# If $2$ altitudes of a triangle are $9$ and $40$ then find the minimum perimeter

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 $$31. Hence I concluded that the minimum perimeter was $$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.

• A) Do you understand why 83 is not attainable? B) Are you making the assumption that the side lengths must be integers, or is that given? – Calvin Lin Nov 24 '20 at 3:32
• @Calvin Lin A) no, that's exactly what I don't understand B) yes, I forgot to mention that in the question; edited now – l1mbo Nov 24 '20 at 3:33
• A) Actually $a \geq 9$. B) In order for $a_{min} = 9$, what other conditions must hold? – Calvin Lin Nov 24 '20 at 3:34
• @Calvin Lin I understood now, fixing two sides and having a fixed altitude fixes the third side, thx for help – l1mbo Nov 24 '20 at 3:44

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)