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XiaNaphryz
LATIN, MATRIPEDICABUS, DO YOU SPEAK IT
(03-20-2009, 09:08 PM)
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Taken from here. Just saw this link from a co-worker, I'll have to give it a try later tonight if I find time.

EDIT: And because there seems to be a lot of confusion to this (either through people not reading the OP fully or just not understanding the term itself), the key point to the problem is that you can ONLY use elementary geometry to solve this. Nothing beyond what's listed here, so no trig, no algebra beyond the few equations listed, no matrices, etc:

Elementary Geometry

Here is everything you need to know to solve the above problems.

Lines and Angles: When two lines intersect, opposite angles are equal and the sum of adjacent angles is 180 degrees. When two parallel lines are intersected by a third line, the corresponding angles of the two intersections are equal.

Triangles: The sum of the interior angles of a triangle is 180 degrees. An isosceles triangle has two equal sides and the two angles opposite those sides are equal. An equilateral triangle has all sides equal and all angles equal. A right triangle has one angle equal to 90 degrees. Two triangles are called similar if they have the same angles (same shape). Two triangles are called congruent if they have the same angles and the same sides (same shape and size).

* Side-Angle-Side (SAS): Two triangles are congruent if a pair of corresponding sides and the included angle are equal.
* Side-Side-Side (SSS): Two triangles are congruent if their corresponding sides are equal.
* Angle-Side-Angle (ASA): Two triangles are congruent if a pair of corresponding angles and the included side are equal.
* Angle-Angle (AA): Two triangles are similar if a pair of corresponding angles are equal.

So a sample solution for this type of problem could look like:

EXAMPLE PROBLEM: Prove that if the two diagonals in a quadrilateral bisect each other, then the quadrilateral is a parallelogram.



PROOF WRITTEN IN TWO-COLUMN FORM:

Code:

Argument 	                                                       Reason why
1. The two lines marked with one brown little line are congruent. 	1. The two diagonals bisect (given).
2. The two lines marked with two brown little lines are congruent. 	2. The two diagonals bisect (given).
3. The two angles marked with blue lines are congruent. 	        3. They are vertical angles.
4. The two yellow triangles are congruent. 	                        4. SAS theorem and 1, 2, and 3. 
5. The angles A and A' are congruent. 	                                5. The two yellow triangles are congruent.
6. The angles A' and A'' are congruent. 	                        6. They are vertical angles.
7. The angles A and A'' are congruent.                                	7. 5 and 6 together.
8. The lines that form bottom and top of the quadrilateral are parallel.8. 7 and the theorem that says that corresponding angles being the same is equivalent to lines being parallel.
9. The lines that form the two sides of the quadrilateral are parallel. 9. Repeat steps 1-8 using the two white triangles.
10. The quadrilateral is a parallelogram. 	                       10. 8 and 9 together.
PROOF IN PARAGRAPH FORM:

Since the diagonals are bisecting each other, the line segments marked with one little line are equal, and similarly the line segments marked with double little lines. The two angles marked with dark blue line are equal, being vertical angles. It follows from SAS congruence theorem that the two yellow triangles are congruent.

Since they are congruent, angles A and A' have the same measure. And, angles A' and A'' are the same because they are vertical angles. So since A and A' are the same, and A' and A'' are the same, it follows that angles A and A'' are the same.

But this is equivalent to the two lines that form the top and bottom of the quadrilateral being parallel.

An identical argument using the two white triangles instead of the two yellow ones proves that the two sides of the quadrilateral are parallel.

So the quadrilateral is a parallelogram.

Here's the actual problem:

Using only elementary geometry, determine angle x. Provide a step-by-step proof.

You may only use elementary geometry, such as the fact that the angles of a triangle add up to 180 degrees and the basic congruent triangle rules (side-angle-side, etc.). You may not use more advanced trigonomery, such as the law of sines, the law of cosines, etc.
There is a review of elementary geometry below above.



This is the hardest problem I have ever seen that is, in a sense, easy. It really can be done using only elementary geometry. This is not a trick question.

Here is a very small hint. Here is a small hint.

-----

Tips for Writing Proofs

Proofs may be written informally using plain English. Just be sure to include all the steps in your reasoning, or at least all the key steps. Providing a diagram is very helpful but not required. You can draw a diagram on the computer or you can draw it on paper and then scan it or photograph it with a digital camera. Name each point you use with a letter. If you don't provide a diagram, you will need to describe the named points with words (ex., say "the intersection of AE and DB is G "). Identify lines with two letters (ex., say "line AB" or simply "AB"). Identify triangles with three letters (ex., say "triangle ABC" or "tri ABC" or simply "ABC"). Identify angles with three letters, vertex in the middle (ex., say "angle ABC" or "ang ABC" or "<ABC" or simply "ABC"). It is helpful to number your steps to make it easy to refer back to earlier steps.

