2010年全美数学竞赛（AMC）竞赛
某三角形三边长为连续正整数．最短边占周长$$30 \%$$．则最长边长度是？．
美国AMC8->Knowledge Point->Geometry->Shapes with Straight Sides->Geometric Theorem 课内体系->知识点->三角形
某三角形三边长为连续正整数．最短边占周长$$30 \%$$．则最长边长度是？  设$$n$$，$$n+1$$，和$$n+2$$是三角形三边的长度．那么三角形的周长是$$n+(n+1)+(n+2)=3n+3$$利用最小边的长度是周长的$$30 \%$$这个事实，可以得出这样的结论：  $$n=0.3(3n+3)\Rightarrow 0.9n+0.9\Rightarrow 0.1n=0.9\Rightarrow n=9$$．最长边是$$n+2=11$$．  故选$$\text{E}$$．  由于最短边的长度是整数，等于周界的$$\frac{3}{10}$$，因此周界必须是$$10$$的倍数．将前面的两个整数添加到每个答案选择中，我们可以看到$$11+10+9=30$$．  故选$$\text{E}$$．  Let $$n$$, $$n+1$$, and $$n+2$$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $$n+\left( n+1 \right)+\left( n+2 \right)=3n+3$$. Usin g the fact that the length of the smallest side is $$30 \%$$ of the perimeter, it follows that:  $$n=0.3\left( 3n +3 \right)\Rightarrow n=0.9n+0.9\Rightarrow 0.1n=0.9\Rightarrow n=9$$. The longest side is then $$n+2-11$$. Thus, answer choice $$\left( \text{E} \right)11$$ is correct.  Since the length of the shortest side is a whole number and is equal to $$\frac{3}{10}$$ of the perimeter, it follows that the perimeter must be a multiple of $$10$$. Adding the two previous integers to each answer choice, we see that $$11+10+9=30$$. Thus, answer choice $$\left( \text{E} \right)11$$ is correct. 