Integer length altitudes of a scalene triangle $ABC$

Question

It is given that two of the altitudes of a scalene triangle $ABC$ have length $4$ and $12$. If the length of the third altitude is also an integer, find it's maximum value.

What I Tried: I have no good idea for this. Here is a picture for an example in Geogebra :-

I have to find the maximum value of $AG$ . In this example I suppose, the value of $CD$ can be increased more if we don't take an obtuse triangle, but all I am trying to do here is hit and trial.

Moreover, in this case $AG$ is not an integer (unless you consider it to be $4$) , so how do you make sure that the other altitude is an integer too?

Can anyone help me with this?

Answer 1

$$ \triangle = \dfrac{1}{2}a\cdot4 = \dfrac{1}{2}b\cdot12 = \dfrac{1}{2}c\cdot h$$

$$ {a = \dfrac{\triangle}{2}, b = \dfrac{\triangle}{6}, c = \dfrac{2\triangle}{h}} $$

Now use

$$ a-b \lt c \lt a+b$$

I got $$\boxed{h_{\text{max}}=5}$$

Answer 2

Consider the area of the triangle.

Let $AB = x$. Then the triangle has area $6x$ and thus $AC$ is equal to $3x$.

By triangle inequality, $BC > 2x$.

Since the area is $6x$, this implies that the remaining altitude cannot exceed $6$.

This argument also leads to the triangle inequality for altitudes:

$$\frac1 {h_a} < \frac1{h_b} + \frac1{h_c} \text{ , etc.}$$

To show the existence of a triangle with altitudes $12, 4, 5$, we take note of the corresponding side lengths $x, 3x, 2.4x$ and use Heron's Formula to obtain the corresponding $x$. The lengths will then form a (constructible) triangle with the desired altitudes.