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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestSuppose 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 \(a=\tfrac{1}{\sqrt{3}}\).

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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 \[a=\frac{1}{\sqrt{3}},~b=\frac{2}{3},s=\frac{2\pi}{3\sqrt{3}}\]

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