If $2$ altitudes of a triangle with integer side lengths are $9$ and $40$ units in length, then find the minimum possible perimeter of the triangle
Since the altitude is the shortest distance from a vertex to the opposite side, I got $a>9$ and $b>40$, where $a$ and $b$ are two of the sides of the triangle. For minimum perimeter I took $a_{min}=10,b=41_{min}$ which gave me $31<c<51 \implies c_{min}=32$. Hence I concluded that the minimum perimeter was $83$. However, the answer given is $90$. I suspect this is the right angled triangle of $(9,40,41)$ but I couldn't see why that would be the one with the smallest perimeter.
Best Answer
I want to add something to Calvin Lin's comment to make it more rigorous.
By equating area we have $$40a=9b$$ Now since $a,b$ are integers and $\gcd(9,40)=1$ we must have $a=9k,b=40k$ for some natural $k$. If $k=1$ we see that the condition is satisfied and $a+b+c=90$.If $k\ge 2$ then we have $a+b+c>a+b=49k\ge 98$ hance minimum of $a+b+c$ is indeed $90$.
Note:Just because $a_{min}=9$ we cant say that the minimum value of $a+b+c$ occurs when this occurs(since a,b,c are not completely independent variables)