Triangle ABC has altitudes AD, BE, and CF. If AD=14, BE=16, and CF is a positive integer, then find the largest possible value of CF.

Ive tried drawing things out but don't know how to figure it out. If anyone can help that would be great.

surfz Aug 14, 2024

#2**+1 **

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 \( \boxed{112} \).

kittykat Aug 16, 2024