STEP I 1990 question 1 solution

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STEP I 1990 question 1 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1990 question 1 solution

Let $\varphi = \angle CDY$ . Then

$\theta + \alpha + \varphi = \pi$ (1)

Consider the lengths of the altitudes dropped

from Y to CD, from Y to AB and from X to AD.

They are

$r\sin\varphi$ , $r\sin\alpha$ and $r\sin\theta$ respectively.

Thus the areas of the triangles DCY, DYX, DXA are $\frac{1}{2}r^2\sin\varphi$ , $\frac{1}{2}r^2\sin\alpha$ and $\frac{1}{2}r^2\sin\theta$ respectively, and the area of the quadrilateral is

$\displaystyle K = \frac{r^2}{2}\left(\sin \theta + \sin \alpha + \sin \varphi\right)$ (2)

When $\alpha$ is fixed, and we wish to maximise this area:

$\displaystyle K = \frac{r^2}{2}\left(\sin \theta + \sin \alpha + \sin (\pi - \alpha - \theta)\right)$

$\displayste \frac{dK}{d\theta} = \frac{r^2}{2}\left(\cos \theta - \cos (\pi - \alpha - \theta)\right)$

Setting the derivative to zero, we get

$\cos \theta - \cos (\pi - \alpha - \theta) = 0$

Now, both $\theta$ and $\pi - \alpha - \theta$ clearly belong in the interval $(0, \pi)$ (the latter because it is the angle marked $\varphi$ ), and thus

$\theta = \pi - \alpha - \theta$

$\displaystyle \theta = \frac{1}{2}(\pi - \alpha)$

which is the maximum. This can be seen by working out the second derivative,

$\displaystyle \frac{d^2K}{d\theta^2} = \frac{r^2}{2}\left(- \sin \theta - \sin (\pi - \alpha - \theta)\right)$

which is clearly negative at $\theta = \frac{1}{2}(\pi - \alpha)$ .

Note that in this case, the maximum area is given by

$\displaystyle K_{max} = \frac{r^2}{2}\left(2\sin \frac{\pi - \alpha}{2} + \sin \alpha\right)$ (3)

If we wish to find the maximum when both $\alpha$ and $\theta$ vary, we can proceed by finding the value of $\alpha$ for which the $K_{max}$ is at a maximum. Thus,

$\displaystyle \frac{dK_{max}}{d\alpha} = \frac{r^2}{2}\left(-\cos \frac{\pi - \alpha}{2} + \cos \alpha}\right)$

Setting the derivative to zero:

$\displaystyle -\cos \frac{\pi - \alpha}{2} + \cos \alpha = 0$

Now, both $\frac{1}{2}(\pi - \alpha)$ and $\alpha$ are in $(0, \pi)$ , and thus we can write

$\displaystyle - \frac{\pi - \alpha}{2} + \alpha = 0$

from which follows

$\displaystyle \alpha = \frac{\pi}{3}$ (4)

Again, to make that that this is a maximum, let us take the second derivative:

$\displaystyle \frac{d^2 K_{max}}{d\alpha^2} = \frac{r^2}{2}\left( -\frac{1}{2} \sin \frac{\pi - \alpha}{2} - \sin \alpha\right)$

which again can be seen to be negative, thus ensuring that (4) indeed gives us the maximum K_max .

Now, insertion into (3) gives us the maximum area as

$\displaystyle K = 3\frac{r^2}{2}\sin \frac{\pi}{3}$ , i.e.

$\displaystyle K = \frac{3\sqrt{3}r^2}{4}$

Just for clarification, note that this happens when

$\displaystyle \theta = \alpha = \frac{\pi}{3}$ .

Alternate Solution

Now you could stop reading here, but for the extra educational benefit or something, I couldn't resist pointing out that this question follows very easily from Jensen's inequality, introduced to us avid STEP-solvers in 2007 II/7:

Let f(x) be a concave function on some interval and let $x_1, x_2, \ldots, x_n$ lie in that interval. Jensen's inequality states that

$\displaystyle \frac{1}{n}\sum_{k=1}^n f(x_k) \leq f\left(\frac{1}{n}\sum_{k=1}^n x_k\right)$

and that equality holds if and only if $x_1 = x_ 2 = \cdots = x_n$

Now, in 2007 II/7, it was proven that $\sin x$ was a concave function on $(0, \pi)$ .

Setting, in Jensen's inequality, n = 3 , $x_1 = \theta$ , $x_2 = \alpha$ , $x_3 = \varphi$ , and using (2) and (1) we get

$\displaystyle K = \frac{3r^2}{2} \cdot \frac{1}{3}\left(\sin \alpha + \sin \theta + \sin \varphi\right) \leq \frac{3r^2}{2}\sin \frac{\pi}{3}$

with equality iff $\theta = \alpha = \varphi$ .

This immediately gives us the maximum.

Solution by ukgea.

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