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The minimum of the expression $AP\cdot AG+BP\cdot BG+CP\cdot CG$ is achieved when P coincides with the centroid G of triangle ABC, and this minimum equals $AG^2+BG^2+CG^2=\frac13 \left( a^2+b^2+c^2\right) $. Proof. Using vectors and their scalar products, we have $UV\cdot UW\geq \overrightarrow{UV}\cdot \overrightarrow{UW}$ for any three points U, V and W, and equality holds if and only if the point W lies on the ray UV. Thus, $AP\cdot AG+BP\cdot BG+CP\cdot CG\geq \overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}$, with equality holding if and only if the point G lies on the ray AP, on the ray BP and on the ray CP simultaneously, i. e. if and only if the point P coincides with the point G. Now, $\overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}$ $=\left( \overrightarrow{AG}+\overrightarrow{GP}\right) \cdot \overrightarrow{AG}+\left( \overrightarrow{BG}+\overrightarrow{GP}\right) \cdot \overrightarrow{BG}+\left( \overrightarrow{CG}+\overrightarrow{GP}\right) \cdot \overrightarrow{CG}$ $=\left( \overrightarrow{AG}^2+\overrightarrow{BG}^2+\overrightarrow{CG}^2\right) +\overrightarrow{GP}\cdot \left( \overrightarrow{AG}+\overrightarrow{BG}+\overrightarrow{CG}\right) $. Now, since the point G is the centroid of triangle ABC, we have $\overrightarrow{AG}+\overrightarrow{BG}+\overrightarrow{CG}=\overrightarrow{0}$, and thus $\overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}=\left( \overrightarrow{AG}^2+\overrightarrow{BG}^2+\overrightarrow{CG}^2\right) +\overrightarrow{GP}\cdot \overrightarrow{0}$ $=\overrightarrow{AG}^2+\overrightarrow{BG}^2+\overrightarrow{CG}^2=AG^2+BG^2+CG^2$. After the well-known formula for the median lengths in a triangle, the length $m_a$ of the median of triangle ABC issuing from the vertex A satisfies the equation $4m_a^2=2b^2+2c^2-a^2$. Since the point G is the centroid of triangle ABC and thus divides all medians in the ratio 2 : 1, we have $AG=\frac23 m_a$, and thus $AG^2=\left( \frac23 m_a\right) ^2=\frac{4m_a^2}{9}=\frac{2b^2+2c^2-a^2}{9}$. Similarly, $BG^2=\frac{2c^2+2a^2-b^2}{9}$; $CG^2=\frac{2a^2+2b^2-c^2}{9}$, and thus $AG^2+BG^2+CG^2=\frac{2b^2+2c^2-a^2}{9}+\frac{2c^2+2a^2-b^2}{9}+\frac{2a^2+2b^2-c^2}{9}=\frac13 \left( a^2+b^2+c^2\right) $. Hence, we have $\overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}=\frac13 \left( a^2+b^2+c^2\right) $. Now, since $AP\cdot AG+BP\cdot BG+CP\cdot CG\geq \overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}$, it follows that $AP\cdot AG+BP\cdot BG+CP\cdot CG\geq \frac13 \left( a^2+b^2+c^2\right) $, with equality if and only if the point P coincides with the point G. In other words, the minimum of the expression $AP\cdot AG+BP\cdot BG+CP\cdot CG$ is achieved for P = G and equals $\frac13 \left( a^2+b^2+c^2\right) $. Proof complete. Darij

We can prove this one by this lemma : If P and Q are abitrary points then $\sum \sin{BQC}.PA \geq \sum \sin{BQC}QA$ Plug Q = G.

orl wrote: Let $ABC$ be a triangle with centroid $G$. Determine, with proof, the position of the point $P$ in the plane of $ABC$ such that $AP{\cdot}AG+BP{\cdot}BG+CP{\cdot}CG$ is a minimum, and express this minimum value in terms of the side lengths of $ABC$. Here is my solution to that and i would like someone to tell me please if its ok : We use complex numbers. Put the triagle on complex plane with the origin 0(0,0) on G. Also let z, a , b , c the complex numbers for the points P, A , B , C.Then we have : $AP\cdot AG+BP\cdot BG+CP\cdot CG=\mid(z-a)a\mid+\mid(z-b)b\mid+\mid(z-c)c\mid\geq\mid(a+b+c)z-(a^{2}+b^{2}+c^{2})\mid=\mid a^{2}+b^{2}+c^{2}\mid=\frac{4}{9}(m_{a}^{2}+m_{b}^{2}+m_{c}^{2})=\frac{1}{3}(AB^{2}+BC^{2}+CA^{2})$

