Poniższy zbiór zadań został wykonany w ramach projektu Mazowiecki program stypendialny dla uczniów szczególnie uzdolnionych - najlepsza inwestycja w człowieka w roku szkolnym 2018/2019. Składają się na niego wybrane zadania z Internetowego Koła Matematycznego, rozwiązane i przetłumaczone na język angielski. Są to zestawy zadań na poziomie Olimpiady Matematycznej Juniorów, mające przygotować młodzież do podejmowania trudnych matematycznych wyzwań, w szczególności do startu w ogólnopolskiej Olimpiadzie Matematycznej. Praca wykonana pod kierunkiem mgr Adrianny Żołnierczuk. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II
Worksheet 6, excercise 1 Find all trios (x,y,z) of real numbers that are solutions to the equation: Answer: Let us transform this equation: 5(x 2 + y 2 + z 2 ) = 4(xy + yz + xz). 5(x 2 + y 2 + z 2 ) 4(xy + yz + xz) = 0, 4x 2 + y 2 4xy + 4y 2 + z 2 4yz + 4z 2 + x 2 4xz = 0, (2x y) 2 + (2y z) 2 + (2z x) 2 = 0. Since the square of the real number is always non-negative, total of squares is equal to zero if and only if all the numbers are zeroes, so (2x y) 2 = 0, (2y z) 2 = 0 and (2z x) 2 = 0, therefore (2x y) = 0 (2y z) = 0 (2z x) = 0. Then we have: 2x y = 2y z, which means 2x = 3y z, so x = 3y z. 2 Now 2y z = 2z 3y z, that is 4y 2z = 4z 3y + z and 7y = 7z, so y = z. 2 Then instead of 2x y = 2y z we can write 2x z = 2z z, which gives us 2x = 2z, so x = z = y. We have 2z x = 0 and z = x, so x = 0 = z = y. The only trio (x,y,z) fulfilling this equation is (x, y, z) = (0, 0, 0). Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 2
Worksheet 6, excercise 5 Let p be a prime number. For which p isn t divisible by 360? p 4 5p 2 + 4 Answer: Let us transform equivalently the expression: p 4 5p 2 + 4 = (p 4 4p 2 + 4) p 2 = (p 2 2) 2 p 2 = (p 2 2 + p)(p 2 2 p) = = ((p 2)(p + 1))((p + 2)(p 1)) = (p 2)(p 1)(p + 1)(p + 2). These are (along with p) 5 consecutive integers, so one of them is divisible by 5. Let us notice that 360 = 2 3 3 2 5, so if p is divisible by 5, none of the numbers p 2, p 1, p + 1, p + 2 is divisible by 5. Then (p 2)(p 1)(p + 1)(p + 2) is divisible neither by 5, nor 360. If p isn t divisible by 5, then (p 2)(p 1)(p + 1)(p + 2) is divisible by 5. We need to consider if (p 2)(p 1)(p + 1)(p + 2) is also divisible by 2 3 and 3 2. For p = 2k + 1 we have p 1 divisible by 2 and p + 1 divisible by 2, so either p 1 or p + 1 is divisible by 4, therefore their product is divisible by 2 3. For p = 2k we have p = 2, because 2 is the only even prime number. Then (p 2)(p 1)(p + 1)(p + 2) = 0 1 3 4 = 0, this product is divisible by 2 3. There is still a question: is (p 2)(p 1)(p + 1)(p + 2) divisible by 3 2? If p isn t divisible by 3 at least 2 out of these 4 numbers (p 2), (p 1), (p + 1), (p + 2) are divisible by 3, so the whole product is divisible by 9. For p = 3, (p 2)(p 1)(p + 1)(p + 2) = 1 2 4 5 = 40 and this is obviously not divisible by 360. Ergo, p 4 5p 2 + 4 isn t divisible by 360 only when p = 3 or p = 5. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 3
