Triangle

To find the largest possible value of $CF$, we'll use the relationship between the altitudes in a triangle and its area. The area $\Delta$ of a triangle can be expressed using any of its altitudes:

 

$$\Delta = \frac{1}{2} \times \text{base} \times \text{height}$$

 

For triangle $ABC$, we have the following relationships:

 

- $\Delta = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times CA \times BE = \frac{1}{2} \times AB \times CF$

 

Let the side lengths be $a = BC$, $b = CA$, and $c = AB$. Then:

 

$$\Delta = \frac{1}{2} \times a \times 12 = \frac{1}{2} \times b \times 16 = \frac{1}{2} \times c \times CF$$

 

This simplifies to:

 

$$\Delta = 6a = 8b = \frac{1}{2} c \times CF$$

 

Equating these expressions:

 

$$6a = 8b \quad \text{and} \quad 6a = \frac{1}{2} c \times CF$$

 

### Step 1: Solve for $a$ and $b$

From $6a = 8b$:

 

$$\frac{a}{b} = \frac{8}{6} = \frac{4}{3}$$

 

Thus, $a = \frac{4}{3}b$.

 

 

### Step 2: Substitute into $6a = \frac{1}{2} c \times CF$

Substitute $a = \frac{4}{3}b$ into $6a = \frac{1}{2} c \times CF$:

 

$$6 \times \frac{4}{3}b = \frac{1}{2} c \times CF$$

 

Simplifying:

 

$$8b = \frac{1}{2} c \times CF \quad \text{so} \quad 16b = c \times CF$$

 

### Step 3: Find the Maximum Value of $CF$

We need to maximize $CF$, which is a positive integer. Since $16b = c \times CF$, and $c$ and $CF$ are integers, $CF$ is maximized when $c$ is minimized.

 

Given the relationship $6a = 8b$, $a = \frac{4}{3}b$, and $c = \frac{16b}{CF}$, the smallest integer value of $c$ occurs when $CF$ is as large as possible.

 

If $CF = 16$, then:

 

$$16b = c \times 16 \quad \text{so} \quad c = b$$

 

Since $c$ is minimized and $CF$ is maximized at $16$, this is the largest possible value for $CF$.

 

Thus, the largest possible value of $CF$ is $\boxed{24}$.