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The problem is as follows:

A buggy is set to cross over a snowy terrain. The driver seat has a rectangular window featured in the diagram from below. When the driver activates the mechanism for cleaning the window from the snow, the wiper spins $120^{\circ}$ using the point $M$ as a pivot. Assuming the rotation is clockwise. Find the area covered by the windshield wiper in the buggy. Assume the wiper has a T shape and the rectangular window is $ABCD$ where $BC=2AB=2AE=2AM=1\,m$. Also assume AE is perpendicular to $MN$ and $N$ is midpoint between $AE$.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\left(\frac{\pi}{6}-\frac{\sqrt{3}}{16}\right)\,m^{2}\\ 2.&\left(\frac{\pi}{16}-\frac{\sqrt{3}}{6}\right)\,m^{2}\\ 3.&\left(\frac{\pi}{4}-\frac{\sqrt{3}}{8}\right)\,m^{2}\\ 4.&\left(\frac{\pi}{16}-\frac{\sqrt{3}}{16}\right)\,m^{2}\\ \end{array}$

So far the only thing which I could come up with was the diagram from below:

However I don't know how to find such area. I remember how to calculate the area of a half circle sector by means of the equation:

$A=\frac{\alpha}{360}\pi r^2$

Assuming $\alpha$ is an angle given in sexagesimal units. But in this case the figure doesn't help much. How exactly should I find that weird surface.

Can someone help me with a method relying euclidean geometry or something along precalculus?. I think integrals can be used but I am not looking for such sort of answer.

Please include a drawing in the answer because for me it is not easy to spot how to find such problematic area. Does it require some sort of construction?.

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2 Answers

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The area we want to find out is the one surrounded by red in the figure above. This is the worst windshield wiper ever.

To verify this, we know that since $\overline{AE}\perp\overline{MN}$, the inner circle — with center $M$ and radius $\overline{MN}$ — will never be reached. Same with the two blue areas at the left and right bottom corner.

I assume that it was the inaccurate figure you have that misled you. Anyway, here's my solution. The length and area unit below are $m$ and $m^2$, respectively.

Let's find out the length of $\overline{MN}$ first by noting that $\triangle ANM$ is a right triangle with $\angle AMN=30^{\circ}$. Therefore $$\overline{AM}=0.5\Longrightarrow \overline{MN}=\frac{\sqrt3}4$$

Now let's compute the red area. Let it be $S$.

$$\begin{align}S&=\frac{1}2\cdot (0.5)^2\pi-\frac{120}{360}\cdot \left(\frac{\sqrt3}4\right)^2\pi-2\triangle ANM\\ &=\frac{\pi}8-\frac{\pi}{16}-2\cdot \frac{1}2\cdot \frac{1}4\cdot\frac{\sqrt3}4\\&=\color{red}{\frac{\pi}{16}-\frac{\sqrt3}{16}} \end{align}$$

which is the final answer.

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Based on Student1058's answer I'm posting a revisited version of my attempt solved properly with some annotations which may help somebody in a similar situation.

Typically if you are inside a car and look at the area swept by a windshield wiper you'll notice it generates an area of a semicircle. This happens because the swept is generated by the rotation of the arm which is covered by a cleaning foam or a rubber blade which doesn't cover all the arm of the wiper. This is better understood in the graphic from below:

However in this sort of problem it seems a bit challenging and at the same part confusing if you rush to believe that the segment perpendicular forming the $T$ (meaning $MN \perp AE$) also acts as a normal wiper. But this is not meant as the correct interpretation of the problem as it indicates, only the area which is swept by the arm $AE$. (See images below)

This jumps into the question, how do you find such area?. The best method is to build a tangential circle enclosing the inner part of the wiper. In geometrical terms, this is translated as drawing a circle tangential to point $N$.

The next thing we want to do is to make another circle but this time, one which touches the points $A$ and $E$ (yes similar to the tv channel).

In this case. This happens because the angle on $\angle NAM = 30^{\circ}$.

Because:

Since $AM=\frac{1}{2}\,m$ and $N$ is midpoint of $AM$ then $AN=NE=\frac{1}{4}\,m$

Since $\triangle ANM$ is a right triangle whose sides are:

$NA=\frac{1}{4}$

$\textrm{hypotenuse}=\frac{1}{2}$

Then this proportion lead us to indicate it is a $30-60-90$ Pythagorean triangle, more properly is a special right triangle.

Where the sides are in a proportion of:

$\textrm{hypotenuse}= 2k$

$\textrm{side opposing 30 degrees}= k$

$\textrm{side opposing 60 degrees}= k\sqrt 3$

To obtain $NM=\frac{\sqrt{3}}{4}$ and since $\triangle ANM \cong$ to $\triangle ENM$ this led us to indicate $\angle\,ENM= 30^{\circ}$ and $NM=\textrm{is bisector and median hence a mediatrix}$.

Therefore the $\angle AEM = \angle EMA = 60^{\circ}$ thus making our $\triangle\,AEM$ isosceles into an equilateral triangle. Thus perfectly fitting one-third of the semicircle. Referring to the semicircle between points $AD$.

Had this angle not occured then it would follow the same strategy, you can always make two circles, both tangential to the inner part from where is about to be rotated and the other one touching both ends of the segment, the only difference will lie on the slope of the segment.

Since it is not easy to spot how the arm generates the area. I made succesive stops along the way as the wiper moves touching both cirles. The image from below showcases this motion.

As now it is proved that we must only focus in the areas covered first by the arc between $AN$ and the ring between $NN$ and finally from the arc between $NE$.

From a visual inspection we can notice that since arc $AN$ equals to the arc $NE$ their swept areas will be the same.

Thus we proceed to calculate such areas:

The easiest part is the ring:

This comes from:

$A=\frac{r^2\theta}{2}$

where $\theta=\textrm{given in radian units}$

therefore: $As \frac{120^{\circ}}{180}\pi=\frac{2\pi}{3}$

$A_{NN}=\frac{(r_1^2-r_2^2)\frac{2\pi}{3}}{2}=\frac{(\frac{1}{2}^2-\frac{\sqrt{3}}{4}^2)\frac{2\pi}{3}}{2}$

$A_{NN}=\frac{\pi}{48}$

Now we will focus in the other two arcs:

This can be obtained by subtracting our right triangle from the circle sector on $NAM$

Hence:

$A_{AN}=\left[\frac{r_{AM}^2\theta_{\angle\,NMA}}{2}\right]-A_{\triangle\,NAM}$

since: $\theta_{\angle\,NMA}=\frac{30\pi}{180}=\frac{\pi}{6}$

This means:

$A_{AN}=\left[\frac{\frac{1}{2}^2}{2}\cdot\frac{\pi}{6}\right]-\frac{\frac{1}{4}\cdot\frac{\sqrt{3}}{4}}{2}$

$A_{AN}=\frac{\pi}{48}-\frac{\sqrt{3}}{32}$

Since $A_{AE}=A_{AN}$

Notice here we're referring as $AN$ and $NE$ as the arcs.

The total area would be:

$A_{\textrm{swept area}}=A_{NN}+2A_{AN}=\frac{\pi}{48}+2\left(\frac{\pi}{48}-\frac{\sqrt{3}}{32}\right)$

Finally becoming into:

$A_{\textrm{swept area}}=\frac{\pi}{16}-\frac{\sqrt{3}}{16}\,m$

Which corresponds to the fourth option.

This explanation is the most complete I could write. Which hope may help in any similar cases.

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