AMC12 2006 B

AMC12 2006 B · Q10

AMC12 2006 B · Q10. It mainly tests 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，第三条边的长度为 15。这个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

If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: \[\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5\] Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$. Then, the first side has length $3*7=21$, the second side has length $7$, the third side has length $15$, and so the perimeter is $21+7+15=43 \Rightarrow \boxed{\text {(A)}}$.

若第二条边长为 x，则第一条边长为 3x，第三条边长为 15。由三角形不等式可得：\[\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5\] 由于要使周长最大，我们取最大的整数 x。若 $x<7.5$，则 $x=7$。此时第一条边长为 $3*7=21$，第二条边长为 $7$，第三条边长为 $15$，所以周长为 $21+7+15=43 \Rightarrow \boxed{\text {(A)}}$。

Topics

GE.002 Triangles (properties)