I've been using this problem to demonstrate techniques for solving geometric inequalities. Here's another solution. At the minimum, $P$ will be inside the triangle. For any value of $AP$, let us write the expression under consideration as a function of $\theta=\angle PAB$ using the Cosine Rule: \[ AG\times AP+BG\sqrt{AB^2+AP^2-2AB\times AP\cos\theta}+CG\sqrt{AC^2+AP^2-2AC\times AP\cos(\angle A-\theta)}. \] Differentiating, we find that when this is minimised \[ BG\frac{2AB\times AP\sin\theta}{2\sqrt{AB^2+AP^2-2AB\times AP\cos\theta}}=CG\frac{2AC\times AP\sin(\angle A-\theta)}{2\sqrt{AC^2+AP^2-2AC\times AP\cos(\angle A-\theta)}}. \] Cancelling and simplifying back the denominators using the Cosine Rule, \[ \frac{BG\times AB\sin\angle PAB}{BP}=\frac{CG\times AC\sin\angle PAC}{CP} \] or, simplifying using the Sine Rule in triangles $ABP$ and $ACP$, \[ \frac{BG}{\sin \angle APC}=\frac{CG}{\sin \angle APB}. \] By symmetry, we find that the condition $\sin \angle APB: \sin \angle APC: \sin \angle BPC = CG: BG: AG$ must hold at the point $P$ which minimises the expression. As the three angles sum to $360^o$, this condition in fact uniquely determines them, and hence the point $P$, so that this condition is not only necessary for $P$ to minimize the expression, but also sufficient. It is therefore enough to prove that when $P$ is set to $G$ this condition is satisfied. This last bit is not difficult. Extending $AG$ to meet $BC$ gives the midpoint $M$ of $BC$. Applying the Sine Rule twice, $BG\sin \angle BGA=BA\sin \angle BAG =BM\sin\angle BMA$. As $BM=CM$, $\sin\angle BMA=\sin\angle CMA$, it follows that $BG\sin \angle BGA= CG\sin\angle CGA$. By symmetry the whole set of ratio equalities holds.

Let $A_1$, $B_1$, and $C_1$ be the midpoints of $BC$, $AC$, and $AB$, respectively. Let the line through $C$ parallel to $BG$ hit $AA'$ at $X$. $BGCX$ is a parallelogram, so $CX = GB$ and $GX = 2 GA_1 = AG$. Let $AG$ intersect the circumcircle of $\triangle BCG$ at $A'$. $\angle CBA' = \angle CGA' = \angle CGX$, and $\angle BA'C = 180 - \angle BGC = 180 - (180 - \angle GCX) = \angle GCX$, so $\triangle BCA'$ is similar to $\triangle GCX$. Hence, there exists a positive real number $k$ such that $BC = k GX = kGA$, $CA' = k GC$, and $A'B = k CX = kGB$. By Ptolemy's inequality, $PB \cdot CA' + PC \cdot BA' \geq PA' \cdot BC$, so $k PB \cdot GC + k PC \cdot GB \geq k PA' \cdot GA$. Dividing by $k$ and adding $PA \cdot GA$ to both sides yields $PA \cdot GA + PB \cdot GC + PC \cdot GB \geq (PA' + AP) \cdot GA$. By the triangle inequality, $PA' + AP \geq AA'$, so $PA \cdot GA + PB \cdot GC + PC \cdot GB \geq AA' \cdot GA$, with equality holding if and only if $A$, $P$, and $A'$ are collinear and $PBA'C$ is cyclic, i.e., if and only if $P = G$. Hence, the minimum value of the $AP \cdot AG + BP \cdot BG + CP \cdot CG$ is $GA^2 + GB^2 + GC^2= \frac{a^2+b^2+c^2}{3}.$

