Q.How many chords can be drawn through 21 points on a circle?

Step 1 — Chord of contact $L=0$.

For $S \equiv x^2+y^2+8x+10y-8=0$ we have $g=4,\ f=5,\ c=-8$. The chord of contact of the point $(5,1)$ is

Step 2 — Pole of $L=0$ w.r.t. $S'$.

For $S' \equiv x^2+y^2+4y-21=0$ we have $g'=0,\ f'=2,\ c'=-21$. The polar of a point $(x_1,y_1)$ is

This must coincide with $9x+6y+17=0$, so

From $\dfrac{y_1+2}{6}=\dfrac{2y_1-21}{17}$: $\;17(y_1+2)=6(2y_1-21)\Rightarrow 5y_1=-160\Rightarrow y_1=-32.$

Then $\dfrac{y_1+2}{6}=\dfrac{-30}{6}=-5$, so $x_1=9(-5)=-45.$

Step 3. Pole $=(-45,-32)=(-k,-h)\Rightarrow k=45,\ h=32,\ k+h=77.$

We start with the equation

where $x$ and $y$ are positive integers. The key trick is to clear denominators and rearrange into a factored form — this turns a rational equation into a divisor-counting problem.

1. Clear denominators and rearrange Multiply both sides by $2025xy$:

Bring all terms to one side:

2. Complete the product Add $2025^2$ to both sides to factor:

The left side factors as $(x - 2025)(y - 2025)$. So:

3. Interpret the equation

   Let $a = x - 2025$ and $b = y - 2025$. Then $a$ and $b$ are integers. Since $x, y > 0$, we need $a > -2025$ and $b > -2025$, but more importantly, for positive $x, y$, both $a$ and $b$ must be positive? Not necessarily — but if either $a$ or $b$ were negative, the product $ab = 2025^2$ (positive) would force both to be negative. Then $x = 2025 + a$ would be less than 2025, but still positive if $a > -2025$. However, we can check: if $a$ is negative, then $b$ is also negative, and $x$ and $y$ are still positive as long as $a > -2025$. But the problem asks for positive integral solutions — and every pair $(a, b)$ of integers whose product is $2025^2$ gives a valid $(x, y)$ because $2025 + a$ and $2025 + b$ will be positive for all divisors $a$ of $2025^2$ (since the smallest negative divisor is $-2025^2$, giving $x = 2025 - 2025^2 < 0$, which is invalid). So we must restrict to $a$ and $b$ such that $x, y > 0$.

   Actually, the clean way: Since $2025^2 > 0$, $a$ and $b$ have the same sign. If both are negative, then $x = 2025 + a < 2025$, but could still be positive if $a > -2025$. The negative divisors of $2025^2$ that are greater than $-2025$ are exactly the negatives of the positive divisors less than $2025$. But note: $2025^2$'s smallest positive divisor is $1$, so $-1$ gives $x = 2024 > 0$. So negative divisors also yield positive $x, y$ as long as they are not too large in magnitude. However, the standard approach counts all integer divisor pairs $(a, b)$ with $ab = 2025^2$, because for every divisor $d$ of $2025^2$, setting $a = d$, $b = 2025^2/d$ gives $x = 2025 + d$, $y = 2025 + 2025^2/d$, which are automatically positive since $d$ can be negative? Wait: if $d$ is negative, $2025 + d$ might be positive or negative. But the classic result: the number of positive integer solutions $(x, y)$ equals the number of positive divisors of $2025^2$. Why? Because we require $x, y > 0$, and if $a$ is a positive divisor, then $b$ is positive, so $x, y > 2025 > 0$. If $a$ is negative, then $b$ is negative, and $x = 2025 + a$ could be positive only if $|a| < 2025$. But the negative divisors of $2025^2$ that satisfy this are exactly the negatives of the positive divisors less than $2025$. However, these give the same $(x, y)$ pairs as the positive divisor pairs? Let's check: if $a = -d$ (with $d>0$), then $b = -2025^2/d$, and $x = 2025 - d$, $y = 2025 - 2025^2/d$. For this to be positive, we need $d < 2025$ and $2025^2/d < 2025$ i.e. $d > 2025$. Both cannot hold simultaneously unless $d = 2025$, which gives $x = 0$, not positive. So negative divisors do not yield positive $x, y$ except possibly when $d = 2025$? That gives $x=0$, invalid. So indeed, only positive divisors $a, b$ work. Thus we count positive divisor pairs.

   Tip

   The transformation $(x-a)(y-a)=a^2$ is a standard trick for equations of the form $1/x + 1/y = 1/a$. The number of positive solutions is exactly the number of positive divisors of $a^2$.

