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Divide a quadrilateral into equal areas

Today let us consider the following interesting compass-and-straightedge construction problem:
Dividing quadrilateral problem. Given a quadrilateral $ABCD$ and four points $M_1$, $M_2$, $M_3$, $M_4$, in this order, on $AB$. Using compass and straightedge, show how to construct four points $N_1$, $N_2$, $N_3$, $N_4$ on $CD$ so that the four line segments $M_1 N_1$, $M_2 N_2$, $M_3 N_3$ and $M_4 N_4$ divide the quadrilateral into 5 small quadrilaterals of equal area.

Let us now spend a few minutes to see how we would tackle this problem.





In mathematics, when we face a problem and we do not know where to start, it is always helpful to look at some special cases first, or to simplify the problem. Sometimes, a technique that we use to solve a simplified problem may help us to tackle the problem in its general form.

Let us now try to simplify this problem. This problem wants us to divide a quadrilateral into 5 pieces of equal area. So we ask ourselves this question: "is it possible to just divide the quadrilateral into 2 pieces in the same manner?".



Dividing a quadrilateral into 2 equal areas

Dividing a quadrilateral into 2 equal areas. Given a quadrilateral $ABCD$ and a point $M$ on $AB$. Using compass and straightedge, show how to construct a point $N$ on $CD$ so that the line segment $M N$ divides the quadrilateral into 2 parts of equal area.

In this simplified problem, we need to specify a point $N$ such that the two quadrilaterals $AMND$ and $BMNC$ have the same area $$s(AMND) = s(BMNC) .$$ These two quadrilaterals share a common edge $MN$, the two edges $AM$ and $BM$ are on the same line, and the two edges $DN$ and $CN$ are also on the same line. If somehow we can transform these quadrilaterals into triangles of the same area then it may be easier to deal with.

Suppose we make the following constructions:
  • Through $A$ draw a line parallel to $MD$ which meets $CD$ at $P$.
  • Through $B$ draw a line parallel to $MC$ which meets $CD$ at $Q$.

Since $AP$ is parallel to $MD$, we have $s(AMD) = s(PMD)$, and it follows that $$s(AMND) = s(MND) + s(AMD) = s(MND) + s(PMD) = s(MNP).$$

Similarly, since $BQ$ is parallel to $MC$, we have $s(BMC) = s(QMC)$, and thus, $$s(BMNC) =  s(MNC) + s(BMC) = s(MNC) + s(QMC) = s(MNQ).$$

We have transformed the two quadrilaterals $AMND$ and $BMNC$ into two triangles $MNP$ and $MNQ$. So the two quadrilaterals $AMND$ and $BMNC$ have the same area if and only if $$s(MNP) = s(MNQ),$$ that is when $N$ is the midpoint of $PQ$.

We have now solved the problem. The required point $N$ can be constructed as follows:
  • Draw a line through $A$ parallel to $MD$ which cuts $CD$ at $P$.
  • Draw a line through $B$ parallel to $MC$ which cuts $CD$ at $Q$.
  • Construct the midpoint $N$ of the line segment $PQ$.

Having settled the problem of dividing a quadrilateral into 2 parts, let us now move on to the problem of dividing a quadrilateral into 3 parts.




Dividing a quadrilateral into 3 equal areas

Dividing a quadrilateral into 3 equal areas. Given a quadrilateral $ABCD$ and two points $M_1$, $M_2$ on $AB$. Using compass and straightedge, construct two points $N_1$ and $N_2$ on $CD$ so that the two line segments $M_1 N_1$ and $M_2 N_2$ divide the quadrilateral into 3 parts of equal area.

We can see that the technique used in the previous problem will work very well here as well. We will transform the quadrilaterals $AM_1 N_1 D$ and $BM_1 N_1 C$ into triangles as follows.
  • Through $A$ draw a line parallel to $M_1 D$ which meets $CD$ at $P$.
  • Through $B$ draw a line parallel to $M_1 C$ which meets $CD$ at $Q$.

As before, we have $$s(A M_1 N_1 D) = s(M_1 N_1 P), ~~~s(B M_1 N_1 C) = s(M_1 N_1 Q).$$ So $$s(M_1 N_1 P) = s(A M_1 N_1 D) = \frac{1}{2} s(B M_1 N_1 C) = \frac{1}{2} s(M_1 N_1 Q),$$ and it follows that $$N_1 P = \frac{1}{2} N_1 Q.$$

The point $N_1$ is then determined. To construct the point $N_1$, we need to divide the line segment $PQ$ into three equal parts.

There are many ways to divide a line segment into three equal parts. Here is one example:
  • Draw an arbitrary line through $P$ and subsequently, on this line, construct the points $U$, $V$, $W$ so that $PU = UV = VW$.
  • Connect $Q$ with $W$.
  • Through $U$ draw a line parallel to $QW$ which then cuts $PQ$ at $N_1$.
Once the point $N_1$ is determined then the point $N_2$ can also be constructed. Indeed, to construct the point $N_2$, we only need to solve the problem of dividing the quadrilateral $M_1 B C N_1$ into 2 equal parts, which we already know how to!

By now, you probably know how to solve a general problem! 



The general problem 

Dividing a quadrilateral into $n$ equal areas. Given a quadrilateral $ABCD$ and $n$ points $M_1, \dots, M_n$, in this order, on $AB$. Using compass and straightedge, construct $n$ points $N_1, \dots, N_n$ on $CD$ so that the $n$ line segments $M_1 N_1, \dots, M_n N_n$ divide the quadrilateral into $n+1$ parts of equal area.


First, we need to construct the point $N_1$ so that $$s(A M_1 N_1 D) = \frac{1}{n} s(B M_1 N_1 C),$$ once we have $N_1$, the problem of "dividing $ABCD$ into $n+1$ parts" is reduced to the problem of "dividing the quadrilateral $M_1 BC N_1$ into $n$ parts"!



Let us stop here for now.  Today, we have solved an interesting compass-and-straightedge construction problem. We also learn that when we solve a problem, it is helpful to simplify it first, solving for simple cases will help us to understand the problem better and may give us hint on how to tackle the problem in its general form.

See you again in the next post.



Homework.


1. Given a quadrilateral $ABCD$ and a point $M$ on $AB$. Suppose $r$ is a rational number on the interval $(0,1)$. Using compass and straightedge, construct a point $N$ on $CD$ so that $$s(AMND)= r ~s(ABCD).$$
Prove that the point $N$ exists if and only if $$s(AMD) \leq r ~s(ABCD) \leq s(AMCD).$$

2. Using the problem 1 above to derive another solution for the general problem of dividing a quadrilateral into $n$ equal areas by observing that $$s(A M_1 N_1 D) = \frac{1}{n+1} s(ABCD), ~~s(A M_2 N_2 D)= \frac{2}{n+1} s(ABCD), \dots$$

Find the necessary and sufficient condition for the existence of the points $N_1, \dots, N_n$.





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