Bài 3. CÁC HỆ THỨC LƯỢNG TRONG TAM GIÁC VÀ GIẢI TAM GIÁC

\(\overrightarrow{AB}.\overrightarrow{CB}+\overrightarrow{AC}.\overrightarrow{BC}=12\)

\(\Leftrightarrow\overrightarrow{BC}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)=12\)

\(\Leftrightarrow\overrightarrow{BC}.\overrightarrow{BC}=12\)

\(\Rightarrow BC^2=12\Rightarrow BC=2\sqrt{3}\)

Gọi G là trọng tâm tam giác \(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\)

\(\overrightarrow{MA}^2+\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MA}.\overrightarrow{MC}=0\)

\(\Leftrightarrow\overrightarrow{MA}\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)=0\)

\(\Leftrightarrow\overrightarrow{MA}\left(\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\right)=0\)

\(\Leftrightarrow3\overrightarrow{MA}.\overrightarrow{MG}=0\)

\(\Rightarrow\) M thuộc đường tròn đường kính AG

Bán kính: \(R=\dfrac{1}{2}AG=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{6}\)

\(\widehat{ABC}=180^0-\left(30^0+75^0\right)=75^0\)

\(\Rightarrow\Delta ABC\) cân tại A \(\Rightarrow AB=AC=6\)

\(S_{ABC}=\dfrac{1}{2}AB.AC.sinA=\dfrac{1}{2}.6.6.sin30^0=9\)

A đúng Vì:

Trong 1 tam giác ta luôn có :

\(b-c< a\Rightarrow\left(b-c\right)^2< a^2\Rightarrow b^2+c^2-a^2< 2bc\Rightarrow\dfrac{b^2+c^2}{4}-\dfrac{a^2}{4}< \dfrac{2bc}{4}\Rightarrow\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}< \dfrac{b^2+c^2+2bc}{4}=\dfrac{\left(b+c\right)^2}{4}\Rightarrow\sqrt{\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}}< \dfrac{b+c}{2}\) Mà \(m_a=\sqrt{\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}}\Rightarrow m_a< \dfrac{b+c}{2}\)

\(\overrightarrow{BM}=\dfrac{1}{3}\overrightarrow{MC}=\dfrac{1}{3}\left(\overrightarrow{MB}+\overrightarrow{BC}\right)\Rightarrow\overrightarrow{BM}=\dfrac{1}{4}\overrightarrow{BC}\)

\(k\overrightarrow{AN}=\overrightarrow{CN}=\overrightarrow{CA}+\overrightarrow{AN}\Rightarrow\left(1-k\right)\overrightarrow{AN}=\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}\)

\(\Rightarrow\overrightarrow{AN}=\dfrac{1}{1-k}\overrightarrow{AB}+\dfrac{1}{1-k}\overrightarrow{AD}\)

\(\overrightarrow{AM}.\overrightarrow{DN}=0\Leftrightarrow\left(\overrightarrow{AB}+\overrightarrow{BM}\right)\left(\overrightarrow{DA}+\overrightarrow{AN}\right)=0\)

\(\Leftrightarrow\left(\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AD}\right)\left(\dfrac{1}{1-k}\overrightarrow{AB}+\dfrac{k}{1-k}\overrightarrow{AD}\right)=0\)

\(\Rightarrow\dfrac{1}{1-k}AB^2+\dfrac{k}{4\left(1-k\right)}AD^2=0\)

\(\Leftrightarrow\dfrac{1}{1-k}+\dfrac{k}{4\left(1-k\right)}=0\Leftrightarrow k=-4\)

Đáp án B

\(S_{ABN}=3S_{ANC}\) , mà \(S_{ABN}+S_{ANC}=S_{ABC}\)

\(\Rightarrow S_{ANC}=\dfrac{1}{4}S_{ABC}\Rightarrow\overrightarrow{NC}=\dfrac{1}{4}\overrightarrow{BC}\)

Gọi \(N\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{NC}=\left(-1-x;-2-y\right)\\\overrightarrow{BC}=\left(-3;-5\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-1-x=-\dfrac{3}{4}\\-2-y=-\dfrac{5}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=-\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow N\left(-\dfrac{1}{4};-\dfrac{3}{4}\right)\)

Gọi G là trọng tâm tam giác\(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)

Đặt \(P=MA^2+MB^2+MC^2=\left(\overrightarrow{MG}+\overrightarrow{GA}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GB}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GC}\right)^2\)

\(=3MG^2+GA^2+GB^2+GC^2+2\overrightarrow{MG}\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)\)

\(=3MG^2+GA^2+GB^2+GC^2\)

Do  \(GA^2+GB^2+GC^2\) ko đổi nên \(P_{min}\) khi \(MG_{min}\Leftrightarrow M\) là chân đường vuông góc hạ từ G xuống BC

\(\Rightarrow\dfrac{CM}{BC}=\dfrac{2}{3}\Rightarrow\dfrac{BM}{BC}=\dfrac{1}{3}\)

\(\Rightarrow\dfrac{S_{ABM}}{S_{ABC}}=\dfrac{1}{3}\)