The Cow Problem

G

greatwhiteshark

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A cow is tethered to one corner of a square barn, 10 feet by 10 feet, with a rope 100 feet long. What is the maximum grazing area of the cow?
 
In the diagram the cow is tethered to the northeast corner of a square barn, 10 feet by 10 feet, with a rope 100 feet long. The maximum grazing area of the cow is three-quarters of the area of a circle with radius 100 plus one-quarter of a circle with radius 90.
http://img156.echo.cx/img156/1421/cow6tj.gif
 
I have to disagree. Unless you see something I'm missing it's a little worse than that.
I drew the barn at P(0,0), P(10,0), P(0,10) and P(10,10)
Yes, 3/4 of a 100' circle.
The two 90' circles around the next two corners meet at P(X,Y) on x=y
The 90' circle around P(10,0) has an angle so that
sin(T) = Y/90
Cos(T) = (X-10)/90
Since x = y and sin² + cos² = 1
(X/90)²+((X-10)/90)² = 1
That gives X which in turn gives T radians.
The area of that segment is pi*r²*T/(2pi) = 90²T/2
You have two of them to add in.
Next there is a triangle P(0,10), P(10,0), P(X,Y) to be added.
That is straight forward, but includes triangle P(10,0), P(0,10), P(10,10) which must then be subtracted.
 
Why ever must anything be subtracted?
The question is: “What is the maximum grazing area of the cow?”
The fact that the beast can overgraze an area twice does not change the maximum possible greasing area.
We are not finding the area with integrals, nor or we doing probability.
Therefore, there is no need to worry about over counting.

Frankly, I do not follow you argument where you say “You have two of them to add in.” In the diagram it is perfectly clear the there are two cases for second circles, each of radius 90. One has center at (0,10) and the other center is at (10,0). I have redrawn the diagram.
http://img193.echo.cx/img193/645/cow29rt.gif
It is clear that the two circles intersect in y=x.
But I see no additional triangular regions to consider.
Again it is just a matter of over grazing.
The grazing area outside the ‘red’ three-quarters is one-quarter of a circle with radius 90.
 
Agreed on the 3/4 100' circle.

Consider only the 90' circles from the first corner each way. Each direction, the 90' rope sweeps some angle to reach the point of intersection of the two 90' circles on x = y. That the center of the 90' circles is NOT on x = y, is significant. It forces the cow to sweep an angle just greater than 45º to get to the point of intersection. Let's call the angle 'q'.

tan(q) = [10+5*(sqrt(161)-1)]/(sqrt(161)-1) = 1.171107
q = 49.506220º

Therefore, it looks like we have 2*q/360º times the area of the 90' circle, just greater than 25% (27.50345%)

Just one more thing. Are we SURE the point of intersection is more than 80' from the barn corner at the origin?

sqrt(2*(sqrt(161)-1)) = 82.650724 Yes, we are sure.
 
Tkhunny: did you look at my second diagram?
Do you agree that the question is about the maximum POSSIBLE grazing area?
Not the area of overgrazing.
 
I think the question is about counting all area that can be reached by the cow, the total grazing area. Count each blade of grass only once.

I see it this way - How I wish I could post a drawing.

Area 1) 3/4 of the 100' circle.
Area 2) 13.751725% of one 90' circle. (See above)
Area 3) 13.751725% of the other 90' circle (See above)

Since this is already grater than the solution using 2*12.5% of the 90' circle, I thought I was done. Looking more closely, I see there is another chunk.

Area 4) Isosceles Triangle whose base is a diagonal across the barn (connecting the corners where the 90' circles are centered). The apex is the distant point of intersection of the to 90' circles. The area is 50*(sqrt(161)-2) less half the barn, about 484.428877 ft<sup>2</sup>.

All told, I get 31045.149 ft<sup>2</sup>, which is almost 99% of what there would have been, had there been no barn.

One more way, just for fun, I broke out the integral calculus and calculated the area between the 100' and the two 90' circles. It's not quite symmetrical parallel to the y-axis. It would be in polar coordinates. This gives the 100' circle less the barn less the calculated area -- 31145.15183 ft<sup>2</sup>. I'm quite disappointed that this is even more than I calculated above, but I'm tired of looking at it.
 
“How I wish I could post a drawing.”
WELL USE http://www.imageshack.us/ , it is free!

But what about my second drawing is wrong?
The question is about the total grazing area possible.
It is a classical contest problem: easy to solve.
Why over complicate it?
 
I give...what am I doing wrong?

Ta-da! That wasn't so hard.
 
TKH has the angle to the intersection right. I wanted shark to finish it so I left everything in equations.
I think I agree with everything TKH says up the area 4. The center of the base is P(5,5) and I make X=Y=68.44 so we have a height of
(68.44-5)*sqrt(2) and a base of 10sqrt(2)
area = (1/2)*10sqrt(2)*63.44*sqrt(2) = 634.4
Minus 1/2 the barn = 50 gives
584.4 for the triangle.
 
There it is. Height of isosceles triangle is 50*sqrt(161), giving

Area 1) + Area 2) + Area 3) + Area 4) =

(3/4)*pi*100<sup>2</sup> + .2750345*pi*90<sup>2</sup> + [50*sqrt(161) - 50] =

31145.149333 ft<sup>2</sup>

From the integral, 31145.151483 ft<sup>2</sup>

Now we're talking.
 
Pheeew! TKH and G agree except for the typo saying "Height of" instead of "Area of".
I don't recognize the 50sqrt(161) formula but we both end up with 634.4... so it must be good. Its nice to be vindicated.

Now we're done talking.
------------------
Gene
 
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