By law of sines on $\triangle BGL$, we have $\frac{BG}{\sin\angle BLG}=\frac{BL}{\sin\angle BGK}$. Similarly, in $\triangle CGL$ we have $\frac{CG}{\sin\angle CLG}=\frac{CL}{\sin\angle CGK}$. However, $L$ is the midpoint of $Bc$ and we have $\sin\angle BLG=\sin\angle CLG$, so $\frac{BG}{CG}=\frac{\sin\angle CGK}{\sin\angle BGK}$. Now, since $\angle BOK=2\angle BGK$, with $O$ the circle's center, we have $BK=2R\sin\angle BGK$, where $R$ is the radius of the circle. Similarly, $CK=2R\sin\angle CGK$, so $\frac{CK}{BK}=\frac{BG}{CG}$, and hence $\frac{BG}{CK}=\frac{CG}{BK}$. Similarly we have $\frac{AG}{BG}=\frac{\sin\angle BGN}{\sin\angle AGN}=\frac{\sin\angle BGN}{\sin CGK}$ since they are opposite angles. Since quadrilateral $BGCK$ is cyclic, we have $\angle BKC=\angle BGN\implies BC=2R\sin\angle BKC=2R\sin\angle BGN$. It follows that $\frac{BC}{CK}=\frac{\sin\angle BGN}{\sin\angle CGK}=\frac{AG}{BG}$, so $\frac{BG}{CK}=\frac{AG}{BC}$ and hence $BC:CK:BK=AG:BG:CG$. Now, by Ptolemy's inequality on quadrilateral $PBKC$, we have $(PK)(BC)\le (BP)(CK)+(CP)(BK)$. Hence $(PK)(AG)\le (BP)(BG)+(CP)(CG)$. Therefore, \[AG(AP+PK)\le (AP)(AG)+(BG)(BG)+(CP)(CG)\] and it follows that $(AK)(AG)\le (AP)(AG)+(BP)(BG)+(CP)(CG)$, and equality holds if and only if $P$ lies on the circle between $B$ and $C$, which will give Ptolemy's equality, and $P$ lies on $AK$, for triangle equality. Thus the minimum is achieved when $P=G$. Finally, by the law of cosines, we have \[\begin{cases}AB^{2}=AL^2+BL^2-2(BL)(AL)\cos\angle ALB \\ AC^2=AL^2+CL^2-2(CL)(AL)\cos\angle ALC\end{cases}\] Adding the two equations yields \[AB^2+AC^2=2AL^2+\frac{BC^2}{2},\] so \[AG^2=\frac{4}{9}AL^2 = \frac{2AB^2+2AC^2-BC^2}{9}\] It follows that \[\text{min}\{AG^2+BG^2+CG^2\}=\frac{AB^2+BC^2+CA^2}{3}\] $\Box$

Note that $|AP|\cdot |AG|\ge (A-P)\cdot (A-G)$, where the right hand side is a vector dot product. Thus if $S$ is the desired sum, \begin{align*} S&\ge (A-P)\cdot(A-G)+(B-P)\cdot(B-G)+(C-P)\cdot(C-G)\\ &=A\cdot (A-P)+B\cdot (B-P)+C\cdot (C-P)-P\cdot (A+B+C-3G)\\ &=A\cdot (A-P)+B\cdot (B-P)+C\cdot (C-P)-P\cdot 0\\ &=A\cdot (A-P)+B\cdot (B-P)+C\cdot (C-P)-G\cdot 0\\ &=A\cdot (A-P)+B\cdot (B-P)+C\cdot (C-P)-G\cdot (A+B+C-3G)\\ &=(A-G)\cdot(A-G)+(B-G)\cdot(B-G)+(C-G)\cdot(C-G)\\ &=|AG|^2+|BG|^2+|CG|^2 \end{align*}So $S$ is minimized when $P=G$ (and equality can only hold if $A,P,G$, $B,P,G$, and $C,P,G$ are collinear triples, i.e. if $P=G$). By well known-formulas the minimum value of $S$ can also be expressed as $\frac{1}{3}(a^2+b^2+c^2)$.

Let $X,Y,Z$ be the midpoints of $BC,CA,AB$. By the Law of Cosines on $APX$, we have that $PX^2=AP^2+AX^2-2\cdot AP\cdot AX\cdot\cos\angle PAX$, so $AP\cdot AX=\frac{AP^2+AX^2-PX^2}{2\cdot\cos\angle PAX}\leq\frac{AP^2+AX^2-PX^2}{2}$. By the length of the median formula on $BPC$, we have that $PX^2=\frac{BP^2}{2}+\frac{CP^2}{2}-\frac{BC^2}{4}$, so $AP\cdot AX\leq\frac{AP^2+AX^2-\frac{BP^2}{2}-\frac{CP^2}{2}+\frac{BC^2}{4}}{2}$. Summing through the other vertices, we have that $AP\cdot AG+BP\cdot BG+CP\cdot CG=\frac{2}{3}(AP\cdot AX+BP\cdot BY+CP\cdot CZ)\leq \frac{AX^2+BY^2+CZ^2+\frac{BC^2+CA^2+AB^2}{4}}{2}=\frac13(AB^2+BC^2+CA^2)$ by well-known formulas. Equality holds iff $\angle PAX=\angle PBY=\angle PCZ=0^\circ$, or iff $P=G$. The only problem is the possibility that (WLOG) $\angle PAX\geq90^\circ$. This is easily resolved by some weak inequalities that are too annoying to type.

