1972 AMC 12 第 23 题

Question

能包含图中由

3

个单位正方形组成的对称图形的最小圆半径为：

The radius of the smallest circle containing the symmetric figure composed of

3

unit squares shown is:

\sqrt2

\sqrt{1.25}

1.25

\dfrac{5\sqrt{17}}{16}

以上皆非

None of these

利用对称性，把圆心放在图形的竖直对称轴上

By symmetry, place the circle’s center on the vertical axis of the figure

令圆心到下方外角点与上方外角点的距离相等

Equate its distances to a lower outer corner and an upper outer corner

把底边中点置于

O=(0,0)

。一个下方外角点为

A=(1,0)

，一个上方外角点为

B=(\frac{1}{2},2)

。由对称性，最小外接圆的圆心为

P=(0,k)

。在最优位置，两类外角点都在圆上，所以

1+k^2=\left(\frac12\right)^2+(2-k)^2\text{。}

解得

k=\frac{13}{16}

。因此

\begin{aligned} r^2&=1+\left(\frac{13}{16}\right)^2 =\frac{425}{256},\\ r&=\frac{5\sqrt{17}}{16} \end{aligned}\text{。}

因此，正确答案是 D。

Put the midpoint of the bottom edge at

O=(0,0).

A lower outer corner is

A=(1,0),

and an upper outer corner is

B=(\frac{1}{2},2).

By symmetry the center of the smallest enclosing circle is

P=(0,k).

At the optimum both types of outer corner lie on the circle, so

1+k^2=\left(\frac12\right)^2+(2-k)^2.

This gives

k=\frac{13}{16}.

Hence

\begin{aligned} r^2&=1+\left(\frac{13}{16}\right)^2 =\frac{425}{256},\\ r&=\frac{5\sqrt{17}}{16}. \end{aligned}

Therefore, the correct answer is D.

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