1959 AMC 12 第 26 题

Question

一个等腰三角形的底边长为

\sqrt2

。两条腰上的中线互相垂直。该三角形的面积为：

The base of an isosceles triangle is

\sqrt2.

The medians to the legs intersect each other at right angles. The area of the triangle is:

1.5

2

2.5

3.5

4

将底边两端点关于原点对称放置，并把第三个顶点置于底边的垂直平分线上

Place the base endpoints symmetrically about the origin and the third vertex on the perpendicular bisector

写出两条中线的方向向量，并令其点积为零

Write direction vectors for the two medians and set their dot product equal to zero

将底边两端点置于

A=(-\frac{\sqrt2}{2},0)

和

B=(\frac{\sqrt2}{2},0)

，并设第三个顶点为

C=(0,h)

。从

A

出发的中线方向为

\left(\frac{3\sqrt2}{4},\frac h2\right)\text{，}

从

B

出发的中线方向为

\left(-\frac{3\sqrt2}{4},\frac h2\right)\text{。}

两者点积为零，所以

-\frac98+\frac{h^2}{4}=0\text{，}

且

h=\frac{3\sqrt2}{2}

。因此面积为

\frac12\cdot\sqrt2\cdot\frac{3\sqrt2}{2}=\frac32\text{。}

因此，正确答案是 A。

Put the base endpoints at

A=(-\frac{\sqrt2}{2},0)

and

B=(\frac{\sqrt2}{2},0),

and let the third vertex be

C=(0,h).

The median from

A

has direction

\left(\frac{3\sqrt2}{4},\frac h2\right),

while the median from

B

has direction

\left(-\frac{3\sqrt2}{4},\frac h2\right).

Their dot product is zero, so

-\frac98+\frac{h^2}{4}=0,

and

h=\frac{3\sqrt2}{2}.

Thus the area is

\frac12\cdot\sqrt2\cdot\frac{3\sqrt2}{2}=\frac32.

Therefore, the correct answer is A.

其他年份的第 26 题