AMC12 2011 B

AMC12 2011 B · Q22

AMC12 2011 B · Q22. It mainly tests Triangles (properties), Area & perimeter.

Let $T_1$ be a triangle with side lengths $2011$, $2012$, and $2013$. For $n \geq 1$, if $T_n = \triangle ABC$ and $D, E$, and $F$ are the points of tangency of the incircle of $\triangle ABC$ to the sides $AB$, $BC$, and $AC$, respectively, then $T_{n+1}$ is a triangle with side lengths $AD, BE$, and $CF$, if it exists. What is the perimeter of the last triangle in the sequence $\left(T_n\right)$?

设 $T_1$ 是一个边长为 $2011$, $2012$, $2013$ 的三角形。对 $n \geq 1$，若 $T_n = \triangle ABC$，且 $D, E, F$ 分别是 $\triangle ABC$ 的内切圆与边 $AB$, $BC$, $AC$ 的切点，则（若存在）$T_{n+1}$ 是一边长为 $AD, BE, CF$ 的三角形。序列 $\left(T_n\right)$ 中最后一个三角形的周长是多少？

(A) \frac{1509}{8} \frac{1509}{8}

(B) \frac{1509}{32} \frac{1509}{32}

(D) \frac{1509}{128} \frac{1509}{128}

(E) \frac{1509}{256} \frac{1509}{256}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer: (D) Let $AB = c$, $BC = a$, and $AC = b$ Then $AD = AF$, $BE = BD$ and $CF = CE$ Then $a = BE + CF$, $b = AD + CF$, $c = AD + BE$ Hence: Note that $a + 1 = b$ and $a - 1 = c$ for $n = 1$, I claim that it is true for all $n$, assume for induction that it is true for some $n$, then Furthermore, the average for the sides is decreased by a factor of 2 each time. So $T_n$ is a triangle with side length $\frac{2012}{2^{n- 1}} - 1$, $\frac{2012}{2^{n-1}}$, $\frac{2012}{2^{n-1}} + 1$ and the perimeter of such $T_n$ is $\frac{(3)(2012)}{2^{n-1}}$ Now we need to find when $T_n$ fails the triangle inequality. So we need to find the last $n$ such that $\frac{2012}{2^{n-1}} > 2$ For $n = 10$, perimeter is $\frac{(3)(2012)}{2^{9}} = \frac{1509}{2^7} = \frac{1509}{128}$

答案：(D) 设 $AB = c$, $BC = a$, $AC = b$。则 $AD = AF$, $BE = BD$, 且 $CF = CE$。于是 $a = BE + CF$, $b = AD + CF$, $c = AD + BE$。因此：注意当 $n = 1$ 时有 $a + 1 = b$ 且 $a - 1 = c$。我断言对所有 $n$ 都成立；假设对某个 $n$ 成立，用归纳法可得。此外，每次三边的平均值都会缩小为原来的一半。所以 $T_n$ 的边长为 $\frac{2012}{2^{n- 1}} - 1$, $\frac{2012}{2^{n-1}}$, $\frac{2012}{2^{n-1}} + 1$。这样的 $T_n$ 的周长为 $\frac{(3)(2012)}{2^{n-1}}$。现在需要找出 $T_n$ 何时不再满足三角不等式。也就是找出最大的 $n$ 使得 $\frac{2012}{2^{n-1}} > 2$。当 $n = 10$ 时，周长为 $\frac{(3)(2012)}{2^{9}} = \frac{1509}{2^7} = \frac{1509}{128}$。

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