Let $A = (5,12)$, $B = (0,0)$, and $C = (14,0)$. For a point $P$ in the plane, the minimum value of $PA^2 + PB^2 + PC^2$ can be expressed in the form $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Answer:

45

Step-by-step explanation:

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To find the minimum value of the sum of the squares of distances, we can use calculus. The minimum value can be expressed as $233/9$.

To find the minimum value of $PA^2 + PB^2 + PC^2$, we need to find the point $P$ that minimizes the sum of the squares of the distances from $P$ to $A$, $B$, and $C$. Let's denote the coordinates of $P$ as $(x, y)$. Using the distance formula, we can find the expressions for the squares of the distances:

1. 
   
2.  
3. $PA^2 = (x - 5)^2 + (y - 12)^2$
4. 
   
5.  
6. $PB^2 = x^2 + y^2$
7. 
   
8.  
9. $PC^2 = (x - 14)^2 + y^2$

The sum of these expressions is $PA^2 + PB^2 + PC^2$:

$PA^2 + PB^2 + PC^2 = (x - 5)^2 + (y - 12)^2 + x^2 + y^2 + (x - 14)^2 + y^2$

Simplifying the expression:

$PA^2 + PB^2 + PC^2 = 3x^2 + 3y^2 - 38x - 24y + 365$

To find the minimum value, we can use calculus. Taking the partial derivatives of this expression with respect to $x$ and $y$ and setting them to zero, we can find the critical points. The coordinates of the point $P$ that minimizes the sum of the squares of the distances are $(x, y) = (13/3, 8/3)$. Plugging these values into the expression, we get:

$PA^2 + PB^2 + PC^2 = (13/3)^2 + (8/3)^2 = 233/9$

Therefore, the minimum value can be expressed as $233/9$, and $m + n = 233 + 9 = 242$.

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