A Geometry Exercise

This post was inspired by a problem I came across when studying for a math contest, namely problem 23 in the 2007 Cayley Contest.

I think this is an interesting problem because the solution is not so obvious, and there are several incorrect approaches that I’ve tried (and given up on). I’ve modified the problem slightly to make it more interesting.

Here we have a rectangle, $ABCD$ . It’s rotated to the right by $\theta$ degrees, using $A$ as the pivot. This forms $AEFG$ as the new rectangle. We are given $x$ and $y$ , we are asked to find the area of the shaded region.

Can you come up with a general formula for the area of the shaded region, using the given variables $x$ , $y$ , and $\theta$ ?

Solution

I will use the notation $(ABC)$ to denote the area of triangle $\triangle ABC$ .

The solution is to draw a line parallel to $BC$ through $E$ , the blue line. This splits the lower shaded area into two triangles: $\triangle AXE$ and $\triangle EYH$ , and a rectangle: $BCYX$ .

Since $ABCD$ is the same as $AEFG$ , the top shaded area is the same as the bottom shaded area.

Let’s find the area of triangle $\triangle AXE$ .

The angle $\angle BAE$ is equal to $\theta$ . Of course, $AE$ is equal to $y$ . Using some basic trigonometry, $AX = y \cos \theta$ , $XE = y \sin \theta$ , and $BX = y - y \cos \theta$ . Now we know the area of triangle $\triangle AXE$ :

$\begin{array}{rcl} (AXE) &=& \frac{1}{2} (y \cos \theta) (y \sin \theta) \\ &=& \frac{1}{4} y^2 \sin(2 \theta)\end{array}$

Next we find the area of $\triangle EYH$ . $EY = x - y \sin \theta$ , and using trigonometry again, $HY = \tan \theta (x - y \sin \theta)$ .

$\begin{array}{rcl} (EYH) &=& \frac{1}{2} (x - y \sin \theta) [\tan \theta (x - y \sin \theta)] \\ &=& \frac{1}{2} (x - y\sin \theta)^2 \tan \theta \end{array}$

Finally, for the rectangle $BCYX$ :

$\begin{array}{rcl} (BCYX) &=& x (y - y \cos \theta) \end{array}$

Now we add all this up and simplify ( $T$ is total):

$\begin{array}{rcl} T &=& 2[(AXE) + (EYH) + (BCYX)] \\ &=& 2[\frac{1}{4} y^2 \sin(2 \theta) + \frac{1}{2}(x - y \sin \theta)^2 \tan \theta + x (y - y \cos \theta)] \\ &=& \tan \theta (x^2+y^2) - 2xy (\sec \theta - 1)\end{array}$

It’s rather tedious to get from the second to the last step, but Wolfram Alpha could do it for me.

Notice that we assume that $EF$ intersects $DC$ . If $\theta$ is large enough, then $EF$ would intersect $AD$ , and this formula would no longer work.

Of course it would also fail if $y$ is greater than $x$ , so my formula works for only a rather limited range of values.

This entry was posted on Monday, February 22nd, 2010 at 5:12 am and is filed under Mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to A Geometry Exercise

nschoe says:

February 24, 2010 at 12:01 pm

Hi,
Nice exercice, I remember I had something a bit similar, but it was a rhombus which was rotated. I’ll blog it if I find it again.
Well, sorry to ask such idiot thing, but I have recently written a Maths post on my own blog, but I really got pissed of with formatting, putting operations in the center of the page so that they appear “pretty”, but your way is definitely better, can I ask you how you produced your “format” (the operations your wrote are dramatically more “eyes-candy” than mine), is it LaTeX-generated ?
Thanks for your answer.

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luckytoilet says:

February 24, 2010 at 3:14 pm

Yea, it’s from Latex. WordPress supports latex rendering out of the box, and it’s really essential for any math blog posts. You should give it a try if you’re into math blogging.

Reply
nschoe says:

February 26, 2010 at 6:19 am

Hi, thanks. LaTex .. I’ll try it. It seems … convenient for Maths stuff.

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