Today we will look at two problems that seem to be unrelated. The first one is a beautiful geometry problem about finding shortest path and the other one is about a property of an ellipse.
But first, let us introduce the ellipse. An ellipse is drawn below.
To draw an ellipse, we need to specify two points $F_1$ and $F_2$, they are called the two focus points of the ellipse, and we need to specify a length $\ell$. The sum of the distances from any point on the ellipse to the two focus points is always equal to $\ell$. That is, for any point $P$ on the ellipse, $PF_1 + PF_2 = \ell$.
We can make an ellipse drawing tool from a piece of thick paper and a sewing thread as follows.
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| making ellipse drawing tool from a piece of thick paper and a sewing thread |
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| drawing an ellipse |
So now we already know what an ellipse is, let us look at the first problem. Here we have a straight line $t$ and two points $A$ and $B$ lying on the same side of the line. The problem is to find a point $M$ on the line so that the path from $A$ to $M$ and back to $B$ is the shortest.
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| Find $M$ so that $MA + MB$ is minimum |
Problem 1: Given a line $t$ and two points $A$ and $B$ on the same side of $t$. Find a point $M$ on the line $t$ such that $MA + MB$ is minimum.
Can you guess where that point $M$ is? My prediction is that there are three possibilities (see the figure below). The point $M$ may be the point $A'$ -- the intersection point where we draw from $A$ a straight line perpendicular to $t$, the point $M$ could be $B'$, or it could be a point $C$ lying somewhere between $A'$ and $B'$.
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| prediction: $M$ could be $A'$, $B'$, or a point $C$ lying in the middle |
We can use a ruler and measure the following three distances
- $A'A+ A'B$
- $B'A + B'B$
- $CA + CB$
to see which one is the smallest.
How are you going? Have you found out which one is the smallest? My measure showed that $CA + CB$ is the smallest. But where is exactly the point $C$ between $A'$ and $B'$ so that $CA + CB$ is minimum?
Let us look at another way. Suppose that we have already found a point $M$ so that $MA + MB$ is minimum, and this minimum value is $MA + MB = \ell$. Let us draw an ellipse that passes though $M$ and that has two focus points $A$ and $B$ as in the figure below. We observe that
- for any point $P$ on the ellipse, $PA + PB = MA + MB = \ell$;
- on the other hand, for any point $P$ on the line $t$, $PA + PB \geq MA + MB = \ell$.
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| for any point $P$ on the ellipse: $PA+PB=\ell$, but for any point $P$ on the line $t$: $PA+PB \geq \ell$ |
This observation leads us to conclude that the ellipse has to be tangent to the line $t$ at the point $M$. Why is that? It is because if the ellipse is not tangent to $t$ then it will cut the line $t$ at two points $M$ and $N$. As in the following figure, we pick an arbitrary point $X$ between $M$ and $N$, then
$$XA + XB < XA + XU + UB = AU + UB = \ell = MA + MB.$$
So $XA + XB < MA + MB$, this is contradictory to our assumption that $MA + MB$ is minimum.
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| if the ellipse is not tangent to the line $t$ then we can find a point $X$ better than $M$ |
Even though we haven't found a way to specify the point $M$ on the line $t$ so that $MA+MB$ is minimum, but our analysis shows that if $M$ is such a point then the ellipse that passes through $M$ and that has two focus points $A$ and $B$ must be tangent to $t$ at the point $M$. The line $t$ is called a tangent line of the ellipse.
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| if $MA + MB$ is minimum then $t$ is a tangent line to the ellipse |
We still don't know how to construct the point $M$. Let us look at some special cases. If $B$ is on the line $t$, then the point $M$ must be $M=B$. If the point $B$ approaches very near the line $t$ then the point $M$ must also be very near $B$ to have $MA + MB$ minimal.
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| if $B$ is near $t$ then $M$ should be near $B$ so that $MA + MB$ is minimal |
Even though we are given that $A$ and $B$ are two points on the same side of the line $t$, let us consider what would happen if $B$ is on the other side of the line. Obviously, if $B$ is on the other side of the line then $MA + MB$ is minimum when $M$ is the intersection point of $AB$ and the line $t$.
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| if $B$ is on the other side of the line then $MA + MB$ is minimum when $M$ is the intersection point of $AB$ and $t$ |
Well, can we construct a point on the other side of the line that have the same role as $B$? Yes we can! Let $K$ be the reflection point of $B$ over the line $t$, then for any point $P$ on the line $t$ we have $PB = PK$. So $PA + PB = PA + PK$ and $PA + PK$ is minimum when $P$ is the intersection point of $AK$ and $t$. And the problem is solved!
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| $PA + PB = PA + PK$ |
Solution to problem 1: Let $K$ be the reflection point of $B$ over the line $t$. For any point $P$ on the line $t$, we have $PB = PK$, and thus, $PA + PB = PA + PK \geq AK$. Therefore, the point $M$ such that $MA + MB$ is minimum is the intersection point of $AK$ and $t$.
Problem 2: Given an ellipse with two focus points $A$ and $B$. Through a point $M$ on the ellipse, draw a tangent line as in the figure below. Prove that $\angle xMA = \angle yMB$.
Thank to our above analysis of the problem 1, problem 2 becomes quite simple. To solve problem 2, we don't need to write equation for the ellipse, nor do we need to know the equation for the tangent line.
Solution to problem 2: As shown in the following figure, by the above analysis, we know that the point $M$ where the ellipse touches the line is the point $M$ on the line such that $MA + MB$ achieves its minimum. Let $K$ be the reflection point of $B$ over the tangent line, then $M$ is the intersection point of $AK$ and the tangent line. So we have $\angle yMB = \angle yMK = \angle xMA$ and the problem is proved.
That's all for now. Today, we learn that by analysing a problem in different angles we may achieve some interesting results that may seem unrelated to our original problem. As in our problem 1, it is a problem about finding shortest path, but our analysis has led us to a beautiful property of the ellipse.
If you know how to write the equation of an ellipse and its tangent lines, it is an interesting problem to find a direct proof of the problem 2 using these equations.
Homework.
1. Given an angle $xOy$ and two points $A$ and $B$ inside that angle. Find a point $X$ on $Ox$ and a point $Y$ on $Oy$ so that $AX + XY + YB$ is minimum.
2. Given a triangle $ABC$. Find a point $M$ so that $MA + MB + MC$ is minimum.
