1
$\begingroup$

The problem is as follows:

The diagonals of a quadrilateral measure $8\,cm$ and $10\,cm$. Using this information find the most probable perimeter of this quadrilateral indicated as a range.

The alternatives given in my book are:

$\begin{array}{ll} 1.&\textrm{Between 14 cm and 16 cm}\\ 2.&\textrm{Between 18 cm and 36 cm}\\ 3.&\textrm{Between 38 cm and 40 cm}\\ 4.&\textrm{Between 42 cm and 50 cm}\\ 5.&\textrm{Between 12 cm and 14 cm}\\ \end{array}$

In this problem I don't know what sort of strategy can be used?. Does it exist a relationship between the diagonals of a quadrilateral?.

So far what it come to my mind was to use the triangle inequality which states that the sum of the two sides of a triangle is greater than the third side.

Given a quadrilateral ABCD then this means the diagonals are from $AC$ and from $BD$.

$AC<AB+BC$

$BD<BC+CD$

$AC<CD+AD$

$BD<AB+AD$

Given that the diagonals are AC and BD then its sum must be less than: By summing the expressions given:

$2AC+2BD<2AB+2BC+2CD+2AD$

$AC+BD<AB+BC+CD+AD$

This must be the upper bound in the perimeter of the quadrilateral. But how about the lower bound?. Well, for this part I'm using the reverse triangle inequality which states that any side is greater than the difference of the other two sides provided one is greater than the other.

In this case I'm assuming $AB>BC$ and $BC>CD$ and $CD>AD$ and $AB>AD$

Then this would mean:

$AC>AB-BC$

BD>BC-CD

$AC>CD-AD$

$BD>AB-AD$

The thing here is that you may not get the perimeter directly.

Suming these:

$2AC+2BD>2AB-2AD$

$AC+BD>AB-AD$

But we know that we're assuming $BD>AC$.

Then summing both gets:

$2BD>AB-AD$

The perimeter is $AB+BC+CD+AD$

But that's where I'm stuck. Can someone help me here to finish this proof?. I think I'm near but I don't know how to use those inequalities to get to the answer. Can someone help me?.

By the way, the official answer from my book states to be the second option. Which it kinds of confirms my initial finding for the upper bound. But it doesn't make sense for the lower bound. Shouldn't it be less?. Therefore, can someone help me?

$\endgroup$
1
$\begingroup$

What you found is actually the lower bound for the perimeter:

$$18 = AC + BD < AB + BC + CD + DA$$

To find the upper bound, we must assume that $ABCD$ is convex. Otherwise the perimeter can be as large as we want (consider a very long arrow-shape.)

If $ABCD$ is convex, the diagonals $AC$ and $BD$ intersect within the quadrilateral at some point $E$. Then we have, by triangle inequality:

$$AE + BE > AB$$$$BE + CE > BC$$$$CE + DE > CD$$$$DE + AE > DA$$

Summing them up, we have:

$$2 (AE + BE + CE + DE) = 2(AC + BD) = 36 > AB + BC + CD + DA$$

hence the perimeter will be bounded between $18$ cm and $36$ cm (or the sum and twice the sum of the diagonals).

$\endgroup$
1
  • $\begingroup$ Actually any perimeter less than $20$ is impossible. Just consider the vertices at the two ends of the longer diagonal. $\endgroup$ – David K Nov 24 '20 at 4:26
1
$\begingroup$

You found a lower bound, but not the lower bound.

Instead of adding your four inequalities all together, just add the ones that have $AC$ on the left. Make a separate sum of inequalities with $BD$ on the left. The results are \begin{align} 2AC &< AB + BC + CD + AD, \\ 2BD &< AB + BC + CD + AD. \end{align} That is, the perimeter must be greater than twice the diagonal, for each diagonal. In short, $$ AB + BC + CD + AD > 2 \max \{ AC, BD \}. $$

Since in your example, $\max \{ AC, BD \} = 10,$ the minimum perimeter is $20.$

To put this intuitively, to get around the quadrilateral once starting at $A$ you must first travel from $A$ to $C$, which is a path at least $10$ units long since $A$ and $C$ are $10$ units apart, and then you must get back from $C$ to $A$, which is at least another $10$ units.

There is no upper bound if we allow quadrilaterals that are not convex. But if the quadrilateral is convex then the diagonals intersect. We maximize the perimeter by intersecting the diagonals very near an end of each diagonal at a very shallow angle; in the limit we have the two diagonals collinear and joined at one endpoint, so that two vertices (say $B$ and $C$) are coincident, so that you go from $A$ to $B$ (distance $8$, since $B=C,$ assuming WLOG that $AC$ is the short diagonal), from $B$ to $C$ (distance $0$), from $C$ to $D$ (distance $10$), and then $D$ to $A$ (distance $18$), for a total perimeter $36$.

In conclusion,

$$ 20 < \mathrm{perimeter} < 36. $$

This is not exactly the same as answer 2, which has a slightly larger range, but answer 2 is the only one that covers the actual minimum and maximum; all the other answers give perimeters that are much too short or much too long. In fact every other answer is giving a range of numbers for which we know the quadrilateral cannot have any of those values as its perimeter.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.