AMC10 2019 B

AMC10 2019 B · Q4

AMC10 2019 B · Q4. It mainly tests Linear equations, Sequences & recursion (algebra).

All lines with equation $ax+by=c$ such that $a,b,c$ form an arithmetic progression pass through a common point. What are the coordinates of that point?

所有方程 $ax+by=c$ 使得 $a,b,c$ 构成等差数列的直线都经过一个公共点。该点的坐标是什么？

(A) (-1,2) (-1,2)

(B) (0,1) (0,1)

(D) (1,0) (1,0)

(E) (1,2) (1,2)

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): If $a, b, c$ form an arithmetic progression, then $a=b-d$ and $c=b+d$ for some number $d$. Then the given linear equation becomes $(b-d)x+by=b+d$, which is equivalent to $b(x+y-1)-d(x+1)=0$. This will hold for all values of $b$ and $d$ if and only if $x+y-1=0$ and $x+1=0$, which means $x=-1$ and $y=2$. Thus $(-1,2)$ is the unique point through which all such lines pass.

答案 (A)：若 $a,b,c$ 构成等差数列，则存在某个数 $d$ 使得 $a=b-d$ 且 $c=b+d$。则所给线性方程变为 $(b-d)x+by=b+d$，这等价于 $b(x+y-1)-d(x+1)=0$。当且仅当 $x+y-1=0$ 且 $x+1=0$ 时，上式对所有 $b$ 与 $d$ 的取值都成立，这意味着 $x=-1$ 且 $y=2$。因此 $(-1,2)$ 是所有这类直线都经过的唯一点。

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