claim that ψ(2 ab + 1) ≥ ψ(2 a + 1) + ψ(2 b + 1). Note that 2 ab + 1 2 a + 1 = b  i=1  b i  (−2 a − 1) i−1 ≡ b −  b 2  (2 a + 1) (mod (2 a + 1) 2 ), so if a prime p divides 2 a + 1 exactly k ≥ 1 times, then p divides 2 ab + 1 either k times (if p doesn’t divide b) or k + 1 times (if p divides b). In any case p divides 2 ab + 1 at most twice as many times as p divides 2 a + 1. The same is true for prime factors of 2 b + 1. As in the first solution, 2 ab + 1 > (2 a + 1) 2 (2 b + 1) 2 , so, in light of the above, 2 ab + 1 must have a prime factor dividing neither 2 a + 1 nor 2 b + 1. Clearly 2 ab +1 is divisible by lcm[2 a +1, 2 b +1]. Because 2 ab +1 has a prime factor not dividing lcm[2 a + 1, 2 b + 1], we have ψ(2 ab + 1) ≥ ψ(lcm[2 a + 1, 2 b + 1]) + 1 = ψ(2 a + 1) + ψ(2 b + 1) −ψ(gcd(2 a + 1, 2 b + 1)) + 1 = ψ(2 a + 1) + ψ(2 b + 1) −ψ(3) + 1 = ψ(2 a + 1) + ψ(2 b + 1), completing the induction. 1.4. Let n be an integer greater than 1. Let a 1 , a 2 , . . . , a n , b 1 , b 2 , . . . , b n be nonnegative real numbers with a 1 a 2 ···a n = b 1 b 2 ···b n , b 1 + b 2 + ···+ b n = 1, and  1≤i<j≤n |a i − a j | ≤  1≤i<j≤n |b i − b j |. Determine the maximum value of a 1 + a 2 + ···+ a n . Solution: The answer is n−1, which can be obtained by setting a 1 = 0 and a 2 = ···a n = 1, and b 1 = b 2 = ··· = b n−1 = 0 and 22 b n = 0. Now we show that n  i=1 a i = a 1 + a 2 + ···+ a n ≤ n −1. (∗) Without loss of generality, we assume that a 1 ≤ a 2 ≤ ··· ≤ a n and b 1 ≤ b 2 ≤ ··· ≤ b n . If a 1 = 0, then n  i=1 a i ≤  1≤i<j≤n |a i − a j | ≤  1≤i<j≤n |b i − b j | = n  i=1 (2i −n − 1)b i ≤ (n −1) n  i=1 b i = n −1, establishing the inequality (∗). If a 1 > 0 and n = 2. Then the given inequality becomes a 2 −a 1 ≤ b 2 −b 1 . It follows that (a 1 +a 2 ) 2 = 4a 1 a 2 +(a 1 −a 2 ) 2 ≤ 4a 1 a 2 +(b 2 −b 1 ) 2 = 4b 1 b 2 +(b 2 −b 1 ) 2 = (b 2 +1b 1 ) 2 = 1, establishing the inequality (∗). Now we assume that a 1 > 0 and n > 2. We approach indirectly by assuming that the inequality (∗) is false; that is n  i=1 = a 1 + a 2 + ···+ a n > n −1. (∗ ′ ) We have  1≤i<j≤n |a i − a j | = n  1=1 (2i −n − 1)a i = n  i=1 a i + n  1=1 (2i −n − 2)a i > n −1 + n  i=1 (2i −n − 2)a i . 23 and  1≤i<j≤n |b i − b j | = n  1=1 (2i −n − 1)b i = (n −1) n  i=1 b i − 2 n  1=1 (n −i)b i = n −1 − 2 n−1  i=1 (n −i)b i . Since  1≤i<j≤n |a i −a j | =  1≤i<j≤n |b i −b j |, combining the last two inequalities gives 2 n−1  i=1 (n −i)b i < n  i=1 (n + 2 − 2i)a i = na 1 + (n −2)a 2 + ···+ (2 − n)a n = na 1 − (n −2)(a n − a 2 ) −(n −4)(a n−1 − a 3 ) −··· ≤ na 1 . Consequently, by AM-GM inequality, we obtain na 1 > 2 n−1  i=1 (n −i)b i ≥ 2(n −1) n−1  (n −1)!b 1 b 2 ···b n−1 = 2(n −1) n−1  (n −1)!a 1 a 2 ···a n b n > 2(n −1) n−1  (n −1)!a 1 a 2 ···a n ≥ 2(n −1)a 1 n−1  (n −1)!a n . Therefore, by noting n ≥ 3, a n <  n 2(n−1)  n−1 (n −1)! < n −1 n , implying that  n i=1 a i ≤ na n < n − 1, contradicts to our assumption (∗ ′ ). Thus our assumption was wrong and the desired inequality (∗) holds. 24 Note: One of the key to this solution is always considering the equality case of the inequality (∗) by taking out terms a i −a j for 1 < i < j ≤ n. 25 Some Important Ideas You Should Know F1.1. To clip a convex n-gon means to choose a pair of consecutive sides AB, BC and to replace them by the three segments AM, MN , and NC, where M is the midpoint of AB and N is the midpoint of BC. In other words, one cuts off the triangle MBN to obtain a convex (n + 1)-gon. A regular hexagon P 6 of area 1 is clipped to obtain a heptagon P 7 . Then P 7 is clipped (in one of the seven possible ways) to obtain an octagon P 8 , and so on. Prove that no matter how the clippings are done, the area of P n is greater than 1/3, for all n ≥ 6. Solution: The key observation is that for any side S of of P 6 , there is some sub-segment of S that is a side of P n . (This is easily proved by induction on n.) Thus P n has a vertex on each side of P 6 . Since P n is convex, it contains a hexagon Q with (at least) one vertex on each side of P 6 . (The hexagon may be