HMMT 二月 2010 · 冲刺赛 · 第 33 题
HMMT February 2010 — Guts Round — Problem 33
题目详情
- [ 21 ] Let a = 3, and for n > 1, let a be the largest real number such that 1 n ( ) 2 2 4 a + a = 10 a a − 9 . n − 1 n n − 1 n What is the largest positive integer less than a ? 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND Note: Problems 34 and 36 are estimation problems. You will receive a number of points (between 0 and 25) based on how close your answer is to the correct answer. Throughout, we will let C denote the correct answer and A denote your answer. If a described scoring method would ever assign a negative number of points, you will receive zero points instead.
解析
- [ 21 ] Let a = 3, and for n > 1, let a be the largest real number such that 1 n ( ) 2 2 4 a + a = 10 a a − 9 . n − 1 n n − 1 n What is the largest positive integer less than a ? 8 √ 1 3+ 5 Answer: 335 Let t be the larger real such that a = t + . Then t = . We claim that n n n 1 t 2 n t = 2 t . Writing the recurrence as a quadratic polynomial in a , we have: n n − 1 n 2 2 4 a − 10 a a + 4 a + 9 = 0 . n − 1 n n n − 1 √ 5 3 2 Using the quadratic formula, we see that a = a + a − 4. (We ignore the negative square n n − 1 n − 1 4 4 1 root, since a is the largest real number satisfying the polynomial.) Substituting t + for a , n n − 1 n − 1 t n − 1 √ √ 1 2 2 we see that a − 4 = t − 2 + , so we have: 2 n − 1 n − 1 t n − 1 √ ( ) ( ) 2 5 1 3 1 1 a = t + + t − = 2 t + n n − 1 n − 1 n − 1 4 t 4 t 2 t n − 1 n − 1 n − 1 √ 128(3+ 5) 2 √ so t = 2 t , as claimed. Then a = + . The second term is vanishingly small, n n − 1 8 2 128(3+ 5) √ √ so b a c = b 64(3 + 5) c . We approximate 5 to two decimal places as 2 . 24, making this expression 8 √ b 335 . 36 c = 335. Since our value of 5 is correct to within 0 . 005, the decimal is correct to within 0 . 32, which means the final answer is exact. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th 13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUND