1962 AMC 12 第 39 题

Question

一个各边不等的三角形，其两条中线

AN

和

BP

的长度分别为

3

英寸和

6

英寸。其面积为

3\sqrt{15}

平方英寸。以英寸为单位，第三条中线的长度为：

The medians

AN

and

BP

of a triangle with unequal sides are, respectively,

3

inches and

6

inches long. Its area is

3\sqrt{15}

square inches. The length of the third median, in inches, is:

4

3\sqrt3

3\sqrt6

6\sqrt3

6\sqrt6

答案：C

知识点：中线（几何）余弦定理面积

难度评级：2090

小提示：

三条中线可组成一个三角形的三边，该三角形面积是原三角形面积的四分之三

The three medians form the side lengths of a triangle whose area is three-fourths the original area

大提示：

利用两条已知中线和该面积求出两种可能的夹角，再排除使两条中线相等的情形

Use the two known median lengths and that area to find the two possible included angles, then reject the case that makes two medians equal

解答：

以三条中线为边所组成的三角形，其面积为

\frac34\left(3\sqrt{15}\right)=\frac{9\sqrt{15}}4\text{。}

若

\theta

是边长

3

与

6

的夹角，则

9\sin\theta=\frac{9\sqrt{15}}4\text{，}

所以

\cos\theta=\pm\frac{1}{4}

。由余弦定理，第三条中线

m

满足

m^2=3^2+6^2-2(3)(6)\cos\theta\text{，}

得到

m^2=36

或

54

。若

m=6

，则两条中线相等，从而两条边也相等。由于原三角形各边不等，故

m=\sqrt{54}=3\sqrt6

。

所以正确答案是 C。

The triangle whose sides are the three medians has area

\frac34\left(3\sqrt{15}\right)=\frac{9\sqrt{15}}4.

If

\theta

is the included angle between its sides

3

and

6,

then

9\sin\theta=\frac{9\sqrt{15}}4,

so

\cos\theta=\pm\frac{1}{4}.

By the law of cosines, the third median

m

satisfies

m^2=3^2+6^2-2(3)(6)\cos\theta,

giving

m^2=36

or

54.

The value

m=6

would make two medians, and hence two sides, equal. Because the triangle has unequal sides,

m=\sqrt{54}=3\sqrt6.

Therefore, the correct answer is C.

1962 AMC 12 第 39 题

答案：C

难度评级：2090

小提示：

大提示：

解答：

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