Enrichment - Free Printable

Let's solve each question step by step based on the Fano Plane shown in the image and the axioms provided.

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Understanding the Fano Plane

The Fano Plane is a finite projective plane with:
- 7 points: labeled 1 through 7
- 7 lines: represented as curves or straight lines, labeled ℓ₁ to ℓ₇
- Each line contains exactly 3 points
- Each pair of points lies on exactly one line
- Any two lines intersect at exactly one point
- There are no parallel lines
- Every set of four points has at least one that is not collinear with the other three

This structure satisfies the axioms of a projective plane, which differ from Euclidean geometry.

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Question 1: What line contains the points 1 and 2? And 7? 5 and 6? Are there any violations of the first axiom?

#### First Axiom (Projective Plane):
> Two distinct points in $ P' $ are only on one line.

So we need to find which line passes through:
- Points 1 and 2
- Points 7 and 5
- Points 5 and 6

From the diagram:

- Line ℓ₁ connects points 1, 2, and 4 → so it contains 1 and 2 ✔
- Line ℓ₂ connects points 1, 3, 7 → contains 1 and 7
- Line ℓ₃ connects points 2, 5, 6 → contains 5 and 6

Now check if these pairs lie on only one line:
- 1 and 2 → only on ℓ₁ ✔
- 7 and 5 → Let’s see: Is there a line with both 7 and 5?

Look at lines:
- ℓ₂: 1, 3, 7
- ℓ₄: 3, 5, 6
- ℓ₅: 4, 5, 7 → yes! So 5 and 7 are on ℓ₅

Is there another line with 5 and 7? No. Only ℓ₅ → ✔

Similarly, 5 and 6 → ℓ₃ → only line with both → ✔

✔ No violations of the first axiom — every pair of points lies on exactly one line.

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Answer to Question 1:

- The line containing points 1 and 2 is ℓ₁
- The line containing points 7 and 5 is ℓ₅
- The line containing points 5 and 6 is ℓ₃
- No violations of the first axiom.

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Question 2: Which point is the intersection of line 1 and line 2? Line 4 and line 2? Line 6 and line 7? Are there any violations of the second axiom?

#### Second Axiom:
> Any two lines from $ L' $ intersect in a unique point (there are no parallel lines)

So we need to find intersections:

- Line ℓ₁ and ℓ₂:
- ℓ₁: 1, 2, 4
- ℓ₂: 1, 3, 7
- Common point: 1 → Intersection = point 1

- Line ℓ₄ and ℓ₂:
- ℓ₄: 3, 5, 6
- ℓ₂: 1, 3, 7
- Common point: 3 → Intersection = point 3

- Line ℓ₆ and ℓ₇:
- ℓ₆: 4, 5, 7
- ℓ₇: 2, 4, 6
- Common point: 4 → Intersection = point 4

Now, check if any two lines do not intersect — i.e., are parallel?
In this Fano Plane, all lines intersect at exactly one point.

For example:
- ℓ₁ and ℓ₃: ℓ₁ = {1,2,4}, ℓ₃ = {2,5,6} → intersect at 2
- ℓ₁ and ℓ₅: ℓ₁ = {1,2,4}, ℓ₅ = {4,5,7} → intersect at 4
- All pairs intersect.

✔ No violations of the second axiom — every pair of lines intersects at exactly one point.

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Answer to Question 2:

- Intersection of ℓ₁ and ℓ₂ → point 1
- Intersection of ℓ₄ and ℓ₂ → point 3
- Intersection of ℓ₆ and ℓ₇ → point 4
- No violations of the second axiom.

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Question 3: Why is it that regardless of which four points you choose, one point will not be on the same line as the other three?

We are told in the third axiom:
> There is at least one set of four points in $ P' $ such that no three are collinear.

But here, the question says: "regardless of which four points you choose, one point will not be on the same line as the other three."

Let’s think about this.

In the Fano Plane:
- There are only 7 points.
- Each line has exactly 3 points.
- Total number of lines: 7
- Each pair of points determines a unique line.

Now suppose we pick any four points.

Can all four be collinear? No — because no line has more than 3 points.

So, no line contains 4 points, meaning that in any group of 4 points, at least one is off the line defined by the other three.

But wait — could three of them be on a line, and the fourth not?

Yes — but that's what we’re trying to show.

Let’s suppose we pick any four points.

There are only 7 points total, and only 7 lines.

Suppose we pick four points: A, B, C, D.

Can A, B, C be collinear? Possibly.

Then D must not be on that line — otherwise, it would be part of a 4-point line, which doesn't exist.

But even if A, B, C are on a line, D might still be on a different line with some of them.

But the key is: no line has more than 3 points, so no four points can be collinear.

Therefore, in any set of four points, it is impossible for all four to be on the same line.

Hence, at least one of them is not on the line defined by the other three.

