Difference between revisions of "2001 AMC 10 Problems/Problem 4"
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<math> \textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6 </math> | <math> \textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6 </math> | ||
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| − | + | == Solution 1 == | |
[[File:circle-triangle problem.PNG]] | [[File:circle-triangle problem.PNG]] | ||
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Therefore, <math> 2 \times 3 = \boxed{\textbf{(E) }6} </math>. | Therefore, <math> 2 \times 3 = \boxed{\textbf{(E) }6} </math>. | ||
| − | === Solution | + | == Solution 2 == |
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| + | We know that the maximum amount of points that a circle and a line segment can intersect is <math>2</math>. Therefore, because there are <math>3</math> line segments in a triangle, the maximum amount of points of intersection is <math>2 \times 3 = \boxed{\textbf{(E) }6}</math>. | ||
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| + | ==Video Solution by Daily Dose of Math== | ||
| + | |||
| + | https://youtu.be/OrbG-toqEvg?si=rWBUWTuRN5v2WbRV | ||
| − | + | ~Thesmartgreekmathdude | |
== See Also == | == See Also == | ||
Latest revision as of 15:10, 15 July 2024
Problem
What is the maximum number of possible points of intersection of a circle and a triangle?
Solution 1
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 
points, but what if the circle was bigger? Then, each side would intersect the circle at 2 points.
Therefore, 
.
Solution 2
We know that the maximum amount of points that a circle and a line segment can intersect is 
. Therefore, because there are line segments in a triangle, the maximum amount of points of intersection is .
Video Solution by Daily Dose of Math
https://youtu.be/OrbG-toqEvg?si=rWBUWTuRN5v2WbRV
~Thesmartgreekmathdude
See Also
| 2001 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
