Question

1. Tính độ dài phân giác trong AD của \(\Delta ABC\) theo \(a=BC;b=CA;c=AB;\alpha=\widehat{BAC}\)

2. Cho \(\Delta ABC,G\) là trọng tâm và M tùy ý.

CM: \(MA^2+MB^2+MC^2=3MG^2+\dfrac{1}{3}\left(a^2+b^2+c^2\right)\)

3. Cho \(\Delta ABC\), tìm max \(P=cosA+cosB+cosC\)

4. Cho \(\Delta ABC\), tìm min \(Q=cos2A+cos2B+cos2C\)

5. Cho \(\Delta ABC\), điểm M tùy ý. Tìm min \(F=\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MB}.\overrightarrow{MC}+\overrightarrow{MC}.\overrightarrow{MA}\)

6. CM: \(F=cos2A+cos2B-cos2C\le\dfrac{3}{2}\)

7. Tứ giác ABCD nội tiếp \(\left(O;R\right)\).

Tìm \(M\in\left(O;R\right)\) sao cho \(F=MA^2+MB^2+MC^2-3MD^2\) đạt min, max

Answer 1

1.

\(\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AB}+\dfrac{c}{b+c}\overrightarrow{BC}=\dfrac{\left(b+c\right)\overrightarrow{AB}+c\overrightarrow{BC}}{b+c}=\dfrac{b\overrightarrow{AB}+c\overrightarrow{AC}}{b+c}\)

\(\Rightarrow AD^2=\dfrac{\left(b\overrightarrow{AB}+c\overrightarrow{AC}\right)^2}{\left(b+c\right)^2}=\dfrac{2b^2c^2+2b^2c^2.cosA}{\left(b+c\right)^2}=\dfrac{2b^2c^2\left(1+cos\alpha\right)}{\left(b+c\right)^2}\)

\(\Rightarrow AD=\dfrac{bc\sqrt{2+2cos\alpha}}{b+c}\)

Answer 2

2.

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

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

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

\(=3MG^2+\dfrac{4}{9}\left(AM^2+MB^2+MC^2\right)\)

\(=3MG^2+\dfrac{4}{9}\left(\dfrac{2b^2+2c^2-a^2}{4}+\dfrac{2a^2+2c^2-b^2}{4}+\dfrac{2a^2+2b^2-c^2}{4}\right)\)

\(=3MG^2+\dfrac{4}{9}.\dfrac{3}{4}\left(a^2+b^2+c^2\right)\)

\(=3MG^2+\dfrac{1}{3}\left(a^2+b^2+c^2\right)\)

Answer 3

3.

Hình vẽ:

Đặt các vecto đơn vị \(\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}\) cùng hướng \(\overrightarrow{AB};\overrightarrow{BC};\overrightarrow{CA}\)

Khi đó \(\left(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right)^2=3-2\left(cosA+cosB+cosC\right)=3-2P\)

\(\Rightarrow3-2P=\left(\overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}\right)^2\ge0\Rightarrow P\le\dfrac{3}{2}\)

\(maxP=\dfrac{3}{2}\Leftrightarrow\Delta ABC\) đều

Answer 4

4. 

Gọi O là tâm đường tròn ngoại tiếp tam giác ABC

\(\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)^2=OA^2+OB^2+OC^2+2\left(\overrightarrow{OA}.\overrightarrow{OB}+\overrightarrow{OB}.\overrightarrow{OC}+\overrightarrow{OC}.\overrightarrow{OA}\right)\)

\(=3OA^2+2OA^2\left(cosAOB+cosBOC+cosCOA\right)\)

\(=3OA^2+2OA^2\left(cos2A+cos2B+cos2C\right)\)

\(\Rightarrow cos2A+cos2B+cos2C=\dfrac{\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)^2-3OA^2}{2OA^2}\ge-\dfrac{3OA^2}{2OA^2}=-\dfrac{3}{2}\)

\(minQ=-\dfrac{3}{2}\Leftrightarrow\Delta ABC\) đều

Answer 5

5.

Gọi G là trọng tâm tam giác ABC

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

\(\Leftrightarrow9MG^2=\left(\overrightarrow{MG}+\overrightarrow{GA}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GB}\right)^2+\left(\overrightarrow{MG}+\overrightarrow{GC}\right)^2+2F\)

\(\Leftrightarrow9MG^2=3MG^2+GA^2+GB^2+GC^2+2\overrightarrow{MG}\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)+2F\)

\(\Leftrightarrow F=\dfrac{6MG^2-\left(GA^2+GB^2+GC^2\right)}{2}\ge-\dfrac{1}{2}\left(GA^2+GB^2+GC^2\right)\)

\(minF=-\dfrac{1}{2}\left(GA^2+GB^2+GC^2\right)\Leftrightarrow M\equiv G\)

Answer 6

7.

Gọi G là trọng tâm tam giác ABC

\(F=\left(\overrightarrow{MO}+\overrightarrow{OA}\right)^2+\left(\overrightarrow{MO}+\overrightarrow{OB}\right)^2+\left(\overrightarrow{MO}+\overrightarrow{OC}\right)^2-3\left(\overrightarrow{MO}+\overrightarrow{OD}\right)^2\)

\(=3MO^2+\left(OA^2+OB^2+OC^2\right)+2\overrightarrow{MO}\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)-3MO^2-3OD^2-6\overrightarrow{MO}.\overrightarrow{OD}\)

\(=3MO^2+3R^2+2\overrightarrow{MO}.3\overrightarrow{OG}-3MO^2-3R^2-6\overrightarrow{MO}.\overrightarrow{OD}\)

\(=6\overrightarrow{MO}\left(\overrightarrow{OG}-\overrightarrow{OD}\right)=6\overrightarrow{MO}.\overrightarrow{DG}=6R.DG.cos\left(\overrightarrow{MO},\overrightarrow{DG}\right)\)

Ta có \(-1\le cos\alpha\le1\left(0^o\le\alpha\le180^o\right)\)

\(maxF=6R.OD\Leftrightarrow cos\left(\overrightarrow{MO},\overrightarrow{DG}\right)=1\Leftrightarrow\overrightarrow{MO},\overrightarrow{DG}\) cùng hướng

\(minF=-6R.OD\Leftrightarrow cos\left(\overrightarrow{MO},\overrightarrow{DG}\right)=-1\Leftrightarrow\overrightarrow{MO},\overrightarrow{DG}\) ngược hướng