HMMT 二月 2010 · 冲刺赛 · 第 33 题
HMMT February 2010 — Guts Round — Problem 33
题目详情
英文原题
- [ 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 nn − 1 n
What is the largest positive integer less than a ?
13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUNDth
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 thecorrect answer and A denote your answer. If a described scoring method would ever assign a negativenumber of points, you will receive zero points instead.
解析
英文解析
- [ 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 nn − 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 thatn n n 1
t 2
t = 2 t . Writing the recurrence as a quadratic polynomial in a , we have:nn n − 1 n
2 2
4 a − 10 a a + 4 a + 9 = 0 .
n − 1 nn n − 1
√
5 3
Using the quadratic formula, we see that a = a + a − 4. (We ignore the negative square 2
n n − 1
n − 1
4 4
root, since a is the largest real number satisfying the polynomial.) Substituting t + for a ,1
n n − 1 n − 1
n − 1 t
√
√
2 21
we see that a − 4 = t − 2 + , so we have:
n − 1 n − 12
n − 1 t
√
( ) ( )
5 1 3 1 12
a = t + + t − = 2 t +
n n − 1 n − 1 n − 1
4 t 4 t 2 tn − 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
128(3+ 5)2
√ √
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.
13 ANNUAL HARVARD-MIT MATHEMATICS TOURNAMENT, 20 FEBRUARY 2010 — GUTS ROUNDth