Geometry

Given a triangle $ABC$ with altitudes $AD = 12$, $BE = 16$, and $CF = h$, where $h$ is a positive integer, we are to find the largest possible value of $CF$.

 

We use the property that the product of the altitudes of a triangle is proportional to its area:

$$A = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times AC \times BE = \frac{1}{2} \times AB \times CF$$

 

First, express the area $A$ in terms of $a$, $b$, and $c$, the lengths of the sides opposite the respective altitudes:

$$A = \frac{1}{2} \times a \times 12 = \frac{1}{2} \times b \times 16 = \frac{1}{2} \times c \times h$$

 

Thus, we have:

$$a \times 12 = b \times 16 = c \times h$$

 

Let $K$ be the constant of proportionality. Then:

$$a \times 12 = K \quad \text{(1)}$$

$$b \times 16 = K \quad \text{(2)}$$

$$c \times h = K \quad \text{(3)}$$

 

From (1) and (2):

$$a \times 12 = b \times 16$$

 

Solving for $b$:

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

 

From (1) and (3):

$$c \times h = a \times 12$$

 

Solving for $c$:

$$c = \frac{a \times 12}{h}$$

 

For $a$, $b$, and $c$ to form a valid triangle, the triangle inequality must be satisfied:

$$a + b > c, \quad b + c > a, \quad \text{and} \quad c + a > b$$

 

Substituting $b = \frac{3}{4}a$ and $c = \frac{12a}{h}$:

$$a + \frac{3}{4}a > \frac{12a}{h}$$

 

Simplifying:

$$\frac{7a}{4} > \frac{12a}{h}$$

 

Cancel out $a$ (assuming $a \neq 0$):

$$\frac{7}{4} > \frac{12}{h}$$

 

Solving for $h$:

$$h > \frac{12 \times 4}{7} = \frac{48}{7} \approx 6.857$$

 

Since $h$ must be an integer, the minimum possible value for $h$ is 7. We test larger values:

 

Next, test if $h = 7, 8, 9, \ldots$:

 

### For $h = 7$:

$$c = \frac{12a}{7}$$

 

Checking triangle inequality with $a + \frac{3}{4}a > \frac{12a}{7}$:

$$\frac{7}{4} > \frac{12}{7} \quad \text{(True)}$$

 

Other inequalities are tested similarly and hold true:

$$\frac{3}{4}a + \frac{12a}{7} > a \implies \frac{55}{28} > 1 \quad \text{(True)}$$

 

This checks out, hence we test for higher $h$.

 

### For $h = 8$:

$$c = \frac{12a}{8} = \frac{3a}{2}$$

 

$$a + \frac{3}{4}a > \frac{3a}{2} \implies \frac{7}{4}a > \frac{3}{2}a \quad \text{(True)}$$

 

Thus higher $h$:

 

### For $h = 24$:

 

The largest altitude $CF$ satisfying all inequalities is 24. 

 

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