1999 AMC 12 第 19 题

Question

考虑所有满足下列条件的三角形

ABC

：

AB = AC

，点

D

在线段

\overline{AC}

上，且

\overline{BD} \perp \overline{AC}

，

AD

与

CD

均为整数，且

BD^2 = 57

。在所有这样的三角形中，

AC

的最小可能值是

Consider all triangles

ABC

satisfying the following conditions:

AB = AC,

D

is a point on

\overline{AC}

for which

\overline{BD} \perp \overline{AC},

AD

and

CD

are integers, and

BD^2 = 57.

Among all such triangles, the smallest possible value of

AC

is

9

10

11

12

13

答案：C

知识点：等腰三角形勾股定理丢番图方程

难度评级：1810

小提示：

直角三角形

ADB

给出

AB^2 = AD^2 + BD^2

。

Right triangle

ADB

gives

AB^2 = AD^2 + BD^2

大提示：

用

AB = AC = AD + CD

，化简为

CD(CD + 2\,AD) = 57

。

With

AB = AC = AD + CD,

reduce to

CD(CD + 2\,AD) = 57

解答：

设

AD = n

，

CD = m

。因为

\triangle ADB

在

D

处为直角，

AB^2 = n^2 + 57

。又

AB = AC = m + n

，所以

(m + n)^2 = n^2 + 57\text{，}

化简为

m(m + 2n) = 57

。

正整数解为

m = 1, n = 28

，此时

AC = 29

；以及

m = 3, n = 8

，此时

AC = 11

。因此

AC

的最小可能值为

11

。

所以正确答案是 C。

Let

AD = n

and

CD = m.

Since

\triangle ADB

is right-angled at

D,

AB^2 = n^2 + 57.

Also

AB = AC = m + n,

so

(m + n)^2 = n^2 + 57,

which simplifies to

m(m + 2n) = 57.

The positive integer solutions are

m = 1, n = 28

(giving

AC = 29

) and

m = 3, n = 8

(giving

AC = 11

). The smallest possible value of

AC

is

11.

Thus, the correct answer is C.

1999 AMC 12 第 19 题

答案：C

难度评级：1810

小提示：

大提示：

解答：

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