$PA^2\sin A+PB^2\sin B+PC^2\sin C$ is minimum if P is the incenter.

Let $ABC$ be a triangle and $P$ is a point in the plane of the triangle $ABC$.If $a,b,c$ are the lengths of sides $BC,CA,AB$ opposite to angles $A,B,C$ respectively then prove that

1. $PA^2\sin A+PB^2\sin B+PC^2\sin C$ is minimum if P is the incenter.
   

2. $PA^2\tan A+PB^2\tan B+PC^2\tan C$ is minimum if P is the orthocenter.
   

3. $PA^2+PB^2+PC^2$ is minimum if P is the centroid.
   

4. $PA^2\sin 2A+PB^2\sin 2B+PC^2\sin 2C$ is minimum if P is the circumcenter.
   

How do i have to formulate this expression into a form that is differentiable wrt a single variable or is there some different hidden approach will be used here.Please guide me,how should i solve this question?

Solution 1:

Let us use the following :

For a point $P$ inside $\triangle{ABC}$, $$\vec{OP}=\frac{[\triangle{PBC}]\cdot\vec{OA}+[\triangle{PCA}]\cdot\vec{OB}+[\triangle{PAB}]\cdot\vec{OC}}{[\triangle{ABC}]}\tag 1$$ where $[\triangle{PBC}]$ is the area of $\triangle{PBC}$.

Proof :

$\qquad\qquad\qquad\qquad$

Since $$\vec{OA'}=\frac{A'C\cdot \vec{OB}+BA'\cdot\vec{OC}}{A'C+BA'}=\frac{[\triangle{PCA}]\cdot\vec{OB}+[\triangle{PAB}]\cdot\vec{OC}}{[\triangle{PCA}]+[\triangle{PAB}]},$$ we have $$\begin{align}\vec{OP}&=\frac{PA'\cdot\vec{OA}+AP\cdot\vec{OA'}}{PA'+AP}\\&=\frac{[\triangle{PBC}]\cdot\vec{OA}+([\triangle{PAB}]+[\triangle{PCA}])\cdot\vec{OA'}}{[\triangle{PBC}]+([\triangle{PAB}]+[\triangle{PCA}])}\\&=\frac{[\triangle{PBC}]\cdot\vec{OA}+[\triangle{PCA}]\cdot\vec{OB}+[\triangle{PAB}]\cdot\vec{OC}}{[\triangle{PBC}]+[\triangle{PCA}]+[\triangle{PAB}]}\\&=\frac{[\triangle{PBC}]\cdot\vec{OA}+[\triangle{PCA}]\cdot\vec{OB}+[\triangle{PAB}]\cdot\vec{OC}}{[\triangle{ABC}]}\end{align}$$ (By the way, $(1)$ holds for any point $P$ if we define the area of a "flipped" triangle as negative.)

For 1.

Let $I$ be the incenter.

Since $[\triangle{IBC}]:[\triangle{ICA}]:[\triangle{IAB}]=a:b:c$, by the law of sines, we have

$$\begin{align}\vec{PI}&=\frac{a\vec{PA}+b\vec{PB}+c\vec{PC}}{a+b+c}\\&=\frac{R\sin A\vec{PA}+R\sin B\vec{PB}+R\sin C\vec{PC}}{R\sin A+R\sin B+R\sin C}\\&=\frac{\sin A\vec{PA}+\sin B\vec{PB}+\sin C\vec{PC}}{\sin A+\sin B+\sin C}\end{align}$$where $R$ is the circumradius of $\triangle{ABC}$.

So, setting $k=\sin A+\sin B+\sin C$ gives $$\vec{PI}=\frac{\sin A\vec{PA}+\sin B(\vec{AB}-\vec{AP})+\sin C(\vec{AC}-\vec{AP})}{k}=\frac{k\vec{PA}+\sin B\vec{AB}+\sin C\vec{AC}}{k}$$ $$\Rightarrow \vec{PA}=\vec{PI}-\frac{\sin B}{k}\vec{AB}-\frac{\sin C}{k}\vec{AC}.$$

Similarly, we have $$\vec{PB}=\vec{PI}-\frac{\sin A}{k}\vec{BA}-\frac{\sin C}{k}\vec{BC},\quad\vec{PC}=\vec{PI}-\frac{\sin B}{k}\vec{CB}-\frac{\sin A}{k}\vec{CA}.$$

