In a right triangle, the perimeter is equal to 30. How many integer values can the hypotenuse take? (Answer:2)
I did:
$a+b+h = 30\rightarrow a+b = h-30\\
a^2+b^2 = h^2 \rightarrow h = \sqrt{a^2+b^2}\\
a-b<h<a+b \rightarrow a-b < h < \sqrt{a^2+b^2}-30...\\
\text{I didn't find other relationships...}
 $
 A: Lower bound: use AM-QM inequality:
$$\frac{h^2}{2}=\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2=\frac{(30-h)^2}{4}$$
Solving the quadratic inequality for positive $h$ will give to you $h\ge 30(\sqrt{2}-1)= 12.42\ldots$ (lowest $h$ is achieved when the triangle is isosceles)
Upper bound: use $a+b>h$
$30=a+b+h>h+h=2h$
Therefore $h<15$
So the integral values of $h$ are $13$ and $14$
You can verify by solving $a+b=30-h$ and $a^2+b^2=h^2$
For $h=13$ you will get the well-known Pythagorean triplet $5,12,13$
For $h=14$ you will get $8-\sqrt{34},8+\sqrt{34},14$
A: The sides of a general triangle are the roots of the cubic equation
\begin{align} 
 x^3-2\rho x^2+(\rho^2+r^2+4rR)x-4\rho r R&=0
 \tag{1}\label{1}
\end{align}
where $\rho,r$ and $R$ are the semiperimeter, the inradius and
the circumradius, respectively, of the corresponding triangle.
We are given that $\rho=\tfrac{30}2=15$ and also,
since $c$ is the hypotenuse, $c=2R$. So, \eqref{1} transforms to
\begin{align} 
c(r+c-15)^2&=0
\tag{2}\label{2}
\\
\text{or just}\quad
c&=15-r
\tag{3}\label{3}
,
\end{align}
so the integer values of $c$ are defined by integer values of $r$.
Also, since for any valid non-degenerate non-equilateral triangle
\begin{align}
r&\in(0,\tfrac R2)
\tag{4}\label{4}
,
\end{align}
we must have
\begin{align} 
15-r&> 4r
\tag{5}\label{5}
\\
\text{or }\quad
r&<3
\tag{6}\label{6}
\end{align}
and we have only two options left:
\begin{align}
r_1&=1
,\quad
c_1=14
\tag{7}\label{7}
\\
\text{and }\quad
r_2&=2
,\quad
c_2=13
\tag{8}\label{8}
.
\end{align}
So, the answer is indeed, $2$, we don't even have to find the other side lengths.
However, it's trivial, since by factoring \eqref{1} they can be found
as a roots of quadratic
\begin{align}
x^2-(30-c) x+450-30c&=0
\tag{9}\label{9}
,
\end{align}
hence for $c_1=14$ two other sides are
$8\pm\sqrt{34}$
and for
$c_2=13$ two other sides are
$5$ and $12$.

Edit
Using geometric approach, when $c=|AB|$ is fixed, the third point $C$
must be located at the intersection of the circle centered at
$O$, the midpoint of $AB$, and the ellipse focused at $A,B$
for which $|AC|+|BC|=30-|AB|$. These are only two possible solutions, that corresponds to $c_1$ and $c_2$.
\begin{align}
\begin{array}{ccc} 
R            & \text{major semi-axis} & \text{minor semi-axis} 
\\ \hline
7 & 8 & \sqrt{15}
\\ 
6.5 & 8.5 & \sqrt{30}
\\ \hline
\end{array}
\end{align}


A: The sides of a right triangle with integer sides are of the form
$a(m^2-n^2), 2amn, a(m^2+n^2)$
for integer $a, m, n$
with $m > n$.
We therefore want
$p
=2a(m^2+mn)
=2am(m+n)$
where $p=30$ here,
or
$p/2
=am(m+n)$.
For p=30,
possible values for $a$ are
1, 3, 5, 15.
For these,
$m(m+n)
=15, 5, 3, 1
$.
If
$m(m+n)=15$
with $m > n$,
$m=3, n=2,$
sides $=5, 12, 13$.
If
$m(m+n)=5, 3, 1$,
no solutions.
Here are bounds on
$m$ and $n$
with $m > n$
and
$m(m+n)=r$
in terms of $r$.
If
$m(m+n)=r$,
$n
=\dfrac{r}{m}-m
$
so,
since $m > n$,
$\dfrac{r}{m}-m
\lt m$
or
$r < 2m^2$
or
$m > \sqrt{r/2}
$
and $m | r$.
Also,
if
$m(m+n)=r$,
$m^2+mn-r=0$,
$m
=\dfrac{-n+\sqrt{n^2+4r}}{2}
$.
Since $m > n$,
$-n+\sqrt{n^2+4r}
\gt 2n$
or
$n^2+4r > 4n^2
$
or
$n < \sqrt{4r/3}
$.
Similarly,
$r \ge m(m+1)$
or
$4r+1 \ge 4m(m+1)+1
=(2m+1)^2
$
so
$m \le \sqrt{4r+1}/2
=\sqrt{r+\frac14}
\lt \sqrt{r}+\frac12
$.
A: There are indeed not enough restrictions. We can write $\cos(\theta) = \frac a h$, $\sin(\theta) = \frac b h$, hence $h\cos(\theta)+h\sin(\theta)+h=30$, or $\cos(\theta)+\sin(\theta)+1 = \frac{30}h$.
$\cos(\theta)+\sin(\theta)$ is equal to $\sqrt 2\sin(\theta +\frac{\pi}4)$, so we get $\sqrt 2 \sin(\theta +\frac{\pi}4) = \frac2 {\sqrt 2}(\frac{30}h-1)$ or $\theta = \sin^{-1}(\frac2 {\sqrt 2}(\frac{30}h-1))-\frac{pi}4$, which has solutions for more than two values of $h$.
If we have the requirement that all sides are positive integers, there's only one solution.
All Pythogorean triples are of the form $2kmn$, $k(m^2-n^2)$, $k(m^2+n^2)$ for some positive integers $k,n,m$. $k$ has to be a factor of $30$, so that gives $1,2,3,5,6,10,30$. The smallest Pythogorean triplet is $3,4,5$, which adds up to $12$, for $k=2.5$, which doesn't work because $k$ needs to be an integer. All larger $k$ are prohibited (a larger $k$ would means that the reduced triples is smaller, but $3,4,5$ is smallest). The only smaller $k$ is $1$, so we now know that any solution must be of the form $2mn$, $m^2-n^2$, $m^2+n^2$. Adding those together, we get $2m^2+2mn=2m(m+n)$, thus $m(m+n)=15$. The only factorizations of $15$ are $15\cdot1$ and $5\cdot3$. The factorization $15\cdot1$ would require $m=1$, which doesn't work ($m>n$, so if $m=1$, $n=0$, but $m$ and $n$ both have to be positive), so that leaves $5\cdot3$. Since $m+n>m$, the $3$ must be $m$ and the $5$ is $m+n$, so $n=2$. That gives $5,12,13$.
