[Guest]

I have got a square field with one cow. My cow lied 13 meters from one of the corner posts of the field, 17 meters from the corner post diagonally opposite that one, and 20 meters from a third corner post. Can someone tell me how big (perimeter) a field I own?

You may assume the field is flat, and the distance of the cow from the corner poles is measured from the center of mass of the cow -- you know what I mean.

Edited February 24, 2008 by brhan

[Guest]

ill try it...

cow is not sitting in middle of field or at least along the line of the diagional of the first 2 stated corners.

nope its doin my head in..

thats all from me

i first thought 30m diag but the sq thing cuts that out..

[bonanova]

By construction the field is about 19 km on a side.

Perimeter is about 76 km and area about 360 km².

You have a formula for it?

[Guest]

I did some maths like these when i was in high school, but now i forgot everything.

[Guest]

From the figure:

AO²+OC²=BO²+OD² .... (Simple Pythagoras Theorem)...(iv)

=> AO = 10 m

Now, Area of the square = sum of the area of 4 triangles

a²=Ar(ABO)+Ar(BOC)+Ar(COD)+A(DOA)

Let, f(a)=4a² - [{(30²-a²)(a²-4²)}^1/2+{(23²-a²)(a²-3²)}^1/2+{(30²-a²)(a²-10²)}^1/2+{(37²-a²)(a²-3²)}^1/2]

where 10<a<23 (else f(a) would not be real)

Solving for f(a)=0

a=20.42296 m

Hence, Perimeter = 81.69184 m

[Guest]

From the figure:

AO²+OC²=BO²+OD² .... (Simple Pythagoras Theorem)...(iv)

=> AO = 10 m

Now, Area of the square = sum of the area of 4 triangles

a²=Ar(ABO)+Ar(BOC)+Ar(COD)+A(DOA)

Let, f(a)=4a² - [{(30²-a²)(a²-4²)}^1/2+{(23²-a²)(a²-3²)}^1/2+{(30²-a²)(a²-10²)}^1/2+{(37²-a²)(a²-3²)}^1/2]

where 10<a<23 (else f(a) would not be real)

Solving for f(a)=0

a=20.42296 m

Hence, Perimeter = 81.69184 m

The approach looks correct to me. My comment would be the B and C poles must be in diagonally opposite position.

I have got a square field with one cow. My cow lied 13 meters from one of the corner posts of the field, 17 meters from the corner post diagonally opposite that one, and 20 meters from a third corner post. Can someone tell me how big (perimeter) a field I own?

[Guest]

The approach looks correct to me. My comment would be the B and C poles must be in diagonally opposite position.

I have got a square field with one cow. My cow lied 13 meters from one of the corner posts of the field, 17 meters from the corner post diagonally opposite that one, and 20 meters from a third corner post. Can someone tell me how big (perimeter) a field I own?

My mistake .. Didn't read "that one"!!

AO = 7.615773 m

f(a) = 4a²-[{(20.615773²-a²)(a²-5.38423²)}^1/2+(24.615773²-a²)(a²-9.38423²)}^1/2+(33²-a²)(a²-7²)}^1/2+(37²-a²)(a²-3²)}^1/2

9.38423<a<24.615773

a = 19.20937 m

Perimeter = 76.83749 m

But, I think my square was better, at least AO was 10 m, making calculations a bit easy

[bonanova]

My mistake .. Didn't read "that one"!!

AO = 7.615773 m

f(a) = 4a²-[{(20.615773²-a²)(a²-5.38423²)}^1/2+(24.615773²-a²)(a²-9.38423²)}^1/2+(33²-a²)(a²-7²)}^1/2+(37²-a²)(a²-3²)}^1/2

9.38423<a<24.615773

a = 19.20937 m

Perimeter = 76.83749 m

But, I think my square was better, at least AO was 10 m, making calculations a bit easy

Nice job Aatif,

My ruler and compass solution first tried 18.5m side [too small] then 20m [too big].

I interpolated to 19m, which was a tad small but within a thickness or two of my pencil line, so I stopped there.

Why it didn't occur to me to drop perpendiculars from the cow to the fence I don't know.

brhan, good puzzle.

[EventHorizon]

perimeter = 12 * sqrt(41)

[EventHorizon]

perimeter = 12 * sqrt(41)

I first set one side of the field to the length of the other side of the field

x + y = a + b

then put everything in terms of x

x + sqrt(x^2 + 17^2 - 20^2) = sqrt(20^2 - x^2) + sqrt(13^2 - x^2)

after (a whole lot of) simplification this becomes

0 = 3649x^4 - 979600x^2 + 64000000

Solving for x results in...

x = sqrt(6400/41)

One side length then equals

sqrt(13^2 - (6400/41)) + sqrt(20^2 - (6400/41))

= 23/sqrt(41) + 100/sqrt(41) = 123/sqrt(41) = 3*sqrt(41)

Multiplying by four gives 12 * sqrt(41).

