AMC10 2006 B

AMC10 2006 B · Q10

AMC10 2006 B · Q10. It mainly tests Linear inequalities, Triangles (properties).

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?

在一个具有整数边长的三角形中，一条边是第二条边的三倍，第三条边长为15。该三角形的最大可能周长是多少？

(A) 43 43

(B) 44 44

(D) 46 46

(E) 47 47

Answer

Correct choice: (A)

正确答案：（A）

Solution

Let the sides of the triangle have lengths $x$, $3x$, and $15$. The Triangle Inequality implies that $3x < x + 15$, so $x < 7.5$. Because $x$ is an integer, the greatest possible perimeter is $7 + 21 + 15 = 43$.

设三角形边长为$x$、$3x$和15。三角不等式表明$3x < x + 15$，因此$x < 7.5$。因为$x$是整数，最大可能周长为$7 + 21 + 15 = 43$。

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