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# How does one solve this problem?

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Points A, B, and C are given in the coordinate plane. There exists a point Q and a constant k such that for any point P,
PA² + PB² + PC² = 3PQ² + k.
If A = (4,-4), B = (3,8), and C = (-1, 2), then find the constant k.

Aug 9, 2024

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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:

$PA^2 = (x^2 - 8x + 16) + (y^2 + 8y + 16)$

$PB^2 = (x^2 - 6x + 9) + (y^2 - 16y + 64)$

$PC^2 = (x^2 + 2x + 1) + (y^2 - 4y + 4)$

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

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:

$k = 194 - 24 = 170$

The constant $$k$$ is $$\boxed{170}$$.