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| the Fermat's problem: find $M$ so that $MA + MB + MC$ is minimum |
The interesting part is that we have used a property of the ellipse to work out the solution. Our analysis shows that the point $M$, if it exists, can only be at four possible locations. It is either the point $A$, $B$, $C$, or the point specified by $\angle AMB = \angle BMC = \angle CMA = 120^{o}$.
Let us emphasize that, in the above statement, we say that if this point $M$ exists then it has to be at four possible positions. However, we have not shown that such a point $M$, for which $MA + MB + MC$ is minimal, exists. So with mathematical arguments, we have to be extra careful.
Today, we will solve the Fermat problem in a "more conventional" way! We will look at two separate cases.
The first case is when all the angles of the triangle $ABC$ are smaller than $120^{o}$. We will show that the point $M$ that we are looking for is specified by $\angle AMB = \angle BMC = \angle CMA = 120^{o}$.
The second case is when one of the angles of the triangle $ABC$ is greater than or equal to $120^{o}$. For example, when $\angle A \geq 120^{o}$, we will show that the point $M$ is $A$.
Case 1: $\angle A, \angle B, \angle C < 120^{o}$.
Draw outside the triangle $ABC$ three equilateral triangles $ABC'$, $BCA'$ and $CAB'$.
Take an arbitrary point $P$. Use $A$ as the center, make a $60^{o}$ rotation so that the triangle $APB$ will be rotated to the triangle $AQC'$. By this rotation, we have obtained two things. Firstly, two triangles $APB$ and $AQC'$ are equal triangles, and secondly, the triangle $APQ$ is an equilateral triangle. So we have $PB = QC'$ and $PA = PQ$.
Using rotation, we transform $PA + PB + PC = CP + PQ + QC'$. So $PA + PB + PC$ will be minimal when it is equal to $CC'$, it happens when the four points $C$, $P$, $Q$ and $C'$, in this order, lying on the same straight line.
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| $PA = PQ$, $PB = QC'$, so $PA + PB + PC = CP + PQ + QC' \geq CC'$ |
To make $C$, $P$, $Q$ lie on a straight line we must have $\angle CPA = 180^{o} - \angle APQ = 120^{o}$.
Similarly, to have $P$, $Q$, $C'$ lying on a straight line we must have $\angle AQC' = 180^{o} - \angle PQA = 120^{o}$, and so $\angle APB = \angle AQC' = 120^{o}$.
Since $\angle APQ = 60^{o}$, if $C$, $P$, $Q$, $C'$ lie on a straight line then $P$ must be the intersection point of $CC'$ with the circumcircle of the equilateral triangle $ABC'$.
In summary, we have shown that $PA + PB + PC \geq CC'$ and the necessary condition for $PA + PB + PC = CC'$ is that $\angle CPA = \angle APB = 120^{o}$ and the point $P$ is the intersection point of $CC'$ with the circumcircle of the equilateral triangle $ABC'$.
To complete the proof, we construct the point $M$ such that $\angle CMA = \angle AMB = \angle BMC = 120^{o}$ and we show that $MA + MB + MC = CC'$, thus showing $M$ is the point that we are looking for.
Indeed, if we take $M$ as the intersection point of $CC'$ with the circumcircle of the equilateral triangle $ABC'$ then $\angle AMB = 120^{o}$ and $\angle AMC' = 60^{o}$, and thus, $\angle AMC = 120^{o}$. Let $N$ be the point on $CC'$ so that $AMN$ is an equilateral triangle. By the rotation of $60^{o}$ at center $A$, the triangle $AMB$ will be rotated to the triangle $ANC'$. Thus, the point $M$ satisfies the equality $MA + MB + MC = MN + NC' + MC = CC'$.
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| At the Fermat's point $M$ we have $MA + MB + MC = CC'$ |
We have constructed the point $M$ such that $\angle AMC = \angle AMB = \angle BMC = 120^{o}$ and showed that $MA + MB + MC = CC'$. Thus, $M$ is our solution point.
By similarity, $M$ must also lie on $AA'$ and $BB'$ and we have $AA' = BB' = CC'$, and also the three circumcircles of the equilateral triangles $ABC'$, $BCA'$, $CAB'$ must meet at $M$.
We remark that the condition that $\angle A, \angle B, \angle C < 120^{o}$ is a necessary condition to ensure that the four points $C$, $M$, $N$ and $C'$ lie on a straight line in this order. If one of the angles of the triangle $ABC$ is greater than $120^{o}$ then the points $C$, $M$, $N$ and $C'$ lie on a straight line, but not in this order (see the figure below).
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| if one of the angles of the triangle $ABC$ is greater than $120^{o}$ then $C$, $M$, $N$ and $C'$ lie on a straight line, but not in this order |
Case 2: $\angle A \geq 120^{o}$.
Here we will sketch the proof, the details is left for the readers.
The triangle $AQC'$ is obtained from the triangle $APB$ by a rotation so $PB = QC'$. The triangle $APQ$ is an equilateral triangle, so $PA = PQ$. Therefore, $PA + PB + PC = CP + PQ + QC'$.
Next, we need to prove that $CP + PQ + QC' \geq CA + AC' = CA + AB$ and $CP + PQ + QC'$ is minimum when $P=A$.
If you need some hints to prove the inequality $CP + PQ + QC' \geq CA + AC'$, please read this post "Solve for special cases first!".
Let us stop here for now. If you know other ways to solve the Fermat's problem please post it here so we all can share. See you again in the next post!
Homework. Given a square $ABCD$. Find on the four sides of the square four points $X$, $Y$, $Z$, $T$ so that $XY + YZ + ZT + TX$ is minimum.
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| $PA = PQ$, $PB = QC'$, so $PA + PB + PC = CP + PQ + QC' \geq CA + AC' = CA + AB$ |
The triangle $AQC'$ is obtained from the triangle $APB$ by a rotation so $PB = QC'$. The triangle $APQ$ is an equilateral triangle, so $PA = PQ$. Therefore, $PA + PB + PC = CP + PQ + QC'$.
Next, we need to prove that $CP + PQ + QC' \geq CA + AC' = CA + AB$ and $CP + PQ + QC'$ is minimum when $P=A$.
If you need some hints to prove the inequality $CP + PQ + QC' \geq CA + AC'$, please read this post "Solve for special cases first!".
Let us stop here for now. If you know other ways to solve the Fermat's problem please post it here so we all can share. See you again in the next post!
Homework. Given a square $ABCD$. Find on the four sides of the square four points $X$, $Y$, $Z$, $T$ so that $XY + YZ + ZT + TX$ is minimum.