2
$\begingroup$

Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.

The three vertices of the triangle would be $(a,0), (x,y), (x, -y)$.

The area of the triangle by Heron's formula is $$A^2 = (x-a)^2y^2 = (x-a)^2b^2\left( 1- \dfrac{x^2}{a^2}\right) \tag{1}.$$

Hence $$\dfrac{dA}{dx} = 0 \implies (x-a)^2 \left( x + \dfrac{a}2\right) = 0.$$

We have minimum at $x = a$ and maximum at $x = -\frac{a}{2}$.

Substituting back in $(1)$ and taking square roots on both the sides gives $$A = \dfrac{\sqrt{3}ab}{4}$$.

The given answer is $3A$.

What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene ?

$\endgroup$
  • $\begingroup$ How do you know the vertex on the major axis lies on the line of symmetry of the triangle? $\endgroup$ – Shuri2060 Aug 11 '17 at 20:42
  • $\begingroup$ @Shuri2060 Otherwise the triangle won't be isosceles. $\endgroup$ – user8277998 Aug 11 '17 at 20:46
  • $\begingroup$ Isosceles triangles can be inscribed without that condition. Take the triangle $(a,0),(-a,0),(0,b)$, for example. $\endgroup$ – Shuri2060 Aug 11 '17 at 20:48
  • 1
    $\begingroup$ @123 If that is the case, then you (or the original problem) perhaps need to make it clear which vertex of the isosceles triangle is referred to. A triangle has three vertices. The example above has a vertex at one of those points. $\endgroup$ – Shuri2060 Aug 11 '17 at 20:53
  • 1
    $\begingroup$ The problem states the triangle's [unique] "vertex" lies at $(a,0)$, so all is clear. $\endgroup$ – David G. Stork Aug 11 '17 at 20:59
1
$\begingroup$

Everything is correct until you use the value of $x$ to calculate the area. You should have $$A^2=(-\frac{3a}{2})^2b^2(1-\frac 14)$$ which will give you the correct answer $$A=\frac{3\sqrt{3}}{4}ab$$

$\endgroup$
  • $\begingroup$ Oh, I realise my mistake, I took $x = a/2$ thinking that length can't be negative. I was so wrong ... . $\endgroup$ – user8277998 Aug 11 '17 at 20:47
2
$\begingroup$

By AM-GM $$S_{\Delta}=(a-x)y=(a-x)b\sqrt{1-\frac{x^2}{a^2}}=ab\left(1-\frac{x}{a}\right)\sqrt{1-\frac{x^2}{a^2}}=$$ $$=\frac{ab}{\sqrt3}\cdot\sqrt{\left(1-\frac{x}{a}\right)^3\left(3+\frac{3x}{a}\right)}\leq\frac{ab}{\sqrt3}\cdot\sqrt{\left(\frac{3\left(1-\frac{x}{a}\right)+3+\frac{3x}{a}}{4}\right)^4}=\frac{3\sqrt3ab}{4}.$$ The equality occurs for $1-\frac{x}{a}=3+\frac{3x}{a}$, which says that the equality indeed occurs,

which says that $\frac{3\sqrt3ab}{4}$ is a maximal value.

Done!

$\endgroup$
  • $\begingroup$ Really cool method but I feel like this would be impossible to do without knowing the answer in the first place. $\endgroup$ – user8277998 Aug 12 '17 at 6:45
  • 1
    $\begingroup$ @123 I don't agree with you! We need just to delete $\frac{x}{a}$ inside the root and we made it by AM-GM. By the way, I did not know the answer before. $\endgroup$ – Michael Rozenberg Aug 12 '17 at 6:49
1
$\begingroup$

The given answer is correct. Your mistake is in the formula for the area, which should be $A/2 = 1/2\ {\rm base} \times {\rm height}$ for each half half triangle or

$A = (a-x) b \sqrt{1 - x^2/a^2}$

for the full triangle.

The derivative is then:

$${d A \over d x} = -\frac{b (a+2 x) \sqrt{1-\frac{x^2}{a^2}}}{a+x} .$$

Set it to $0$ to find $x = -a/2$.

Then compute the area:

$$A = {3 \sqrt{3} \over 4} a b$$

enter image description here

$\endgroup$
1
$\begingroup$

We perform an orthogonal projection to map the ellipse to the unit circle.

Let the maximal area of our isosceles triangle be $\mathcal{A}$ which we wish to find, and let $\triangle ABC$ be the isosceles triangle with maximal area inscribed in our unit circle. Since orthogonal projections preserve area ratios, $\triangle ABC$ is the projection of the triangle with area $\mathcal{A}$. Note that since we wish to maximise the area, $\triangle ABC$ is simply an equilateral triangle with area $\frac{3\sqrt{3}}{4}$. Hence, by preservation of area ratios,

$$\begin{align*} \frac{\mathcal{A}}{\text{ Area of ellipse}} &=\frac{[ABC]}{\text{ Area of circle}} \\ \implies \mathcal{A} &=\frac{3\sqrt{3}}{4\pi}\cdot \pi ab \\ &= \frac{3ab\sqrt{3}}{4} \end{align*}$$which is our answer.

$\endgroup$

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.