AMC12 2022 A

AMC12 2022 A · Q25

AMC12 2022 A · Q25. It mainly tests Factoring, Circle theorems.

A circle with integer radius $r$ is centered at $(r, r)$. Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$, $b_i$, and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$. What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$?

半径为整数 $r$ 的圆心在 $(r, r)$。不同的长度为 $c_i$ 的线段连接点 $(0, a_i)$ 到 $(b_i, 0)$ 对于 $1 \le i \le 14$，且与圆相切，其中 $a_i$、$b_i$ 和 $c_i$ 均为正整数，且 $c_1 \le c_2 \le \cdots \le c_{14}$。对于 $r$ 的最小可能值，求比例 $\frac{c_{14}}{c_1}$？

(A) \frac{21}{5} \frac{21}{5}

(B) \frac{85}{13} \frac{85}{13}

(D) \frac{39}{5} \frac{39}{5}

(E) 17 17

Answer

Correct choice: (E)

正确答案：（E）

Solution

Suppose that with a pair $(a_i,b_i)$ the circle is an excircle. Then notice that the hypotenuse must be $(r-x)+(r-y)$, so it must be the case that \[a_i^2+b_i^2=(2r-a_i-b_i)^2.\] Similarly, if with a pair $(a_i,b_i)$ the circle is an incircle, the hypotenuse must be $(x-r)+(y-r)$, leading to the same equation. Notice that this equation can be simplified through SFFT to \[(a_i-2r)(b_i-2r)=2r^2.\] Thus, we want the smallest $r$ such that this equation has at least $14$ distinct pairs $(a_i,b_i)$ for which this holds. The obvious choice to check is $r=6$. In this case, since $2r^2=2^3\cdot 3^2$ has $12$ positive factors, we get $12$ pairs, and we get another two if the factors are $-8,-9$ or vice versa. One can check that for smaller values of $r$, we do not even get close to $14$ possible pairs. When $r=6$, the smallest possible $c$-value is clearly when the factors are negative. When this occurs, $a_i=4, b_i=3$ (or vice versa), so the mimimal $c$ is $5$. The largest possible $c$-value occurs when the largest of $a_i$ and $b_i$ are maximized. This occurs when the factors are $72$ and $1$, leading to $a_i=84, b_i=13$ (or vice-versa), leading to a maximal $c$ of $85$. Hence the answer is $\frac{85}5=\boxed{17}$.

假设对于一对 $(a_i,b_i)$，圆是外切圆。那么注意斜边必须是 $(r-x)+(r-y)$，所以必须有 \[a_i^2+b_i^2=(2r-a_i-b_i)^2.\] 类似地，如果对于一对 $(a_i,b_i)$ 圆是内切圆，斜边必须是 $(x-r)+(y-r)$，导致相同方程。注意此方程通过 SFFT 可简化为 \[(a_i-2r)(b_i-2r)=2r^2.\] 因此，我们要找最小 $r$ 使得此方程有至少 $14$ 个不同的对 $(a_i,b_i)$。明显选择检查 $r=6$。在这种情况下，因为 $2r^2=2^3\cdot 3^2$ 有 $12$ 个正因子，我们得到 $12$ 个对，如果因子是 $-8,-9$ 或反之则再得两个。可以检查对于更小的 $r$ 值，我们甚至远达不到 $14$ 个可能对。当 $r=6$ 时，最小 $c$ 值显然当因子为负时发生。此时 $a_i=4, b_i=3$（或反之），所以最小 $c$ 是 $5$。最大可能 $c$ 值发生在 $a_i$ 和 $b_i$ 中最大的被最大化时。这发生在因子为 $72$ 和 $1$ 时，导致 $a_i=84, b_i=13$（或反之），导致最大 $c$ 为 $85$。因此答案是 $\frac{85}5=\boxed{17}$。

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