-----

Please don't search the the web for the answer -- that's cheating. You will only deprive yourself of many hours of delicious frustration.

I did not invent these problems. After I first read problem 1, I worked on it for many hours over several days before I eventually figured it out. A couple of years later I came back to the problem, but I had forgotten my proof. It took me many hours to figure it out again! Problem 2 also took me many hours to solve.

How hard are these problems? Any teenage student and some younger students can understand the proof, but very very few are able to discover the proof on their own. Of the hundreds of people that have emailed me, I'd estimate only one or two percent (mostly math professionals and college students) have solved it without significant hints. (The hints given above are not significant hints.) Most people who think they have found the solution are wrong.

These problems have been published in many places. Problem 2 first appeared here: Langley, "A Problem", Mathematical Gazette, 1922. Dr. Gary Gruber says his high school teacher showed him problem 1 in about 1955. Tom Rike says problem 1 first appeared in print here: Harry Schor, The New York State Mathematics Teachers' Journal, 1974. It also appeared here: "Problem 134", Eureka (now Crux Mathematicorum), 1976. Dr. Gruber popularized problem 1 in several papers (such as "The Genius Test") which appeared in newspapers throughout the 1990s (Universal Press Syndicate and Los Angeles Times Syndicate). That's where I discovered it.

Last edited by XiaNaphryz; 03-20-2009 at 10:32 PM.
Domino Theory
343i Test
(03-20-2009, 09:09 PM)
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kk
Nerevar
they call me "Man Gravy".
(03-20-2009, 09:10 PM)
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um, isn't it literally just a combination of "all the angles of a triangle add up to 180 degrees" and "the two angles of a straight line add up to 180 degrees", then you've just got a big polynomial function to solve? Am I missing something there?
Skittleguy
Ring a Bell for me
(03-20-2009, 09:11 PM)
Seriously? People find this hard?
FromTheFuture
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(03-20-2009, 09:13 PM)
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Wait, i'm confused. I'm pretty sure this is easy.
Like the hat?
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(03-20-2009, 09:14 PM)
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40 degrees?
Yaweee
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(03-20-2009, 09:15 PM)
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Originally Posted by Nerevar

um, isn't it literally just a combination of "all the angles of a triangle add up to 180 degrees" and "the two angles of a straight line add up to 180 degrees", then you've just got a big polynomial function to solve? Am I missing something there?

No, that's it. 4 equations, 4 unknowns, easy shit. It took me about a minute to get it here, and chugging things through should only take a few more.
levious
That throwing stick stunt of yours has boomeranged on us.
(03-20-2009, 09:15 PM)
I thought proofs were more complicated than just explaining your steps in plain english... otherwise I don't think I would have grown to hate proofs.
layzie1989
Member
(03-20-2009, 09:16 PM)
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i really don't feel like writing anything down right now...
RiskyChris
Banned
(03-20-2009, 09:16 PM)
This is so easy, why are there 2000 words in the OP.
daw840
Member
(03-20-2009, 09:17 PM)
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Uh, yeah. If I had a little time to sit here and figure this out it would be really simple. Why is this considered hard?
XiaNaphryz
LATIN, MATRIPEDICABUS, DO YOU SPEAK IT
(03-20-2009, 09:17 PM)
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Originally Posted by Yaweee

No, that's it. 4 equations, 4 unknowns, easy shit.

I think the trick is that you have to answer it as a geometric proof, no equations.
RiskyChris
Banned
(03-20-2009, 09:18 PM)

Originally Posted by XiaNaphryz

I think the trick is that you have to answer it as a geometric proof, no equations.

Not a proof!
zoku88
Member
(03-20-2009, 09:18 PM)
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Originally Posted by XiaNaphryz

I think the trick is that you have to answer it as a geometric proof, no equations.

???

If those equations are derived from geometric reasoning, that should be sufficient for a proof.
Yaweee
Member
(03-20-2009, 09:19 PM)
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Originally Posted by XiaNaphryz

I think the trick is that you have to answer it as a geometric proof, no equations.

Proofs are equations, just written with a slightly different formalism that would actually make things shorter, if you know the right path of logic to take.
Metalic Sand
who is Emo-Beas?
(03-20-2009, 09:19 PM)
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Think my brain just exploded, Yes im horrible at math.
XiaNaphryz
LATIN, MATRIPEDICABUS, DO YOU SPEAK IT
(03-20-2009, 09:21 PM)
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Originally Posted by Yaweee

Proofs are equations, just written with a slightly different formalism that would actually make things shorter, if you know the right path of logic to take.

