2
$\begingroup$

MOP 1997: Let $Q$ be a quadrilateral whose side lengths are $a,b,c,d$ in that order. Show that the area of $Q$ does not exceed $(ac+bd)/2$.

My solution: Without loss of generality, let $a$ be the longest side. I first prove that the area cannot exceed (ab+cd)/2. Drawing a diagonal, we can divide the quadrilateral into two triangles with sides are $a$ and $b$, and $c$ and $d$. The triangle with sides $a$ and $b$ can have maximum area if the angle between them is $\pi/2$. In this case, the area between them will be $\frac{1}{2}ab$. Similarly, the area of the other triangle can have a maximum of $\frac{1}{2}cd$. Adding them up, we get that the maximum area of the quadrilateral can be $\frac{1}{2}(ab+cd)$.

Now if this maximum is achieved, which means that the angle between $a$ and $b$, and $c$ and $d$ are both $\pi/2$, by drawing a diagram, we can convince ourselves that $a=c$. This means that $a$ and $b$ are both the longest sides in the quadirlateral. Hence, by the rearrangement inequality, we have $(ab+cd)/2\leq (ac+bd)/2$. Hence, the area of the quadrilateral is less than or equal to $(ac+bd)/2$.

Is the reasoning given above correct? Does there exist a better proof?

$\endgroup$
  • 1
    $\begingroup$ What's the MOP? Did the question come from here? $\endgroup$ – Viktor Glombik Mar 9 at 20:18
  • $\begingroup$ @ViktorGlombik- It did. It's the very first question in the book $\endgroup$ – Anju George Mar 9 at 20:19
  • $\begingroup$ You claim that 'by drawing a diagram, we can convince ourselves that $a=c$'. Are you sure? $\endgroup$ – Dr. Mathva Mar 9 at 20:25
  • $\begingroup$ @Dr.Mathva- Not sure anymore. It seems I didn't consider the possibility of the angle between $b$ and $c$ being greater than $\pi/2$. $\endgroup$ – Anju George Mar 9 at 20:29
  • $\begingroup$ You're right when you claim that the area is maximized when the angle in between is $\pi/2$ $\endgroup$ – Dr. Mathva Mar 9 at 20:30
3
$\begingroup$

Let $d_1$ and $d_2$ be diagonals of the quadrilateral.

Thus, $$S=\frac{1}{2}d_1d_2\sin\measuredangle(d_1,d_2)\leq\frac{1}{2}d_1d_2$$ and by the Ptolemy $$ac+bd\geq d_1d_2\geq2S.$$

$\endgroup$
  • 2
    $\begingroup$ For clarity, the inequality, not the equation for cyclic quadrilaterals. $\endgroup$ – J.G. Mar 9 at 20:35
  • $\begingroup$ @J.G. All these named the Ptolemy's theorem. $\endgroup$ – Michael Rozenberg Mar 9 at 20:37
  • 2
    $\begingroup$ @MichaelRozenberg I don't think you're wrong. I just thought those hyperlinks would help some readers. $\endgroup$ – J.G. Mar 9 at 21:26
  • 1
    $\begingroup$ @Jean Marie Yes, you are right! I know, it's not so good, to say this, but I really think so. Because for the concave quadrilateral $ABCD$ the idea with replacing $\Delta ABC$ on $\Delta CBA$ not always works. $\endgroup$ – Michael Rozenberg Mar 9 at 22:16
  • 1
    $\begingroup$ I must say as well that I have learned a lot from your mastering of inequalities ! $\endgroup$ – Jean Marie Mar 9 at 22:18
1
$\begingroup$

Alternatively (and I know that this solution isn't diagrammatic):

As you claimed, the area of a triangle given two sides $a,b$ is maximized when $\gamma=90°$ since you can express the area of a triangle as $$A=\frac{a·b·\sin\gamma}{2}$$ and $\sin\gamma\leq1$ with equality if $\gamma=90°$. Thus the area of the quadrilateral, given the sides $a,b,c,d$ is maximized when this quadrilateral is cyclic. The area $K$ of cyclic quadrilaterals can be expressed as
$$K=\frac{1}{2}·(ac+bd)·\sin\theta\leq \frac{1}{2}·(ac+bd)·1$$

$\endgroup$

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.