2022 AMC 12B 第 19 题

Question

在

\triangle ABC

中，中线

\overline{AD}

和

\overline{BE}

相交于

G

，且

\triangle AGE

是等边三角形。那么

\cos(C)

可写为

\dfrac{m\sqrt p}{n}

，其中

m

和

n

是互质正整数，

p

是不被任何质数平方整除的正整数。求

m + n + p

。

In

\triangle ABC

medians

\overline{AD}

and

\overline{BE}

intersect at

G

and

\triangle AGE

is equilateral. Then

\cos(C)

can be written as

\dfrac{m\sqrt p}{n},

where

m

and

n

are relatively prime positive integers and

p

is a positive integer not divisible by the square of any prime. What is

m + n + p?

44

48

52

56

60

答案：A

知识点：中线（几何）重心余弦定理

难度评级：2020

小提示：

重心把

AD

分开，使

AG = \tfrac23 AD

，且

GE = \tfrac13 BE

；同时

AE = \tfrac12 AC

。

The centroid splits

AD

so

AG = \tfrac23 AD,

and

GE = \tfrac13 BE;

also

AE = \tfrac12 AC

大提示：

令这三段相等，并用中线长度公式联系边长，再用余弦定理。

Set all three equal and use the median-length formula to relate the sides, then apply the law of cosines

解答：

令

a = BC

、

b = CA

、

c = AB

。由于

E

是

AC

的中点，

AE = \tfrac{b}{2}

。重心性质给出

AG = \tfrac23 m_a

和

GE = \tfrac13 m_b

，其中

m_a, m_b

分别是从

A

和

B

出发的中线长。

等边三角形

\triangle AGE

说明

AG = GE = AE

。由

\tfrac23 m_a = \tfrac{b}{2}

得

m_a = \tfrac34 b

，结合

m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}

得

2c^2 - a^2 = \tfrac{b^2}{4}

。由

\tfrac13 m_b = \tfrac{b}{2}

得

m_b = \tfrac32 b

，所以

a^2 + c^2 = 5b^2

。

解这两个关系得

c^2 = \tfrac{7b^2}{4}

、

a^2 = \tfrac{13b^2}{4}

。取

b = 2

，则

a^2 = 13

，

c^2 = 7

，于是

\begin{aligned} \cos C &= \dfrac{a^2 + b^2 - c^2}{2ab} \\ &= \dfrac{13 + 4 - 7}{2 \cdot \sqrt{13} \cdot 2} \\ &= \dfrac{5\sqrt{13}}{26} \end{aligned}\text{。}

因此

m + n + p = 5 + 26 + 13 = 44

。

所以正确答案是 A。

Let

a = BC,

b = CA,

c = AB.

Since

E

is the midpoint of

AC,

AE = \tfrac{b}{2}.

The centroid gives

AG = \tfrac23 m_a

and

GE = \tfrac13 m_b,

where

m_a, m_b

are the medians from

A

and

B.

Equilateral

\triangle AGE

means

AG = GE = AE.

From

\tfrac23 m_a = \tfrac{b}{2}

we get

m_a = \tfrac34 b,

which with

m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4}

gives

2c^2 - a^2 = \tfrac{b^2}{4}.

From

\tfrac13 m_b = \tfrac{b}{2}

we get

m_b = \tfrac32 b,

giving

a^2 + c^2 = 5b^2.

Solving,

c^2 = \tfrac{7b^2}{4}

and

a^2 = \tfrac{13b^2}{4}.

Taking

b = 2

gives

a^2 = 13,

c^2 = 7,

so

\begin{aligned} \cos C &= \dfrac{a^2 + b^2 - c^2}{2ab} \\ &= \dfrac{13 + 4 - 7}{2 \cdot \sqrt{13} \cdot 2} \\ &= \dfrac{5\sqrt{13}}{26}. \end{aligned}

Then

m + n + p = 5 + 26 + 13 = 44.

Thus, the correct answer is A.

2022 AMC 12B 第 19 题

答案：A

难度评级：2020

小提示：

大提示：

解答：

其他年份的第 19 题