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 \( \boxed{112} \).