Difference between revisions of "Fermat point"

Latest revision as of 16:07, 29 December 2019

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.

@@ Line 1: / Line 1: @@
-The '''Fermat point''' (also called the Torricelli point) of a triangle <math>\triangle ABC</math> is a point <math>P</math> which has the minimum total distance to three [[vertices]] (i.e., <math>AP+BP+CP</math>).
+The '''Fermat point''' (also called the Torricelli point) of a triangle <math>\triangle ABC</math> (with no angle more than <math>120^{\circ}</math> is a point <math>P</math> which has the minimum total distance to three [[vertices]] (i.e., <math>AP+BP+CP</math>).
 <asy>
@@ Line 16: / Line 16: @@
 ==Construction==
 A method to find the point is to construct three equilateral triangles out of the three sides from <math>\triangle ABC</math>, 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 <math>P</math> must lie inside the triangle <math>\triangle ABC</math>. Otherwise, we suppose that WLOG, <math>P</math> and <math>A</math> are on opposite sides of <math>BC</math>. Then, consider <math>P'</math> the reflection of <math>P</math> about <math>BC</math>. Note <math>PB=P'B</math>, <math>PC=P'C</math>, and <math>PA>P'A</math>, so thus <math>P</math> is not the Fermat Point.
+Suppose that <math>\angle A</math> was acute. Consider the <math>60^{\circ}</math> rotation about <math>\angle A</math>. For any point <math>X</math>, let the image of this point be <math>X'</math>. Then, we see that <math>AC=AC'</math> and <math>\angle CAC'=60^{\circ}</math>. so <math>ACC'</math> is equilateral. Now, consider the point <math>P</math> inside the triangle. Then, <math>AP=AP'</math> and <math>\angle PAP'=60^{\circ}</math>, so <math>AP=PP'</math>. Thus, we get that <math>AP+BP+CP=BP+PP'+C'P'</math> as <math>CP=C'P'</math>.
+Now, WLOG let <math>\angle B\leq\angle C</math>. We have that <math>\angle BCC'\leq 180^{\circ}</math>. We note that <math>AP+BP+CP=BP+PP'+C'P'\geq BC'</math> with equality if and only if <math>P,P'\in BC'</math>.
+This means that <math>\angle APC'=60^{\circ}</math> (as then <math>P'\in CP</math>). Thus, we see that <math>APCC'</math> is cyclic, so thus as <math>P</math> lies on <math>BC'</math>, we see that <math>P</math> is the intersection of the circumcircle of <math>ACC'</math> and <math>BC'</math> (not <math>C'</math>). Thus, note that as <math>\angle APC'=\angle CPC'=60^{\circ}</math>, <math>\angle APC=120^{\circ}</math>. Similarly, <math>\angle CPB=180^{\circ}-\angle CPC'=120^{\circ}</math>. Thus, we have found the Fermat Point.
+===Vector Proof (Due to Titu Andreescu and Oleg Mushkarov)===
+We will let our origin be the point <math>P</math> with <math>\angle APB=\angle BPC=\angle CPA=120^{\circ}</math>.
+Consider the point in the plane <math>X</math>. Let <math>a=\vec{A},b=\vec{B},c=\vec{C},x=\vec{X}</math> and <math>i,j,k</math> the unit vectors along <math>a,b,c</math>. Then, <math>|a|=a\cdot i=(a-x)\cdot i+x\cdot i\leq |a-x|+x\cdot i</math>. Similarly, <math>|b|=b\cdot j=(b-x)\cdot j+x\cdot j\leq |b-x|+x\cdot j</math> and <math>|c|=c\cdot k=(c-x)\cdot k+x\cdot k\leq |c-x|+x\cdot k</math>. Noting that <math>i+j+k=\vec{0}</math> and adding, we see that <math>|a|+|b|+|c|\leq |a-x|+|b-x|+|c-x|</math>, or <math>AP+BP+CP\leq AX+BX+CX</math>. Thus, the origin or point <math>P</math> is the desired point.
+==Generalizations==
+There are two main generalizations:
+===Weighted Generalization===
+The problem goes as following: which point <math>P</math> minimizes <math>m\cdot AP+n\cdot BP+p\cdot CP</math>, where <math>m,n,p</math> are positive reals?
+===Polygon Generalization===
+The problem goes as following: for the polygon <math>A_1A_2\cdots A_n</math>, which point <math>P</math> minimizes <math>A_1P+A_2P+\cdots+A_nP</math>?
+Using the second solution, it is easy to see the point is the point <math>P</math> where the unit vectors to the vertices sum to <math>0</math>. For a quadrilateral, it is the intersection of the diagonals.
 ==See Also==

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