April wrote: The inradius of triangle $ ABC$ is $ 1$ and the side lengths of $ ABC$ are all integers. Prove that triangle $ ABC$ is right-angled. We have that $ S=p.r$ where $ S$ is area, $ p$ is semiperimeter and $ r$ is inradius of a given triangle. From Heron's formula we have: $ S=\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)}$ Thus $ 1=r=\frac{S}{p}=\sqrt{\frac{\left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)}{\left(\frac{a+b+c}{2}\right)}}$ or $ \left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=\frac{a+b+c}{2}$ From triangle inequality we have that $ a+b>c$ but since sides of the triangle are integers, we can write $ a+b=c+x$ where $ x$ is positive integer. Similarly, we get that $ b+c=a+y$ and $ c+a=b+z$, with $ y$ and $ z$ positive integers. Also we have that $ x+y=2b$, $ y+z=2c$ and $ z+x=2a$ thus $ x,y,z$ are all odd or all even. Form equation above we have: $ \left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=\frac{a+b+c}{2}$ is equivalent to $ \frac{xyz}{8}=\frac{x+y+z}{2}$ or $ xyz=4(x+y+z)$ Since $ x,y,z$ all cannot be odd hence they are all even. We get equation $ x_{1}y_{1}z_{1}=x_{1}+y_{1}+z_{1}$ We can easily get (maybe it is well known) that the only solution in positive integers of this equation is $ (x_{1},y_{1},z_{1})=(1;2;3)$ and all possible permutations. Hence we get that $ (x,y,z)=(2,4,6)$ and solving the following system of equations $ a+b=c+2$ $ b+c=a+4$ $ c+a=b+6$ we get that sides of a triangle are $ (a,b,c)=(3;4;5)$ but these are the sides of a right triangle.

I like this problem and your solution Quote: We can easily get (maybe it is well known) that the only solution in positive integers of this equation... Can you show me your proof please ? But are you sure isn't $ \left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=(\frac{a+b+c}{2})^{2}$ ??

zweig wrote: Quote: We can easily get (maybe it is well known) that the only solution in positive integers of this equation... Can you show me your proof please ? $ a+b+c=abc \Rightarrow (c-1)(ab-1)+(a-1)(b-1)=2$ After checking three cases we get solutions. zweig wrote: But are you sure isn't $ \left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=(\frac{a+b+c}{2})^{2}$ ?? There shouldn't be a square. After squaring both sides of the equation $ S=r.p$ we get: $ \left(\frac{a+b+c}{2}\right)\left(\frac{a+b-c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{c+a-b}{2}\right)=\left(\frac{a+b+c}{2}\right)^{2}$ and after cancelling $ \frac{a+b+c}{2}$ we get the result.

This problem was also proposed to 2006 Catalunya Mathematical Olympiad (problem 3): http://www.xtec.es/recursos/mates/aqui/olimp42/olimp42.pdf in english, the problem says 3. Prove that the unique triangle with integer sides and such that area is equal to semiperimeter, is the one with sides 3,4 and 5.

The double area of this triangle is equal to $ pr = p = a+b+c$, where a,b,c are the sides. So, S=p. After substituting $ a+b-c = x$, $ b+c-a = y$ and $ a+c-b = z$, by Heron we get: $ x+y+z = xyz$, where $ x,y,z$ are all positive integers. Wlog, $ x\leq y\leq z$. Then, $ xyz = x+y+z\leq 3z$, or $ xy\leq 3$, where $ x,y\geq 1$. From here we get that the sides of this triangle are 3,4 and 5. As $ 3^{2}+4^{2}= 5^{2}$, this triangle is right angled, Q.E.D.

Wow, What a undoubtable solution

The area is equal to the semiperimeter, so the semiperimeter is an integer. By Heron's, this is equivalent to: $$\sqrt{s(s-a)(s-b)(s-c)}=s\Leftrightarrow (s-a)(s-b)(s-c)=s.$$Let $x=s-a,y=s-b,z=s-c$. Note that $\{x,y,z\}\subseteq\mathbb Z$, and $xyz=x+y+z$. WLOG, assume that $x\ge y\ge z$, then $xyz\le3z$, so $yz\le3$. Case 1: $yz=1$ Then $x=x+y+z$ which is absurd. Case 2: $yz=2$ Then $x=y+z=3$ since $y=2$ and $z=1$ must hold. These together imply $s=x+y+z=6$, so $(a,b,c)=(3,4,5)$ and permutations, which are right-angled. Case 3: $yz=3$ Then $2x=y+z=4$, so $x=2$, a contradiction with our ordering assumption.