So, no matter which four points you choose, you cannot have all four on one line.

Thus, one point will not be on the same line as the other three.

But note: It might be that three are collinear, and the fourth is not — that's fine.

Or no three are collinear — also possible.

But in every case, not all four are collinear, so at least one is off the line of the others.

Wait — the phrasing is: "one point will not be on the same line as the other three"

That means: for any four points, there exists one point that is not on the line formed by the other three.

But is this always true?

Let’s test.

Suppose we pick four points where three are collinear, and the fourth is not.

Say: 1, 2, 4 (on ℓ₁), and 3.

Now, consider whether 3 is on the line of 1, 2, 4 → no.

But now, is 1 on the line of 2, 4, 3?

- Line of 2, 4, 3? Do they lie on a line?
- ℓ₁: 1,2,4 → includes 2,4
- ℓ₂: 1,3,7 → includes 3
- ℓ₃: 2,5,6 → includes 2
- ℓ₄: 3,5,6 → includes 3
- ℓ₅: 4,5,7 → includes 4
- ℓ₆: 4,5,7 → includes 4
- ℓ₇: 2,4,6 → includes 2,4

Wait — is there a line with 2, 3, 4?

Check:
- ℓ₁: 1,2,4 → no 3
- ℓ₂: 1,3,7 → no 2,4
- ℓ₃: 2,5,6 → no 3,4
- ℓ₄: 3,5,6 → no 2,4
- ℓ₅: 4,5,7 → no 2,3
- ℓ₆: 4,5,7 → no 2,3
- ℓ₇: 2,4,6 → no 3

So no line contains 2, 3, 4 → so they are not collinear

So in the set {1,2,3,4}, no three are collinear?

Wait — 1,2,4 are on ℓ₁ → so 1,2,4 are collinear.

So in this set: 1,2,4 are collinear; 3 is not.

So 3 is not on the line of 1,2,4 → so 3 is not on the same line as the other three.

Now, is 1 on the line of 2,3,4? No — no such line.

Similarly, 2 not on line of 1,3,4? No such line.

4 not on line of 1,2,3? No such line.

So in this case, each point is off the line of the other three.

But the statement says: "one point will not be on the same line as the other three."

So it's sufficient that at least one is not.

But actually, more than one may not be.

But the key is: can we find four points where one of them IS on the line of the other three?

Yes — for example: take points 1,2,4, and 5.

- 1,2,4 are on ℓ₁
- 5 is not on ℓ₁

So 5 is not on the line of 1,2,4 → okay.

Now take points 1,3,7, and 2.

- 1,3,7 are on ℓ₂
- 2 is not on ℓ₂ → so 2 is not on line of 1,3,7

Now try: Can we find a set where one point is on the line of the other three?

But wait — any three points define a line (axiom 1). But four points cannot be collinear.

So if we pick any four points, then no line contains all four, so at least one point is not on the line defined by the other three.

But could it be that three of them are collinear, and the fourth is not?

Yes — e.g., 1,2,4, and 3.

Then 1,2,4 are on ℓ₁, but 3 is not on ℓ₁.

So 3 is not on the line of 1,2,4.

But is 1 on the line of 2,3,4? No — no such line.

So again, no point is on the line of the other three.

But suppose we pick four points where three are collinear, and the fourth is on a different line with two of them?

Example: Take points 1,2,4 (on ℓ₁), and 5.

- 1,2,4 on ℓ₁
- 5 is not on ℓ₁
- Is 5 on line of 1,2,4? No
- Is 1 on line of 2,4,5? Check: 2,4,5 — is there a line with them?

- ℓ₇: 2,4,6 → no 5
- ℓ₃: 2,5,6 → no 4
- ℓ₅: 4,5,7 → no 2

So no line contains 2,4,5 → so 1 is not on that line.

Similarly, none of the points are on the line of the other three.

But let’s pick a better example.

Take points: 1, 2, 5, 6

- ℓ₃: 2,5,6 → so 2,5,6 are collinear
- 1 is not on ℓ₃
- So 1 is not on the line of 2,5,6

Now, is 2 on the line of 1,5,6? No such line → no

So again, 1 is not on the line of the other three

But suppose we pick four points where one is on the line of the other three?

But since no line has more than 3 points, no four points can be collinear, so no point can be on the line of the other three unless those three are collinear and the fourth is on that line — but that would require four collinear points, which is impossible.

Therefore, no four points are collinear, so in any set of four points, it is impossible for one point to be on the line defined by the other three, unless the other three are collinear and the fourth is on that line — but that would make four collinear points, which is impossible.

Wait — contradiction?

Let’s clarify.

Suppose three points A,B,C are collinear (on line L).

Then any fourth point D is not on L → so D is not on the line of A,B,C.

But is A on the line of B,C,D?

Only if B,C,D are collinear and A is on that line.

