Question

The number of triangle(s) whose side lengths are consecutive integers and one of the angle is twice as large as another, is
(correct answer + 2, wrong answer - 0.50)

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

In $△ABC$, let
$\angle B=x,\angle A=2x\Rightarrow \angle C=180-3x$,
Now,
$\sin C=\sin 3x=\sin 3B\Rightarrow \sin A=\sin 2x=\sin 2B\Rightarrow {\sin }^{2}A={\sin }^{2}2B\Rightarrow {\sin }^{2}A=2\sin B\cos B\sin 2B\Rightarrow {\sin }^{2}A=\sin B(\sin B+\sin 3B)\Rightarrow {\sin }^{2}A=\sin B(\sin B+\sin C)\Rightarrow a^2=b(b+c)\Rightarrow a^2-b^2=bc\cdots \left(1\right)$

When $b$ is shortest,
Case $1:a$ is the largest, then
$(b+2{)}^2-b^2=b(b+1)\Rightarrow 2(2b+2)=b(b+1)\Rightarrow 4(b+1)=b(b+1)\Rightarrow b=4$
The triangle sides are $4,5,6$

Case $2:c$ is the largest, then
$(b+1{)}^2-b^2=b(b+2)\Rightarrow 2b+1=b^2+2b\Rightarrow b=1$
The triangle sides are $1,2,3,$ which is not possible.

When $c$ is the shortest,
Only one possible side squence
$c,b,a\to c,c+1,c+2$
Using equation $\left(1\right)$,
$(c+2{)}^2-(c+1{)}^2=(c+1)c\Rightarrow 2c+3=c^2+c\Rightarrow c^2-c-3=0$
Solution is non integers.

Hence, only one such triangle is possible.