"Farmer with a piece of land"-Question

DanteQ

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Oct 12, 2008
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Hi, I'm new to the forum, but was wondering if you guys could help me out.

The problem:
mapfa7.png


There is this farmer, who has a piece of land, which is bordered of by 2 fences. (the black lines). Somewhere within this piece of land is a drinking place for his cows, (named W). He wants to divide his land in two parts. A huge piece of land to the right of a new third fence (brown line) and a small triangular piece to the left of the third fence (brown line). The triangular piece of land is only for one cow, so it's probably gonna be much to large, that's why he wants to make it as small as possible.

The Question:How does he have to place the brown fench, to make the area of the triangle as small as possible, but still have it cross through W, so both parts can still have acces to the water.

What I already have:
I have already proven, that to get the smallest area, you have to make a triangle, with W in the exact middle of the brown Fence. (so the situation drawn in the example isn't far of of the best angle the fence can make).

What I still need, and can't figure out:
An Geometric way to exactly draw this fence, knowing that W has to be in the exact middle of the brown line.

I hope the problem is understandable to you. (english isn't my native language, and describing situations doesn't make it easier... lol).
 
My interpretation is:

The location of W is fixed - that is the well has already been dug. (We do not have choice of re-digging the hole).

W is the middle point (bi-secting) of the potential dividing lines.

Are both of those statements correct?
 
The location of W is indeed the well, and has already been digged, but we just don't know where. I need a general way to derive a line L (brown) for any location of W.

map2gp8.png


Including my proof of L1=L2 leads to the smallest area A, I need:

A way to create a line L which is cut into two equally sized lines L1 and L2 by W, and which reaches until the black lines.

Hope this makes it clearer for you
 
Can you forget about the farmer/cow story and simply state the "givens";
is the angle where the black lines meet given?
are the coordinates of point W given?

Could it be worded simply (as example; O = origin):
straight line OA is infinitely long and is situated in the top right quadrant;
it forms an angle of D degrees with the x-axis;
point W is situated at (X,Y);
point P is on OA, point Q is on the x-axis such that PWQ is a straight line;
what is the shortest length of PQ such that PW = QW?

D, X and Y would be givens.
 
Denis said:
Can you forget about the farmer/cow story and simply state the "givens";
is the angle where the black lines meet given?
are the coordinates of point W given?

Could it be worded simply (as example; O = origin):
straight line OA is infinitely long and is situated in the top right quadrant;
it forms an angle of D degrees with the x-axis;
point W is situated at (X,Y);
point P is on OA, point Q is on the x-axis such that PWQ is a straight line;
what is the shortest length of PQ such that PW = QW?

D, X and Y would be givens.

map3uf9.png

You seem to understand most of the problem, All you say is correct, and you can even say that there is only one length of PQ such that PW = QW.
There are none given values, but you can use d, X and Y as variables.
My teacher has informed us it is also possible to create this line by using Geometric drawing methodes, like using bi-section, altitudes or medians etc.
(maybe even circels??? 2 points equaly far away from W sound a lot like a middelpoint W with two points on the circel...)
 
Construct two lines through point W, one parallel to each of the two lines you are starting with.

The two new lines will intersect the original lines at two points. Call them B and C.

Draw a third new line through these points, B and C, of intersection.

Finally construct another line through W, but parallel to line BC. That should do it. Do you know why?
 
An easy way:
draw line WV parallel to OQ, V on OP
draw line WR perpendicular to OQ, R on OQ
join VR: you get triangle ROV with height = Y, RO = X, angle VOR = d
solve for OV; then OP = twice OV, and you're done.

In other words, OP = 2Y / SIN(d)
 
wjm11 said:
Construct two lines through point W, one parallel to each of the two lines you are starting with.

The two new lines will intersect the original lines at two points. Call them B and C.

Draw a third new line through these points, B and C, of intersection.

Finally construct another line through W, but parallel to line BC. That should do it. Do you know why?

So you mean this:

map4lq9.png


It seems to work!

The way I see it:
Because W is the middle of PQ, if you draw two lines parralel to the given black lines, These lines (BW and CW) will cross the blacklines OP and OQ precisely in the middle. This creates a length CB of half the size of PQ at half the distance with W, so if you draw a parralel line (parralel with BC I mean) through W, it will be precisely twice as big and it will be located at 2*(1/2)OP = OP and 2*(1/2)OQ = OQ, so it wil be the asked line PQ .

Or you could say you divided the complete triangle in 4 equal triangles, so OBC is equaly shaped as OQP.

Does that sound right?

An easy way:
draw line WV parallel to OQ, V on OP
draw line WR perpendicular to OQ, R on OQ
join VR: you get triangle ROV with height = Y, RO = X, angle VOR = d
solve for OV; then OP = twice OV, and you're done.

In other words, OP = 2Y / SIN(d)

This also seems to be true, with the same technique as above, only now its calculated instead of drawn.
The problem was preffered to be drawn, but still , Thanx alot! It helps me out greatly!
 
DanteQ said:
[
An easy way:
draw line WV parallel to OQ, V on OP
draw line WR perpendicular to OQ, R on OQ
join VR: you get triangle ROV with height = Y, RO = X, angle VOR = d
solve for OV; then OP = twice OV, and you're done.

In other words, OP = 2Y / SIN(d)
This also seems to be true, with the same technique as above, only now its calculated instead of drawn.
The problem was preffered to be drawn, but still , Thanx alot! It helps me out greatly!

C'mon Dante: since you have length of OP, then you've found P's location,
so simply mark off P on OA, then draw line from P to x-axis going through W,
meeting the x-axis at point Q.

HOWEVER, if you prefer getting the coordinates of P and Q, just as easy:
let k = Y TAN(90 - d)
P coordinates = (2k, 2Y)
Q coordinates = (2X - 2k, 0)
 
Denis said:
DanteQ said:
[
An easy way:
draw line WV parallel to OQ, V on OP
draw line WR perpendicular to OQ, R on OQ
join VR: you get triangle ROV with height = Y, RO = X, angle VOR = d
solve for OV; then OP = twice OV, and you're done.

In other words, OP = 2Y / SIN(d)
This also seems to be true, with the same technique as above, only now its calculated instead of drawn.
The problem was preffered to be drawn, but still , Thanx alot! It helps me out greatly!

C'mon Dante: since you have length of OP, then you've found P's location,
so simply mark off P on OA, then draw line from P to x-axis going through W,
meeting the x-axis at point Q.

HOWEVER, if you prefer getting the coordinates of P and Q, just as easy:
let k = Y TAN(90 - d)
P coordinates = (2k, 2Y)
Q coordinates = (2X - 2k, 0)

I already understood what you meant, and what the implications of those statements were... I just replied to thank you :D
 
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