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\(\dfrac{h_b}{h_a^2}+\dfrac{h_c}{h_b^2}+\dfrac{h_a}{h_c^2}=\dfrac{\dfrac{2S_{ABC}}{b}}{\dfrac{4S_{ABC}^2}{a^2}}+\dfrac{\dfrac{2S_{ABC}}{c}}{\dfrac{4S^2_{ABC}}{b^2}}+\dfrac{\dfrac{2S_{ABC}}{a}}{\dfrac{4S_{ABC}^2}{c^2}}\)
\(=\dfrac{a^2}{2bS_{ABC}}+\dfrac{b^2}{2cS_{ABC}}+\dfrac{c^2}{2aS_{ABC}}\)
\(=\dfrac{1}{2S_{ABC}}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\)
\(\ge\dfrac{1}{2.\dfrac{a+b+c}{2}r}.\dfrac{\left(a+b+c\right)^2}{a+b+c}=\dfrac{1}{r}\)
Hình như có dấu = chứ nhỉ
Đẳng thức xảy ra khi tam giác ABC đều
\(a=2b-2c\Rightarrow sinA.2R=2sinB.2R-2sinC.2R\)
\(\Rightarrow sinA=2sinB-2sinC\)
\(ah_a=bh_b=ch_c\Rightarrow\left(2b-2c\right)h_a=bh_b=ch_c\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{h_a}=\dfrac{2b-2c}{b}.\dfrac{1}{h_b}\\\dfrac{1}{h_a}=\dfrac{2b-2c}{c}.\dfrac{1}{h_c}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{h_a}=\dfrac{1}{h_b}-\dfrac{1}{h_c}+\left(\dfrac{b}{c.h_c}-\dfrac{c}{b.h_b}\right)\)
Câu này đề sai tiếp, biểu thức \(\dfrac{b}{c.h_c}-\dfrac{c}{b.h_b}\) kia không thể bằng 0
a: Ta có: \(S_{\Delta ABC} = \frac{1}{2} AB \cdot AC \cdot \sin A\)
=>\(S^2 = \frac{1}{4} AB^2 \cdot AC^2 \cdot \sin^2 A\)
\(=\frac{1}{4}AB^2\cdot AC^2(1-\cos^2A)=\frac{1}{2^2}\left[AB^2\cdot AC^2-(AB\cdot AC\cdot\cos A)^2\right]\)
=>\(S^2 = \frac{1}{4} \left[ AB^2 \cdot AC^2 - \left(\vec{AB} \cdot \vec{AC}\right)^2 \right]\)
=>\(S=\frac12\cdot\sqrt{AB^2\cdot AC^2-\left(\overrightarrow{AB}\cdot\overrightarrow{AC}\right)^2}\)
b:Ta có: \(S = \frac{1}{2} a \cdot h_a = \frac{1}{2} b \cdot h_b = \frac{1}{2} c \cdot h_c\)
=>\(a = \frac{2S}{h_a}, \quad b = \frac{2S}{h_b}, \quad c = \frac{2S}{h_c}\)
b+c=2a
=>\(\frac{2S}{h_b}+\frac{2S}{h_c}=2\cdot\left(\frac{2S}{h_a}\right)\)
=>\(\frac{1}{h_b}+\frac{1}{h_c}=\frac{2}{h_a}\)
c: Xét ΔABC có \(m_{a};m_{b};m_{c}\) lần lượt là độ dài ba đường trung tuyến ứng với các cạnh BC,AC,AB
nên ta có:
\(m_{a}^2=\frac{2(b^2 + c^2) - a^2}{4};m_{b}^2=\frac{2(a^2 + c^2) - b^2}{4};m_{c}^2=\frac{2(a^2 + b^2) - c^2}{4}\)
\(m_b^2 + m_c^2 = \frac{2a^2 + 2c^2 - b^2 + 2a^2 + 2b^2 - c^2}{4} = \frac{4a^2 + b^2 + c^2}{4}\)
Ta có: \(m_b^2 + m_c^2 = 5m_a^2\)
=>\(\frac{4a^2 + b^2 + c^2}{4}=5\cdot\frac{2(b^2 + c^2) - a^2}{4}\)
=>\(4a^2+b^2+c^2=10(b^2+c^2)-5a^2\)
=>\(4a^2+5a^2=10(b^2+c^2)-(b^2+c^2)\)
=>\(9a^2=9(b^2+c^2)\)
=>\(a^2=b^2+c^2\)
=>ΔABC vuông tại A
Đặt \(AB=c;BC=a\)
\(S=\frac{1}{2}ah_a=\frac{1}{2}bh_b=\frac{1}{2}ch_c\Rightarrow ah_a=bh_b=ch_c=2S\)
\(\Rightarrow\left\{{}\begin{matrix}h_a=\frac{2S}{a}\\h_b=\frac{2S}{b}\\h_c=\frac{2S}{c}\end{matrix}\right.\) \(\Rightarrow\frac{2S}{b}=\frac{2S}{a}+\frac{2S}{c}\Rightarrow\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\Rightarrow\frac{b}{a}+\frac{b}{c}=1\)
\(cosB=\frac{7}{8}=\frac{a^2+c^2-b^2}{2ac}\Leftrightarrow b^2=a^2+c^2-\frac{7}{4}ac\)
\(\Leftrightarrow\left(\frac{a}{b}\right)^2+\left(\frac{c}{b}\right)^2-\frac{7}{4}\left(\frac{a}{b}\right)\left(\frac{c}{b}\right)=1\)
Đặt \(\left\{{}\begin{matrix}\frac{a}{b}=x>0\\\frac{c}{b}=y>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=1\\x^2+y^2-\frac{7}{4}xy=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=xy\\x^2+y^2-\frac{7}{4}xy=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=xy\\\left(x+y\right)^2-\frac{15}{4}xy=1\end{matrix}\right.\)
\(\Leftrightarrow\left(xy\right)^2-\frac{15}{4}xy-1=0\) \(\Rightarrow\left[{}\begin{matrix}xy=4\\xy=-\frac{1}{4}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=4\\xy=4\end{matrix}\right.\) \(\Rightarrow x=y=2\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{a}{b}=2\\\frac{c}{b}=2\end{matrix}\right.\) \(\Rightarrow a=c=2b\)
\(\Rightarrow p=\frac{a+b+c}{2}=\frac{5b}{2}\) \(\Rightarrow S=\sqrt{p\left(p-b\right)\left(p-2b\right)\left(p-2b\right)}=\frac{b^2\sqrt{15}}{4}\)