Rhombus in Rectangle
Take any rectangle ABCD such that AB > BC. The point P is on AB
and Q is on CD. Show that there is exactly one position of P and Q
such that APCQ is a rhombus.
Age
14 to 16
Challenge level
Problem
Take any rectangle $ABCD$ such that $AB > BC$. The point $P$ is on $AB$ and $Q$ is on $CD$. Show that there is exactly one position of $P$ and $Q$ such that $APCQ$ is a rhombus.
Show that if the rectangle has the proportions of A4 paper ($AB=BC$ $\sqrt 2$) then the ratio of the areas of the rhombus and the rectangle is $3:4$.
Show also that, by choosing a suitable rectangle, the ratio of the area of the rhombus to the area of the rectangle can take any value strictly between $\frac{1}{2}$ and $1$.
Getting Started
Call the base of the rectangle $b$ and the height $h$. If the distance from A to P is $x$, can you write an equation linking $b$, $h$ and $x$ in order to make $AP=PQ$? How many possible values can x take in your equation?
Student Solutions
Another Tough Nut! Take any rectangle $ABCD$ such that $AB > BC$ and say the lengths of $AB$ and $CD$ are $S$ and $s$ respectively. The point $P$ is on $AB$ and $Q$ is on $CD$. For $APCQ$ to be a rhombus, the lengths $AP$ and $PC$ must be equal. Consider the point $P$ coinciding with $A$ (such that $AP=0$) and then $P$ moving along $AB$ so that the length $AP$ increases continuously from $0$ to $S$ while the length of $PC$ decreases continuously from $\sqrt{S^2 + s^2}$ to $s$. As $AP < PC$ initially (when P is at A) and $AP > PC$ finally (when $P$ is at $B$) there must be one point at which $AP = PC$. Similarly there is exactly one position of $Q$ such that $CQ = QA$ making $APCQ$ into a rhombus.
Now take $AP = PC = x$ than you can use Pythagoras' Theorem to find $x$ in terms of $S$ and $s$ so that you can find the ratio of the areas of the areas of the rhombus and the rectangle.
Teachers' Resources
Why do this problem?
This problem allows students to gain an understanding of a geometrical situation by using an algebraic representation. It can be approached numerically at first and then generalised. The problem can be used to practise expansion of brackets, changing the subject of a formula, fractions and surds, as well as the application of Pythagoras' theorem.Possible approach
Start with a particular rectangle, for example 6 units by 4
units. If I positioned$Q$ 1 unit from$D$, would it be a rhombus?
Using Pythagoras, show that $AQ$ and $QC$ would not be equal.By
calling the distance $QC$ $x$, students could try to write down an
equation which must be true for the shape to be a rhombus, that is
for $AQ=QC$. By rearranging to make $x$ the subject, students
should be able to justify the statement that there is only one
possible value of $x$.
It may be necessary to try other numerical examples before
generalising, but once a general form is found linking $x$ with the
base and height of the rectangle, the rest of the problem can be
tackled.
Key questions
What must be true about the lengths $DQ$ and $PB$?
What must be true about the lengths $AQ$ and $QC$?
What happens to the length of $AQ$ as I move $Q$ further from
$D$?