AMC12 2014 B

AMC12 2014 B · Q13

AMC12 2014 B · Q13. It mainly tests Quadratic equations, Triangles (properties).

Real numbers $a$ and $b$ are chosen with $1 < a < b$ such that no triangle with positive area has side lengths $1, a,$ and $b$ or $\frac{1}{b}, \frac{1}{a},$ and $1$. What is the smallest possible value of $b$?

选择实数$a$和$b$，满足$1 < a < b$，使得没有正面积三角形具有边长$1, a,$和$b$或$\frac{1}{b}, \frac{1}{a},$和$1$。$b$的最小可能值是多少？

(A) $3 + \frac{\sqrt{3}}{2}$ $3 + \frac{\sqrt{3}}{2}$

(B) $\frac{5}{2}$ $\frac{5}{2}$

(D) $3 + \frac{\sqrt{6}}{2}$ $3 + \frac{\sqrt{6}}{2}$

(E) 3 3

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): There is a triangle with side lengths $1$, $a$, and $b$ if and only if $a>b-1$. There is a triangle with side lengths $\frac1b$, $\frac1a$, and $1$ if and only if $\frac1a>1-\frac1b$, that is, $a<\frac{b}{b-1}$. Therefore there are no such triangles if and only if $b-1\ge a\ge \frac{b}{b-1}$. The smallest possible value of $b$ satisfies $b-1=\frac{b}{b-1}$, or $b^2-3b+1=0$. The solution with $b>1$ is $\frac12(3+\sqrt5)$. The corresponding value of $a$ is $\frac12(1+\sqrt5)$.

答案（C）：当且仅当 $a>b-1$ 时，存在边长为 $1$、$a$、$b$ 的三角形。当且仅当 $\frac1a>1-\frac1b$（即 $a<\frac{b}{b-1}$）时，存在边长为 $\frac1b$、$\frac1a$、$1$ 的三角形。因此，当且仅当 $b-1\ge a\ge \frac{b}{b-1}$ 时，不存在这样的三角形。$b$ 的最小可能值满足 $b-1=\frac{b}{b-1}$，即 $b^2-3b+1=0$。取满足 $b>1$ 的解为 $\frac12(3+\sqrt5)$。对应的 $a$ 为 $\frac12(1+\sqrt5)$。

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