The lines L1, L2, ..., L20 are distinct. For n = 1, 2, 3, ..., 10 all the lines L2n-1 are parallel t

Q: Wait, why do we subtract $\binom{10}{2}$ for the parallel lines? Shouldn't we subtract something else?

Think of it this way: the formula $\binom{20}{2}$ assumes every pair of lines intersects. For the 10 parallel lines, we assumed they intersected at $\binom{10}{2}$ distinct points. But parallel lines never intersect! So, those $\binom{10}{2}$ points are 'ghost' intersections that we counted but don't exist. We must remove them entirely.

Q: I'm confused about the concurrent lines. Why is the loss $\binom{10}{2} - 1$ and not just $\binom{10}{2}$?

Great question! When we calculate $\binom{10}{2}$ for the concurrent lines, we are counting the number of pairs. In a general position, each pair would create a unique intersection point. However, because these lines are concurrent, all those pairs intersect at the *same* point $P$. We counted $\binom{10}{2}$ points, but they all represent just one physical location. We need to remove the $\binom{10}{2}$ 'extra' points and add back the $1$ point that actually exists.

Q: What if the point $P$ lies on one of the parallel lines? Does that change the answer?

In JEE Advanced problems, unless specified, we assume the 'maximum' number of intersection points. If $P$ were to lie on one of the parallel lines, it would create *more* intersection points (the intersection of the concurrent line and the parallel line). To maximize the count, we assume $P$ is not on any of the parallel lines.

Question

The lines $L_1, L_2, \dots, L_{20}$ are distinct. For $n = 1, 2, 3, \dots, 10$ all the lines $L_{2n-1}$ are parallel to each other and all the lines $L_{2n}$ pass through a given point $P$. The maximum number of points of intersection of pairs of lines from the set $\{L_1, L_2, \dots, L_{20}\}$ is equal to :

Sigma Percentile · Accepted Answer

### The Geometry of Order and Chaos

Welcome, future engineer. Today, we are not just solving a problem about lines; we are exploring the delicate balance between order and chaos in geometry. Imagine you are standing in a vast, empty plane with twenty distinct lines at your disposal.

If you were to scatter them randomly, they would create a complex web of intersections. However, this problem imposes structure and rules. Our objective is to find the maximum number of intersection points possible within these constraints.

### Phase 1: The Ideal World of General Position

Let us begin by ignoring the constraints. Imagine a world where every line is in "general position," meaning no two lines are parallel and no three lines are concurrent.

In this ideal, chaotic world, every single pair of lines intersects at exactly one unique point. If we have $20$ lines, the number of pairs is determined by the combination formula $\binom{20}{2}$.

Using the combination formula, we calculate:

$$\binom{20}{2} = \frac{20 	imes 19}{2 	imes 1} = 190$$

So, in our ideal world, we would have $190$ intersection points. We must now account for the reality of the specific constraints provided.

### Phase 2: The Parallelism Trap

Consider the first set of lines: the odd-indexed lines $L_1, L_3, \dots, L_{19}$. There are $10$ of them, and they are all parallel.

In our initial calculation of $190$, we assumed that every pair of these lines intersected. However, parallel lines never meet. Therefore, every pair of lines chosen from this set of $10$ represents an intersection point that we counted in our $190$ but which does not exist in reality.

How many such "ghost" intersections are there? We use the combination formula for the $10$ parallel lines:

$$\binom{10}{2} = \frac{10 	imes 9}{2 	imes 1} = 45$$

We have overcounted by $45$ points. Subtracting these from our total, we are left with $190 - 45 = 145$ points.

### Phase 3: The Concurrency Collapse

Now, let us turn our attention to the second set: the even-indexed lines $L_2, L_4, \dots, L_{20}$. There are $10$ of them, and they are all concurrent, meaning they all pass through a single point $P$.

In our initial calculation of $190$, we assumed that every pair of these lines intersected at a unique point. In reality, all these pairs intersect at the same location $P$.

We counted $\binom{10}{2} = 45$ intersections for these lines, but only $1$ point actually exists. This means we have overcounted by $45 - 1 = 44$ points. We must subtract these $44$ "extra" points from our running total.

### The Final Synthesis

We have navigated the chaos of the general position, corrected for the parallel lines that refuse to meet, and adjusted for the concurrent lines that meet too often. Let us assemble the final result:

$$	ext{Total Points} = (	ext{Theoretical Maximum}) - (	ext{Loss from Parallelism}) - (	ext{Loss from Concurrency})$$

Substituting our values:

$$	ext{Max Points} = 190 - 45 - 44$$

$$	ext{Max Points} = 101$$

There we have it. **$101$ points**. It is a beautiful, precise number that emerges from the interplay of these constraints.

Remember, in JEE Advanced, the math is rarely just about the calculation; it is about visualizing the geometry. When you see parallel lines, think "subtraction of pairs." When you see concurrent lines, think "collapse of points." Keep this mindset, and you will master any geometry problem that comes your way.

Enter Numerical Value:

Visualized Solution

Identifying the Two Families

The Theoretical Maximum

Calculating $^{20} C_{2}$

Loss Due to Parallelism

Calculating Parallel Loss

Loss Due to Concurrency

Calculating Concurrent Loss

The Final Calculation

The Sigma Insight: Combinations and Selection

The Geometry of Order and Chaos

Phase 1

Phase 2

Phase 3

The Final Synthesis

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