1972 AMC 12 第 34 题

Question

Dick 年龄的三倍加上 Tom 的年龄，等于 Harry 年龄的两倍。Harry 年龄立方的两倍，等于 Dick 年龄立方的三倍加上 Tom 年龄的立方。他们三人的年龄两两互质。三人年龄的平方和为：

Three times Dick’s age plus Tom’s age equals twice Harry’s age. Double the cube of Harry’s age is equal to three times the cube of Dick’s age added to the cube of Tom’s age. Their respective ages are relatively prime to each other. The sum of the squares of their ages is:

42

46

122

290

326

答案：A

知识点：年龄问题立方和与立方差丢番图方程最大公约数

难度评级：2270

小提示：

设三人的年龄为

D

、

T

和

H

，并把一次方程改写为

2(H-D)=D+T

Let the ages be

D,

T,

and

H,

and rewrite the linear equation as

2(H-D)=D+T

大提示：

改写三次方程后，分别分解两个立方差

Factor both differences of cubes after rewriting the cubic equation

解答：

方程为

\begin{aligned} 3D+T&=2H,\\ 2H^3&=3D^3+T^3 \end{aligned}\text{。}

将它们改写为

2(H-D)=D+T

和

\begin{gathered} 2(H-D)(H^2+HD+D^2)\\ =(D+T)\\ \qquad\cdot(D^2-DT+T^2) \end{gathered}\text{。}

约去相等的正因子，得

\begin{aligned} H^2+HD+DT-T^2&=0,\\ (H+T)(H+D-T)&=0 \end{aligned}\text{，}

所以

T=H+D

。一次方程进一步给出

H=4D

，两两互质条件迫使

(D,H,T)=(1,4,5)\text{。}

所求平方和为

1^2+4^2+5^2=42

。

因此，正确答案是 A。

The equations are

\begin{aligned} 3D+T&=2H,\\ 2H^3&=3D^3+T^3. \end{aligned}

Rewrite them as

2(H-D)=D+T

and

\begin{gathered} 2(H-D)(H^2+HD+D^2)\\ =(D+T)\\ \qquad\cdot(D^2-DT+T^2). \end{gathered}

Canceling the equal positive factors gives

\begin{aligned} H^2+HD+DT-T^2&=0,\\ (H+T)(H+D-T)&=0, \end{aligned}

so

T=H+D.

The linear equation then gives

H=4D,

and pairwise relative primality forces

(D,H,T)=(1,4,5).

The requested sum is

1^2+4^2+5^2=42.

Therefore, the correct answer is A.

1972 AMC 12 第 34 题

答案：A

难度评级：2270

小提示：

大提示：

解答：

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