So, $$\small\begin{align}&PA^2\sin A+PB^2\sin B+PC^2\sin C\\&=\left(\vec{PI}-\frac{\sin B}{k}\vec{AB}-\frac{\sin C}{k}\vec{AC}\right)^2\sin A+\left(\vec{PI}-\frac{\sin A}{k}\vec{BA}-\frac{\sin C}{k}\vec{BC}\right)^2\sin B+\left(\vec{PI}-\frac{\sin B}{k}\vec{CB}-\frac{\sin A}{k}\vec{CA}\right)^2\sin C\\&=kPI^2+\left(-2\sin A\sin B(\vec{AB}+\vec{BA})-2\sin B\sin C(\vec{BC}+\vec{CB})-2\sin C\sin A(\vec{AC}+\vec{CA})\right)\cdot\frac{\vec{PI}}{k}+s\\&=(\sin A+\sin B+\sin C)PI^2+s\end{align}$$ where $s$ is a constant independent from $P$.

So, $PA^2\sin A+PB^2\sin B+PC^2\sin C$ is minimum if $P=I$.

For 2.

Let $H$ be the orthocenter.

Since $[\triangle{HCA}]:[\triangle{HAB}]=A'C:BA'=\frac{AA'}{\tan C}:\frac{AA'}{\tan B}=\tan B:\tan C$ where $A'$ is on the side $BC$ such that $AA'$ is perpendicular to $BC$, we have $$[\triangle{HBC}]:[\triangle{HCA}]:[\triangle{HAB}]=\tan A:\tan B:\tan C.$$

So, $$\vec{PH}=\frac{\tan A\vec{PA}+\tan B\vec{PB}+\tan C\vec{PC}}{\tan A+\tan B+\tan C}.$$

By the similar argument as the one in the first, we can have $$PA^2\tan A+PB^2\tan B+PC^2\tan C=(\tan A+\tan B+\tan C)PH^2+t$$ where $t$ is a constant independent from $P$.

Thus, $PA^2\tan A+PB^2\tan B+PC^2\tan C$ is minimum if $P=H$.

For 3.

Let $G$ be the centroid.

We have $$\vec{PG}=\frac{\vec{PA}+\vec{PB}+\vec{PC}}{3}.$$

So, $$\vec{PA}=\vec{PG}-\frac{\vec{AB}+\vec{AC}}{3},\quad\vec{PB}=\vec{PG}-\frac{\vec{BA}+\vec{BC}}{3},\quad\vec{PC}=\vec{PG}-\frac{\vec{CB}+\vec{CA}}{3}.$$

Hence, we have $$\begin{align}&PA^2+PB^2+PC^2\\&=\left(\vec{PG}-\frac{\vec{AB}+\vec{AC}}{3}\right)^2+\left(\vec{PG}-\frac{\vec{BA}+\vec{BC}}{3}\right)^2+\left(\vec{PG}-\frac{\vec{CB}+\vec{CA}}{3}\right)^2\\&=3PG^2+u\end{align}$$ where $u$ is a constant independent from $P$.

Thus, $PA^2+PB^2+PC^2$ is minimum if $P=G$.

For 4.

Let $O$ be the circumcenter.

Since $$\begin{align}[\triangle{OBC}]:[\triangle{OCA}]:[\triangle{OAB}]&=\frac{R^2\sin(2A)}{2}:\frac{R^2\sin(2B)}{2}:\frac{R^2\sin(2C)}{2}\\&=\sin(2A):\sin(2B):\sin(2C),\end{align}$$ we have $$\vec{PO}=\frac{\sin(2A)\vec{PA}+\sin(2B)\vec{PB}+\sin(2C)\vec{PC}}{\sin(2A)+\sin(2B)+\sin(2C)}.$$

By the similar argument as the one in the first, we have $$PA^2\sin(2A)+PB^2\sin(2B)+PC^2\sin(2C)=(\sin(2A)+\sin(2B)+\sin(2C))PO^2+v$$ where $v$ is a constant independent from $P$.

Thus, $PA^2\sin(2A)+PB^2\sin(2B)+PC^2\sin(2C)$ is minimum if $P=O$.

Solution 2:

For 1, let $X,Y,Z$ be the projections of $P$ onto the sides of triangle $ABC$. Note that by the extended law of sines, $YZ=AP\sin A$, etc. Therefore our expression $$ AP^2\sin A+BP^2\sin B+CP^2\sin C=AP\cdot YZ+BP\cdot ZX+CP\cdot XY\ge 2[AYPZ]+2[BZPX]+2[CXPY]=2[ABC] $$ with equality when $P$ is the incenter, as then $AP\perp YZ$, etc.

Solution 3:

The third part is trivialised by the following identity $$PA^2+PB^2+PC^2=PG^2+AG^2+BG^2+CG^2$$ where G is the centroid.