2015 AIME II 第 6 题

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6.

Steve 对 Jon 说:“我在想一个多项式,它的根全都是正整数。这个多项式形如 P(x)=2x32ax2P(x) = 2x^3 - 2ax^2 +(a281)xc+ (a^2 - 81)x - c,其中 aacc 是正整数。你能告诉我 aacc 的值吗?”

经过一些计算后,Jon 说:“这样的多项式不止一个。”

Steve 说:“你说得对。这是 aa 的值。”他写下一个正整数并问:“你能告诉我 cc 的值吗?”

Jon 说:“cc 仍然有两个可能值。”

求这两个可能的 cc 值之和。

Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form P(x)=2x32ax2P(x) = 2x^3 - 2ax^2 +(a281)xc+ (a^2 - 81)x - c for some positive integers aa and c.c. Can you tell me the values of aa and c?c?"

After some calculations, Jon says, "There is more than one such polynomial."

Steve says, "You're right. Here is the value of a.a." He writes down a positive integer and asks, "Can you tell me the value of c?c?"

Jon says, "There are still two possible values of c.c."

Find the sum of the two possible values of c.c.

答案:440
知识点:韦达定理多项式逻辑推理分类讨论
难度评级:2500
解答:

除以 22 后,根 rstr \le s \le t 满足 r+s+t=ar + s + t = ars+rt+st=a2812rs + rt + st = \frac{a^2 - 81}{2},以及 rst=c2rst = \frac{c}{2}。因此 r2+s2+t2=(r+s+t)22(rs+rt+st)=a2(a281)=81. \begin{aligned} &r^2 + s^2 + t^2 \\ &= (r + s + t)^2 - 2(rs + rt + st) \\ &= a^2 - (a^2 - 81) \\ &= 81. \end{aligned}

平方和为 8181 的正整数三元组为 (1,4,8)(1, 4, 8)(4,4,7)(4, 4, 7), 和 (3,6,6)(3, 6, 6),对应的 a=r+s+ta = r + s + t 分别为 13131515, 和 1515。 因为知道 aa 后 Jon 仍有两个选择,所以 a=15a = 15,两个多项式来自 (4,4,7)(4, 4, 7)(3,6,6)(3, 6, 6)

对应的 c=2rstc = 2rst 分别为 2447=2242 \cdot 4 \cdot 4 \cdot 7 = 2242366=2162 \cdot 3 \cdot 6 \cdot 6 = 216,和为 224+216=440224 + 216 = 440

Dividing by 2,2, the roots rstr \le s \le t satisfy r+s+t=a,r + s + t = a, rs+rt+st=a2812,rs + rt + st = \frac{a^2 - 81}{2}, and rst=c2.rst = \frac{c}{2}. Therefore r2+s2+t2=(r+s+t)22(rs+rt+st)=a2(a281)=81. \begin{aligned} &r^2 + s^2 + t^2 \\ &= (r + s + t)^2 - 2(rs + rt + st) \\ &= a^2 - (a^2 - 81) \\ &= 81. \end{aligned}

The triples of positive integers whose squares sum to 8181 are (1,4,8),(1, 4, 8), (4,4,7),(4, 4, 7), and (3,6,6),(3, 6, 6), with a=r+s+ta = r + s + t equal to 13,13, 15,15, and 15.15. Since knowing aa still left Jon two choices, a=15,a = 15, and the two polynomials come from (4,4,7)(4, 4, 7) and (3,6,6).(3, 6, 6).

The corresponding values of c=2rstc = 2rst are 2447=2242 \cdot 4 \cdot 4 \cdot 7 = 224 and 2366=216,2 \cdot 3 \cdot 6 \cdot 6 = 216, with sum 224+216=440.224 + 216 = 440.

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