# Folding a rectangular paper and finding the area of the triangle so formed.

Given a rectangular sheet of paper ABCD such that the lengths of AB and AD are respectively 7 and 3 cms.Suppose B' and D' are two points on AB and AD respectively such that if the paper is folded along B'D' then A falls on A' on the side DC. Determine the maximum possible area of the triangle AB'D'.

I discovered quite a few basic facts that everybody can, but cannot actually make any progress. Please help.

• I sugest " A falls on A' on the side DC. " should be " A falls on A' on the side BC. " Jan 7, 2015 at 13:22

My suggestion is to compute AD' and AB' as a function of AD and $\alpha$. $\Delta AOD' \sim \Delta ADA' \to AD' \times AD = AO\times AA'$

and note that $AA' = \frac{AD}{\cos \alpha}$ and $AO = \frac{1}{2}AA'$

$\therefore AD' = \frac{AA'^2}{2AD}=\frac{AD}{2\cos^2 \alpha}$

$AB' = \frac{AD'}{\tan \alpha} = \frac{AD}{2\cos^2 \alpha} \times \frac{\cos \alpha}{\sin \alpha}=\frac{AD}{\sin 2\alpha}$

$S_{AB'D'} = \frac{1}{2}AB' \times AD' = \frac{AD^2}{2\cos \alpha \sin 2\alpha}$

Since AD is constant, maximizing $S_{AB'D'}$ is to maximize the expression of $\alpha$. You can find the bound of $\alpha$ by moving (1) D' to D and (2) B' to B combining with the assumption 3:7 given. Then, the min or max could be easily obtained.  Rectangle $$ABCD$$ is drawn above with vertex $$A$$ at the origin. Let $$DA' = x$$ then slope of $$AA' = \dfrac{3}{x}$$ and therefore, slope of $$B'D' = - \dfrac{x}{3}$$ and we have that $$B'D'$$ passes through $$( \dfrac{x}{2}, \dfrac{3}{2} )$$, therefore, its equation is

$$y' = \dfrac{3}{2} - \dfrac{x}{3} (x' - \dfrac{x}{2} )$$

$$x'$$ intercept is $$\dfrac{x}{2} + \dfrac{9}{2x}$$ and $$y'$$ intercept is $$\dfrac{3}{2} + \dfrac{x^2}{6}$$

We must have the $$x'$$ intercept in $$[0, 7]$$, i.e.

$$0 \le x + \dfrac{9}{x} \le 14$$

So that

$$0 \le x^2 + 9 \le 14 x$$

And finaly $$x^2 - 14 x + 9 \le 0$$, which implies that $$x \gt 0.67544$$

And we also have the condition of the $$y'$$ intercept, namely,

$$AD' = \dfrac{3}{2} + \dfrac{x^2}{6} \le 3$$

From which, we must have $$x \le 3$$

Area of $$\triangle AB'D' = [AB'D'] = \dfrac{1}{2} \left(\dfrac{x}{2} + \dfrac{9}{2x} \right) \left(\dfrac{3}{2} + \dfrac{x^2}{6} \right)$$

And this simplifies to

$$[AB'D'] = S(x) = \dfrac{1}{2} \left( \dfrac{3}{2} x + \dfrac{1}{12} x^3 + \dfrac{27}{4x} \right) = \dfrac{1}{8} \left( 6 x + \dfrac{x^3}{3} + \dfrac{27}{x} \right)$$

Differentiating this expression with respect to $$x$$ and equating to zero, gives us,

$$6 + x^2 - 27/x^2 = 0$$

Multiplying through by $$x^2$$,

$$x^4 + 6 x^2 - 27 = 0$$

$$\Rightarrow (x^2 + 9)(x^2 - 3) = 0$$

And therefore, $$x = \sqrt{3}$$ gives the critical value for the area. To identify it as a local maximum or local minimum, we need to check the sign of second derivative.

$$S''(x) = \dfrac{1}{8} ( 2 x + \dfrac{54}{x^3} ) \gt 0$$

So, at $$x = \sqrt{3}$$ we have a local minimum.

In addition, we have to check the value of $$S(x)$$ at the end points $$x = 0.67544$$ and $$x = 3$$. We have

$$S(0.67544) = 5.5161$$

$$S (\sqrt{3}) = 3.4641$$

$$S(3) = 4.5$$

Therefore, $$x = \sqrt{3}$$ corresponds to the minimum area of $$\triangle AB'D'$$ where it is defined.