Worksheet 7, excercise 1 We have such numbers a, b and c that a + b + c = 20 and 1 + 1 + 1 = 1. Find the sum: a+b b+c c+a 4 w = Answer: Let us note that c a + b + a b + c + 1 (a + b + c)( a + b + 1 b + c + 1 c + a ) = a + b + c a + b b c + a. + a + b + c b + c + a + b + c a + c = c a + b + a + b a + b + a b + c + b + c b + c + b c + a + c + a c + a = 1 + 1 + 1 + w = w + 3, but on the other hand Then w = 5 3 = 2. 1 (a + b + c)( a + b + 1 b + c + 1 c + a ) = 20 1 4 = 5. = Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 4
Worksheet 7, excercise 4 Let us take positive numbers a, b, c, d, such that a b c d and a + b + c + d 1. Prove that a 2 + 3b 2 + 5c 2 + 7d 2 1. Answer: Both a + b + c + d and 1 are positive, so we can square sides of this inequality: (a + b + c + d) 2 1 2 = 1, a 2 + b 2 + c 2 + d 2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd 1. There is a b and b > 0, so ab b 2, therefore 2ab 2b 2. Similarly 2ac c 2, 2ad 2d 2, 2bc 2c 2, 2bd 2d 2 and 2cd 2d 2. After adding up all these inequations we get: 2b 2 + 4c 2 + 6d 2 2(ab + ac + ad + bc + bd + cd). Now we can make use of all our conclusions and get a 2 + 3b 2 + 5c 2 + 7d 2 = a 2 + b 2 + c 2 + d 2 + 2b 2 + 4c 2 + 6d 2 QED a 2 + b 2 + c 2 + d 2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd 1. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 5
Worksheet 8, excercise 1 Show that for every trio of real numbers (a, b, c) there is an equation: w = (a b b a )(b c c b )(c a a c ) = 0. Answer: 1) When a, b or c = 0: Without the loss of generality, we can assume that a = 0, then (a b b a ) = 0 b b 0 = 0 0 = 0, so w = 0 (b c c b )(c a a c ) = 0. 2) Neither a, b nor c = 0: Then at least two of the numbers (a, b, c) are both positive or negative. Without the loss of generality we can assume that these two numbers are a and b. 2a) a and b are positive a = a and b = b, so a b b a = ab ba = 0, therefore w = 0 (b c c b )(c a a c ) = 0. 2b) a and b are negative a = a and b = b, so a b b a = ab ( ba) = ab + ab = 0, therefore w = 0 (b c c b )(c a a c ) = 0. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 6
Worksheet 8, excercise 2 Find all natural numbers n that meet the inequalities 2010 < n 10 < 2011. n + 10 n 10 Answer: Obviously 2010 n+10 > 0 and so are the other sides of this inequation, so we can square them and get: 2010 2 20112 < n 10 < n + 10 n 10. Since n is a natural number, n 0, therefore n+10 > 0, ergo we can multiply our inequation by n+10: 2010 2 < (n 10)(n + 10) < 2011 2, 2010 2 < n 2 100 < 2011 2, 2010 2 + 100 < n 2 < 2011 2 + 100, 2010 2 < 2010 2 + 100 < n 2 < 2011 2 + 100 < 2012 2. So the only natural number meeting this inequality is n 2 = 2011 2, n = 2011. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 7
Worksheet 8, excercise 4 Find, whether there exist 5 consecutive integer numbers, which squared and summed give a squared integer number. Answer: Let n be an integer number. The question is: is there an integer number a fulfilling the following equation: (a 2) 2 + (a 1) 2 + (a) 2 + (a + 1) 2 + (a + 2) 2 = n 2? If so, there must be (a 2 + 4 4a) + (a 2 + 1 2a) + a 2 + (a 2 + 1 + 2a) + (a 2 + 4 + 4a) = n 2, in other words: 5a 2 + 10 = n 2. Let s find out what are the possible remainders of squared integer number divided by 4: 1) if n 0 (mod 4), then n 2 0 (mod 