1951 AMC 12 第 32 题

Question

若

\triangle ABC

内接于以

\overline{AB}

为直径的半圆，则

\overline{AC}+\overline{BC}

必须：

If

\triangle ABC

is inscribed in a semicircle whose diameter is

\overline{AB},

then

\overline{AC}+\overline{BC}

must be:

等于

\overline{AB}

Equal to

\overline{AB}

等于

\overline{AB}\sqrt2

Equal to

\overline{AB}\sqrt2

\ge \overline{AB}\sqrt2

\le \overline{AB}\sqrt2

\overline{AB}^{\,2}

直径所对的圆周角是直角

The angle subtending the diameter is a right angle

对两条直角边

a,b

，比较

(a+b)^2

与

2(a^2+b^2)

For legs

a,b

, compare

(a+b)^2

with

2(a^2+b^2)

由泰勒斯定理，

AC

和

BC

是以

AB

为斜边的直角三角形的两条直角边。因此

\begin{aligned} (AC+BC)^2 &\le2(AC^2+BC^2)\\ &=2AB^2 \end{aligned}\text{。}

两边取正平方根，得到

AC+BC\le AB\sqrt2

；当三角形为等腰直角三角形时取等号。

因此，正确答案是 D。

By Thales’ theorem,

AC

and

BC

are the legs of a right triangle with hypotenuse

AB.

Therefore

\begin{aligned} (AC+BC)^2 &\le2(AC^2+BC^2)\\ &=2AB^2. \end{aligned}

Taking positive square roots gives

AC+BC\le AB\sqrt2,

with equality for an isosceles right triangle.

Thus, the correct answer is D.

其他年份的第 32 题