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Given that $R$ is the circumradius of $\triangle ABC$, and $\cos A=\frac1{2R}$, $\cos B=\frac1{R}$ and $\cos C= \frac3{2R}$. Then would the $\triangle ABC$ be unique? If so how easily we may find its area or otherwise the maximum possible area?

Any help is appreciated :)

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3 Answers 3

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Since $$ \cos(B)=2\cos(A)\tag{1} $$ and $$ \cos(C)=3\cos(A)\tag{2} $$ and furthermore, since $A+B+C=\pi$, $$ \cos(C)=-\cos(A+B)\tag{3} $$ Using $(1)$ and $(2)$ in $(3)$ says $$ 3\cos(A)=\sin(A)\sin(B)-2\cos^2(A)\tag{4} $$ Moving $2\cos^2(A)$ to the left and squaring yields $$ \begin{align} 4\cos^4(A)+12\cos^3(A)+9\cos^2(A)&=(1-\cos^2(A))(1-4\cos^2(A))\\ 12\cos^3(A)+14\cos^2(A)-1&=0\\ \cos(A)&=0.24312617957188074493\tag{5} \end{align} $$ Since $R=\dfrac1{2\cos(A)}$ and the area is $2R^2\sin(A)\sin(B)\sin(C)$, we get $$ \begin{align} \text{Area} &=\frac{\sqrt{1-\cos^2(A)}\sqrt{1-4\cos^2(A)}\sqrt{1-9\cos^2(A)}}{2\cos^2(A)}\\ &=4.90482198456130394784\tag{6} \end{align} $$

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By law of sines, we have $$\sin A=\frac{a}{2R}\Rightarrow 1-\left(\frac{1}{2R}\right)^2=\frac{a^2}{4R^2}\Rightarrow a=\sqrt{4R^2-1}.$$ Also, we have $$b=\sqrt{4R^2-4},\ \ \ \ c=\sqrt{4R^2-9}.$$

Then, the area $S$ is $$S=\frac 12bc\sin A=\frac{\sqrt{(4R^2-1)(4R^2-4)(4R^2-9)}}{4R}.$$

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  • $\begingroup$ Ah nice, thanks :) $\endgroup$
    – Sawarnik
    Sep 26, 2014 at 18:01
  • $\begingroup$ @Sawarnik: You are welcome:) $\endgroup$
    – mathlove
    Sep 26, 2014 at 18:03
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Note $A+B+C=\pi,\cos {(A+B+C)}=-1 \implies 1-\cos(A)\sin(B)\sin(C)-\sin(A)\cos(B)\sin(C)-\sin(A)\sin(B)\cos(C)+\cos(A)\cos(B)\cos(C)=0$

let $D=2R \implies D \ge 3 $, we have $\sin A= \sqrt{1-\cos^2 A}=\dfrac{\sqrt{D^2-1^2}}{D},\sin B=\dfrac{\sqrt{D^2-2^2}}{D},\sin C=\dfrac{\sqrt{D^2-3^2}}{D}$

$D^3+1*2*3=3\sqrt{D^2-1^2}\sqrt{D^2-2^2}+2\sqrt{D^2-1^2}\sqrt{D^2-3^2}+1*\sqrt{D^2-3^2}\sqrt{D^2-2^2} \iff (2D+12)*\sqrt{(D^2-9)}\sqrt{(D^2-4)}=D^4-12D^2+12D+72 \iff D^3-14D-12=0 \iff D=4.1131$

which means the triangle is unique.

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