3
$\begingroup$

I have an equilateral triangle with a fixed side-length $x$. What is the largest area of a rectangle that can be put inside the triangle?

$\endgroup$
  • $\begingroup$ Largest in what sense? Longest single side, longest perimeter, biggest area, etc? $\endgroup$ – adfriedman Oct 28 '17 at 16:48
  • $\begingroup$ My bad, it is biggest area $\endgroup$ – Pavel Straka Oct 28 '17 at 17:08
3
$\begingroup$

Consider the situation as in the image..

enter image description here

Then the side $b$ depends on $a$ by the formula $b = \tan(60°) \frac{x-a}{2} = \sqrt{3}\frac{x-a}{2}$, so the area of the rectangle is $S = ab = a\sqrt{3}\frac{x-a}{2}$.

Now look for the maximum of $S$ as a function of $a$ (calculate the first derivative and find the zero point or notice it is a parabola, so its maximum is in the middle of zero-points). In both cases you get that the maximum is in the point $a = \frac{x}{2}$, hence the maximum area is $\frac{x}{2}\sqrt{3}\frac{x-\frac{x}{2}}{2}=\frac{x^2\sqrt{3}}{8}$.

Note that the area of the entire triangle $ \frac{x^2\sqrt{3}}{4}$, so the maximum rectangle area is $1/2$ the triangle area.

$\endgroup$
1
$\begingroup$

Let $a$ be the length of the side of the rectangle, which placed on the side of our triangle, and $b$ be the length of another side of the rectangle.

Thus, $$\frac{\frac{x\sqrt3}{2}-b}{\frac{x\sqrt3}{2}}=\frac{a}{x},$$ which gives $$b=\frac{\sqrt3}{2}(x-a).$$ Id est, by AM-GM we obtain: $$S_{rectangle}=ab=\frac{\sqrt{3}}{2}a(x-a)\leq\frac{\sqrt3}{2}\left(\frac{a+x-a}{2}\right)^2=\frac{x^2\sqrt3}{8}.$$ The equality occurs for $$a=x-a,$$ which says that we got a maximal value.

$\endgroup$

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.