[Guest]

I first set one side of the field to the length of the other side of the field

x + y = a + b

then put everything in terms of x

x + sqrt(x^2 + 17^2 - 20^2) = sqrt(20^2 - x^2) + sqrt(13^2 - x^2)

after (a whole lot of) simplification this becomes

0 = 3649x^4 - 979600x^2 + 64000000

Solving for x results in...

x = sqrt(6400/41)

One side length then equals

sqrt(13^2 - (6400/41)) + sqrt(20^2 - (6400/41))

= 23/sqrt(41) + 100/sqrt(41) = 123/sqrt(41) = 3*sqrt(41)

Multiplying by four gives 12 * sqrt(41).

It would appear that you used the Pythagorean theorem to solve for the lengths of the sides. That only works for right triangles.

[EventHorizon]

It would appear that you used the Pythagorean theorem to solve for the lengths of the sides. That only works for right triangles.

They are right triangles. I guess I should have posted a picture too....I thought aatif's would be sufficient.

[Guest]

They are right triangles. I guess I should have posted a picture too....I thought aatif's would be sufficient.

The triangles with side lengths (17,20,x) and (13,20,x) are not right triangles, as your solution implies. Also, you used very different naming than aatif, and aatif's picture is not accurate, since he swapped the legs with lengths 17 and 20.

[Guest]

The triangles with side lengths (17,20,x) and (13,20,x) are not right triangles, as your solution implies. Also, you used very different naming than aatif, and aatif's picture is not accurate, since he swapped the legs with lengths 17 and 20.

Nope, EventHorizon's solution doesn't imply that (17,20,x) and (13,20,x) are right angle triangles. It just uses the fact that:

AO²+CO²=BO²+DO² ... (i) (Naming the square from top left corner clockwise as ABCD, with the cow tied at O)

(i)

and, 13²-x²=AO²-y² ...(ii)

You can use (i) and (ii) to write y =f(x) = sqrt(x² + 17² - 20²)

And thus his results, which are correct.

Edited February 26, 2008 by Aatif

[bonanova]

Hold on... There's two fields. And the cow is loose!']The cow is at [x,y] and is the center of 13, 17 and 20 m radius circles.

The corners of the field are [0,0], [0,a], [a,a], [a,0]; three corners lie one on each circle.

The origin is on the 13 m circle.

[1] x² + y² = 13²

[2] (a-x)² + y² = 20²

[3] (a-x)² + (a-y)² = 17²

[2]-[1]: (a-x)² - x² = 20² - 13² = 231 --> x = (a² - 231) / 2a

[2]-[3]: y² - (a-y)² = 20² - 17² = 111 --> y = (a² + 111) / 2a

let A = a² [area]

x² = (A²-462A+53361) / 4A

y² = (A²+222A+12231) / 4A

x² + y² = (2A²-240A+65682) / 4A = 13² = 169

A² - 458A + 32841 = 0

A = 89, 369.

Both solutions are physically realizable.

A1 = 89 m² -> a1 = 9.434 m -> perimeter1 = 37.736 m

Cow is at x1 = -71/a1 = -7.526 m and y1 = 100/a1 = 10.6 m

Negative x and y>a mean the cow is outside the field.

Bad cow.

A2 = 369 m² -> a2 = 19.209 m -> perimeter2 = 76.837 m

Cow is at x2 = 69/a2 = 3.592 m and y2 = 240/a2 = 12.494 m

This cow is inside the field.

Good cow.

[Guest]

Nope, EventHorizon's solution doesn't imply that (17,20,x) and (13,20,x) are right angle triangles. It just uses the fact that:

AO²+CO²=BO²+DO²

Sorry if I'm just a bit slow here, but I fail to see how this statement would be true unless any of the triangles comprised of O and any two of {A,B,C,D} are right triangles, which they are not. What if the cow were lying very close to one of the four corners? Let's say the three provided lengths were:

AO=1, BO=19, and CO=20. At a glance it would be evident that DO would have to be somewhat close to 19 as well. According to your statement, that would mean 1²+20²=19²+19². I think the solution both you and EventHorizon provided is flawed, but I certainly could be missing something.

[Guest]

Sorry if I'm just a bit slow here, but I fail to see how this statement would be true unless any of the triangles comprised of O and any two of {A,B,C,D} are right triangles, which they are not. What if the cow were lying very close to one of the four corners? Let's say the three provided lengths were:

AO=1, BO=19, and CO=20. At a glance it would be evident that DO would have to be somewhat close to 19 as well. According to your statement, that would mean 1²+20²=19²+19². I think the solution both you and EventHorizon provided is flawed, but I certainly could be missing something.

Duh Puck,

Yes, the bigger triangles AOB, and BOC are not right triangles. But these guys divided the square into eight right triangles.