This solution was found with prior knowledge of what the answer was, but oh well, I still think it's pretty nice. We first present a lemma. LEMMA: In any triangle $\triangle ABC$, we have \[AP^2+BP^2+CP^2=AG^2+BG^2+CG^2+3GP^2.\] Proof. We use physics. Consider the set of mass points $\Sigma = \{1A,1B,1C\}$. Let $J_{P,\Sigma}$ denote the moment of inertia of $P$ about $\Sigma$. Note that $G$ is trivially the center of mass of $\Sigma$. Now recall that by the Parallel Axis Theorem \[J_{P,\Sigma} = J_{G,\Sigma}+3GP^2,\]where the $3$ is the total mass of the system $\Sigma$. Hence by the definition of moment of inertia we get the desired equality. $\blacksquare$ Now note that by Law of Cosines we can rewrite this as \[AP^2+BP^2+CP^2 = AG^2+BG^2+CG^2+\sum_{cyc}\left(AP^2+AG^2-2AP\cdot AG\cos\angle PAG\right).\]This rearranges to \[\sum_{cyc}AP\cdot AG\cos\angle PAG = AG^2+BG^2+CG^2.\]Finally, note that $AP\cdot AG\cos\angle PAG\leq AP\cdot AG$ and similar, so we end up with \[AP\cdot AG+BP\cdot BG+CP\cdot CG\geq AG^2+BG^2+CG^2,\]with equality occurring when $P\equiv G$. As others have pointed out above, this can be rewritten as $\tfrac13(a^2+b^2+c^2)$.

Any barybash ? By totally thoughtless barybash, reduced the problem to given $a,b,c$ sides of a triangle, find all triples $(x,y,z)$ of reals such that the following is minimized, $$ \sum_{\sigma} \frac{xy}{(x+y+z)^2} c_{xy} + \sum_{\sigma} \frac{x}{x+y+z} d_x $$, where $$c_{xy} = 6a^4 + a^2 \sum_{\sigma} a^2 $$$$ d_x = 6c^2b^2-2(b^2+c^2)(\sum_{\sigma} a^2)$$(and cyclic permutations)

