2014 AMC 12A 第 25 题

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25.

抛物线 PP 的焦点为 (0,0)(0,0),并经过点 (4,3)(4,3)(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+3y1000|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+3y1000?|4x+3y|\le1000?

3838

4040

4242

4444

4646

答案:B
知识点:抛物线格点丢番图方程
难度评级:2650
小提示:

(0,0)(0,0)(4,3)(4,3)(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 4y3x+25=04y-3x+25=0;参数化格点后,把 4x+3y|4x+3y| 化为 50u+251000|50u+25|\le1000

The directrix is 4y3x+25=04y-3x+25=0; parametrize the lattice points and reduce 4x+3y|4x+3y| to 50u+251000|50u+25|\le1000

解答:

因为 (0,0)(0,0)A=(4,3)A=(4,3)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 4y3x+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(4y3x))=25\big(25+2(4y-3x)\big)。令 4x+3y=5s4x+3y=5s,可推出 ss55 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+251000|4x+3y|=|50u+25|\le1000 等价于 2u+139|2u+1|\le39,即 20u19-20\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 4y3x+25=0.4y-3x+25=0.

Equating distances to focus and directrix gives (4x+3y)2(4x+3y)^2 =25(25+2(4y3x)).=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+251000|4x+3y|=|50u+25|\le1000 iff 2u+139,|2u+1|\le39, i.e. 20u19.-20\le u\le19. That gives 4040 lattice points.

Thus, the correct answer is B.

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