I begin

Geometry Level 4

$\begin{cases} a^{2}+b^{2}+c^{2}=50 \\ a^{3}+b^{3}+c^{3}=216 \\ ab+bc+ca=47 \end{cases}$

For $\triangle ABC$ with sides $a,b$ and $c$ , let $R$ and $r$ denote the radius of circumcircle and incircle of $\triangle ABC$ . If the values of $a,b$ and $c$ satisfy the system of equations above, find the value of $R \times r$ .

The answer is 2.5.

1 solution

Siddharth Singh
Jun 20, 2015

$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$

$(a+b+c)^{2}=50+2(47)$

$(a+b+c)^{2}=144$

$(a+b+c)=12$ -(1)

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-(ab+bc+ca))$

$216-3abc=12*(50-47)$

$3abc=180$

$abc=60$ -(2)

$\frac{abc}{R}=r(\frac{a+b+c}{2})$

$\frac{60}{R}=r(\frac{12}{2})$

$R*r=\frac{60}{12*2}=\boxed { 2.5 }$

Answer should be 10 (Acc. to your solution)

$\large{\frac{60}{R}=r\frac{12}{2} \\ R\times r = \frac{60\times 2}{12}=10}$

Department 8 - 5 years, 7 months ago

I begin

The answer is 2.5.

1 solution

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