Lấy O nằm trong tam giác ABC. Các tia AO, BO, CO cắt BC, AC, AB lần lượt tại P, Q, R. Tìm vị trí của O để AP/AO + BQ/OB + CR/OC đạt GTNN
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Đặt S OBC=S1, S OAC=S2, S OAB=S3, S=S ABC
Kẻ AH vuông góc BC< OK vuông góc BC
=>OK//AH
OP/AP=OK/AH=1/2*OK*BC/1/2*AH*CB=S1/S
=>\(\dfrac{AP-OP}{AP}=\dfrac{S-S_1}{S}\)
=>\(\dfrac{OA}{AP}=\dfrac{S_2+S_3}{S}\)
Cmtương tự, ta được: \(\dfrac{OB}{BQ}=\dfrac{S_1+S_3}{S};\dfrac{OC}{CR}=\dfrac{S_1+S_2}{S}\)
=>\(\dfrac{OA}{AP}+\dfrac{OB}{BQ}+\dfrac{OC}{CR}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mai mình nghĩ cho cái này thay nọ thay kia, áp dụng ta lét ( lấy B làm đỉnh ) gợi ý là vậy chứ chưa giải ra :v
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2:
a: HM là đường trung bình của ΔEBC
=>EH=HB
KM là đường trug bình của ΔFBC
=>FK=KC
ΔAHM có EO//HM
=>AE/AH=AO/AM
ΔAKM có KM//FO
nên AF/AK=AO/AM
=>AE/AH=AF/AK
=>EF//HK
b: ΔAHM có EO//HM
=>MA/MO=HA/HE
=>MA/MO=HA/HB
ΔAKM có FO//KM
=>MA/MO=KA/KF=KA/KC
=>HA/HB=KA/KC
=>HK//BC
=>EF//BC
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{OA}{AD}=\frac{S_{AOB}}{S_{ABD}}=\frac{S_{AOC}}{S_{ACD}}=\frac{S_{AOB}+S_{AOC}}{SABC}\)
Tương tự rồi cộng lại ta đc
\(\frac{OA}{AD}+\frac{OB}{BE}+\frac{OC}{CF}=\frac{2\left(S_{AOB}+S_{BOC}+S_{COA}\right)}{S_{ABC}}=2\)
Bài Giải
Đặt SBOC=x2,SAOC=y2,SAOB=z2 ⇒SABC=SBOC+SAOC+SAOB=x2+y2+z2
Ta có : ADOD =SABCSBOC =AO+ODOD =1+AOOD =x2+y2+z2x2 =1+y2+z2x2
⇒AOOD =y2+z2x2 ⇒√AOOD =√y2+z2x2 =√y2+z2x
Tương tự ta có √OBOE =√x2+z2y2 =√x2+z2y ;√OCOF =√x2+y2z2 =√x2+y2z
⇒P=√x2+y2z +√y2+z2x +√x2+z2y ≥x+y√2z +y+z√2x +x+z√2y
=1√2 [(xy +yx )+(yz +zy )+(xz +zx )]≥1√2 (2+2+2)=3√2
Dấu "=" xảy ra khi x=y=z⇒SBOC=SAOC=SAOB=13 SABC
⇒ODOA =OEOB =OFOC =13 ⇒O là trọng tâm của tam giác ABC
Vậy MinP=3√2 khi O là trọng tâm của tam giác ABC
The two triangles BAP and BAO have the same height from B, so we have: \(\dfrac{S_{BAP}}{S_{BAO}}=\dfrac{AP}{AO}\)
Similarly, we have: \(\dfrac{S_{CAP}}{S_{CAO}}=\dfrac{AP}{AO}\), from that, we have: \(\dfrac{AP}{AO}=\dfrac{S_{BAP}}{S_{BAO}}=\dfrac{S_{CAP}}{S_{CAO}}=\dfrac{S_{BAP}+S_{CAP}}{S_{BAO}+S_{CAO}}=\dfrac{S_{ABC}}{S_{BAO}+S_{CAO}}\)
Thus, we also have \(\dfrac{BQ}{OB}=\dfrac{S_{ABC}}{S_{BOC}+S_{AOB}}\); \(\dfrac{CR}{OC}=\dfrac{S_{ABC}}{S_{BOC}+S_{AOC}}\)
So we get: \(\dfrac{AP}{AO}+\dfrac{BQ}{OB}+\dfrac{CR}{OC}=\dfrac{S_{ABC}}{S_{COA}+S_{AOB}}+\dfrac{S_{ABC}}{S_{AOB}+S_{BOC}}\)\(+\dfrac{S_{ABC}}{S_{BOC}+S_{AOC}}\)
If \(S_{BOC}=a;S_{COA}=b;S_{AOB}=c\left(a,b,c>0\right)\), then \(P=\dfrac{AP}{AO}+\dfrac{BQ}{OB}+\dfrac{CR}{OC}=\dfrac{S_{ABC}}{b+c}+\dfrac{S_{ABC}}{c+a}+\dfrac{S_{ABC}}{a+b}\)
\(=S_{ABC}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
We have already had the inequality: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\) (This is true with all of the positive real number \(x,y,z\). If you don't know about this, please check it on the Internet) \(P\ge S_{ABC}\left(\dfrac{9}{a+b+b+c+c+a}\right)=S_{ABC}.\dfrac{9}{2\left(a+b+c\right)}\)\(=S_{ABC}.\dfrac{9}{2S_{ABC}}=\dfrac{9}{2}\) (vì \(a+b+c=S_{BOC}+S_{COA}+S_{AOB}=S_{ABC}\))
In conclusion, the minimum value of \(\dfrac{AP}{AO}+\dfrac{BQ}{OB}+\dfrac{CR}{OC}\) is \(\dfrac{9}{2}\), happens when \(a=b=c=\dfrac{1}{3}S_{ABC}\) or \(S_{BOC}=S_{COA}=S_{AOC}=\dfrac{1}{3}S_{ABC}\)
Consider \(S_{BOC}=\dfrac{1}{3}S_{ABC}\Leftrightarrow\dfrac{S_{BOC}}{S_{ABC}}=\dfrac{1}{3}\)
We have \(\dfrac{S_{BOP}}{S_{ABP}}=\dfrac{PO}{PA}\) and \(\dfrac{S_{COP}}{S_{ACP}}=\dfrac{PO}{PA}\)
Therefore, we have \(\dfrac{PO}{PA}=\dfrac{S_{BOP}}{S_{ABP}}=\dfrac{S_{COP}}{S_{ACP}}=\dfrac{S_{BOP}+S_{COP}}{S_{ABP}+S_{ACP}}=\dfrac{S_{BOC}}{S_{ABC}}=\dfrac{1}{3}\)
Similarly, we have \(\dfrac{OQ}{BQ}=\dfrac{1}{3};\dfrac{OR}{CR}=\dfrac{1}{3}\)
These means O is the centroid of the triangle ABC.
So in order to minimize the value of \(\dfrac{AP}{AO}+\dfrac{BQ}{OB}+\dfrac{CR}{OC}\), O must be the centroid of the triangle ABC.