I meant a proof you would do for geometry class (and ONLY geometry class, as I can't ever remember doing this sort of thing in any other math course I've taken since then). Stuff like:

AC = BC
BD = CD
Draw a line parallel to AB, called DE
Because AC = BC and BD is || to DE, we can state yadda yadda yadda
123rl
Member
(03-20-2009, 09:22 PM)
So what's the answer? I worked it out but very quickly and may have missed something obvious (not that I was ever good with geometry :D)

Is it 40 degrees?
MisterHero
Super Member
(03-20-2009, 09:23 PM)
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luckily i kept notes in my graphing calculator
zoku88
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(03-20-2009, 09:23 PM)
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Originally Posted by XiaNaphryz

I meant a proof you would do for geometry class (and ONLY geometry class, as I can't ever remember doing this sort of thing in any other math course I've taken since then). Stuff like:

AC = BC
BD = CD
Draw a line parallel to AB, called DE
Because AC = BC and BD is || to DE, we can state yadda yadda yadda

the 'we can state" part could easily by followed by an equation, though.... which is what I meant when I said, "it should be okay if you state the geometric meaning of the equations"
GaimeGuy
Volunteer Deputy Campaign Director, Obama for America '16
(03-20-2009, 09:24 PM)
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This is supposed to be hard?
XiaNaphryz
LATIN, MATRIPEDICABUS, DO YOU SPEAK IT
(03-20-2009, 09:24 PM)
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A quick Google search produced this link as having the answer. I only peeked at the first few steps to verify the link, but I want to see if I can figure this out on my own. :D

Originally Posted by zoku88

the 'we can state" part could easily by followed by an equation, though.... which is what I meant when I said, "it should be okay if you state the geometric meaning of the equations"

Equations aren't really used in elementary geometry proofs though, at least in terms of stuff like solving for polynomials and whatnot. It's mainly using established equations to fill in variables and make comparisons, then deriving the angles from that.
Last edited by XiaNaphryz; 03-20-2009 at 09:27 PM.
Ela Hadrun
Probably plays more games than you
(03-20-2009, 09:24 PM)
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Yeah... this is so easy it actually makes me miss geometry

goddamn I love geometry it is so much more real than stupid calculus

maybe if I had taken physics calc would make sense

but geometry = best
Timedog
good credit (by proxy)
(03-20-2009, 09:28 PM)
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this algebraic systems of equations crap or whatever you guys are talking about is not simple geometry.
zoku88
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(03-20-2009, 09:28 PM)
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Originally Posted by XiaNaphryz

Equations aren't really used in elementary geometry proofs though.

I think people use whatever methods are easiest for them for proofs, as long as it stands logically. I mean, for this proof, whatever logical statements you use can be easily turned into a linear algebra problem. I mean, they're equivalent. And because the equations are derived geometrically, it would still be a geometric proof. Although, maybe they're not used in elementary geometry classes.
XiaNaphryz
LATIN, MATRIPEDICABUS, DO YOU SPEAK IT
(03-20-2009, 09:30 PM)
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Originally Posted by zoku88

I think people use what methods are easiest for them for proofs, as long as it stands logically. I mean, for this proof, whatever logical statements you use can be easily turned into a linear algebra problem.

Again, the whole setup for this problem is to solve it using elementary geometry proofs. It's purposely limiting the methods you can use to solve this - no equations to solve for unknowns, no trig, etc. It says right at the beginning: Using only elementary geometry.
~Kinggi~
FIND ME AN ESCORT
NO SHARP KNEEEEEEES
(03-20-2009, 09:31 PM)
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40
scottnak
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(03-20-2009, 09:31 PM)
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Originally Posted by zoku88

I think people use what methods are easiest for them for proofs, as long as it stands logically. I mean, for this proof, whatever logical statements you use can be easily turned into a linear algebra problem. I mean, they're equivalent.

But that's the point of this whole problem.
It's EASY to get the answer.
But HARD to do so keeping it a GEOMETRIC problem.

A challenge indeed...
zoku88
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(03-20-2009, 09:33 PM)
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Originally Posted by scottnak

But that's the point of this whole problem.
It's EASY to get the answer.
But HARD to do so keeping it a GEOMETRIC problem.

A challenge indeed...

Uhm. The algebra is just a representation of the geometry. I'm trying to tell you that there is no difference. Whatever proof you construct will still be algebraic.

Originally Posted by TheRagnCajun

So instead of solving a set of equations, you just sub one equation into another and plug in the values at the end.

