AMC12 2014 A

AMC12 2014 A · Q25

AMC12 2014 A · Q25. It mainly tests Systems of equations, Coordinate geometry.

The parabola $P$ has focus $(0,0)$ and goes through the points $(4,3)$ and $(-4,-3)$. For how many points $(x,y)\in P$ with integer coordinates is it true that $|4x+3y|\leq 1000$?

抛物线 $P$ 的焦点为 $(0,0)$，并经过点 $(4,3)$ 和 $(-4,-3)$。有几个点 $(x,y)\in P$ 满足整数坐标且 $|4x+3y|\leq 1000$？

(A) 38 38

(B) 40 40

(D) 44 44

(E) 46\qquad 46\qquad

Answer

Correct choice: (B)

正确答案：（B）

Solution

The parabola is symmetric through $y=- \frac{4}{3}x$, and the common distance is $5$, so the directrix is the line through $(1,7)$ and $(-7,1)$, which is the line \[3x-4y = -25.\] Using the point-line distance formula, the parabola is the locus \[x^2+y^2 = \frac{\left\lvert 3x-4y+25 \right\rvert^2}{3^2+4^2}\] which rearranges to $(4x+3y)^2 = 25(6x-8y+25)$. Let $m = 4x+3y \in \mathbb Z$, $\left\lvert m \right\rvert \le 1000$. Put $m = 25k$ to obtain \[25k^2 = 6x-8y+25\]\[25k = 4x+3y.\] and accordingly we find by solving the system that $x = \frac{1}{2} (3k^2-3) + 4k$ and $y = -2k^2+3k+2$. One can show that the values of $k$ that make $(x,y)$ an integer pair are precisely odd integers $k$.* For $\left\lvert 25k \right\rvert \le 1000$ this is $k= -39,-37,-35,\dots,39$, so $40$ values work and the answer is $\boxed{\textbf{(B)}}$.

抛物线关于直线 $y=- \frac{4}{3}x$ 对称，公共距离为 $5$，故准线是通过 $(1,7)$ 和 $(-7,1)$ 的直线，即 \[3x-4y = -25.\] 使用点到直线距离公式，抛物线为轨迹 \[x^2+y^2 = \frac{\left\lvert 3x-4y+25 \right\rvert^2}{3^2+4^2}\]，整理得 $(4x+3y)^2 = 25(6x-8y+25)$。令 $m = 4x+3y \in \mathbb Z$，$\left\lvert m \right\rvert \le 1000$。置 $m = 25k$，得 \[25k^2 = 6x-8y+25\]\[25k = 4x+3y.\] 解方程组得 $x = \frac{1}{2} (3k^2-3) + 4k$，$y = -2k^2+3k+2$。可以证明使 $(x,y)$ 为整数对的 $k$ 恰为奇整数。$\left\lvert 25k \right\rvert \le 1000$ 时，$k= -39,-37,-35,\dots,39$，共 $40$ 个值，故答案为 $\boxed{\textbf{(B)}}$。

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