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.

making ellipse drawing tool from a piece of thick paper and a sewing thread |

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.

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 $B'$.

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 $t$, $PA + PB \geq MA + MB = \ell$.

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.

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.

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.

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 $t$.

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!

$PA + PB = PA + PK$ |

**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.

## No comments:

## Post a Comment