bader

Problem 1:

We can leverage the concept of sum and product of roots in quadratic equations.

Since x = 4 is a root, we know that f(4) = 0. This translates to:

16a + 4b + a = 0

Combining like terms, we get: 17a + 4b = 0 (Equation 1)

We are also given that f(x) = ax^2 + bx + a. Since a ≠ 0, we can rewrite f(x) by factoring out a:

f(x) = a(x^2 + bx/a + 1/a)

We know that one root is x = 4. This means that (x - 4) is a factor of f(x). Let's rewrite f(x) using this information:

f(x) = a(x - 4)(x + r) for some constant r (since the product of the roots is cr)

Equating both ways of expressing f(x), we get:

a(x^2 + bx/a + 1/a) = a(x - 4)(x + r)

Expanding both sides:

ax^2 + bx + a = ax^2 - (4a + r)x + 4ar

Equating the coefficients of like terms on both sides, we get:

b = - (4a + r) (Equation 2)

Now we have two independent equations (Equation 1 and Equation 2) with unknowns a, b, and r. We can use the fact that one root is x = 4 (reflected in Equation 1) to solve for the other root.

Since we're looking for the other root, let's try to eliminate a and b from the equations and solve for r.

From Equation 1: b = -17a/4

Substitute this into Equation 2:

-17a/4 = - (4a + r)

Solve for r:

17a/4 = 4a + r

r = 17a/4 - 4a

r = a/4

Now that we know r (product of roots), we can find the other root itself. Since the product of roots (cr) is also equal to f(4) (which is 0 in this case), we have:

(4)(other root) = 0

Since a ≠ 0, this implies the other root must be:

Other root = 0

In conclusion, the other root of the equation f(x) = 0 is x = 0.

Given the problem's conditions, we need to determine the area of triangle \(ABC\). Let's start by defining the required variables and constraints.

First, we know:

- \(AX = 13\)

- \(BX = 10\)

- \(CX = 4\)

We also know that the circumradii of triangles \(ABX\) and \(ACX\) are equal. Let \(R\) be the common circumradius.

### Step 1: Applying the Extended Law of Sines

The extended law of sines for any triangle \( \triangle ABX \) states:

\[
R = \frac{AX \cdot BX \cdot CX}{4 \cdot K}
\]

where \( K \) is the area of the triangle.

For triangle \( ABX \):

\[
R = \frac{AB \cdot BX}{2 \cdot K_{ABX}} = \frac{13 \cdot 10}{2 \cdot K_{ABX}}
\]

Thus,

\[
K_{ABX} = \frac{13 \cdot 10}{2 \cdot R}
\]

For triangle \( ACX \):

\[
R = \frac{AC \cdot CX}{2 \cdot K_{ACX}} = \frac{13 \cdot 4}{2 \cdot K_{ACX}}
\]

Thus,

\[
K_{ACX} = \frac{13 \cdot 4}{2 \cdot R}
\]

Given that both triangles share the same radius \( R \), equate their areas relative to \( R \):

\[
\frac{13 \cdot 10}{2 \cdot R} = \frac{13 \cdot 4}{2 \cdot R}
\]

### Step 2: Solving for Common Areas

Simplifying the equations, we find:

\[R = 12\]

### Step 3: Ratio and Heron's Formula
 
Let's reconsider a simplified geometrical context:

\[
\text{Total sum of lengths } AX + BX + CX = 13 + 10 + 4 = 27
\]

The sides of triangle are \(AB = 17\), \(BC = 14\), and \(CA = 13\). Applying Heron's formula:

\[
s = \frac{17 + 14 + 13}{2} = 22
\]

\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{22(22-17)(22-14)(22-13)}
\]

\[
\text{Area} = \sqrt{22 \cdot 5 \cdot 8 \cdot 9} = \sqrt{7920} = 30 \sqrt{11}
\]

Thus, the area of triangle \(ABC\) is:
\[
\boxed{30\sqrt{11}}
\]

There are 8 candies to distribute, and the twins must get the same amount. So, let's say they each get x candies (where x can be any number from 0 to 4).

This leaves 8 - 2x candies for the remaining 3 children. We don't care about the order we give candy to these 3 children, so we can focus on how many candies each gets. There are 3 + (8-2x) - 1 = 9 - 2x candies to distribute among these 3 children, where everyone gets at least 1 (because 9-2x >= 3).

We can use stars and bars to solve this kind of problem. Imagine we have 9 - 2x stars and 2 indistinguishable bars. Placing the bars divides the stars into 3 groups representing the candy allocation for the 3 children (excluding the twins).

For example, if the twins each get 2 candies (x = 2), we have 5 stars and 2 bars. Here, ||** represents 2 candies for the first child, 0 candies for the second child, and 3 candies for the third child.

The number of arrangements of stars and bars is the same as the number of ways to distribute the candies. This can be calculated using the formula for combinations: nCr = n! / (r! * (n-r)!), where n is the total number of objects (stars + bars) and r is the number of bars.

In our case, n = (9 - 2x) + 2 = 11 - 2x and r = 2. So, the number of arrangements for a given value of x is:

(11 - 2x)! / (2! * (11 - 2x - 2)!) = (11 - 2x)! / (2! * (9 - 2x)!)

Now we need to consider all the possible values of x (from 0 to 4) and add up the number of arrangements for each case. This gives us the total number of ways to distribute the candy.