degenerate, as some of its vertices may coincide.) Let P 6 = A 1 A 2 A 3 A 4 A 5 A 6 , and let Q = B 1 B 2 B 3 B 4 B 5 B 6 , with B i on A i A i+1 (indices are considered modulo 6). The side B i B i+1 of Q is entirely contained in triangle A i A i+1 A i+2 , so Q encloses the smaller regular hexagon R (shaded in the diagram below) whose sides are the central thirds of the segments A i A i+2 , 1 ≤ i ≤ 6. The area of R is 1/3, as can be seen from the fact that its side length is 1/ √ 3 times the side length of P 6 , or from a dissection argument (count the small equilateral triangles and halves thereof in the diagram below). Thus [P n ] ≥ [Q] ≥ [R] = 1/3. We obtain strict inequality by observing that mathcalP n is strictly larger that Q: if n = 6, this is obvious; if n > 6, then P n cannot equal Q because P n has more sides. Note: With a little more work, one could improve 1/3 to 1/2. The minimal area of a hexagon Q with one vertex on each side of P 6 is in fact 1/2, attained when the vertices of Q coincide in pairs at every other vertex of P 6 , so the hexagon Q degenerates into an equilateral triangle. If the conditions of the problem were changed so that the cut-points could be anywhere within adjacent segments instead of just at the midpoints, then the best possible bound would be 1/2. 26 F1.2. Find all integers n for which a regular hexagon can be divided into n parallelograms of equal area. F1.3. Let n be a positive integer greater than or equal to three. (a) Let C be a circle with center O and radius 1, and let S 1 , S 2 , . . . S n be n semicircles with center O and radius 1. Circle C is covered by semicircles S 1 , S 2 , . . . , and S n . Prove that one can find indices i 1 , i 2 , and i 3 such that sphere C is covered by S i 1 , S i 2 , and S i 3 . (b) Let S be a sphere with center O and radius 1, and let H 1 , H 2 , . . . H n+1 be n + 1 hemispheres with center O and radius 1. Sphere S is covered by hemispheres H 1 , H 2 , . . . , and H n . Prove that one can find indices i 1 , i 2 , i 3 , i 4 such that sphere S is covered by H i 1 , H i 2 , H i 3 , H i 4 . F1.4. Let P be a 1000-sided regular polygon. Some of its diagonals were drawn to obtain a triangulation the polygon P . (The region inside P is cut into triangular regions, and the diagonals drawn only intersect at the vertices of P .) Let n be the number of different lengths of the drawn diagonals. Determine the minimum value of n. (By Tiankai Liu) The answer is 10. This can be achieved by using diagonals corresponding to arc lengths of 400, 200, 120, 80, 40, 24, 16, 8, 4, and 2. First we can draw a regular pentagon with sides subtending arcs of length 200; the interior of this pentagon can be triangulated with diagonals subtending arcs of length 400. Then we divide each the remaining five congruent 201-gons into five congruent 41-gons using diagonals subtending arcs of length 40; the interiors of the hexagons thus formed can be triangulated by diagonals subtending arcs of lengths 80 and 120. Proceeding in a sim- ilar manner, We divide each of the 25 congruent 41-gons into five nonagons using diagonals subtending arcs of length 8; the interiors of the hexagons thus formed can be triangulated by diagonals subtending arcs of lengths 16 and 24. Each of the remaining nonagons can be triangulated in a straightforward way with diagonals subtending arcs of lengths 8, 4, and 2. Suppose, for the sake of contradiction, that the answer is less than 10. Let the arc lengths corresponding to the different [...]... are +1 Otherwise, the entries of the center row, starting from the center entry and working toward either side, must be +1, 1, 1, +1, 1, 1, +1, , 1, 1, +1, 1, 1 But then the width of the matrix is 2n +1 − 1 ≡ 2 (mod 3), which is impossible It follows that the center row must contain only +1 s, and likewise for the center column We may now observe that each of the four remaining (2n − 1) ×... 