Question

[ngunhat]

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

Answer 1

[Lê Song Phương]

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.