# Quadrilateral with maximum area inside a semicircle [closed]

The quadrilateral $ABCD$ inside a semicircle with radius 1 has a maximum area. Calculate that maximum area. Clues to solution include the observation that the vertices of such quadrilateral must be on perimeter of semicircle.

## closed as off-topic by Carl Mummert, Claude Leibovici, user228113, Harish Chandra Rajpoot, hardmathFeb 19 '16 at 11:46

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• If this is a puzzle that you already have a solution for, wouldn't Puzzling.SE be more appropriate? – DylanSp Feb 17 '16 at 20:56
• You could reflect the answer in the diameter to give a simpler question – Henry Feb 17 '16 at 21:00
• Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Feb 18 '16 at 9:47
• @HamidRezaEbrahimi It's less about explaining the parameters of the problem and more about explaining why you want to solve it. Is this a.) a textbook problem you're looking for help on, b.) something you're casually researching and exploring, c.) something you're posing as a challenge to the community, or d.) something else? – DylanSp Feb 18 '16 at 13:15
• It's not clear to me, though, what the point is of posting a question to which you know an answer. If your actual question is, does anyone know a proof that's better than the one I'm posting, then that's what you should ask. If you're not actually asking a question, well, consider starting a blog, instead. – Gerry Myerson Feb 18 '16 at 22:05

## 1 Answer

First we note that all of the vertices of such quadrilateral must be on perimeter of semicircle:

For triangle $ABC$ having maximum area implies that B must be right in the middle of arc $AC$ (to have a maximum height).Similarly arcs $BC$ and $CD$ should be equal.
Suppose that the angle of arc $AB$ is $\alpha(0\le\alpha\le\frac{\pi}{3})$. Then

$$\begin{split} S(ABCD) &=S(AOB)+S(BOC)+S(COD)-S(AOD) \\ &=3S(AOB)-S(AOD)\\ &=\frac32\sin(\alpha)-\frac12\sin(3\alpha)\\ &=\frac32\sin(\alpha)-\frac12(3\sin(\alpha)-4\sin^3(\alpha))\\ &=2\sin^3(\alpha)\\ &\le 2 \sin^3(\frac{\pi}{3})\\ &=2(\frac{\sqrt 3}{2})^3\\ &=\frac{3\sqrt3}{4} \end{split}$$
So the maximum area of such quadrilateral inside a semicircle is $\frac{3\sqrt3}{4}$.

• If someone believes an answer has low scientific value , at least he/she should provide a better one. – Hamid Reza Ebrahimi Feb 20 '16 at 3:39
• Ah, we may be approaching the source of your misunderstanding. Scientific value is NOT the only factor users take into account when deciding how to vote. You are not the only user who has been surprised by the fact that people's views of what the site should be about also affect their voting. There are several heated discussions in Meta around this theme. BTW, a nice way of reasoning! I figured this out when I was training for IMO about 35 years ago, so it's a bit old hat for me. Surely some youngsters will benefit from reading this. – Jyrki Lahtonen Feb 20 '16 at 6:56