# Fermat point

The Fermat point (also called the Torricelli point) of a triangle $\triangle ABC$ (with no angle more than $120^{\circ}$ is a point $P$ which has the minimum total distance to three vertices (i.e., $AP+BP+CP$).

$[asy] pair A=(2,4), B=(1,1), C=(6,1); pair pAB=rotate(60,B)*A, pCA=rotate(60,A)*C; path PC=pAB--C, PB=pCA--B; D(MP("A",A,2N)--MP("B",B,SW)--MP("C",C,SE)--cycle,green); D(C--pCA--A--pAB--B,red+dashed); DPA(PC^^PB,lightblue); D(MP("C'",pAB,NW),orange); D(MP("B'",pCA,NE),orange); D(A); D(B); D(C); D(MP("P",IP(PC,PB),E),blue); [/asy]$

## Construction

A method to find the point is to construct three equilateral triangles out of the three sides from $\triangle ABC$, then connect each new vertex to each opposite vertex, as these three lines will concur at first Fermat point.

## Proofs

We shall present a standard triangle inequality proof as well as a less-known vector proof:

### Geometric Proof

First, we shall note that $P$ must lie inside the triangle $\triangle ABC$. Otherwise, we suppose that WLOG, $P$ and $A$ are on opposite sides of $BC$. Then, consider $P'$ the reflection of $P$ about $BC$. Note $PB=P'B$, $PC=P'C$, and $PA>P'A$, so thus $P$ is not the Fermat Point.

Suppose that $\angle A$ was acute. Consider the $60^{\circ}$ rotation about $\angle A$. For any point $X$, let the image of this point be $X'$. Then, we see that $AC=AC'$ and $\angle CAC'=60^{\circ}$. so $ACC'$ is equilateral. Now, consider the point $P$ inside the triangle. Then, $AP=AP'$ and $\angle PAP'=60^{\circ}$, so $AP=PP'$. Thus, we get that $AP+BP+CP=BP+PP'+C'P'$ as $CP=C'P'$.

Now, WLOG let $\angle B\leq\angle C$. We have that $\angle BCC'\leq 180^{\circ}$. We note that $AP+BP+CP=BP+PP'+C'P'\geq BC'$ with equality if and only if $P,P'\in BC'$.

This means that $\angle APC'=60^{\circ}$ (as then $P'\in CP$). Thus, we see that $APCC'$ is cyclic, so thus as $P$ lies on $BC'$, we see that $P$ is the intersection of the circumcircle of $ACC'$ and $BC'$ (not $C'$). Thus, note that as $\angle APC'=\angle CPC'=60^{\circ}$, $\angle APC=120^{\circ}$. Similarly, $\angle CPB=180^{\circ}-\angle CPC'=120^{\circ}$. Thus, we have found the Fermat Point.

### Vector Proof (Due to Titu Andreescu and Oleg Mushkarov)

We will let our origin be the point $P$ with $\angle APB=\angle BPC=\angle CPA=120^{\circ}$.

Consider the point in the plane $X$. Let $a=\vec{A},b=\vec{B},c=\vec{C},x=\vec{X}$ and $i,j,k$ the unit vectors along $a,b,c$. Then, $|a|=a\cdot i=(a-x)\cdot i+x\cdot i\leq |a-x|+x\cdot i$. Similarly, $|b|=b\cdot j=(b-x)\cdot j+x\cdot j\leq |b-x|+x\cdot j$ and $|c|=c\cdot k=(c-x)\cdot k+x\cdot k\leq |c-x|+x\cdot k$. Noting that $i+j+k=\vec{0}$ and adding, we see that $|a|+|b|+|c|\leq |a-x|+|b-x|+|c-x|$, or $AP+BP+CP\leq AX+BX+CX$. Thus, the origin or point $P$ is the desired point.

## Generalizations

There are two main generalizations:

### Weighted Generalization

The problem goes as following: which point $P$ minimizes $m\cdot AP+n\cdot BP+p\cdot CP$, where $m,n,p$ are positive reals?

### Polygon Generalization

The problem goes as following: for the polygon $A_1A_2\cdots A_n$, which point $P$ minimizes $A_1P+A_2P+\cdots+A_nP$?

Using the second solution, it is easy to see the point is the point $P$ where the unit vectors to the vertices sum to $0$. For a quadrilateral, it is the intersection of the diagonals.