It's a right triangle again!

Knowing that all Pythagorean triplets $(a,b,c)$ can be expressed as:

$a = p^2 - q^2$ , $b = 2pq$ , $c = p^2 + q^2$ ; $(p,q \in \mathbb{N}, p > q)$

and solving for the altitude in question gives:

$\sin \theta = \frac{h}{2pq} = \frac{p^{2} - q^{2}}{p^{2} + q^{2}} \Rightarrow h = \frac{(2pq)(p^{2} - q^{2})}{p^{2} + q^{2}}$ (where $\theta$ is one of the acute angles).

Now substitution of these expressions gives:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{h} = \frac{1}{p^{2} - q^{2}} + \frac{1}{2pq} + \frac{p^{2} + q^{2}}{(2pq)(p^{2} - q^{2})} = 1$ ;

or $(2pq) + (p^2 - q^2) + (p^2 + q^2) = (2pq)(p^2 - q^2);$

or $2p(p+q) = 2pq(p+q)(p-q)$ ;

or $1 = q(p-q)$ ;

or $\boxed{p =2, q = 1}$ .

This result shows that the 3-4-5 Pythagorean triangle is the only solution that satisfies $\frac{1}{a} + \frac{1}{b} + \frac{1}{h} = 1$ .

Starting with the target condition $1/a+1/b+1/h=1$ and rearranging to solve for $h$ , we see that $h=ab/(ab-a-b)$ .

Now, let's call the hypotenuse $c$ . The area of the triangle, then, equals $0.5ab = 0.5hc$ Solving for $c$ , we get $c=ab/h$ . And plugging in our above definition of $h$ , we see that $c=ab-a-b$ .

By the Pythagorean Theorem, we know $a^2+b^2=c^2$ , and plugging in our new expression for $c$ , we get $a^2+b^2=a^2b^2+a^2+b^2-2a^2b-2ab^2+2ab$ . Simplifying, we get $0=a^2b^2-2a^2b-2ab^2+2ab$ , and since neither $a$ nor $b$ can be $0$ , we can divide through by $ab$ and use SFFT to get $0=(a-2)(b-2)-2$ . Since both $a$ and $b$ have to be integers, the only solutions to this are where one of the factors is 2, and the other is 1. So there is only one triangle that satisfies the target condition, which ends up having $a=3$ and $b=4$ .

$a^2+b^2=c^2,~~~~~~~h*c=ab.\\ \therefore~\dfrac 1 a+\dfrac 1 b+\dfrac 1 h=1,~~~~~~\implies~\dfrac{a+b}{ab}+\dfrac{\sqrt{a^2+b^2}}{ab} = 1.\\ Squaring~ to~ eliminate~ square~ root~ sign~and~ simplifying, ~since~ab~\neq 0, we~get,\\ ab=2(a+b-1),~ reduces ~to~~~\color{#D61F06}{a*\dfrac{b-2}{b-1}=2.......(1)}\\ a, ~b~ are~ integers,~\dfrac{b-2}{b-1},~a ~fraction.~\therefore~let~,~~ \dfrac a {b-1}=n,~a~+tive~integer.\\ So~\color{#D61F06}{n*(b-2)=2} .\\ Let~n=1,~~\implies,~~b=4,~and~a*\dfrac 2 3=2,~that ~is~a=3.\\ If~n=2, ~(1) ~is~ not~ satisfied.\\ If~n>2,~from~(1)~we~see~b~does~not~remain~an~integer.\\ So~ only~one~solution,~~3-4-5.$