首页
AEIS
全国热线
400-880-6200

AMC10每日一题(2001年真题#04)

2001 AMC 10 竞赛试题/第04题


Problem

What is the maximum number of possible points of intersection of a circle and a triangle?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

Solution

Solution 1

Circle-triangle problem.PNG

We can draw a circle and a triangle, such that each side is tangent to the circle. This means that each side would intersect the circle at one point.

You would then have $3$ points, but what if the circle was bigger? Then, each side would intersect the circle at 2 points.

Therefore, $2 \times 3 = \boxed{\textbf{(E) }6}$.

Solution 2

We know that the maximum amount of points that a circle and a line segment can intersect is $2$. Therefore, because there are $3$ line segments in a triangle, the maximum amount of points of intersection is $\boxed{\textbf{(E) }6}$.


 . '文章底图' .
点击关注
课窝考试网(http://www.ikewo.cn)声明
本站凡注明原创和署名的文章,未经课窝考试网许可,不得转载。课窝考试网的部分文章素材来自于网络,版权归原作者所有,仅供学习与研究,如果侵权,请提供版权证明,以便尽快删除。
文章页广告
下一篇
AMC10每日一题(2001年真题#05)

2001 AMC 10 竞赛试题/第05题

2018-05-17
相关文章
课窝公众号
课窝公众号
回到顶部