Putnam_2015_A1

Putnam_2015_A1


WLOG, assume A and B lie on the branch of the hyperbola contained in the first quadrant. Also let the x-coordinate of A be less than that of B.

For the area of triangle APB to be maximised, we require the perpendicular distance of P from AB to be maximal. Therefore, the gradient of the curve at P should be equal to the gradient of AB.
Letting A=(a, a^{-1}), B=(b, b^{-1}) and P=(p, p^{-1}):

\begin{align} \left.\frac{d}{dx}\right|_{x=p}\frac1x&=\frac{b^{-1}-a^{-1}}{b-a}\\ -\frac1{p^2}&=-\frac1{ab}\\ p&=\sqrt{ab} \end{align}


Now, to find the areas bounded by AP and BP, we shall integrate the area under the curve and subtract from the area of the corresponding trapezium:

\begin{align} \text{area bounded by }AP &= ACDP - \int_a^{\sqrt{ab}}\frac1x\,dx\\ &= \frac{{\left(\sqrt{ab}\right)^{-1}}+a^{-1}}2\left(\sqrt{ab} - a\right)-\big[\ln{x}\big]_a^{\sqrt{ab}}\\ &= \frac{\sqrt{ab}+a}{2a\sqrt{ab}}\left(\sqrt{ab} - a\right)-\big[\ln{x}\big]_a^{\sqrt{ab}}\\ &= \frac{ab-a^2}{2a\sqrt{ab}}-\left(\ln{\sqrt{ab}}-\ln{a}\right)\\ &= \frac{b-a}{2\sqrt{ab}}-\ln{\sqrt{\frac{b}a}}\\\\ \text{area bounded by }BP &= PDEB - \int_{\sqrt{ab}}^b\frac1x\,dx\\ &= \frac{b^{-1}+{\left(\sqrt{ab}\right)^{-1}}}2\left(b - \sqrt{ab}\right)-\big[\ln{x}\big]_{\sqrt{ab}}^b\\ &= \frac{b+\sqrt{ab}}{2b\sqrt{ab}}\left(b-\sqrt{ab}\right)-\big[\ln{x}\big]_{\sqrt{ab}}^b\\ &= \frac{b^2-ab}{2b\sqrt{ab}}-\left(\ln{b}-\ln{\sqrt{ab}}\right)\\ &= \frac{b-a}{2\sqrt{ab}}-\ln{\sqrt{\frac{b}a}}\\\\ \end{align}

\therefore\text{area bounded by }AP=\text{area bounded by }BP as desired.

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Nice!!

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