If the sides of a quadrilateral are $a,b,c,d$, prove that the area cannot exceed $(ac+bd)/2$.

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?

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

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 $$ac+bd\geq d_1d_2\geq2S.$$

• For clarity, the inequality, not the equation for cyclic quadrilaterals. – J.G. Mar 9 at 20:35
• @J.G. All these named the Ptolemy's theorem. – Michael Rozenberg Mar 9 at 20:37
• @MichaelRozenberg I don't think you're wrong. I just thought those hyperlinks would help some readers. – J.G. Mar 9 at 21:26
• @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. – Michael Rozenberg Mar 9 at 22:16
• I must say as well that I have learned a lot from your mastering of inequalities ! – Jean Marie Mar 9 at 22:18

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
$$K=\frac{1}{2}·(ac+bd)·\sin\theta\leq \frac{1}{2}·(ac+bd)·1$$