Chủ đề này có 2 trả lời

[Leuleudoraemon]

2. Trong tất cả các tam giác nội tiếp đường tròn (O) bán kính R>0  cho trước, hãy xác định tam giác có diện tích lớn nhất. 

[12345678987654321123456789]

Gọi $ABC$ là tam giác bất kì nội tiếp $(O,R)

$D$ là điểm chính giữa cung $BC$ chứa $A$ và $E$ là trung điểm $BC$

Đặt $OE=x\Rightarrow CE=\sqrt{R^2-x^2}$

Dễ thấy

$$S_{ABC}\leq S_{DBC}=CE\cdot DE=\sqrt{R^2-x^2}\cdot (R+x)=\sqrt{(R+x)^3(R-x)}$$

$$=\frac1{\sqrt3}\sqrt{(R+x)(R+x)(R+x)(3R-3x)}\leq\frac1{\sqrt3}\sqrt{\left(\frac{R+x+R+x+R+x+3R-3x}4\right)^4}$$

$$=\frac1{\sqrt3}\sqrt{\left(\frac{3R}2\right)^4}=\frac{3\sqrt3 R^2}4$$

Vậy

$$Max\;S_{ABC}=\frac{3\sqrt3 R^2}4\Leftrightarrow \left\{\begin{matrix}A\equiv D\\R+x=3R-3x\end{matrix} \right.\Leftrightarrow AB=BC=AC$$

 

*** Cannot compile formula:



\definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666}
\definecolor{xdxdff}{rgb}{0.49019607843137253,0.49019607843137253,1.}
\definecolor{ududff}{rgb}{0.30196078431372547,0.30196078431372547,1.}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm]
\clip(-0.28,-2.36) rectangle (5.74,3.78);
\draw [line width=2.pt] (2.72,0.58) circle (2.6461292485439936cm);
\draw [line width=2.pt] (1.28,2.8)-- (0.8148927000958486,-1.2564547845922895);
\draw [line width=2.pt] (0.8148927000958486,-1.2564547845922895)-- (4.639419939222375,-1.2414903504865396);
\draw [line width=2.pt] (1.28,2.8)-- (4.639419939222375,-1.2414903504865396);
\draw [line width=2.pt] (2.70964642655458,3.2261089931287614)-- (0.8148927000958486,-1.2564547845922895);
\draw [line width=2.pt] (2.70964642655458,3.2261089931287614)-- (4.639419939222375,-1.2414903504865396);
\draw [line width=2.pt] (2.70964642655458,3.2261089931287614)-- (2.7271563196591115,-1.2489725675394145);
\draw (2.24,0.04) node[anchor=north west] {$x$};
\begin{scriptsize}
\draw [fill=ududff] (1.28,2.8) circle (2.5pt);
\draw[color=ududff] (1.42,3.17) node {$A$};
\draw [fill=ududff] (2.72,0.58) circle (2.5pt);
\draw[color=ududff] (2.44,0.65) node {$O$};
\draw [fill=xdxdff] (0.8148927000958486,-1.2564547845922895) circle (2.5pt);
\draw[color=xdxdff] (0.38,-1.13) node {$B$};
\draw [fill=xdxdff] (4.639419939222375,-1.2414903504865396) circle (2.5pt);
\draw[color=xdxdff] (4.88,-1.29) node {$C$};
\draw [fill=uuuuuu] (2.70964642655458,3.2261089931287614) circle (2.0pt);
\draw[color=uuuuuu] (2.84,3.55) node {$D$};
\draw [fill=uuuuuu] (2.7271563196591115,-1.2489725675394145) circle (2.0pt);
\draw[color=uuuuuu] (2.86,-0.91) node {$E$};
\end{scriptsize}
\end{tikzpicture}



*** Error message:
Error: Nothing to show, formula is empty

Bài viết đã được chỉnh sửa nội dung bởi 12345678987654321123456789: 01-04-2018 - 10:25

[conankun]

Gọi $ABC$ là tam giác bất kì nội tiếp $(O,R)
$D$ là điểm chính giữa cung $BC$ chứa $A$ và $E$ là trung điểm $BC$
Đặt $OE=x\Rightarrow CE=\sqrt{R^2-x^2}$
Dễ thấy
$$S_{ABC}\leq S_{DBC}=CE\cdot DE=\sqrt{R^2-x^2}\cdot (R+x)=\sqrt{(R+x)^3(R-x)}$$
$$=\frac1{\sqrt3}\sqrt{(R+x)(R+x)(R+x)(3R-3x)}\leq\frac1{\sqrt3}\sqrt{\left(\frac{R+x+R+x+R+x+3R-3x}4\right)^4}$$
$$=\frac1{\sqrt3}\sqrt{\left(\frac{3R}2\right)^4}=\frac{3\sqrt3 R^2}4$$
Vậy
$$Max\;S_{ABC}=\frac{3\sqrt3 R^2}4\Leftrightarrow \left\{\begin{matrix}A\equiv D\\R+x=3R-3x\end{matrix} \right.\Leftrightarrow AB=BC=AC$$
 

*** Cannot compile formula:
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*** Error message:
Error: Nothing to show, formula is empty

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