Ôn tập cuối năm môn Hình học

Do tam giác ABC vuông tại A và AM là phân giác \(\Rightarrow\widehat{MAC}=\dfrac{1}{2}\widehat{A}=45^0\)

\(tanC=\dfrac{AB}{AC}=\sqrt{3}\Rightarrow C=60^0\)

\(\Rightarrow\widehat{AMC}=180^0-\left(45^0+60^0\right)=75^0\)

Áp dụng định lý hàm sin cho tam giác AMC:

\(\dfrac{AM}{sinC}=\dfrac{AC}{sin\widehat{AMC}}\Rightarrow AM=\dfrac{a.sin60^0}{sin75^0}=a\left(\dfrac{3\sqrt{2}-\sqrt{6}}{2}\right)\)

c: =>(1+cosB)^2/sin^2B=(2a+c)^2/4a^2-c^2

=>\(\dfrac{1+cosB}{1-cosB}=\dfrac{2a+c}{2a-c}\)

=>2ac*cosB=c^2

=>a=b

=>ΔCAB cân tại C

d: \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

=>\(c^2+b^2-a^2=2cb\cdot cos120=-cb\)

=>a^2=c^2+b^2+cb(ĐPCM)

a.

\(\dfrac{cosA}{a}+\dfrac{cosB}{b}+\dfrac{cosC}{c}=\dfrac{a}{bc}\)

\(\Leftrightarrow\dfrac{b^2+c^2-a^2}{2abc}+\dfrac{a^2+c^2-b^2}{2abc}+\dfrac{a^2+b^2-c^2}{2abc}=\dfrac{a}{bc}\)

\(\Leftrightarrow a^2+b^2+c^2=2a^2\)

\(\Leftrightarrow b^2+c^2=a^2\)

\(\Rightarrow\) Tam giác ABC vuông tại A theo Pitago đảo

b.

\(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=2R\Rightarrow\left\{{}\begin{matrix}sinA=\dfrac{a}{2R}\\sinB=\dfrac{b}{2R}\\sinC=\dfrac{c}{2R}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{m_a}=\dfrac{b}{m_b}=\dfrac{c}{m_c}\)

\(\Rightarrow\dfrac{m_a^2}{a^2}=\dfrac{m_b^2}{b^2}=\dfrac{m_c^2}{c^2}\)

\(\Rightarrow\dfrac{2\left(b^2+c^2\right)-a^2}{a^2}=\dfrac{2\left(c^2+a^2\right)-b^2}{b^2}=\dfrac{2\left(a^2+b^2\right)-c^2}{c^2}\)

\(\Rightarrow\dfrac{b^2+c^2}{a^2}=\dfrac{c^2+a^2}{b^2}=\dfrac{a^2+b^2}{c^2}=\dfrac{2\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=2\)

\(\Rightarrow\left\{{}\begin{matrix}b^2+c^2=2a^2\\c^2+a^2=2b^2\\a^2+b^2=2c^2\end{matrix}\right.\)

\(\Rightarrow a^2=b^2=c^2\)

\(\Rightarrow a=b=c\) hay tam giác ABC đều

c.

Do A;B;C là 3 góc của 1 tam giác \(\Rightarrow A+B+C=180^0\)

\(\Rightarrow A+C=180^0-B\)

\(cos\left(A+C\right)+3cosB=1\)

\(\Leftrightarrow cos\left(180^0-B\right)+3cosB=1\)

\(\Leftrightarrow-cosB+3cosB=1\)

\(\Leftrightarrow cosB=\dfrac{1}{2}\)

\(\Rightarrow B=60^0\)

Ta có: t.g AHB = t.g AHC (do H là trực tâm t.g ABC)

=> AB = AC và HB = HC

<=> T.g ABC cân tại A

=> \(\left\{{}\begin{matrix}tanA.\overrightarrow{HA}=\overrightarrow{0}\\tanB.\overrightarrow{HB}=\overrightarrow{0}\\tanC.\overrightarrow{HC}=\overrightarrow{0}\end{matrix}\right.\)

=> đpcm

Đặt \(\overrightarrow{u}=\left(x;\dfrac{1}{y}\right);\overrightarrow{v}=\left(y;\dfrac{1}{z}\right);\overrightarrow{w}=\left(z;\dfrac{1}{x}\right)\)

\(\Rightarrow\left|\overrightarrow{u}\right|=\sqrt{x^2+\dfrac{1}{y^2}}\) ; \(\left|\overrightarrow{v}\right|=\sqrt{y^2+\dfrac{1}{z^2}}\) ; \(\left|\overrightarrow{w}\right|=\sqrt{z^2+\dfrac{1}{x^2}}\)

\(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\left(x+y+z;\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\Rightarrow\left|\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right|=\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)

Áp dụng BĐT vecto:

\(\left|\overrightarrow{u}\right|+\left|\overrightarrow{v}\right|+\left|\overrightarrow{w}\right|\ge\left|\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right|\)

\(\Rightarrow\sqrt{x^2+\dfrac{1}{y^2}}+\sqrt{y^2+\dfrac{1}{z^2}}+\sqrt{z^2+\dfrac{1}{x^2}}\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z^2}\right)}\)

\(\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{9}{x+y+z}\right)^2}=\sqrt{\left(x+y+z\right)^2+\dfrac{1}{\left(x+y+z\right)^2}+\dfrac{80}{\left(x+y+z\right)^2}}\)

\(\ge\sqrt{2\sqrt{\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}}+\dfrac{80}{1}}=\sqrt{82}\)

P=bc/a^2b+a^2c+ac/b^2c+b^2a+ab/c^2a+c^2b

bc/a^2b+a^2c=bc/a^2(b+c)=1/a^2(1/b+1/c)=1/a^2:(1/b+1/c)

Đặt x=1/a; y=1/b; z=1/c

P=x^2/y+z+y^2/x+z+z^2/x+y

\(\left(y+z+x+z+y+x\right)\cdot P>=\left(\sqrt{y+z}\cdot\dfrac{x}{\sqrt{y+z}}+\sqrt{z+x}\cdot\dfrac{y}{\sqrt{z+x}}+\sqrt{x+y}\cdot\dfrac{z}{\sqrt{x+y}}\right)^2\)

=>P>=1/2(x+y+z)=3/2

\(x^2+xy+y^2=\dfrac{3}{4}\left(x+y\right)^2+\dfrac{1}{4}\left(x-y\right)^2\ge\dfrac{3}{4}\left(x+y\right)^2\)

\(\Rightarrow\sqrt{x^2+xy+y^2}\ge\sqrt{\dfrac{3}{4}\left(x+y\right)^2}=\dfrac{\sqrt{3}}{2}\left|x+y\right|\ge\dfrac{\sqrt{3}}{2}\left(x+y\right)\)

Tương tự:

\(\sqrt{y^2+yz+z^2}\ge\dfrac{\sqrt{3}}{2}\left(y+z\right)\)

\(\sqrt{z^2+zx+x^2}\ge\dfrac{\sqrt{3}}{2}\left(z+x\right)\)

Cộng vế:

\(\Rightarrow\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)

Dấu "=" xảy ra khi \(x=y=z\)