4) 2) if n 1 (mod 4), then n 2 1 (mod 4) 3) if n 2 (mod 4), then n 2 4 0 (mod 4) 4) if n 3 (mod 4), then n 2 9 1 (mod 4) So the possible remainders are 1 and 0. Now let s check what remainders of division by 4 can 5a 2 + 10 give: 1) if a 0 (mod 4), then 5a 2 + 10 10 2 (mod 4) 2) if a 1 (mod 4), then 5a 2 + 10 5 + 10 = 15 3 (mod 4) 3) if a 2 (mod 4), then 5a 2 + 10 0 + 10 = 30 2 (mod 4) 4) if a 3 (mod 4), then 5a 2 + 10 45 + 10 = 55 3 (mod 4) So 5a 2 +10 is not any integer squared, ergo there don t exist any 5 consecutive integer numbers, which squared and summed give squared integer number. Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 8
Worksheet 9, excercise 1 Determine all trios of integer numbers fulfilling this system of equations: { a + bc = 2012 ab + c = 2011. Answer: Let s substract the secong equation and transform the expression: then a + bc (ab + c) = 2012 2011 = 1, a ab + bc c = 1, a(1 b) + c(b 1) = 1, a(1 b) c(1 b) = 1, (a c)(1 b) = 1. Since a, b, c are integer numbers, then both (a c) and (1 b) equal 1 or 1. (1) If a c = 1 = 1 b, then 1 b = 1, so b = 0. When a c = 1, we have (1 + c) + 0 c = 2012, so c = 2011 and a = 1 + 2011 = 2012. (2) If a c = 1 = 1 b, then 1 b = 1, co b = 2. When a = 1+c = c 1, we have (c 1)+2 c = 2012, so 3c = 2013, c = 671 and a = c 1 = 671 1 = 670. The only trios fulfilling this system of equations are (a, b, c) = (2012, 0, 2011) and (a, b, c) = (670, 2, 671). Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 9
Worksheet 10, excercise 1 Solve the system of equations: (x + y)(x + y + z) = 72 (y + z)(x + y + z) = 120 (z + x)(x + y + z) = 96. Answer: Let s define a = x + y + z. After adding up these 3 equations we get: Then (x + y) a + (y + z) a + (z + x) a = 72 + 120 + 96 = 288. (x + y + y + z + z + x) a = 288, (2(x + y + z)) a = 288, 2a a = 288, a 2 = 144, So a = 12 or a = 12. If a = 12, then x + y + z = 12. From the first equation we get x + y = 72 x + y + z = 72 12 = 6, z = (x + y + z) (x + y) = 12 6 = 6. Now, from the third equation we get: x + z = 96 x + y + z = 96 12 = 8, x = (x + z) z = 8 6 = 2, y = (x + y + z) x z = 12 2 6 = 4. So one of the solution is (x, y, z)=(2, 4, 6) and the opposite one is (x, y, z)=(-2, -4, -6). Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 10
Worksheet 10, excercise 2 Let a, b be real numbers. Prove that a 2 + b 2 + 1 ab + a + b. Answer: The square of the real number is always non-negative, so (a b) 2 0, therefore a 2 + b 2 2ab, (b 1) 2 0, therefore b 2 + 1 2 2b, (a 1) 2 0, therefore a 2 + 1 2 2a, After adding up all these 3 inequations we get: 2(a 2 + b 2 + 1) 2(ab + a + b). We divide both sides of this inequation by 2: a 2 + b 2 + 1 ab + a + b. QED Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 11
Worksheet 10, excercise 4 Can these two equations coexist at the same time? y(4y 3x) = 2 and x(y x) = 3 Answer: If both of them are true then: y(4y 3x) x(y x) = 2 3 = 1, 4y 2 3xy xy + x 2 = 1, x 2 + (2y) 2 4xy = 1, (x 2y) 2 = 1. That is impossible, because the square of the real number is always nonnegative, so these two equations cannot coexist at the same time. QED Szkoła Podstawowa nr 112 Przymierza Rodzin im. Jana Pawła II 12