3440.13 – A Girl and Her Dog

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)(1). When Mitzi is tied outside the garden, she roams over the area ABDFJGKABDFJGK, which we abbreviate II. When tied inside the garden, the relevant area is ABECHGKABECHGK, 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=68π≈213.6I = \dfrac{2}{3} \pi 10^2 + \dfrac{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,

10sin⁡(120°)=8sin⁡(x)⇝x=arcsin⁡(8sin⁡(120°)10)≈43.9° \dfrac{10}{\sin(120°)} = \dfrac{8}{\sin(x)} \leadsto x = \arcsin \left(\dfrac{8 \sin(120°)}{10} \right) \approx 43.9°

We next find that,

y≈180−120−43.9=16.1,z≈120−2(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 ≈128⋅10⋅sin⁡16.1≈11.1area sector ACK≈87.8360π102=76.6.\begin{aligned}\text{area triangle AEC } &\approx \dfrac{1}{2} 8 \cdot 10 \cdot \sin 16.1 \approx 11.1 \\[1em]\text{area sector ACK} &\approx \dfrac{87.8}{360} \pi 10^2 = 76.6.\end{aligned}

Therefore the internal roaming area for Mitzi is

II≈76.6+2⋅11.1=98.8II \approx 76.6 + 2 \cdot 11.1 = 98.8