That was what I was talking about...
Like the hat?
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(03-20-2009, 09:33 PM)
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Originally Posted by XiaNaphryz

A quick Google search produced this link as having the answer.

guess i was wrong. whoops.
TheRagnCajun
(03-20-2009, 09:33 PM)

Originally Posted by XiaNaphryz

A quick Google search produced this link as having the answer. I only peeked at the first few steps to verify the link, but I want to see if I can figure this out on my own. :D

Equations aren't really used in elementary geometry proofs though, at least in terms of stuff like solving for polynomials and whatnot. It's mainly using established equations to fill in variables and make comparisons, then deriving the angles from that.

So instead of solving a set of equations, you just sub one equation into another and plug in the values at the end. I don't see a difference from a practical sense. The problem looks pretty easy, and I'm pretty sure I can guess what the small hint is. there are triangles inside of triangles
ChoklitReign
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(03-20-2009, 09:37 PM)
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[IMG]http://i42.************/4rvwgk.png[/IMG]
Am I missing something? I'm tired today. θ2 + θ3 = 140 I know, so where do I go? As I said, I'm just tired.
Son of Godzilla
Banned
(03-20-2009, 09:38 PM)
Oh my god I miss simple geometry. Proofs were so neat and cute back then.
Yaweee
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(03-20-2009, 09:39 PM)
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Originally Posted by ChoklitReign

[IMG]http://i42.************/4rvwgk.png[/IMG]
Am I missing something? I'm tired today. θ2 + θ3 = 140 I know, so where do I go? As I said, I'm just tired.

With the way you've drawn it...

t2 + t3 = 140
t1 + t2 = 160
t1 + x = 150
x + t3 = 130

4 equations, 4 unknowns. Solve via your preferred method. Algebra or Matrices are the simplest way to go.

EDIT: Nevermind, that's only 3 LI equations.
Last edited by Yaweee; 03-21-2009 at 01:49 AM.
Superman00
Liverpool01
(03-20-2009, 09:40 PM)

Originally Posted by zoku88

Uhm. The algebra is just a representation of the geometry. I'm trying to tell you that there is no difference. Whatever proof you construct will still be algebraic.

That was what I was talking about...


You can't do any equation, basically you are just using arithmatic. All you are doing is writing the numbers down on the triangle.
~Kinggi~
FIND ME AN ESCORT
NO SHARP KNEEEEEEES
(03-20-2009, 09:41 PM)
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I got mine wrong cause the visual representation is misleading in the OP.

[IMG]http://i42.************/4rvwgk.png[/IMG]


^^ that would have been better.
XiaNaphryz
LATIN, MATRIPEDICABUS, DO YOU SPEAK IT
(03-20-2009, 09:43 PM)
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I don't think a lot of people here remember elementary geometry. Here's some example problems and proofs from a quick Google:



PROBLEM: Prove that if the two diagonals in a quadrilateral bisect each other, then the quadrilateral is a parallelogram.



PROOF WRITTEN IN TWO-COLUMN FORM:

Code:

Argument 	                                                       Reason why
1. The two lines marked with one brown little line are congruent. 	1. The two diagonals bisect (given).
2. The two lines marked with two brown little lines are congruent. 	2. The two diagonals bisect (given).
3. The two angles marked with blue lines are congruent. 	        3. They are vertical angles.
4. The two yellow triangles are congruent. 	                        4. SAS theorem and 1, 2, and 3. 
5. The angles A and A' are congruent. 	                                5. The two yellow triangles are congruent.
6. The angles A' and A'' are congruent. 	                        6. They are vertical angles.
7. The angles A and A'' are congruent.                                	7. 5 and 6 together.
8. The lines that form bottom and top of the quadrilateral are parallel.8. 7 and the theorem that says that corresponding angles being the same is equivalent to lines being parallel.
9. The lines that form the two sides of the quadrilateral are parallel. 9. Repeat steps 1-8 using the two white triangles.
10. The quadrilateral is a parallelogram. 	                       10. 8 and 9 together.

Originally Posted by ~Kinggi~

I got mine wrong cause the visual representation is misleading in the OP.

Hey, it said it wasn't to scale. ;)
Haly
One day I realized that sadness is just another word for not enough coffee.
(03-20-2009, 09:43 PM)
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Definitely not.
half a moon
Member
(03-20-2009, 09:43 PM)
You can solve it numerous ways with advanced math... but that's not the point.
bachikarn
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(03-20-2009, 09:44 PM)
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Originally Posted by Yaweee

With the way you've drawn it...

t2 + t3 = 140
t1 + t2 = 160
t1 + x = 150
x + t3 = 130

4 equations, 4 unknowns. Solve via your preferred method. Algebra or Matrices are the simplest way to go.

