Difference between revisions of "2023 AIME I Problems/Problem 3"

(Solution 1)
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==Solution 1==
 
==Solution 1==
 
We can see that for n lines, there should be <math>\frac{n*(n-1)}{2}</math> points where two lines intersect assuming that there are only points where two lines meet. Thus there should be <math>\frac{40*39}{2}=780</math> points where two lines intersect
 
 
==Solution 2==
 
  
 
In this solution, let <b><math>\boldsymbol{n}</math>-line points</b> be the points where exactly <math>n</math> lines intersect. We wish to find the number of <math>2</math>-line points.
 
In this solution, let <b><math>\boldsymbol{n}</math>-line points</b> be the points where exactly <math>n</math> lines intersect. We wish to find the number of <math>2</math>-line points.
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~MRENTHUSIASM
 
~MRENTHUSIASM
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 +
==Solution 2==
 +
 +
We can see that for n lines, there should be <math>\frac{n*(n-1)}{2}</math> points where two lines intersect assuming that there are only points where two lines meet. Thus there should be <math>\frac{40*39}{2}=780</math> points where two lines intersect
  
 
==Video Solution by TheBeautyofMath==
 
==Video Solution by TheBeautyofMath==

Revision as of 01:58, 26 November 2023

Problem

A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect.

Solution 1

In this solution, let $\boldsymbol{n}$-line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$-line points.

There are $\binom{40}{2}=780$ pairs of lines. Among them:

  • The $3$-line points account for $3\cdot\binom32=9$ pairs of lines.
  • The $4$-line points account for $4\cdot\binom42=24$ pairs of lines.
  • The $5$-line points account for $5\cdot\binom52=50$ pairs of lines.
  • The $6$-line points account for $6\cdot\binom62=90$ pairs of lines.

It follows that the $2$-line points account for $780-9-24-50-90=\boxed{607}$ pairs of lines, where each pair intersect at a single point.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~MRENTHUSIASM

Solution 2

We can see that for n lines, there should be $\frac{n*(n-1)}{2}$ points where two lines intersect assuming that there are only points where two lines meet. Thus there should be $\frac{40*39}{2}=780$ points where two lines intersect

Video Solution by TheBeautyofMath

https://youtu.be/3fC11X0LwV8

~IceMatrix

See also

2023 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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