5 super hard questions....

Suppose we have n distinct lines on a plane. What is the maximum and minimum number of intersections?
Suppose we have n distinct points on a plane. What is the maximum and minimum number of line that they can form?
What is the maximum number of region that n lines can divide a plane into?
Given n collinear points , what is the maximum and minimum distance between the points?
Arrange seven points on a plane such that , for any 3 points , we can find 2 points with distance 1.

#Geometry #QuestionsOrSuggestionsForBrilliant #ProblemSolving #InternationalMathOlympiad(IMO) #Hard

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

Minimum=0, all lines are parallel.For maximum, no line must be parallel or concurrent. Now each of n lines intersects at 1 point. Thus, maximum number of intersections= $\dbinom{n}{2}$
Minimum=1, all points are collinear. For maximum, assume all points on a circle, ie. no 3 of them are collinear, this will give again $\dbinom{n}{2}$
This can be proved by recursion. See complete proof here
Minimum $=0^+$ , maximum $=\infty$ . Either I am not understanding it properly or its vague.

Pranjal Jain - 6 years, 2 months ago

For 1 , 2 , how do u know that it is N 2 ?? And , can you please explain what's that……??

werjohi khhun - 6 years, 2 months ago

How many ways are there to select $2$ objects from $N$ given objects? That's $\dbinom{N}{2}$ . Thus, in $1.$ , I selected $2$ lines out of $n$ to find number of intersections. Similarly, in $2.$ , I selected $2$ points out of $n$ points, to draw a line.

In general, $\dbinom{x}{y}$ is the number of ways to select $y$ objects out of $x$ given objects. For more details, see this. I hope it helps.
Though if you don't know about Permutations as well, I'll recommend you to read a wiki on permuations from here first.
$\dbinom{N}{2}=\dfrac{N(N-1)}{2}$
If you need some more help, feel free to either comment here or at Moderator's Messageboard here

Pranjal Jain - 6 years, 2 months ago

@Pranjal Jain – Oh , i never learnt about that tyep of permutations.... But why isit 2 ?? In what situation we can use 3,4 or 5 ?

Thank you.

werjohi khhun - 6 years, 2 months ago

@Werjohi Khhun – When you are supposed to select $3$ objects out of $N$ objects, use $3$ .

Say, we are given $5$ objects, $(A,B,C,D,E)$

Number of ways to choose $2$ objects: $\dbinom{5}{2}=10$

{(A,B),(A,C),(A,D),(A,E),(B,C),(B,D),(B,E),(C,D),(C,E),(D,E)}

Number of ways to choose $4$ objects: $\dbinom{5}{4}=5$

{(A,B,C,D),(A,B,C,E),(A,B,D,E),(A,C,D,E),(B,C,D,E)}

Number of ways to select $5$ objects: $\dbinom{5}{5}=1$

{(A,B,C,D,E)}

Pranjal Jain - 6 years, 2 months ago

Solution needed.

werjohi khhun - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$