Have you actually tried it? I think they are linearly dependent, so you can't get an answer from just that. Unless I'm doing something wrong.
XiaNaphryz
LATIN, MATRIPEDICABUS, DO YOU SPEAK IT
(03-20-2009, 09:45 PM)
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Originally Posted by Halycon

What the fuck is this.

It takes less than 3 seconds.

Try reading the very first line of the problem, which I've bolded, highlighted, and underlined. I guess that's still not enough.

Nice stealth edit btw. ;P
Armitage
Constantly slobbing some bloke's knob
...Not Groovy
(03-20-2009, 09:46 PM)
I think you guys are underestimating the problem a little bit, it's not just 180 - X - Y = Z the whole way through.
Crayon Shinchan
Aquafina Fanboy
(03-20-2009, 09:47 PM)
Remember, no sin or cosine functions people.

I don't know. I guess I must be maths dumb after 10+ years of not using it.
zoku88
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(03-20-2009, 09:48 PM)
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Originally Posted by XiaNaphryz

I don't think a lot of people here remember elementary geometry

That's just a particular method of proving something via elementary geometry, most commonly associated with high school geometry classes. That approach (a two column proof) is not inheritable more geometric than any other proof using elementary geometry (unless they're crazy and use trig.)


If you wanted, you could make an elementary geometric proof that is a proof by contradiction (which would mean that you probably found the answer beforehand somehow.)


In the end, any proof we come up with will most likely be elementary geometry not because of the structure of our proofs but by the tools we used to come up with our proofs. 'Elementary geometric proof' does not define a structure.
HamPster PamPster
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(03-20-2009, 09:49 PM)
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Originally Posted by Armitage

I think you guys are underestimating the problem a little bit, it's not just 180 - X - Y = Z the whole way through.

And thats why I can't solve it :(

Screw you math - someone post the what math teachers want you to believe xkcd comic - I can't do it because I'm at work not doing math!
TheRagnCajun
(03-20-2009, 09:52 PM)

Originally Posted by XiaNaphryz

When two lines intersect, opposite angles are equal and the sum of adjacent angles is 180 degrees. When two parallel lines are intersected by a third line, the corresponding angles of the two intersections are equal.

Triangles: The sum of the interior angles of a triangle is 180 degrees..


Granted I havent' done a lot of proofs in my day, but This is all you need to solve the problem using 'equations'. I dont' see how it can be cheating when it has clearly been laid out as fair game. Whether you choose to sub in the values right away and solve using a matrix, or just demonstrate the relationship of each angle one by one, you're doing the same thing. One method is just more efficient.
Yaweee
Member
(03-20-2009, 09:54 PM)
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Originally Posted by bachikarn

Have you actually tried it? I think they are linearly dependent, so you can't get an answer from just that. Unless I'm doing something wrong.

Hmm, I can't get it to work out either, or just yet. You might be right about them being linearly dependent. Another equation or two might be needed for the mix.
Last edited by Yaweee; 03-20-2009 at 09:57 PM.
Haly
One day I realized that sadness is just another word for not enough coffee.
(03-20-2009, 09:54 PM)
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Originally Posted by XiaNaphryz

Try reading the very first line of the problem, which I've bolded, highlighted, and underlined. I guess that's still not enough.

Nice stealth edit btw. ;P

You were too fast for me :p
zoku88
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(03-20-2009, 09:56 PM)
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Originally Posted by bachikarn

Have you actually tried it? I think they are linearly dependent, so you can't get an answer from just that. Unless I'm doing something wrong.

I just checked them, they're independent, I believe, unless I wrote something down incorrectly.

With row swaps, should be able to define a matrix with

1 1 0 0
0 1 1 0
0 0 1 1
1 0 0 1

t1 t2 t3 x
XiaNaphryz
LATIN, MATRIPEDICABUS, DO YOU SPEAK IT
(03-20-2009, 09:58 PM)
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Originally Posted by TheRagnCajun

Granted I havent' done a lot of proofs in my day, but This is all you need to solve the problem using 'equations'. I dont' see how it can be cheating when it has clearly been laid out as fair game.

I think this is the key disconnect many of us are having. In the mindset of elementary geometry, what was stated there doesn't necessarily equate to "you can just use equations to solve for it." It's providing you rules to use for your proof as you step through it like in the example two-column proofs I posted above. "Because opposite angles are equal, I can state this unknown is n degrees." When I think of geometry class back in junior high, I don't think equations at all. I think of that style of proof we had to write up.

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