If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled.

Question

If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius My Attempt From sine law, $$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$ So, $$a=2R \sin A$$ $$b=2R \sin B$$ $$c=2R \sin C$$ Then, $$8R^2=a^2+b^2+c^2$$ $$8R^2=4R^2 \sin^2 (A)+ 4R^2 \sin^2 (B) + 4R^2 \sin^2 C$$ $$8R^2=4R^2(\sin^2 (A)+\sin^2 (B) +\sin^2 (C)$$ $$2=\sin^2 (A)+\sin^2 (B)+\sin^2 (C)$$

Answer 1

Since $$4 = 2\sin^2 A + 2\sin^2 B + 2\sin^2 C$$ we have $$(1-2\sin^2 A ) + (1-2\sin^2B) + 2 - 2\sin^2C=0.$$ or $$2\cos^2 C + (\cos(2A) + \cos(2B)) = 0$$ Since $$\cos(2A)+\cos(2B) = 2\cos(A+B) \cos(A-B) = -2\cos C\cos(A-B),$$ we have $$2\cos C(\cos C - \cos(A-B)) = 0$$ Replace $\cos C = -\cos(A+B)$, we get $$ \cos C(\cos(A+B) + \cos(A-B)) = 0$$ equivalently $$\cos A \cos B \cos C = 0.$$ The conclusion follows.

Answer 2

HINT:

Use the equality valid for any $A$, $B$, $C$ with sum $\pi$ $$1-(\cos^2 A + \cos^2 B + \cos^2 C)- 2 \cos A \cos B \cos C = 0$$

ADDED: The equality follows from the following formula valid for any angles $\alpha$, $\beta$, $\gamma$ with sum $2s$

$$1-(\cos^2\alpha + \cos^2 \beta + \cos^2 \gamma)- 2 \cos \alpha \cos \beta \cos \gamma = - 4 \cos s\cos (s-\alpha) \cos (s-\beta) \cos (s - \gamma)$$

Answer 3

If $8R^2=a^2+b^2+c^2$ then putting $$\begin{cases}R=\dfrac{r^2+s^2}{2}\\a=r^2-s^2\\b=2rs\\c=r^2+s^2\end{cases}$$ it is verified the identity $$8(\dfrac{r^2+s^2}{2})^2=(r^2-s)^2+(2rs)^2+(r^2+s^2)^2\iff2(r^2+s^2)^2=2(r^2+s^2)^2$$ This show that $a,b,c$ satisfy the well known parametrics of the Pythagorean triples (when $c$ is the diameter of the circumcircle i.e. $2R=r^2+s^2$).

Answer 4

By the law of cosines, \begin{align} a^2+b^2+c^2 &= 2ab\cos\gamma+ 2bc\cos\alpha+ 2ca\cos\beta \tag{1}\label{1} \end{align}

Using expressions for the area $S$ of triangle \begin{align} S&=\tfrac12ab\sin\gamma= \tfrac12bc\sin\alpha= \tfrac12ca\sin\beta ,\\ S&=2R^2\sin\alpha\sin\beta\sin\gamma , \end{align}

we have \begin{align} a^2+b^2+c^2 &= 2ab\sin\gamma\,\cot\gamma+ 2bc\sin\alpha\,\cot\alpha+ 2ca\sin\beta\,\cot\beta \\ &=4S(\cot\alpha+\cot\beta+\cot\gamma) \\ &=8R^2 \sin\alpha\sin\beta\sin\gamma (\cot\alpha+\cot\beta+\cot\gamma) =8R^2 , \end{align}

\begin{align} \cos\alpha\sin\beta\sin\gamma+ \sin\alpha\cos\beta\sin\gamma+ \sin\alpha\sin\beta\cos\gamma &=1 ,\\ (\cos\alpha\sin\beta+\sin\alpha\cos\beta)\sin\gamma+ \sin\alpha\sin\beta\cos\gamma &=1 ,\\ \sin(\alpha+\beta)\sin(\alpha+\beta) + \sin\alpha\sin\beta\cos\gamma &=1 ,\\ \sin^2(\alpha+\beta) + \sin\alpha\sin\beta\cos\gamma &=1 ,\\ 1-\cos^2(\alpha+\beta) - \sin\alpha\sin\beta\cos(\alpha+\beta) &=1 ,\\ -\cos(\alpha+\beta)(\sin\alpha\sin\beta+\cos(\alpha+\beta)) &=0 ,\\ -\cos(\alpha+\beta)(\sin\alpha\sin\beta+\cos\alpha\cos\beta-\sin\alpha\sin\beta) &=0 ,\\ -\cos(\alpha+\beta)\cos\alpha\cos\beta&=0 ,\\ \cos\alpha\cos\beta\cos\gamma&=0 . \end{align}

Answer 5

$$F=\sin^2A+\sin^2B+\sin^2C=1-\cos^2A+1-(\cos^2B-\sin^2C)$$

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

and $\cos(B+C)=\cos(\pi-A)=?$

$$2-F=\cos^2A+\cos(B+C)\cos(B-C)$$

$$=\cos^2A-\cos A\cos(B-C)$$

$$=\cos A\{\cos A-\cos(B-C)\}$$

$$=-\cos A\{\cos(B+C)+\cos(B-C)\}=?$$