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Show that the maximum area of a rectangle which can be inscribed in a triangle of area $A$ is $\dfrac{A}{2}$.

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I was trying to solve this as an application of maxima/minima, but it becomes a little clumsy.

From the figure, $AH=\frac{x}{\tan A}$ and $BK=\frac{x}{\tan B}$ , where $FH=GK=x$ (say)

$\therefore$ Area $(\Delta)$ of rectangle $FGKH=x\times HK=x(AB-(AH+BK))$

$=x\left(c-\left(\frac{x}{\tan A}+\frac{x}{\tan B}\right)\right)$ $\qquad$$(AB=c)$

Then simplifying $\dfrac{d(\Delta)}{dx}=0$ to get the answer becomes somewhat tiresome.

This answer gives a nice approach for the solution. Is there any simple yet rigorous alternative proof of this proposition?

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Hint:

The hyperbola whose asymptotes are $DA$ and $DC$ and passes through $F$ is tangent to $AC$.

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  • $\begingroup$ You mean asymptote, not axes, right? $\endgroup$ – Hrhm Aug 17 '16 at 17:15
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This is only an expansion of ajotatxe's answer.

First, let's only consider right triangles, with the right angle at the origin, and the two other vertices at $(a,0)$ and $(0,b)$ with $a>0$, $b>0$. Consider the hyperbola $\displaystyle xy=\frac{ab}{4}$. Note that the point $\displaystyle \left(\frac{a}{2},\frac{b}{2}\right)$ is on this hyperbola. Also note that the tangent to the hyperbola at $\displaystyle \left(\frac{a}{2},\frac{b}{2}\right)$ is $m^2y+x=2m$, which is the hypotenuse of the right triangle. Because $xy=c$ is a convex function for positive $x$ and $y$, this means that the hypotenuse (besides the point $\displaystyle \left(\frac{a}{2},\frac{b}{2}\right)$) is in the region $\displaystyle xy<\frac{ab}{4}$. So any other rectangle inscribed inside the rectangle (with one vertex on the origin, one vertex on the hypotenuse, and sides parallel to the axes) can never have area greater than or equal to $\displaystyle\frac{ab}{4}$.

I am confident that you can generalize this to non-right triangles.

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Another way is to minimize the area of the little green triangles. Note the upper triangle CFG is similar to the given triangle ABC. Now slide the little triangle on the right, GKB, along the base so that GK coincides with FH. This creates a new triangle, A(FG)B, which is also similar to the given triangle. The area of these similar triangles is proportional to the square of the base length, and the base of CFG plus the base of A(FG)B equals AB, the base of the given triangle ABC.

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