what are the integral sides possible?

Let a triangle have sides $a$,$b$,$c$ and $c$ is the greatest side , triangle ABC is obtuse in nature having integral sides. Find the smallest perimeter possible Given A=2B where A and B are angles of respective vertices

• Is $\angle B$ the obtuse angle? – Sawarnik Apr 30 '14 at 7:56
• @Sawarnik, as $c$ is the greatest side, $C$ must be the obtuse angle. – Gerry Myerson Apr 30 '14 at 9:46
• @GerryMyerson Oh, right, didn't see that condition. – Sawarnik Apr 30 '14 at 11:05

Basic conditions for enumeration

A brute force approach could enumerate triangles with integral edge lengths in order of increasing perimeter. In other words, examine partitions of increasingly larger numbers into three summands, taking the triangle inequality into account. What we'd need is some easy approach to detect this $A=2B$ condition. From the law of cosines, we get

$$\cos A=\frac{-a^2+b^2+c^2}{2bc}\qquad \cos B=\frac{a^2-b^2+c^2}{2ac}$$

We also know that $\cos A=\cos(2B)=2\cos^2B-1$ so we obtain the equation

$$\frac{-2a^2+b^2+c^2}{2bc}=2\left(\frac{a^2-b^2+c^2}{2ac}\right)^2-1 \\ -a^4c + a^2b^2c + a^2c^3 = a^4b - 2a^2b^3 + b^5 - 2b^3c^2 + bc^4 \\ a^4b - 2a^2b^3 + b^5 + a^4c - a^2b^2c - 2b^3c^2 - a^2c^3 + bc^4 = 0 \\ (b + c)(-a - b + c)(a - b + c)(-a^2 + b^2 + bc) = 0$$

The first factor would cannot be satisfied for positive edge lengths. The second translates to $c=a+b$ which would be a degenerate triangle. Likewise the third, $b=a+c$. So the only condition which remains is

$$a^2-b^2-bc=0\tag{1}$$

Now do a bit of enumeration, in order of increasing perimeter, and you find the following combination which satisfies both this and the triangle inequality:

$$a=15\qquad b=9\qquad c=16$$

However, the resulting triangle is not obtuse. So let's add a condition for that:

$$a^2+b^2<c^2$$

Now you get the solution

$$a=28\qquad b=16\qquad c=33$$ Avoiding the enumeration

If you want to avoid the brute force enumeration, start with this reformulation of $\text{(1)}$:

$$c=\frac{a^2-b^2}{b}\in\mathbb N$$

So $a^2$ must be a multiple of $b$. On the other hand, $a$ itself cannot be a multiple of $b$ because otherwise, $bc$ would also be a multiple of $b^2$ so $c$ would be a multiple of $b$, and you could divide all lengths by $b$ to obtain a smaller solution. So for some $s,t,u\in\mathbb N$ you have

$$a=stu\qquad b=s^2u\qquad c=\frac{s^2t^2u^2-s^4u^2}{s^2u}=(t^2-s^2)u$$

You can see that $u=1$ is required for minimality, otherwise you could again divide all lengths by $u$. So $b=s^2=\gcd(a,b)^2$ is a square number.

The triangle inequality states

\begin{gather*} c=t^2-s^2<st+s^2=a+b \\ t^2-st-2s^2<0 \\ -s<t<2s \tag2 \end{gather*}

An obtuse angle at $c$ turns into

\begin{gather*} a^2+b^2=s^2t^2+s^4<t^4-2t^2s^2+s^4=(t^2-s^2)^2=c^2\\ 3s^2t^2<t^4 \\ 3s^2<t^2 \\ \sqrt3s<t \tag3 \end{gather*}

Taking $\text{(2)}$ and $\text{(3)}$ together, you get $\sqrt3s<t<2s$ which you can turn into the condition $\sqrt3s<2s-1$ since $t$, being an integer, must be at least one smaller than $s$. This in turn becomes $s>\frac1{2-\sqrt3}\approx3.7$ so $s=4$ is the first case where both conditions can be satisfied. You get

$$6.9\approx4\sqrt3<t<8$$

This can be satisfied by $t=7$, leading the result above:

$$a=st=4\cdot7=28\qquad b=s^2=4^2=16\qquad c=t^2-s^2=7^2-4^2=33$$

Larger $s$ lead to larger minimal values for $t$ which together lead to larger perimeter. Therefore the above solution is indeed minimal.

• In the last step isnt there any other way than counting ??? – avz2611 Apr 30 '14 at 15:18
• @user142634: I have found a way to avoid the brute force enumeration. – MvG Apr 30 '14 at 20:07
• amazing solution . Thanks 4 tht – avz2611 May 1 '14 at 13:01