3440.13 – A Girl and Her Dog


Tags:Problem Set 16GeometryGeometry-Euclidean

Masha's dog Mitzi is tied to a corner of a walled garden with the shape of a regular hexagon 8 meters on a side. Mitzi's rope is 10 meters long and Mitzi is tied up outside the garden---usually. What is the area within which Mitzi can roam?

Answer the same question if Mitzi is tied up inside the garden.


Solution

Take a look at the figure below, particularly part (1). When Mitzi is tied outside the garden, she roams over the area ABDFJGK, which we abbreviate II. When tied inside the garden, the relevant area is ABECHGK, which we call IIII.

Area II consists of 2/3 of a circle with radius 10 and 2/6 of a circle of radius 2. Therefore, area II is

I=23π102+26π22=68π213.6.I = \frac{2}{3} \pi 10^2 + \frac{2}{6} \pi 2^2 = 68 \pi \approx 213.6.

This is the easy one.

Area IIII consists of two triangles and a sector of a circle of radius 10. We need the angles x,yx, y and zz shown in part (2) of the figure. Remember: the vertex angle of a regular hexagon is 120°120°.

Now,

10sin120=8sinxx=arcsin(8sin12010)43.9°\frac{10}{\sin 120} = \frac{8}{\sin x} \leadsto x = \arcsin \left(\frac{8 \sin 120}{10} \right) \approx 43.9°

We next find that,

y18012043.9=16.1,z1202(16.1)=87.8. \begin{aligned} y &\approx& 180 - 120 - 43.9 = 16.1, \\ z &\approx& 120 - 2 (16.1) = 87.8. \end{aligned}

We can now find the areas relevant to IIII:

area triangle AEC =12810sin16.111.1area sector ACK87.8360π102=76.6. \begin{aligned} \text{area triangle AEC } &\approx& = \frac{1}{2} 8 \cdot 10 \cdot \sin 16.1 \approx 11.1 \\ \text{area sector ACK} &\approx& \frac{87.8}{360} \pi 10^2 = 76.6. \end{aligned}

Therefore the internal roaming area for Mitzi is

II76.6+211.1=98.8.II \approx 76.6 + 2 \cdot 11.1 = 98.8.