Point $P$ is chosen at random inside triangle $ABC$. If $AB = 5$, $BC = 4$, and $CA = 3$, then what is the probability that $P$ is closer to $C$ than it is to either $A$ or $B$?


To solve the problem of finding the probability that a randomly chosen point $ P $ inside triangle $ ABC $ is closer to vertex $ C $ than to vertices $ A $ or $ B $, we can use the concept of the triangle's **Voronoi diagram** or **medial lines**.

### Step 1: Understanding the Voronoi Regions
The triangle is divided into three regions by the **angle bisectors** of the triangle, where each region corresponds to the set of points closer to a particular vertex than to the others. The goal is to find the area of the region associated with vertex $ C $.

### Step 2: Medial Lines and Their Intersection
The lines that bisect the angles $ \angle A $ and $ \angle B $ intersect at a point on side $ AB $, and similarly for the other vertices. The angle bisectors divide the triangle into three smaller regions, where each region is the locus of points closer to one vertex than to the others. 

### Step 3: Using Area Ratios
Given that the problem asks for the probability that $ P $ is closer to $ C $, this probability is equivalent to the ratio of the area of the region associated with $ C $ to the area of the entire triangle $ ABC $. Since the regions are determined by the angle bisectors, the area of each region is proportional to the length of the opposite side.

In triangle $ ABC $:
- $ AB = c = 5 $
- $ BC = a = 4 $
- $ CA = b = 3 $

The area of the region closer to vertex $ C $ is proportional to the length of side $ AB $ (opposite $ C $). The probability that a point $ P $ chosen at random within triangle $ ABC $ is closer to $ C $ than to either $ A $ or $ B $ is the ratio of $ AB $ to the perimeter of the triangle $ ABC $.

### Step 4: Calculate the Probability
The perimeter $ p $ of triangle $ ABC $ is:
\[
p = AB + BC + CA = 5 + 4 + 3 = 12
\]
Thus, the probability $ P(\text{closer to } C) $ is:
\[
P(\text{closer to } C) = \frac{AB}{p} = \frac{5}{12}
\]

### Final Answer
The probability that a randomly chosen point $ P $ inside triangle $ ABC $ is closer to $ C $ than to either $ A $ or $ B $ is $ \boxed{\frac{5}{12}} $.

Would you like further details on this, or do you have any additional questions? Here are some related problems to explore:

1. How would the probability change if the sides of the triangle were different?
2. What is the probability that the point is closer to vertex $ A $ instead?
3. How would the problem change if the triangle were not acute?
4. Can this problem be generalized to a quadrilateral?
5. What are the implications if the point is chosen on the boundary of the triangle?

**Tip:** Understanding how the area of regions within a triangle can be divided proportionally is useful for solving a variety of geometry problems.

Explore the solution to find the probability that a randomly chosen point P inside triangle ABC is closer to vertex C than to vertices A or B. Understand how to apply the Triangle Inequality Theorem and area ratios in geometry problems.

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