darij grinberg wrote: The minimum of the expression $AP\cdot AG+BP\cdot BG+CP\cdot CG$ is achieved when P coincides with the centroid G of triangle ABC, and this minimum equals $AG^2+BG^2+CG^2=\frac13 \left( a^2+b^2+c^2\right) $. Proof. Using vectors and their scalar products, we have $UV\cdot UW\geq \overrightarrow{UV}\cdot \overrightarrow{UW}$ for any three points U, V and W, and equality holds if and only if the point W lies on the ray UV. Thus, $AP\cdot AG+BP\cdot BG+CP\cdot CG\geq \overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}$, with equality holding if and only if the point G lies on the ray AP, on the ray BP and on the ray CP simultaneously, i. e. if and only if the point P coincides with the point G. Now, $\overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}$ $=\left( \overrightarrow{AG}+\overrightarrow{GP}\right) \cdot \overrightarrow{AG}+\left( \overrightarrow{BG}+\overrightarrow{GP}\right) \cdot \overrightarrow{BG}+\left( \overrightarrow{CG}+\overrightarrow{GP}\right) \cdot \overrightarrow{CG}$ $=\left( \overrightarrow{AG}^2+\overrightarrow{BG}^2+\overrightarrow{CG}^2\right) +\overrightarrow{GP}\cdot \left( \overrightarrow{AG}+\overrightarrow{BG}+\overrightarrow{CG}\right) $. Now, since the point G is the centroid of triangle ABC, we have $\overrightarrow{AG}+\overrightarrow{BG}+\overrightarrow{CG}=\overrightarrow{0}$, and thus $\overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}=\left( \overrightarrow{AG}^2+\overrightarrow{BG}^2+\overrightarrow{CG}^2\right) +\overrightarrow{GP}\cdot \overrightarrow{0}$ $=\overrightarrow{AG}^2+\overrightarrow{BG}^2+\overrightarrow{CG}^2=AG^2+BG^2+CG^2$. After the well-known formula for the median lengths in a triangle, the length $m_a$ of the median of triangle ABC issuing from the vertex A satisfies the equation $4m_a^2=2b^2+2c^2-a^2$. Since the point G is the centroid of triangle ABC and thus divides all medians in the ratio 2 : 1, we have $AG=\frac23 m_a$, and thus $AG^2=\left( \frac23 m_a\right) ^2=\frac{4m_a^2}{9}=\frac{2b^2+2c^2-a^2}{9}$. Similarly, $BG^2=\frac{2c^2+2a^2-b^2}{9}$; $CG^2=\frac{2a^2+2b^2-c^2}{9}$, and thus $AG^2+BG^2+CG^2=\frac{2b^2+2c^2-a^2}{9}+\frac{2c^2+2a^2-b^2}{9}+\frac{2a^2+2b^2-c^2}{9}=\frac13 \left( a^2+b^2+c^2\right) $. Hence, we have $\overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}=\frac13 \left( a^2+b^2+c^2\right) $. Now, since $AP\cdot AG+BP\cdot BG+CP\cdot CG\geq \overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG}$, it follows that $AP\cdot AG+BP\cdot BG+CP\cdot CG\geq \frac13 \left( a^2+b^2+c^2\right) $, with equality if and only if the point P coincides with the point G. In other words, the minimum of the expression $AP\cdot AG+BP\cdot BG+CP\cdot CG$ is achieved for P = G and equals $\frac13 \left( a^2+b^2+c^2\right) $. Proof complete. Darij "$UV\cdot UW\geq \overrightarrow{UV}\cdot \overrightarrow{UW}$" what is this inequality?

A synthetic solution, the motivation came from IMOSL 1998 G4: For any point $P$ in the plane define $f(P) = AP \cdot AG + BP \cdot BG + CP \cdot CG$. Pick $P$ such that $f(P)$ is minimal. We will prove $P \equiv G$. Let $K$ be the isogonal conjugate of $G$ wrt $\triangle ABC$ and let $D,E,F$ be the projection of $K$ onto sides $BC,CA,AB$, respectively. Claim: $P$ must be contained in $\triangle ABC$. Proof: Let $\mathcal R_A$ be the region of all points that lie on the same side of line $BC$ as of point $A$ (including all points on line $BC$). Define $\mathcal R_B,\mathcal R_C$ similarly. If $P \notin \mathcal R_A$, then if we denote by $P'$ the reflection of $P$ in line $BC$, then we have $f(P') < f(P)$, which would contradict the minimality of $f(P)$. Thus $P \in \mathcal R_A$. Similarly, $P \in \mathcal R_B$ and $P \in \mathcal R_C$. Now it is not hard to see that $\mathcal R_A \cap \mathcal R_B \cap \mathcal R_C$ is precisely the set of all points contained in $\triangle ABC$, which proves our claim. $\square$ Claim: $AK \sin A : BK \sin B : CK \sin C = AG : BG : CG$ Proof: It suffices to $\frac{AK}{BK} = \frac{AG}{BG} \cdot \frac{\sin B}{\sin A}$ as other results follows analogously. Let $Z = \overline{CG} \cap \overline{AB}$. Then, \begin{align*} \frac{AK}{BK} &= \frac{\sin \angle ABK}{\sin \angle BAK} = \frac{\sin \angle CBG}{\sin CAG} = \frac{\sin \angle GCB}{\sin \angle GCA} \cdot \frac{ \sin \angle CBG / \sin \angle GCB}{\sin \angle CAG / \sin \angle GCA} \\ &= \frac{\sin \angle BCZ / BZ}{\sin \angle ACZ / AZ} \cdot \frac{CG / CB}{CG / CA} = \frac{\sin B / CZ}{\sin A /CZ} \cdot \frac{AG}{BG} = \frac{AG}{BG} \cdot \frac{\sin B}{\sin A} \end{align*}$\square$ Now by Sine rule we have $$EF = AK \cdot \sin A ~,~ FD = BK \cdot \sin B ~,~ DE = CK \cdot \sin C$$So because of the previous claim, we obtain that there is positive real number $t$ such that $$EF = t \cdot AG ~,~ FD = t \cdot BG ~,~ DE = t \cdot CG$$Now denote by $[\lambda]$ the area of any figure $\lambda$. Then, $$AP \cdot AG + BP \cdot BG + CP \cdot CG = AP \cdot \frac{EF}{t} + BP \cdot \frac{FD}{t} + CP \cdot \frac{DE}{t} \ge \frac{[AEPF]}{t/2} + \frac{[BFPD]}{t/2} + \frac{CDPE}{t/2} = \frac{[ABC]}{t/2}$$With equality iff $\overline{AP} \perp \overline{EF}$ , $\overline{BP} \perp \overline{FD}$ , $\overline{CP} \perp \overline{DE}$ ; which (because of the isogonal conjugate thing) is true iff $P \equiv G$. So we have finally proven $\boxed{P \equiv G}$. Now we want to find the value of $$f(G) = AG^2 + BG^2 + CG^2$$in terms of $a = BC$ , $b = CA$ , $c = AB$. But using appolonius theorem and the fact $G$ divides each median in the ratio $2:1$, we directly obtain that $$\boxed{f(G) = \frac{a^2 + b^2 + c^2}{3}}$$$\blacksquare$ Comment: I think we can most probably avoid proving the first claim, but I am not exactly sure how to handle the configuration issues.

