# Need to find the ellipse of maximum area inscribed in a semicircle.

An ellipse inscribed in a fixed semi circle touches the semi-circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter.

When does the ellipse have the maximum possible area? What is its eccentricity $e$ in that case?

I managed to solve it using a coordinate based approach combined with some calculus and obtained the correct answer, that is, when $e=\sqrt{\dfrac 23}$.

But that solution was too boring and time-consuming, and I don't think that it was what the examiner had in his mind, since this is a question from KVPY 2014 SB which means it is intended to be solved in like, 2 minutes tops on a single piece of paper approximately $200\space cm^2$ in area.

EDIT: The paper area condition seems too restrictive so any synthetic method would suffice.

• Basing this purely on intuition and not rigorous math, the eccentricity of the ellipse should be greater than that of the circle, so the 2 points where the ellipse meets the semi-circle should be the the end-points of major axis of the ellipse. From there, it should be very simple calculus. – sirblobfish Dec 17 '15 at 19:20
• @sirblobfish firstly the eccentricity of a circle is zero so for ellipse it is trivial to say its eccentricity is greater. And it is not possible for those 2 points to be the endpoints of the major axis because then the bounding diameter would need to coincide with the major axis. – najayaz Dec 17 '15 at 20:00
• I used the coordinate based approach combined with calculus too. It took about $600$ cm² of paper area to solve it. Are you interested? – RicardoCruz Dec 18 '15 at 13:33
• here is a 8 min 21 sec video on youtube, a bit long, and uses calculus, and is probably not what you are looking for :) – Mirko Dec 19 '15 at 23:45
• My intuition tells me we would get a right angle if we connect the origin to the two touching points of the ellipse with the semicircle. That is, the ellipse touches the circle at the points where the diagonal $y=x$ and antidiagonal $y=-x$ cross the semicircle. (For the unit circle, points $(\frac{\pm1}{\sqrt2},\frac1{\sqrt2})$. This appears to be confirmed by the answer below calculating $b=\frac{\sqrt{2}}{3}$ and $a=\sqrt{\frac23}$, and the graph I made using these $a$ and $b$. I feel this might give a "geometric" proof, but don't see how to continue. – Mirko Dec 20 '15 at 2:31

Stretch the original figure by a factor $\sqrt{3}$ in vertical direction. The half circle $H$ then becomes a half ellipse $E$, and the isosceles right triangle $ABC$ with $C=(0,\sqrt{2})$ transforms into the equilateral triangle $\Delta:=ABD$ with $D=(0,\sqrt{6})$. The oblique legs of $\Delta$ are tangent to $E$ at their midpoints $M$ and $N$. We now solve the analogous problem with respect to $E$. I claim that the largest (i.e., largest with respect to area) ellipse inscribed in $E$ is the incircle $I$ of $\Delta$. Proof: It is well known that the largest ellipse contained in the equilateral triangle $\Delta$ is $I$. As all feasible ellipses for the modified problem are subsets of $\Delta$ they all have an area $\leq {\rm area}(I)$. Since $I$ itself is a feasible ellipse the claim follows. The radius of $I$ computes to $$r={1\over3}|OD|={\sqrt{6}\over3}\ .$$ Returning to the original problem we see that the largest ellipse inscribed in $H$ has semiaxes $$a=r={\sqrt{6}\over3},\quad b={r\over\sqrt{3}}={\sqrt{2}\over3}\ ,$$ and touches the semicircle at the points $\bigl(\pm{1\over\sqrt{2}}, \>{1\over\sqrt{2}}\bigr)$. The eccentricity of this ellipse comes to $$e:{\sqrt{a^2-b^2}\over a}=\sqrt{{2\over3}}\ .$$

• This looks promising, I'll try to understand it after I get back from class. – najayaz Dec 23 '15 at 10:22
• @ChristianBlatter: A clever and insightful approach. Transforming the whole space from an isosceles, right triangle to an equilateral triangle and thereby transforming the ellipse, which is the actual object under investigation to a simpler object, a circle. ... And within the transformed space the maximality property of the incircle $I$ is well known! :-) – Markus Scheuer Dec 23 '15 at 11:41

Let the ellipse $$\frac{x^2}{a^2}+\frac{(y-b)^2}{b^2}=1\quad(1)$$ and the semicircle $$y=\sqrt{1-x^2}\quad(2).$$ The ellipse is clearly tangent to $x$-axis.

If we substitue the semicircle equation in the ellipse equation, we get: $$(a^2-b^2)y^2-2a^2by+b^2=0 \quad(3)$$ As the semicircle and the ellipse are tangent the determinant of equation $(3)$ must be zero. Therefore $$b=\sqrt{a^2(1-a^2)} \quad(4)$$ But we know that the ellipse area can be calculated by: $$A=\pi a b \quad(5)$$ From $(4)$ and $(5)$, we get: $$A= \pi \sqrt{a^4(1-a^2)} \quad (6)$$ To calculate the maximum of $A$ we can use the AM-GM inequality: $$(\frac{a^2}{2}\frac{a^2}{2}(1-a^2))^{\frac{1}{3}} \leq \frac{\frac{a^2}{2}+\frac{a^2}{2}+(1-a^2)}{3}=\frac{1}{3} \Rightarrow$$ $$a^2a^2(1-a^2)\leq \frac{4}{27}. \quad(7)$$ We know that the equality holds for $$\frac{a^2}{2}=(1-a^2).\quad(8)$$ Therefore $$a=\sqrt{{\frac{2}{3}}}\quad(9)$$ From $(4)$ and $(9)$, we get: $$b=\frac{\sqrt{2}}{3}$$ Finally we get $$e=\sqrt{\frac{2}{3}}.$$

• This is probably the intended solution. I do not see any faster way of doing this. – Jack Tiger Lam Dec 20 '15 at 2:40
• @JackLam I agree that this is probably an intended solution but now I'm looking for one which avoids analytical means. – najayaz Dec 20 '15 at 6:03

I don't know if there is a geometric method quickly providing the solution. But if I had to answer that question in a couple of minutes I would have reasoned as follows.

If $a$ and $b$ are major and minor semi-axes of the ellipse, I expect $b/a\approx 1/2$, that is $e\approx\sqrt3/2$. The four proposed answers are: $1\over\sqrt2$, $1\over2$, $1\over\sqrt3$, $\sqrt2\over\sqrt3$ and the fourth one is the nearest to my estimate.

• Why did you expect b/a ≈ 1/2? What's the thought process behind it? – strawberry-sunshine Oct 27 '18 at 15:23
• Because 1/2 is the height/width ratio for a semicircle. – Intelligenti pauca Oct 27 '18 at 16:19