2024 AMC 12B 第 21 题

Question

三个不同直角三角形的最小角度数之和为

90^\circ

。这三个三角形的边长都是本原勾股数三元组。其中两个是

3

-

4

-

5

和

5

-

12

-

13

。第三个三角形的周长是多少？

The measures of the smallest angles of three different right triangles sum to

90^\circ.

All three triangles have side lengths that are primitive Pythagorean triples. Two of them are

3

-

4

-

5

and

5

-

12

-

13.

What is the perimeter of the third triangle?

40

126

154

176

208

答案：C

知识点：三角恒等式勾股数

难度评级：2130

小提示：

前两个最小角的正切分别为

\tfrac34

和

\tfrac{5}{12}

；第三个满足

\alpha + \beta + \gamma = 90^\circ

，所以

\tan\gamma = \cot(\alpha + \beta)

。

The smallest angles have tangents

\tfrac34

and

\tfrac{5}{12};

the third must satisfy

\alpha + \beta + \gamma = 90^\circ,

so

\tan\gamma = \cot(\alpha + \beta)

大提示：

计算

\tan(\alpha + \beta) = \dfrac{56}{33}

，所以

\tan\gamma = \dfrac{33}{56}

；找出两条直角边为

33

和

56

的本原三元组。

Compute

\tan(\alpha + \beta) = \dfrac{56}{33},

so

\tan\gamma = \dfrac{33}{56};

find the primitive triple with legs

33

and

56

解答：

3

-

4

-

5

与

5

-

12

-

13

三角形的最小角

\alpha, \beta

满足

\tan\alpha = \tfrac34

，

\tan\beta = \tfrac{5}{12}

。

\begin{aligned} &\tan(\alpha + \beta) = \frac{\tfrac34 + \tfrac{5}{12}}{1 - \tfrac34\cdot\tfrac{5}{12}} \\ &= \frac{\tfrac{14}{12}}{\tfrac{33}{48}} \\ &= \frac{56}{33} \end{aligned}\text{。}

第三个最小角

\gamma

满足

\gamma = 90^\circ - (\alpha + \beta)

，所以

\tan\gamma = \dfrac{33}{56}

。直角边为

33

、

56

的直角三角形斜边为

\sqrt{33^2 + 56^2} = \sqrt{4225} = 65

，构成本原三元组，周长为

33 + 56 + 65 = 154

。

所以正确答案是 C。

The smallest angles

\alpha, \beta

of the

3

-

4

-

5

and

5

-

12

-

13

triangles have

\tan\alpha = \tfrac34

and

\tan\beta = \tfrac{5}{12}.

By the tangent addition formula,

\begin{aligned} &\tan(\alpha + \beta) = \frac{\tfrac34 + \tfrac{5}{12}}{1 - \tfrac34\cdot\tfrac{5}{12}} \\ &= \frac{\tfrac{14}{12}}{\tfrac{33}{48}} \\ &= \frac{56}{33}. \end{aligned}

The third smallest angle

\gamma

satisfies

\gamma = 90^\circ - (\alpha + \beta),

so

\tan\gamma = \dfrac{33}{56}.

The right triangle with legs

33

and

56

has hypotenuse

\sqrt{33^2 + 56^2} = \sqrt{4225} = 65,

a primitive triple. Its perimeter is

33 + 56 + 65 = 154.

Thus, the correct answer is C.

2024 AMC 12B 第 21 题

答案：C

难度评级：2130

小提示：

大提示：

解答：

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