4. Factor 2025

   $2025 = 3^4 \cdot 5^2$ because $2025 = 81 \times 25 = 3^4 \times 5^2$.

   Then $2025^2 = (3^4 \cdot 5^2)^2 = 3^8 \cdot 5^4$.

5. Count the divisors

   The number of positive divisors of $3^8 \cdot 5^4$ is $(8+1)(4+1) = 9 \times 5 = 45$.

   Each divisor $d$ gives a distinct solution $(x, y) = (2025 + d,\ 2025 + 2025^2/d)$, and swapping $d$ and $2025^2/d$ gives the symmetric solution, but we count ordered pairs $(x, y)$ — so all 45 divisors yield distinct ordered pairs.

6. Conclusion

   There are 45 positive integral solutions.

Concept

An equation of the form $\tfrac1x+\tfrac1y=\tfrac1a$ can be cleared to a product form $(x-a)(y-a)=a^2$. Each positive divisor of $a^2$ yields exactly one ordered solution.

Step-by-step solution

1. Rearrange: from $\tfrac1x+\tfrac1y=\tfrac1{2025}$, multiply through by $2025\,xy$:

2. Factor by adding $2025^2$ to both sides:

3. Positivity: if $x\le 2025$ then $\tfrac1x\ge\tfrac1{2025}$, forcing $\tfrac1y\le 0$ — impossible. So $x>2025$ and $y>2025$, meaning $x-2025$ and $y-2025$ are positive. Each positive divisor $d$ of $2025^2$ gives $x=2025+d,\ y=2025+\tfrac{2025^2}{d}$, both positive integers.

4. Count divisors: $2025=3^4\cdot5^2$, so $2025^2=3^8\cdot5^4$. The number of positive divisors is

The key idea is simple: once you fix a side of a polygon, any triangle that uses that side is determined by picking any other vertex (not the two endpoints) to be the third corner. So the count is just the number of vertices left after removing A and B.

1. Find the number of sides of the polygon.

The number of diagonals in an $n$-sided polygon is given by the formula

We are told this equals 35:

Solving $n^2 - 3n - 70 = 0$ gives $(n-10)(n+7)=0$, so $n=10$ (since $n$ must be positive).

Thus the polygon has 10 vertices.

2. Interpret the question.

We have two distinct vertices A and B. They are vertices of the polygon, so AB is either a side or a diagonal. The problem asks for the number of triangles formed by three vertices of the polygon that have AB as one of their sides. That means: choose a third vertex C (different from A and B) such that A, B, C form a triangle.

3. Count the possible third vertices.

Since the polygon has 10 vertices total, and A and B are already taken, the remaining vertices are

Any one of these 8 vertices can be chosen as the third vertex C. Each choice gives a distinct triangle with AB as a side.

4. Conclusion.

The number of such triangles is $8$.

Counting all possible triangles

Choosing any $3$ of the $10$ points:

Of the $4$ collinear points, any $3$ do not form a triangle, so remove those degenerate selections:

Number of genuine triangles $= 120 - 4 = 116$.

Triangles with at least one vertex among the $4$ collinear points

Use the complement. Triangles that use none of the $4$ collinear points must take all $3$ vertices from the remaining $6$ points (no three of which are collinear):

Hence triangles having at least one vertex from the $4$ collinear points:

The problem asks for the number of ways to choose 3 girls from 15 seated at a round table, with the condition that the three chosen are not all seated together — meaning they are not three consecutive girls around the circle.

When dealing with circular arrangements, the concept of "consecutive" is different from a line because the circle has no start or end. The first and last positions are adjacent. So we must handle the counting of "three consecutive" selections carefully.

Let’s break it down.

1. Total number of ways to select any 3 girls from 15 This is simply a combination, since order of selection doesn’t matter:

2. Number of ways to select 3 girls who ARE seated together (consecutive)

   On a circle of 15, "three consecutive" means a block of 3 adjacent seats. How many such blocks are there?

   Imagine fixing the first girl of the block. There are 15 possible starting positions (one for each girl). For each starting position, the next two girls clockwise are forced. So there are exactly 15 blocks of three consecutive girls.

   Watch out

   A common mistake is to think there are 13 blocks (as in a line), but on a circle the block starting at position 14 includes positions 14, 15, and 1 — so all 15 starting points are valid.

   Each block corresponds to exactly one selection of 3 girls (the three in that block). So the number of selections where all three are together is:

3. Apply the complement principle We want selections where the three are not all together. So subtract the "together" cases from the total: