Help w/ altitudes

Given that triangle $ABC$ has altitudes $AD = 14$, $BE = 16$, and $CF = h$, where $h$ is a positive integer, we need to find the largest possible value of $h$.

 

### Step 1: Use the area formula with altitudes

The area of triangle $ABC$ can be expressed in terms of the altitudes and the corresponding sides:

 

$$\text{Area} = \frac{1}{2} \times a \times h_a = \frac{1}{2} \times b \times h_b = \frac{1}{2} \times c \times h_c$$

 

where $h_a = AD$, $h_b = BE$, and $h_c = CF = h$, and $a$, $b$, and $c$ are the lengths of the sides opposite the vertices $A$, $B$, and $C$, respectively.

 

### Step 2: Express the sides in terms of the area and altitudes

Let the area of triangle $ABC$ be denoted as $K$. Then:

 

$$K = \frac{1}{2} \times a \times 14 = 7a$$

$$K = \frac{1}{2} \times b \times 16 = 8b$$

$$K = \frac{1}{2} \times c \times h = \frac{1}{2}ch$$

 

### Step 3: Equate the expressions for the area $K$

From the first two equations:

 

$$7a = 8b \quad \Rightarrow \quad a = \frac{8b}{7}$$

 

Using the equation for $K$ involving $h$:

 

$$7a = \frac{1}{2} ch$$

 

Substitute $a = \frac{8b}{7}$ into the equation:

 

$$7\left(\frac{8b}{7}\right) = \frac{1}{2}ch$$

 

Simplify:

 

$$8b = \frac{1}{2}ch$$

 

Multiply both sides by 2:

 

$$16b = ch$$

 

Thus:

 

$$h = \frac{16b}{c}$$

 

### Step 4: Maximize $h$

We aim to maximize $h = \frac{16b}{c}$, where $b$ and $c$ are positive integers.

 

Recall that $K = 7a = 8b = \frac{1}{2}ch$. We want to maximize $h$, so:

 

$$c = \frac{16b}{h}$$

 

To maximize $h$, minimize $c$. Since $b = 7$, $a = 8$, choose $b = 7$ and $c = 1$ which gives:

 

$$h = \frac{16 \times 7}{1} = 112$$

 

However, since $b = \gcd(7,8)$, choose $b = 7$ for simplicity. So:

 

\[
K = 112, h = 1
]

Thus:

 

$$h \leq 112$$

 

### Final Answer:

 

Since $h = 112$, the maximum value for the given maximum number is $\boxed{112}$.