AO²+CO²=BO²+DO² can only hold when the triangles AOB, BOC, COD and DOA are right triangles.

Edited February 26, 2008 by brhan

[Guest]

AO²+CO²=BO²+DO² can only hold when the triangles AOB, BOC, COD and DOA are right triangles.

I have to conclude that their solution is correct, since everything checks out after plugging the answer back in, so now I'm still just waiting for that aha moment to slap me upside the head. AOB, BOC, COD, and DOA are clearly not right triangles. I understand there are eight smaller right triangles. I immediately drew the perpendiculars when I started trying to solve the problem, figuring that there would be an evident relationship between the resultant sides and angles of the eight smaller right triangles. I never found it, but apparently it's there, and I just missed it, but I can't see it based on what you're saying.

[bonanova]

I have to conclude that their solution is correct, since everything checks out after plugging the answer back in, so now I'm still just waiting for that aha moment to slap me upside the head. AOB, BOC, COD, and DOA are clearly not right triangles. I understand there are eight smaller right triangles. I immediately drew the perpendiculars when I started trying to solve the problem, figuring that there would be an evident relationship between the resultant sides and angles of the eight smaller right triangles. I never found it, but apparently it's there, and I just missed it, but I can't see it based on what you're saying.

Check out the two figures and first 3 equations in post #15.

It may help show how the three distances - 13, 17 and 20 - become hypotenuses of relevant triangles.

As a bonus, find brhan's other, hitherto undiscovered, field!

[Guest]

Check out the two figures and first 3 equations in post #15.

It may help show how the three distances - 13, 17 and 20 - become hypotenuses of relevant triangles.

As a bonus, find brhan's other, hitherto undiscovered, field!

Aha! *gets smacked upside the head*

Your solution had already made sense to me (although I did catch a small typo in your 4th and 5th equations: the denominators should be -2A. Fortunately, your very next step was to square it), but it wasn't helping me to understand Aatif and EventHorizon's first step. This was mostly because I had considered the idea early on and incorrectly ruled it out as a possibility. Consider this diagram:

I intuitively concluded that AO² + OC² could not possibly equal BO² + OD², I guess because the combined length of the latter two was clearly much larger. Of course, that doesn't mean the sum of the squares are larger.

Good job to all, and sorry for my harassment.

Edited February 27, 2008 by Duh Puck

[bonanova]

Perhaps all questions are answered on this.

But it occurred to me it's easy to show the sum of squared distances to diagonally opposite corners are the same.

Let horizontal and vertical lines through the cow divide the fences into lengths u and v vertically and w and x horizontally.

Squared distance to one corner is u² + x² and to the diagonally opposite corner, v² + w².

Squared distances to the other two corners are u² + w² and v² + x².

The squared distances in both cases add to u² + x² + w² + v².

[unreality]

Ah I finally got around to this topic i didnt read any posts but brhan's top one, so I didnt check if there were any changes to the riddle that werent edited in, so tell me if there were

I have got a square field with one cow. My cow lied 13 meters from one of the corner posts of the field, 17 meters from the corner post diagonally opposite that one, and 20 meters from a third corner post. Can someone tell me how big (perimeter) a field I own?

You may assume the field is flat, and the distance of the cow from the corner poles is measured from the center of mass of the cow -- you know what I mean.

13+17 = 30 meters on the diagonal

so each side is 30?2 (hopefully everyone can see that square root symbol between the 30 and the 2)

*4 =

120?2 meters

Whatever that is (no calculator on me)

Good riddle, though it's pretty easy (if im doing it right that is)

The 20m-to-the-third-corner-post doesnt matter

[unreality]

Sorry, my answer was incorrect! woops! I used the wrong part of the triangle lol

Redo of my answer:

the OP said it was a square field. Meaning the 30m diagonal cuts it into two right triangles, with the hypotenuse of 30 m, so each side of both right triangles (in other words, one side of the 4 equal sides of the square), would be 15*sqrt(2) meters. Multiply by 4 to get 60*sqrt(2), which is approx 84.853 meters perimeter

[unreality]

Oh I see, nevermind- he doesnt have to be necessarily on the diagonal between the corners. Woops! lol

[Guest]

Oh I see, nevermind- he doesnt have to be necessarily on the diagonal between the corners. Woops! lol

unreality,

Ya, the cow may not be on the diagonal. There are good diagrams on posts #12, 15 and 20. IMHO, they will be very helpful.

Edited March 1, 2008 by brhan

[unreality]

Yeah I was just like "17m from one corner, 13m from the other, means the diagonal is 30m, wow this is an easy problem!" before I actually thought enough to see that it didn't have to be right on the diagonal hehe. I'd rather not look at the pictures or the answers, I'll figure it out myself ;D (if i can, that is. hehe)