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
·
2 years, 11 months ago

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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 \[a=\frac{1}{\sqrt{3}},~b=\frac{2}{3},s=\frac{2\pi}{3\sqrt{3}}\]
–
Aleksey Korobenko
·
2 years, 12 months ago

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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 \(a=\tfrac{1}{\sqrt{3}}\). – Mark Hennings · 2 years, 11 months ago

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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 \[a=\frac{1}{\sqrt{3}},~b=\frac{2}{3},s=\frac{2\pi}{3\sqrt{3}}\] – Aleksey Korobenko · 2 years, 12 months ago

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