But in the Fano Plane, any three points are collinear only if they form a line.

So unless B,C,D are on a line, then A is not on that line.

But in general, most sets of three points are not collinear.

But the key point is:

> In the Fano Plane, any three points are collinear if and only if they form a line.

And since there are only 7 lines, only certain triples are collinear.

Now, the statement says: "Regardless of which four points you choose, one point will not be on the same line as the other three."

This is true, because:

- Suppose we pick four points: A,B,C,D.
- If none of the subsets of three are collinear, then obviously, no point is on the line of the other three.
- If three are collinear, say A,B,C on a line L, then D is not on L → so D is not on the line of A,B,C.
- But could D be on the line of A,B,C? No — because that would mean D is on L → but then A,B,C,D all on L → four points on one line → impossible.

So in all cases, at least one point is not on the line of the other three.

In fact, exactly one or more may not be.

But the statement says: "one point will not be on the same line as the other three" — meaning at least one.

✔ So it's always true.

Also, this is consistent with the third axiom: there is at least one set of four points with no three collinear — but here we're saying every set of four points has at least one point not on the line of the other three.

Actually, stronger: in every set of four points, no point lies on the line of the other three, because that would require four collinear points.

Wait — is that true?

Suppose we have points A,B,C on line L, and D not on L.

Then:
- Is D on the line of A,B,C? No — because that line is L, and D ∉ L.
- Is A on the line of B,C,D? Only if B,C,D are collinear and A is on that line.

But B,C,D may not be collinear.

But suppose B,C,D are collinear — is that possible?

Yes — for example, take points 2,5,6 (on ℓ₃), and 1.

Then 2,5,6 are on ℓ₃, 1 is not.

Is 1 on the line of 2,5,6? No.

Is 2 on the line of 1,5,6?

- 1,5,6 — is there a line with them?

- ℓ₄: 3,5,6 → no 1
- ℓ₅: 4,5,7 → no 1
- ℓ₆: 4,5,7 → no 1
- ℓ₇: 2,4,6 → no 1,5

No line has 1,5,6 → so no.

So 2 is not on line of 1,5,6.

So again, no point is on the line of the other three.

But suppose we pick four points where two different lines share a point, and we try to see if one point lies on the line of the other three.

But due to the small size, it's hard.

But crucially: since no line has more than 3 points, no four points are collinear, so no point can lie on the line defined by three other points unless those three are collinear and the fourth is on that line, which would imply four collinear points — impossible.

Therefore, in any set of four points, no point lies on the line of the other three.

So the statement is true.

But the question says: "one point will not be on the same line as the other three" — which is true, and even stronger: none of them are.

But the reason is: no line contains four points, so no four points are collinear, so no point can be on the line of the other three.

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Answer to Question 3:

Because no line contains more than three points, it is impossible for any four points to be collinear. Therefore, in any set of four points, at least one point (in fact, all four) is not on the line determined by the other three. Hence, one point will not be on the same line as the other three, regardless of the choice.

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Question 4: Is the Fano Plane a projective plane?

Recall the definition of a projective plane requires:

1. Two distinct points determine a unique line.
2. Any two distinct lines intersect in exactly one point.
3. There exist four points, no three of which are collinear.

We have already verified:
- (1) ✔ Every pair of points lies on exactly one line.
- (2) ✔ Every two lines intersect in exactly one point (no parallels).
- (3) ✔ There exists a set of four points with no three collinear — for example, points 1, 2, 3, 5:
- 1,2,4 → on ℓ₁ → but 5 not on it
- 1,3,7 → on ℓ₂ → 5 not on it
- 2,5,6 → on ℓ₃ → 1,3 not on it
- 1,2,3: not on a line (no line contains 1,2,3)
- Similarly, 1,2,5: no line
- So no three of 1,2,3,5 are collinear → ✔

Therefore, all three axioms are satisfied.

✔ Yes, the Fano Plane is a projective plane.

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Answer to Question 4:

Yes, the Fano Plane is a projective plane because it satisfies all three axioms of a projective plane:
1. Every pair of points lies on exactly one line.
2. Every pair of lines intersects in exactly one point.
3. There exists a set of four points with no three collinear.

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✔ Final Answers:

1.
- Line containing 1 and 2: ℓ₁
- Line containing 7 and 5: ℓ₅
- Line containing 5 and 6: ℓ₃
- No violations of the first axiom.

2.
- Intersection of ℓ₁ and ℓ₂: point 1
- Intersection of ℓ₄ and ℓ₂: point 3
- Intersection of ℓ₆ and ℓ₇: point 4
- No violations of the second axiom.

3.
Because no line contains more than three points, no four points are collinear. Thus, in any set of four points, at least one point is not on the line defined by the other three.

4.
Yes, the Fano Plane is a projective plane because it satisfies all three axioms of a projective plane.