AREA of ellipse inside triangle

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`2 \times 3`	$2 \times 3$
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Comments

Let's denote ellipse half-axes as $a$ and $b$ with $b$ being along the altitude. Let's now stretch the triangle along it's altitude $\frac{a}{b}$ times, then the inscribed ellipse will become inscribed circle of radius $a$ and the triangle will become isosceles triangle with a base of 2 and altitude of $\frac{2a}{b}$ . The incircle radius of such triangle is easy to find using Heron's formula to be: $r=\sqrt{1-\frac{2}{\sqrt{1+\frac{4a^2}{b^2}}+1}}$ on the other hand, it's $a$ . Equating the two after some algebra we get: $b=|1-a^2|$ Since $a$ for obvious reasons is smaller than half of the base, we can omit the absolute value sign. The area of the ellipse is: $s=\pi ab=\pi a(1-a^2)$ At maximum its derivative reaches zero: $s'=\pi(1-3a^2)$ Therefore $a=\frac{1}{\sqrt{3}},~b=\frac{2}{3},s=\frac{2\pi}{3\sqrt{3}}$

Aleksey Korobenko - 7 years, 7 months ago

Suppose that the triangle has vertices $(-1,0)$ , $(1,0)$ and $(0,2)$ . The ellipse must be centred somewhere on the $y$ -axis, and must pass through the origin, so must have equation $\frac{x^2}{a^2} + \frac{(y-b)^2}{b^2} \; = \; 1$ where $a,b > 0$ are the semimajor and semiminor axes (in some order). This ellipse must just touch the line $2x+y=2$ . Substituting $x=1-\tfrac12y$ into the equation for the ellipse, we obtain a quadratic equation for $y$ : $(4a^2+b^2)y^2 - 4(b^2 + 2a^2b)y + 4b^2 = 0$ This equation must have just one root, so must have zero discriminant. The discriminant is $16a^2b^2(a^2+b-1)$ so we deduce that $b = 1-a^2$ . We can now proceed like Aleksey to maximise the area when $a=\tfrac{1}{\sqrt{3}}$ .

Mark Hennings - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$