2014 AMC 12A 第 25 题

先试着解答 2014 AMC 12A 第 25 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2014 AMC 12A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

25.

抛物线 PP 的焦点为 (0,0)(0,0),并经过点 (4,3)(4,3) 与 (−4,−3)(-4,-3)。有多少个满足 (x,y)∈P(x,y)\in P、坐标均为整数且 ∣4x+3y∣≤1000|4x+3y|\le1000 的点?
The parabola PP has focus (0,0)(0,0) and goes through the points (4,3)(4,3) and (−4,−3).(-4,-3). For how many points (x,y)∈P(x,y)\in P with integer coordinates is it true that ∣4x+3y∣≤1000?|4x+3y|\le1000?
3838
4040
4242
4444
4646
答案:B
知识点:抛物线格点丢番图方程
小提示:
(0,0)(0,0) 是 (4,3)(4,3) 与 (−4,−3)(-4,-3) 的中点,所以这条弦是通径,准线与它平行
(0,0)(0,0) is the midpoint of (4,3)(4,3) and (−4,−3),(-4,-3), so this chord is the latus rectum and the directrix is parallel to it
大提示:
准线为 4y−3x+25=04y-3x+25=0;参数化格点后,把 ∣4x+3y∣|4x+3y| 化为 ∣50u+25∣≤1000|50u+25|\le1000。
The directrix is 4y−3x+25=04y-3x+25=0; parametrize the lattice points and reduce ∣4x+3y∣|4x+3y| to ∣50u+25∣≤1000|50u+25|\le1000
解答:
因为 (0,0)(0,0) 是 A=(4,3)A=(4,3) 与 B=(−4,−3)B=(-4,-3) 的中点,所以线段 ABAB 是通径。准线与 ABAB 平行,并位于焦点另一侧、相距 55,其方程为 4y−3x+25=04y-3x+25=0。
令点到焦点和准线的距离相等,得到 (4x+3y)2(4x+3y)^2 =25(25+2(4y−3x))=25\big(25+2(4y-3x)\big)。令 4x+3y=5s4x+3y=5s,可推出 ss 是 55 的倍数;再令 s=5ts=5t,可推出 tt 为奇数。写成 t=2u+1t=2u+1 后,所有整数点可表示为 x=−6u2+2u+4,y=8u2+14u+3。 \begin{aligned} x&=-6u^2+2u+4,\\ y&=8u^2+14u+3 \end{aligned}\text{。}
于是 ∣4x+3y∣=∣50u+25∣≤1000|4x+3y|=|50u+25|\le1000 等价于 ∣2u+1∣≤39|2u+1|\le39,即 −20≤u≤19-20\le u\le19。因此共有 4040 个整数坐标点。
所以正确答案是 B。
Since (0,0)(0,0) is the midpoint of A=(4,3)A=(4,3) and B=(−4,−3),B=(-4,-3), the segment ABAB is the latus rectum, so the directrix is parallel to ABAB at distance 55 on the far side, namely 4y−3x+25=0.4y-3x+25=0.
Equating distances to focus and directrix gives (4x+3y)2(4x+3y)^2 =25(25+2(4y−3x)).=25\big(25+2(4y-3x)\big). Writing 4x+3y=5s4x+3y=5s forces ss to be a multiple of 5,5, and s=5ts=5t forces tt odd; with t=2u+1t=2u+1 the integer points are x=−6u2+2u+4,y=8u2+14u+3. \begin{aligned} x&=-6u^2+2u+4,\\ y&=8u^2+14u+3. \end{aligned}
Then ∣4x+3y∣=∣50u+25∣≤1000|4x+3y|=|50u+25|\le1000 iff ∣2u+1∣≤39,|2u+1|\le39, i.e. −20≤u≤19.-20\le u\le19. That gives 4040 lattice points.
Thus, the correct answer is B.
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