Solution with islander7. Let \(\hat\bullet\) denote the unit vector. Let \(f(P)=AP\cdot AG+BP\cdot BG+CP\cdot CG\), which is convex, so at the unique point \(P\) that minimizes \(f\), \[0=\nabla f(P)=AG\cdot\widehat{\overrightarrow{AP}}+BG\cdot\widehat{\overrightarrow{BP}}+CG\cdot\widehat{\overrightarrow{CP}}.\]But \(\nabla f(G)=\vec{AG}+\vec{BG}+\vec{CG}=0\), so \(P=G\).

Hi all. I'm trying to use vectors to solve this Olympiad Geometry problem (I've actually never done this before!) Here's what I have at the moment: We use vectors. Firstly, note that $G = \frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C})$. We can rewrite the desired expression as $|\overrightarrow{P} - \overrightarrow{A}| \cdot |\frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}) - \overrightarrow{A}| + |\overrightarrow{P} - \overrightarrow{B}| \cdot |\frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}) - \overrightarrow{B}| + |\overrightarrow{P} - \overrightarrow{C}| \cdot |\frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}) - \overrightarrow{C}|$. My natural next step would be to expand that out, but it seems like it's going to get extremely ugly. Is there a better way or am I just destined to bash this out and simplify (or do my best to) what I get?

Bumping this. Thanks in advance for any help!

OlympusHero wrote: Bumping this. Thanks in advance for any help!

hydo2332 wrote: "$UV\cdot UW\geq \overrightarrow{UV}\cdot \overrightarrow{UW}$" what is this inequality? Follows from $a \cdot b = |a||b|\cos\theta$ where $\theta$ is the angle between $a$ and $b$. Alternatively its Cauchy-Schwarz.

OlympusHero wrote: Hi all. I'm trying to use vectors to solve this Olympiad Geometry problem (I've actually never done this before!) Here's what I have at the moment: We use vectors. Firstly, note that $G = \frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C})$. We can rewrite the desired expression as $|\overrightarrow{P} - \overrightarrow{A}| \cdot |\frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}) - \overrightarrow{A}| + |\overrightarrow{P} - \overrightarrow{B}| \cdot |\frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}) - \overrightarrow{B}| + |\overrightarrow{P} - \overrightarrow{C}| \cdot |\frac{1}{3}(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}) - \overrightarrow{C}|$. My natural next step would be to expand that out, but it seems like it's going to get extremely ugly. Is there a better way or am I just destined to bash this out and simplify (or do my best to) what I get? OlympusHero wrote: Bumping this. Thanks in advance for any help!

Note that \begin{align*} AP\cdot AG +BP\cdot BG+CP\cdot CG &\ge \overrightarrow{AP}\cdot \overrightarrow{AG}+\overrightarrow{BP}\cdot \overrightarrow{BG}+\overrightarrow{CP}\cdot \overrightarrow{CG} \\ &= (\overrightarrow{AG}+\overrightarrow{GP})\cdot (\overrightarrow{AG}) + (\overrightarrow{BG}+\overrightarrow{GP})\cdot (\overrightarrow{BG}) + (\overrightarrow{CG}+\overrightarrow{GP})\cdot (\overrightarrow{CG}) \\ &= AG^2 + BG^2 + CG^2 \end{align*}with equality if and only if $P=G$.

