Grazing Cow

Geometry Level 4

A cow is tied to a corner (vertex) of a regular hexagonal fenced area of sides $4$ meters long, with a rope of length $10$ meters long, in a grass field. The cow cannot graze inside the fenced area. What is the maximum possible area of grass field in which the cow can graze?

Let $A$ be the area in square meters. Find $\left\lfloor A \right\rfloor$ . where $\left\lfloor ... \right\rfloor$ represents Greatest Integer Function.

This is an old KVPY problem. Try more such problems here .

The answer is 251.

2 solutions

Mohamed Ahmed Abd El-Fattah
Jun 25, 2015

Consider that :

The yellow line represents the connecting rope (10 meters) .
The dark grey regular hexagon represents the hexagonal fenced area (of side length 4 meters) .
The green region represents the reachable area of grass field by the cow .

We should recall that the area of any circular sector $\large = \ \pi (\text{its radius})^2 \times \frac{(\text{its inner angle})^\circ}{360^\circ}$

Now, we can sort and divide the green region into 3 types of circular sectors :

The largest sector of radius $10$ and inner angle $= 360 - 120 = 240^\circ$ , $\therefore \text{its area} = \pi (10)^2 \times \frac{240^\circ}{360^\circ} = \frac{200\pi}{3}$ , and there is only one sector of this type .
The middle sector of radius $6$ and inner angle $= 180 - 120 = 60^\circ$ , $\therefore \text{its area} = \pi (6)^2 \times \frac{60^\circ}{360^\circ} = \frac{18\pi}{3}$ , and there are two sectors of this type .
The smallest sector of radius $2$ and inner angle $= 180 - 120 = 60^\circ$ , $\therefore \text{its area} = \pi (2)^2 \times \frac{60^\circ}{360^\circ} = \frac{2\pi}{3}$ , and there are two sectors of this type too .

Therefore the total area of the green region (A) $= \frac{200\pi}{3} + ( \frac{18\pi}{3} \times 2) + ( \frac{2\pi}{3} \times 2) = \frac{240\pi}{3} = 80 \pi \approx 251.3274\dots$ $\large \therefore \ \ \ \left \lfloor \text{A} \right \rfloor = \boxed{251}$

But the distance b/w any two opposite vertex of hexan will be (4+2+2=8), and cow is tied with a rope of 10 then this hexagon should be entirely inside of circle. Where am i committing mistake? Then total area=100π-24√3.

himanshu mishra - 5 years, 11 months ago

Your mistake, sir, is considering that the whole area of the circle of radius $10$ (except the area of the hexagon) is an available reachable area by the cow .

You should notice that in the case of maximum area : the cow will tend to move along on the circumference of the circle of radius $10$ , but you should also notice that while this , the rope approaches the fenced hexagon and tends to be tangent to one of the hexagon's sides , but it's only $4 \ \text{Meters}$ of the rope which are tangent to the side , and the rest $6 \ \text{Meters}$ of the rope are free and attached to the cow at its end , then it's bent over the vertex of the hexagon .

Then we start as from first again , but this time we exclude the previously calculated area , and consider that the cow moves with the rope in a circle of radius $6$ .

Then this case is repeated once more , but in a circle of radius $(6 - 4 = 2)$ .

I really recommend you to look carefully at the attached image in the post .

Mohamed Ahmed Abd El-Fattah - 5 years, 11 months ago

I too did the same way. Your detailed explanation with beautiful diagram is nice. (+

Niranjan Khanderia - 2 years, 11 months ago

Prashant Kumar Singh
Nov 9, 2014

Consider a hexagon whose upper vertex tied through 10m rope

we get -

(1) a free flow area i.e. area of circular sector of r=10m & θ=240 degree ——> Area= (πr^2)*2/3 = 200π/3 meter square

(2) Now, hexagon shaped fence of side = 4m bound the rope & will allow further free flow area i.e. area of circular sector of r=(10-4)m=6m & θ=60 degree ON BOTH SIDES ——> Area= (πr^2)/6 = 6π meter square x 2

(3) Again, next fence of side = 4m bound the rope & will allow further free flow area i.e. area of circular sector of r=(6-4)m=2m & θ=60 degree ON BOTH SIDES ——> Area= (πr^2)/6 = 2π/3 meter square x 2

So, Required Area = (200π/3) + (6π) 2 + (2π/3) 2 = 80π = 251.43 meter square

So, Ans = 251