learnmgcat

Given points \( A = (4, -4) \), \( B = (3, 8) \), and \( C = (-1, 2) \), and a point \( Q \) such that for any point \( P \) in the plane, the following equation holds:

\[
PA^2 + PB^2 + PC^2 = 3PQ^2 + k
\]

We need to find the constant \( k \).

### Step 1: Write down the expression for the sum of squares of distances

Let \( P = (x, y) \) and \( Q = (x_Q, y_Q) \). The squared distances from \( P \) to the points \( A \), \( B \), and \( C \) are:

\[
PA^2 = (x - 4)^2 + (y + 4)^2
\]

\[
PB^2 = (x - 3)^2 + (y - 8)^2
\]

\[
PC^2 = (x + 1)^2 + (y - 2)^2
\]

The sum of these squared distances is:

\[
PA^2 + PB^2 + PC^2 = \left[(x - 4)^2 + (y + 4)^2\right] + \left[(x - 3)^2 + (y - 8)^2\right] + \left[(x + 1)^2 + (y - 2)^2\right]
\]

Expanding these expressions:

\[
PA^2 = (x^2 - 8x + 16) + (y^2 + 8y + 16)
\]

\[
PB^2 = (x^2 - 6x + 9) + (y^2 - 16y + 64)
\]

\[
PC^2 = (x^2 + 2x + 1) + (y^2 - 4y + 4)
\]

Adding them together:

\[
PA^2 + PB^2 + PC^2 = \left[3x^2 + (-8x - 6x + 2x) + (16 + 9 + 1)\right] + \left[3y^2 + (8y - 16y - 4y) + (16 + 64 + 4)\right]
\]

Simplifying:

\[
PA^2 + PB^2 + PC^2 = 3x^2 - 12x + 26 + 3y^2 - 12y + 84
\]

Thus:

\[
PA^2 + PB^2 + PC^2 = 3(x^2 - 4x + \frac{26}{3}) + 3(y^2 - 4y + 28)
\]

### Step 2: Express the right-hand side

The expression for \( 3PQ^2 \) is:

\[
3PQ^2 = 3\left[(x - x_Q)^2 + (y - y_Q)^2\right] = 3\left[(x^2 - 2xx_Q + x_Q^2) + (y^2 - 2yy_Q + y_Q^2)\right]
\]

Expanding:

\[
3PQ^2 = 3x^2 - 6xx_Q + 3x_Q^2 + 3y^2 - 6yy_Q + 3y_Q^2
\]

### Step 3: Set up the equation

We equate \( PA^2 + PB^2 + PC^2 \) with \( 3PQ^2 + k \):

\[
3x^2 - 12x + 110 + 3y^2 - 12y + 84 = 3x^2 - 6xx_Q + 3x_Q^2 + 3y^2 - 6yy_Q + 3y_Q^2 + k
\]

By comparing coefficients, we get:

\[
-12x = -6xx_Q \quad \text{and} \quad -12y = -6yy_Q
\]

So:

\[
x_Q = 2, \quad y_Q = 2
\]

Now, matching the constant terms:

\[
110 + 84 = 3x_Q^2 + 3y_Q^2 + k
\]

\[
194 = 3(2^2) + 3(2^2) + k = 12 + 12 + k = 24 + k
\]

Thus:

\[
k = 194 - 24 = 170
\]

### Final Answer:

The constant \( k \) is \( \boxed{170} \).

Finding |u+2w|

Understanding the Problem

We are given three pieces of information about complex numbers u and w:

|u| = 5

|w| = 3

|u+w| = 6

We need to find the value of |u+2w|.

Solution Approach

We will use the Law of Cosines for complex numbers to solve this problem.

Applying the Law of Cosines

Let θ be the angle between u and w. Then, we have:

|u+w|^2 = |u|^2 + |w|^2 + 2|u||w|cosθ

Substituting the given values:

6^2 = 5^2 + 3^2 + 253*cosθ

36 = 34 + 30cosθ

cosθ = 1/15

Now, consider |u+2w|^2:

|u+2w|^2 = |u|^2 + (2|w|)^2 + 2|u|(2|w|)cosθ

|u+2w|^2 = 5^2 + (23)^2 + 25*(23)(1/15)

|u+2w|^2 = 25 + 36 + 12

|u+2w|^2 = 73

Therefore, |u+2w| = √73.

So, the value of |u+2w| is √73.

To find the probability that all three pieces of a stick, which is 6 units long, are shorter than 6 units after it is broken at two random points, we can use a geometric interpretation.

1. **Setting Up the Problem:**

- Let the stick start at 0 and end at 6 (it has a total length of \( L = 6 \) units).

- Denote the two break points as \( X_1 \) and \( X_2 \), where \( 0 < X_1 < X_2 < 6 \).

- The resulting pieces of the stick will have lengths:

- Piece 1: from 0 to \( X_1 \) (length \( X_1 \))

- Piece 2: from \( X_1 \) to \( X_2 \) (length \( X_2 - X_1 \))

- Piece 3: from \( X_2 \) to 6 (length \( 6 - X_2 \))

2. **Condition for the Pieces:**

- We need each piece to be less than 6 units in length:

- \( X_1 < 6 \)

- \( X_2 - X_1 < 6 \)

- \( 6 - X_2 < 6 \)

- The inequalities simplify to:

- \( X_1 < 6 \) (which will always be true since \( X_1 < X_2 < 6 \))

- \( X_2 < 6 + X_1 \) (or simply \( X_2 < 6 \) due to initial constraints)

- \( X_2 > 0 \) (will also hold since \( X_2 > X_1 > 0 \))

3. **Valid Conditions:**

- The only actual restriction is that \( X_2 < 6 \) (which is always satisfied because \( 0 < X_1 < X_2 < 6 \)).

- The lengths of the pieces will always be less than the length of the original stick (6 units).

4. **Geometric Approach:**

- We represent the breaking points \( (X_1, X_2) \) in the coordinate plane (in the region where \( 0 < X_1 < X_2 < 6 \)).

- The total area of the triangle formed by these restrictions in this coordinate system is represented by:

\[
\text{Area of valid region} = \frac{1}{2} \times 6 \times 6 = 18
\]

- However, the actual region that must be considered is bounded by the triangle formed by \( (0,0), (6,0), (6,6) \). The area of the square is \( 6 \times 6 = 36 \).

5. **Calculating the Probability:**

- The area of the successful outcomes where all pieces are shorter than 6 units is the same as the area where the inequalities hold. As we analyzed, if we consider the constraints, the condition is satisfied for all relative placements of the breaking points.

- Thus the probability \( P \) of obtaining such a division where each piece is shorter than 6 units is simply:
\[
P = \frac{\text{area of valid choices}}{\text{total area}} = \frac{18}{36} = \frac{1}{2}
\]

Thus, the probability that all three resulting pieces are shorter than 6 units is \(\frac{1}{2}\) or 50%.

(x,y) = (3 + 2*sqrt(2), 7 - 2*sqrt(2))

The minimum value is 5/9.

We put the system in matrix form:

[2 1 0] [t] = 3.15

[0 2 1] [b] = 4.85

[1 2 1] [m] = 5.25

Solving, we get b = 1.40, t = 1.20, m = 1.60.  Therefore, one bagel costs 140 cents.