Câu hỏi của Nguyễn Gia Nghi

Question

CHO TAM GIÁC ABC. TRÊN CẠNH BC,CA,AB LẦN LƯỢT LẤY ĐIỂM D,E,F (KHÁC CÁC ĐỈNH CỦA TAM GIÁC) SAO CHO AD,BE,CF CẮT NHAU TẠI H.CMR:

a.AH/AD+BH/BE+CH/CF=2

b.AH/HD+BH/HE+CH/HF>=6

Luân Đào · Accepted Answer

Đặt SAHB = S1, SAHC = S2, SBHC = S3

a.

$\dfrac{AH}{AD}=\dfrac{S_1}{S_{ABD}}=\dfrac{S_2}{S_{ACD}}=\dfrac{S_1+S_2}{S}$

Tương tự:

$\dfrac{BH}{BE}=\dfrac{S_1+S_3}{S};\dfrac{CH}{CF}=\dfrac{S_2+S_3}{S}$

$\Rightarrow\dfrac{AH}{AD}+\dfrac{BH}{BE}+\dfrac{CH}{CF}=\dfrac{2\left(S_1+S_2+S_3\right)}{S}=\dfrac{2S}{S}=2$

b.

$\dfrac{AH}{HD}=\dfrac{S_1}{S_{BHD}}=\dfrac{S_2}{S_{CHD}}=\dfrac{S_1+S_2}{S_3}$

Tương tự:

$\dfrac{BH}{HE}=\dfrac{S_1+S_3}{S_2};\dfrac{CH}{HF}=\dfrac{S_2+S_3}{S_1}$

$\Rightarrow\dfrac{AH}{HD}+\dfrac{BH}{HE}+\dfrac{CH}{HF}=\dfrac{AD}{HD}+\dfrac{BE}{HE}+\dfrac{BF}{HF}-3$

$=\dfrac{S}{S_1}+\dfrac{S}{S_2}+\dfrac{S}{S_3}-3\ge\dfrac{9S}{S_1+S_2+S_3}-3=\dfrac{9S}{S}-3=6$

Dấu "=" xảy ra khi H là trọng tâm tam giác ABCCâu trả lời: Đặt SAHB = S1, SAHC = S2, SBHC = S3

a.

$\dfrac{AH}{AD}=\dfrac{S_1}{S_{ABD}}=\dfrac{S_2}{S_{ACD}}=\dfrac{S_1+S_2}{S}$

Tương tự:

$\dfrac{BH}{BE}=\dfrac{S_1+S_3}{S};\dfrac{CH}{CF}=\dfrac{S_2+S_3}{S}$

$\Rightarrow\dfrac{AH}{AD}+\dfrac{BH}{BE}+\dfrac{CH}{CF}=\dfrac{2\left(S_1+S_2+S_3\right)}{S}=\dfrac{2S}{S}=2$

b.

$\dfrac{AH}{HD}=\dfrac{S_1}{S_{BHD}}=\dfrac{S_2}{S_{CHD}}=\dfrac{S_1+S_2}{S_3}$

Tương tự:

$\dfrac{BH}{HE}=\dfrac{S_1+S_3}{S_2};\dfrac{CH}{HF}=\dfrac{S_2+S_3}{S_1}$

$\Rightarrow\dfrac{AH}{HD}+\dfrac{BH}{HE}+\dfrac{CH}{HF}=\dfrac{AD}{HD}+\dfrac{BE}{HE}+\dfrac{BF}{HF}-3$

$=\dfrac{S}{S_1}+\dfrac{S}{S_2}+\dfrac{S}{S_3}-3\ge\dfrac{9S}{S_1+S_2+S_3}-3=\dfrac{9S}{S}-3=6$

Dấu "=" xảy ra khi H là trọng tâm tam giác ABC