However, there's a shortcut! We realize that the arrangements for a given value of x will be the same as the arrangements for 4 - x (because the stars and bars are indistinguishable). For example, the arrangements for x = 2 (as shown above) are the same as the arrangements for x = 4 (where ||** becomes **||).

Therefore, we only need to consider values of x from 0 to 2 and then multiply the final result by 2 (to account for the double counting).

So, summing the arrangements for x = 0, 1, and 2:

x = 0: (11 - 0)! / (2! * (9 - 0)!) = 11!/ (2! * 9!) = 55

x = 1: (11 - 2)! / (2! * (9 - 2)!) = 9!/ (2! * 7!) = 36

x = 2: (11 - 4)! / (2! * (9 - 4)!) = 7!/ (2! * 5!) = 21

Total arrangements (considering double counting): 2 * (55 + 36 + 21) = 2 * 112 = 224

However, there's one last detail. We've counted the arrangements where the twins get 0 candy. But since they must get at least one candy each, we need to subtract the arrangements where x = 0 (which is 55).

Therefore, the final number of ways to distribute the candy is 224 - 55 = 169​ ways.

The equation |z - 12| + |z - 5i| = 13 represents the sum of the distances from the complex number z to two fixed points in the complex plane: 12 (on the real number axis) and 5i (on the imaginary number axis).

Here's how to solve for the smallest possible value of ∣z∣ (which represents the distance from the origin):

Triangle Inequality: The sum of the lengths of any two sides of a triangle must be greater than the absolute value of the difference of the remaining side length.

In this case, consider a triangle formed by points z, 12, and 5i. Applying the triangle inequality:

First inequality: |z - 12| + |5i - z| >= |12 - (5i)|

Second inequality: |z - 5i| + |12 - z| >= |5i - 12|

Since both sides of the original equation |z - 12| + |z - 5i| = 13 are positive, we can rewrite it as:

|z - 12| + |z - 5i| = 13

Simplifying the inequalities:

First inequality:

|12 - (5i)| = |-12 + 5i| = sqrt(12^2 + 5^2) = 13 (using the distance formula)

Therefore, |z - 12| + |z - 5i| >= 13 (which is the same as the original equation)

Second inequality:

|5i - 12| = |-12 - 5i| = sqrt((-12)^2 + (-5)^2) = 13 (using the distance formula)

Therefore, |z - 5i| + |12 - z| >= 13

Smallest possible value of |z|:

Since both inequalities we derived involve |z|, the smallest possible value of |z| will occur when both inequalities become equalities.

This means that triangle z - 12 - 5i must be an isosceles triangle with base 12−(−5i)=12+5i and both legs having the same length as ∣z∣.

In other words, point z must be equidistant to both 12 and 5i.

Finding the midpoint:

The midpoint of the line segment connecting 12 and 5i is:

Midpoint = [(12 + 0i) + (0 + 5i)] / 2 = (6 + 2.5i)

Therefore, for the smallest possible value of ∣z∣, point z must coincide with the midpoint, which is (6+2.5i).

Conclusion: The smallest possible value of ∣z∣ is the distance between the origin and the midpoint (6+2.5i), which can be found using the distance formula:

|z| = sqrt(6^2 + (2.5)^2) = sqrt(41)

Therefore, the smallest possible value of z is 6+2.5i​.

Analyzing the game setup, we can see the following:

Landing on a prime number (2 or 3) results in winning that number of dollars.

Landing on a non-prime number (4) results in losing that number of dollars.

We are given the arc size information (120 degrees for prime numbers, 60 degrees for non-prime), but that doesn't directly affect the expected winnings calculation as long as the landing probability on each section is proportional to

its arc size.

Here's how to calculate the expected winnings:

Probability of landing on a prime number:

There are two prime numbers (2 and 3) and their combined arc span is 120 + 120 = 240 degrees.

Since the total degrees is 360 (full circle), the probability of landing on a prime number is (240 degrees / 360 degrees) = 2/3.

Expected value for landing on a prime:

Winning amount for 2 is $2.

Winning amount for 3 is $3.

Expected value when landing on a prime = (2/3) * ($2 + $3) = $4/3

Expected value for landing on a non-prime:

Losing amount for 4 is -$4 (since it's a non-prime).

We can't directly calculate the expected value for non-primes as there's only one number (4) mentioned. There could be other non-prime numbers with different values.

A common approach is to assume an average loss for non-primes. Let's say the average loss for non-primes is also -$4.

Probability of landing on a non-prime:

Probability of non-prime = 1 - (probability of prime) = 1 - (2/3) = 1/3

Overall expected winnings after 1000 games:

Expected winnings = (# of games * (probability of prime * expected value for prime)) + (# of games * (probability of non-prime * expected value for non-prime))

Expected winnings = (1000 games * (2/3 * ($4/3))) + (1000 games * (1/3 * -$4))

Expected winnings = ($1000 * -$4/3) = -$1333.33 (approximately)

You expect to take a loss of 1333.33

We can solve this by realizing that 50¢ can be represented as the sum of 32¢ and 18¢.

For the 32¢ portion, Bina has two 32¢ coins, so there's only 1 choice (use both 32¢ coins).

For the remaining 18¢, Bina can use:

One 16¢ coin and one 2¢ coin (2 choices)

Two 8¢ coins (1 choice)

Nine 2¢ coins (1 choice)

So in total, there are 1 (32¢) + 2 (16¢+2¢) + 1 (8¢+8¢) + 1 (2¢+2¢+...+2¢) = 5 different combinations that add up to 50¢.