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In a triangle, the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $y$. If $x^2-\mathrm{c}^2-y$, where $\mathrm{c}$ is the length of the third side of the triangle, then the circumradius of the triangle is
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The correct answer is:
$\frac{c}{\sqrt{3}}$
Let $\mathrm{a}$ and $\mathrm{b}$ be the lengths of two sides of $\mathrm{a}$ triangle.
$\therefore \quad$ According to the given condition,
$$
\begin{aligned}
& \mathrm{a}+\mathrm{b}=x \text { and } \mathrm{ab}=y \\
& \therefore \quad x^2-\mathrm{c}^2=y \Rightarrow(\mathrm{a}+\mathrm{b})^2-\mathrm{c}^2=\mathrm{ab} \\
& \Rightarrow a^2+b^2+2 a b-c^2=a b \\
& \Rightarrow a^2+b^2-c^2=-a b \\
& \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2}{2 \mathrm{ab}}=-\frac{1}{2} \\
& \Rightarrow \cos \mathrm{C}=\frac{1}{2} \\
& \Rightarrow \mathrm{C}=\frac{2 \pi}{3} \\
& \Rightarrow \text { circumradius }=\frac{\mathrm{c}}{2 \sin \mathrm{C}}=\frac{\mathrm{c}}{2 \sin \left(\frac{2 \pi}{3}\right)}=\frac{\mathrm{c}}{\sqrt{3}} \\
&
\end{aligned}
$$
$\therefore \quad$ According to the given condition,
$$
\begin{aligned}
& \mathrm{a}+\mathrm{b}=x \text { and } \mathrm{ab}=y \\
& \therefore \quad x^2-\mathrm{c}^2=y \Rightarrow(\mathrm{a}+\mathrm{b})^2-\mathrm{c}^2=\mathrm{ab} \\
& \Rightarrow a^2+b^2+2 a b-c^2=a b \\
& \Rightarrow a^2+b^2-c^2=-a b \\
& \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2}{2 \mathrm{ab}}=-\frac{1}{2} \\
& \Rightarrow \cos \mathrm{C}=\frac{1}{2} \\
& \Rightarrow \mathrm{C}=\frac{2 \pi}{3} \\
& \Rightarrow \text { circumradius }=\frac{\mathrm{c}}{2 \sin \mathrm{C}}=\frac{\mathrm{c}}{2 \sin \left(\frac{2 \pi}{3}\right)}=\frac{\mathrm{c}}{\sqrt{3}} \\
&
\end{aligned}
$$
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