In what follows all point names except for $P$, $O$, and $G$ are taken cyclically. Let $D,E,F$ be the feet of the altitudes from $P$ to $AG$, $BG$, $CG$ respectively, and let $O$ be the midpoint of $PG$. We have \begin{align*} \sum_{\text{cyc}} AG \cdot AP &\ge \sum_{\text{cyc}} AG \cdot AD \\ &= \sum_{\text{cyc}} (AO^2 - OG^2) \\ &= \sum_{\text{cyc}} (AG^2 + AO^2 - OG^2 - AG^2) \\ &= \sum_{\text{cyc}} AG^2 - (\vec{GA} \cdot \vec{GO}) \\ &= \vec{0} \cdot \vec{GO} + \sum_{\text{cyc}}AG^2 \\ &= \sum_{\text{cyc}} AG^2 . \end{align*}Thethird to last equality follows from the law of cosines, equality holding if and only if $D=E=F=G.$

not good with vectors but here is how my brain understands it: \[AP\cdot AG\ge AP\cdot AG\cdot \cos\angle PAG=\overrightarrow{AP}\cdot \overrightarrow{AG}=\overrightarrow{AG}\cdot \overrightarrow{AG}+\overrightarrow{AG}\cdot \overrightarrow{GP}=AG^2+\overrightarrow{AG}\cdot \overrightarrow{GP}\]and summing yields \[AP\cdot AG+BP\cdot BG+CP\cdot CG\ge AG^2+BG^2+CG^2\]with equality if $\cos\angle PAG=\cos\angle PBG=\cos\angle PCG=1$ hence $P=G$. Now let $M$ be the midpoint of $BC$. Notice \[2AM^2=AB^2+AC^2-BM^2-CM^2\implies AM^2=\frac{1}{2}b^2+\frac{1}{2}c^2-\frac{1}{4}a^2\implies AG^2=\frac{2}{9}b^2+\frac{2}{9}c^2-\frac{1}{9}a^2.\]Hence adding yields $AG^2+BG^2+CG^2=\frac{1}{3}(a^2+b^2+c^2)$. $\blacksquare$

We claim that $P=G$, and that this minimum value equals $\frac13 (a^2+b^2+c^2)$. We start by defining $A'$ to be the intersection of $AG$ with the circumcircle of $BGC$, distinct from $G$. We apply Ptolemy's Inequality on $BPCA'$ to get that $$PB\cdot CA' + PC \cdot BA' \geq PA' \cdot AG.$$Using the similar triangles formed by the cyclic quadrilateral $BGCA$, we can determine that $CA'=\dfrac{BC\cdot BG}{AG}$ and $BA'=\dfrac{BC\cdot CG}{AG}$. Thus after dividing out $BC$ and multiplying by $AG$, we have that $$BP\cdot BG+CP\cdot CG \geq PA'\cdot AG.$$Now adding $AP \cdot AG$ to both sides, we get that $$AP\cdot AG + BP \cdot BG + CP \cdot CG \geq (AP+PA')\cdot AG.$$ Note that by the triangle inequality, $AP+PA'\geq AA'$, so we have that $$AP\cdot AG + BP \cdot BG + CP \cdot CG \geq AA'\cdot AG.$$ Equality holds only when $P$ lies on $(GBC)$ and on segment $AA'$, which still leaves two options, $G$ and $A'$. However, these same calculations can be performed by defining $B'$ and $C'$, and $P=G$ is the only point that maintains equality in all of these cases. Now we just have to compute $AG^2+BG^2+CG^2$. By Stewart's Theorem we get that $AM^2=\frac12 b^2+\frac12 c^2-\frac14 a^2$, so $AG^2=\left(\frac23AM\right)^2=\frac29b^2+\frac29c^2-\frac19a^2$. We can similarly compute that $BG^2=\frac29a^2+\frac29c^2-\frac19b^2$ and $CG^2=\frac29a^2+\frac29b^2-\frac19c^2$, which gives us that $AG^2+BG^2+CG^2=\frac13 (a^2+b^2+c^2)$, as desired.