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

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### #11

Posted 26 February 2008 - 02:12 AM

### #12

Posted 26 February 2008 - 02:39 AM

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.

### #13

Posted 26 February 2008 - 05:51 AM

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

### #14

Posted 26 February 2008 - 08:34 AM

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: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.

AO

^{2}+CO

^{2}=BO

^{2}+DO

^{2}... (i) (Naming the square from top left corner clockwise as ABCD, with the cow tied at O)

(i)

and, 13

^{2}-x

^{2}=AO

^{2}-y

^{2}...(ii)

You can use (i) and (ii) to write y =f(x) = sqrt(x

^{2}+ 17

^{2}- 20

^{2})

And thus his results, which are correct.

**Edited by Aatif, 26 February 2008 - 08:34 AM.**

### #15

Posted 26 February 2008 - 09:41 AM

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #16

Posted 26 February 2008 - 02:40 PM

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: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:

AO^{2}+CO^{2}=BO^{2}+DO^{2}

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 1

^{2}+20

^{2}=19

^{2}+19

^{2}. I think the solution both you and EventHorizon provided is flawed, but I certainly could be missing something.

### #17

Posted 26 February 2008 - 03:51 PM

Duh Puck,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 1^{2}+20^{2}=19^{2}+19^{2}. I think the solution both you and EventHorizon provided is flawed, but I certainly could be missing something.

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

**eight**right triangles.

AO

^{2}+CO

^{2}=BO

^{2}+DO

^{2}can only hold when the triangles AOB, BOC, COD and DOA are right triangles.

**Edited by brhan, 26 February 2008 - 03:52 PM.**

### #18

Posted 27 February 2008 - 12:21 AM

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.AO

^{2}+CO^{2}=BO^{2}+DO^{2}can only hold when the triangles AOB, BOC, COD and DOA are right triangles.

### #19

Posted 27 February 2008 - 05:24 AM

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

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!

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #20

Posted 27 February 2008 - 05:48 PM

Aha! *gets smacked upside the head*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, findbrhan's other, hitherto undiscovered, field!

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 AO

^{2}+ OC

^{2}could not possibly equal BO

^{2}+ OD

^{2}, 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 by Duh Puck, 27 February 2008 - 05:49 PM.**

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