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    Maximum-area triangle inscribed in an ellipse

    P-BKS15-2B

    Area
    Prelim
    Seen at
    BERKELEY
    Years
    2015
    Status
    draft

    Find the maximum area of all triangles that can be inscribed in an ellipse with semiaxes \(a\) and \(b\).

    Solution

    1. The linear map\[ T(u,v)=(au,bv) \]maps the unit circle onto the ellipse with semiaxes \(a\) and \(b\), and multiplies every triangle area by \(ab\).

    The unit circle has equation \[ u^2+v^2=1. \] Writing \((x,y)=T(u,v)\) gives \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] so \(T\) maps the unit circle onto the given ellipse. Since \[ \det T=ab, \] the absolute area-scaling factor is \(ab\).

    2. Let a nondegenerate triangle inscribed in the unit circle have interior angles \(A,B,C\). Its side lengths opposite these angles are respectively\[ 2\sin A, \qquad 2\sin B, \qquad 2\sin C. \]

    The central angle subtending the side opposite \(A\) is \(2A\). A chord of the unit circle subtending a central angle \(2A\) has length \[ 2\sin A. \] The same argument applies to the other two sides.

    3. The area \(K\) of such a triangle is\[ K=2\sin A\sin B\sin C. \]

    Using the two sides adjacent to the angle \(A\) and step 2, \[ \begin{aligned} K &= \frac12(2\sin B)(2\sin C)\sin A\\ &= 2\sin A\sin B\sin C. \end{aligned} \]

    4. If \(A,B,C>0\) and \(A+B+C=\pi\), then\[ \sin A\sin B\sin C \leq \left(\frac{\sqrt3}{2}\right)^3, \]with equality exactly when\[ A=B=C=\frac\pi3. \]

    On \((0,\pi)\), set \[ h(x)=\log(\sin x). \] Then \[ h''(x)=-\csc^2x<0, \] so \(h\) is strictly concave. Jensen’s inequality gives \[ \frac{h(A)+h(B)+h(C)}3 \leq h\left(\frac{A+B+C}{3}\right) = h\left(\frac\pi3\right). \] Exponentiating yields \[ \sin A\sin B\sin C \leq \sin^3\left(\frac\pi3\right) = \left(\frac{\sqrt3}{2}\right)^3. \] Strict concavity gives equality exactly when \(A=B=C\).

    5. The maximum area of a triangle inscribed in the unit circle is\[ \frac{3\sqrt3}{4}. \]

    By steps 3 and 4, \[ K \leq 2\left(\frac{\sqrt3}{2}\right)^3 = \frac{3\sqrt3}{4}. \] Equality is attained by an equilateral triangle, whose angles are all \(\pi/3\).

    6. Therefore the maximum area of a triangle inscribed in the given ellipse is\[ \boxed{\frac{3\sqrt3}{4}ab}. \]

    By step 1, \(T\) gives a bijection between triangles inscribed in the unit circle and triangles inscribed in the ellipse, multiplying all their areas by \(ab\). Apply the unit-circle maximum from step 5.

    7. Q.E.D.

    Step 6 gives the requested maximum.