Find the maximum area of ellipse that can be inscribed in an isosceles triangel of area $A$ and having one axis along the perpendiculur from the vertex of the triangle to the base.


So isosceles triangle I made was of coordinates $D(\frac{a} {2},p),B(0,0)$ and $C(a, 0)$.

Now Area of triangle =$A$=$\frac{ap} {2}$ and the point E($\frac{a} {2},0)$ as the feet of perpendicular on BC side by the vertex D. So I assumed that this will be the major axis of that ellipse because the area of the ellipse is $\pi\times a\times b$.

But, from here I am not able to do anything further.

Any suggestions or hints?


  • $\begingroup$ @symchdmath Please see now. $\endgroup$ – jayant98 Dec 19 '18 at 10:16

It's hard to solve this problem straightforward. However, there is a trick for you.

Let's say we have found a solution and ellipse has axes $a$ and $b$. Let's scale the picture in the direction of axis $b$ by $a/b$ times. Then, ellipse will turn into a circle. Isosceles triangle will turn into some other isosceles triangle. However, the ratio of their areas will remain the same. Then we ask ourselves the question: what isosceles triangle has the largest incircles (in terms of area ratio).

$$ R=\frac{A_{incircle}}{A_{triangle}} = \frac{\pi r^2}{A_{triangle}}=\frac{\pi(2A_{triangle}/P)^2}{A_{triangle}}=4\pi\frac{A_{triangle}}{P^2}. $$

We used here a formula $A_{triangle}=\frac12 Pr$, that area of triangle is half of perimeter times radius of incircle. So now question is what is the largest area can isosceles triangle have with given perimeter. That is very well known question. The answer is equilateral triangle.

So the framework to find a the ellipse with the largest area is the following: you scale your triangle by some value $k$ to make it equilateral, draw an incircle, and then scale everything back.

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  • 1
    $\begingroup$ Or, is that the area of ellipse =$\frac{(\pi)A} {3\sqrt{3}}$ $\endgroup$ – jayant98 Dec 19 '18 at 14:20

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