46 with: y1 + y2 + yn = 0 y1 + 2y2 + nyn = 0 y1 + 4y2 + n2 yn = 0 y1 + 8y2 + n3 yn = 0 But 1 1 1 1 2 n 1 4 n2 = 0, thus the only solution to this system is y1 = y2 = yn = 0 Therefore, the only solution to the original problem is just (n, 1, 0, , 0, 1) as desired Third Solution: Subtracting the first given equation by the second given equation, the third by the second, and the fourth by the third,... xn 1 = 0 Then the given xk k =1 45 equations become x1 + x2 + xn = n + 2, x1 + 2x2 + nxn = 2n + 2, x1 + 22 x2 + n2 xn = n2 + n + 4 It is not difficult to solve the above system of equations to obtain (x1 , x2 , xn ) = (n, 1, 1) Second Solution: Define y1 = x1 − n, y2 = x2 − 1, yn = xn − 1, and yi = 0 for 3 ≤ i ≤ n − 1 Then we have (y1 + n) + (y2 + 1) + (yn + 1) = n + 2, (y1 + n) + 2(y2 + 1) + n(yn + 1) ... big as the place value of the highest 1 in the binary representation of 10 00 But ai1 , ai2 , ai3 ≤ a9 ≤ 384 < 512 , a contradiction Therefore 10 is indeed the answer 28 MOP 2004 - 2005 Solutions to the Mathematics Olympiad Test 2 2 .1 Let m be a fixed positive integer with m ≥ 3 A set S is good if there are elements s1 , s2 , , sm 1 , sm (not necessarily distinct) in S such that s1 +s2 +· · ·+sm 1 =... Figure 2 .1 U P Y A D B ( B 1) Q X C ( C 1) Z T W Figure 2 .1 First Solution: Consider the dilation T taking 1 to ω2 It is clear that T is the center of T Assume that line DT meet 1 and ω2 again (other than D) at B1 and C1 , respectively Then T(B1 ) = D, T(D) = C1 , and T(B1 D) = DC1 , implying that ∠B1 AD = ∠DAC1 On the other hand, consider a point B ′ moving along omega1 from D to A, line B ′ D meet... q | prp 1 Therefore, q = p and q | pq 1 1 by Fermat’s theorem Applying lemma 2, we have q | pgcd(q 1, rp) − 1 This means that either p | q − 1 or q | pr − 1 However, q | F (r) | prp − 1 = pr − 1 p 1 i=0 r pir ≡ p (mod pr − 1) This means, q cannot divide p − 1 and p | q − 1 Now we know that all prime divisors of F (r) are equivalent to 1 modulo p We need to find one that is not equivalent to 1 modulo... remains to show that any partition {1, 2, , m2 − m − 1} = A ∪ B contains at least one good set Assume the contrary, and assume without loss of generality that 1 ∈ A; then m − 1 ∈ B and (m 1) 2 ∈ A Also, m2 −m 1 = (m 1) 2 +(m−2) = (m 1) 2 +1+ 1+ · · · +1 ∈ B, and m+m+· · ·+m +1 = (m−2)m +1 = (m 1) 2 , so m ∈ B But then m+m+· · ·+m+(m 1) = (m−2)m+(m 1) = m2 − m − 1 is a sum of m − 1 elements of B that is itself... (r), F (s)) = 1 Proof: By lemma 2 gcd(prp − 1, psp − 1) = pp − 1 Therefore, rp sp rp rp 1 1 1 1 gcd( p p 1 , p p 1 ) = 1 Since F (r) | p p 1 and F (s) | p p 1 , p p p p gcd(F (r), F (s)) = 1 Lemma 4 For any prime r = p there is a prime divisor q of F (r) such that p | q − 1 but p2 |q − 1 Proof: Clearly, F (r) > 1 First we will show that any prime divisor q of F (r) is equivalent to 1 modulo p... 19 97 Show that there exists a real number x such that an = ⌊nx⌋ (the greatest integer ≤ nx) for all 1 ≤ n ≤ 19 97 Solution: Any x that lies in all of the half-open intervals In = an an + 1 , n n , n = 1, 2, , 19 97 will have the desired property Let L= max 1 n 19 97 an ap = n p and U = min 1 n 19 97 We shall prove that am + 1 an < , n m an + 1 aq + 1 = n q 41 or, equivalently, man < n(am + 1) (∗) for all... assume that m > n gcd(am − 1, an − 1) = gcd(am − an , an − 1) = gcd(am−n − 1, an − 1) = agcd(m−n,n) = agcd(m,n) as m − n + n < m + n and gcd(m − n, n) = gcd(m, n) This completes the inductive step (prp 1) (p 1) For each prime r = p let F (r) = (pp 1) (pr 1) By lemma 2 gcd(pr − 1, pp − 1) = p − 1 Since both pr − 1 and pp − 1 divide prp − 1, their product divides (prp − 1) (p − 1) Therefore, F (r) is an . 17 MOP 2004 - 2005 Solutions to the Mathematics Olympiad Test 1 1.1. Let n be a positive integer. There are n. ·d(lcm[2 a + 1, 2 b + 1]) ≥ 4 ψ(a)+ψ(b) , completing the induction. Second Solution: Following the notation of the first solution, a stronger claim is that, for any tenebrous integer a, ψ(2 a +. a i 2 , a i 3 ≤ a 9 ≤ 384 < 512, a contradiction. Therefore 10 is indeed the answer. 28 MOP 2004 - 2005 Solutions to the Mathematics Olympiad Test 2 2.1. Let m be a fixed positive integer with m