Find a triangle such that $a,b,c$ are consecutive natural numbers

Question

Find a triangle $∆ABC$ such that the sides are consecutive natural numbers and one of angle $=$ twice other angle (for example $A=2B$ or $2B=A$ or ...)

I have tried many times but I don't got the complete solution. My try is as follows:

The sides are consecutive natural numbers.

So the sides are: $n,n+1,n+2$ and angle $2\alpha,\alpha,π-3\alpha$

Also I know by law of cosines: $a^{2}=b^{2}+c^{2}-2bc\cos A$.

But how do i find relations?

Answer 1

Hint:- apply sin law ( I.e. x/sin(A)= y/sin(B) =z/sin(C) where x,y,z, are side length and A,B,C are angles ) As the sides you have taken already a, a+1,a+2 And angles will be A,2A,π-3A . Now with these data apply sin law , you will get the desired answer ( it will be 4,5,6).

Solution :-.

 - for the shake of convenience let A=x then B= 2x and C=π-3x 

Applying sine law :- a/(sinx) = (a+2)/(sin2x) = (a+1)/(sin(π-3x))

(1.) a/(sinx) =(a+2)/(sin2x) = (a+2)/(2sinxcosx)

                    Implies, (a+2)/a = 2cosx .............(I) 

(2.) a/(sinx) = (a+1)/(sin(π-3x))

     a/(sinx) = (a+1)/(3sinx-4sin³x)

               3-4cos²x = (a+1)/a 

           4cos²x = ((a+1)/a)+1

        ((a+2)/a)²= (2a+1)/a........(using I ) 

   1/a{(a+2)²/a -(2a+1)} =0.          ...(1/a can't be 0 ) 

 ( a²+4a+4-2a²-a)/a  = 0

     -a²+3a+4=0
   a²-3a-4=0
(a-4)(a+1)=0

Implies , a=4 ( other root we won't take as it is -1 not a natural no ) Hence , sides of triangle is 4, (4+1), (4+2) .

Answer 2

I think it's easiest to let the sides be $s-1$, $s$ and $s+1$.

Let the angles be $P$, $Q$ and $R$ with $P>Q>R$.

Then $P$ is opposite the side with length $s+1$, $Q$ is opposite the side with length $s$, and $P$ is opposite the side with length $s-1$.

I think the Cosine Rule is more useful than the Sine Rule, because you can take advantage of the fact that $\cos(2\theta)=2\cos^2(\theta)-1$.

$\cos (P)=\frac{s^2+(s-1)^2-(s+1)^2}{2s(s-1)}=\frac{s^2+s^2-2s+1-s^2-2s-1}{2s(s-1)}=\frac{s^2-4s}{2s(s-1)}=\frac{s(s-4)}{2s(s-1)}=\frac{s-4}{2(s-1)}$

$\cos (Q)=\frac{(s+1)^2+(s-1)^2-s^2}{2(s+1)(s-1)}=\frac{s^2+2s+1+s^2-2s+1-s^2}{2(s+1)(s-1)}=\frac{s^2+2}{2(s+1)(s-1)}$

$\cos (R)=\frac{s^2+(s+1)^2-(s-1)^2}{2s(s-1)}=\frac{s^2+s^2+2s+1-s^2+2s-1}{2s(s+1)}=\frac{s^2+4s}{2s(s+1)}=\frac{s(s+4)}{2s(s+1)}=\frac{s+4}{2(s+1)}$

We are looking for angles $X$ and $Y=2X$. It is clear that $Y>X$ but not obvious which of the three are $X$ and $Y$.

Case 1: $P=Y=2X$ and $Q=X$

$\cos(P)=2\cos^2 (Q)-1$

$\frac{s-4}{2(s-1)}=2\left(\frac{s^2+2}{2(s+1)(s-1)} \right)^2-1$

$\frac{s-4}{2(s-1)}=\frac{(s^2+2)^2}{2(s+1)^2(s-1)^2} - \frac {2(s+1)^2(s-1)^2}{2(s+1)^2(s-1)^2}$

$(s+1)^2(s-1)(s-4)=(s^2+2)^2 -2(s+1)^2(s-1)^2$

$(s^2+2s+1)(s^2-5s+4)=s^4+4s^2+4-2s^4+4s^2-2$

$s^4-3s^3-5s^2+3s+4=s^4+4s^2+4-2s^4+4s^2-2$

$2s^4-3s^3-13s^2+3s+2=0$

No positive integer roots.

Case 2: $P=Y=2X$ and $R=X$

$\cos(P)=2\cos^2 (R)-1$

$\frac{s-4}{2(s-1)}=2\left(\frac{s+4}{2(s+1)} \right)^2-1$

$\frac{s-4}{2(s-1)}=\frac{2(s+4)^2}{4(s+1)^2} -\frac {4(s+1)^2}{4(s+1)^2}$

$(s+1)^2(s-4)=(s+4)^2(s-1) -2(s+1)^2(s-1)$

$(s^2+2s+1)(s-4)=(s^2+8s+16)(s-1) -2(s+1)(s^2-1)$

$s^3-2s^2-7s-4=s^3+7s^2+8s-16 -2s^3-2s^2+2s+2$

$2s^3-7s^2-17s+10=0$

This gives $s=5$ as a positive integer root

Case 3: $Q=Y=2X$ and $R=X$ (to check there isn't another possibility hiding away there)

$\cos(Q)=2\cos^2 (R)-1$

$\frac{s^2+2}{2(s+1)(s-1)}=2\left(\frac{s+4}{2(s+1)} \right)^2-1$

$\frac{s^2+2}{2(s+1)(s-1)}=\frac{(s+4)^2}{2(s+1)^2}-\frac{2(s+1)^2}{2(s+1)^2}$

$(s^2+2)(s+1)=(s+4)^2(s-1)-2(s+1)^2(s-1)$

$s^3+s^2+2s+2=(s^2+8s+16)(s-1)-2(s^2+2s+1)(s-1)$

$s^3+s^2+2s+2=s^3+7s^2+8s-16-2s^3-2s^2+2s+2$

$2s^3-4s^2-8s+16=0$

This gives $s=2$ as another positive integer root.

Case 2 gives us the correct answer that $s=5$, so the sides are 4, 5 and 6 units long.

What does Case 3 mean?

Taking $s=2$ gives us a triangle with sides 1, 2 and 3 units.

Trying to construct such a triangle gives us instead a few lines drawn over the top of each other: angles $P=180^o$, $Q=0^o$, $R=0^o$.

In such a 'triangle' it is true that $Q=2R$... something to ponder.