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edit: woops ok we were confused since the pic wasnt showing.
Ok im taking a look at it right now, this is actually a pretty cool problem.
Currently realizing how much geometry I've forgotten.
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where is the actual question? [edit: ok pic showing now]
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On February 06 2009 08:33 indecision wrote:
where is the actual question?
Solve X
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Label the inside triangle DEF (with D lying on AB and F lying on AC). Construct a line from E to a point we shall call G on DF such that GEF is 50. Use this to motivate another construction of a line from E to AC. The answer quickly follows to be
+ Show Spoiler +
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9070 Posts
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+ Show Spoiler +It's 80 degrees. GL with finding the solution 
edit: i might spoiler stuff.
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United States17042 Posts
On February 06 2009 08:58 Not_Computer wrote:this looks like fun + Show Spoiler +gl
This was how I was going to do it too, and then i scrolled and saw that you had already done it 
You can't really do it by any other method, unless you start arbitrarily assigning side lengths. Pretty simple problem.
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wait what's so hard about it
he says ab=ac, so you know like all but 4 angles already
and then some simple linear algebra and you get the solutions for all the angles
owat
i get x=free variable
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On February 06 2009 09:17 ydg wrote:
wait what's so hard about it
he says ab=ac, so you know like all but 4 angles already
and then some simple linear algebra and you get the solutions for all the angles
owat
i get x=free variable
LOL thats what it makes it a NiceTriangle, try your simple linear algebra
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how come i get X as any integer within 1<X < 110 . . .
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On February 06 2009 09:25 malongo wrote:Show nested quote +On February 06 2009 09:17 ydg wrote:
wait what's so hard about it
he says ab=ac, so you know like all but 4 angles already
and then some simple linear algebra and you get the solutions for all the angles
owat
i get x=free variable LOL thats what it makes it a NiceTriangle, try your simple linear algebra
I tried that before reading any further down the page, and I wound up with a singular matrix. Very interesting, I'll have to think a bit more
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arg I'm stuck with 4 unknown angles
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On February 06 2009 09:59 Zortch wrote:Don't wanna bother to write my solution but here is my answer: + Show Spoiler +
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On February 06 2009 10:30 Avidkeystamper wrote:
Haha ty for the hint, couldn't find the last angles ;D
+ Show Spoiler +
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the drawing is kinda wrong
that 50 degree angle isn't 50 at all
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Label the inside triangle DEF (with D lying on AB and F lying on AC).
I suppose you found out all the angles by yourself except those problematic 4 mentioned before.
Now, if we suppose the triangle ADF is the same (as in, it has the same angles) as the triangle FEC then X will be 80 and all other angles will verify this solution. That means those two triangles (ADF and FEC) really are the same.
since there can be only 1 correct X angle, this is it, 80 degrees. It's kind of a leap of intuition solution, but since there is only 1 correct solution, and you have this one that verifies all the angles, it must be it
^^
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^Wrong.
If you calculate all the angles but the 4 which depend on x (set x=80), the bottom angle of the small triangle comes out as 60 degrees, whereas it should give you 70 degrees (using symmetry and the sum of a triangle's angles being 180 degrees)
Solution below, using only middle-school geometry and simple equations:
+ Show Spoiler [Full solution] +We calculate all angles possible like I said before, and get 4 unknown angles, all of which depend on what x is. From there, we calculate x so that it solves all the equations at once. The unknown angles are x, the left angle of the small triangle (let's call it alpha), and the left and right angles missing from the top triangle (beta and gamma). We know that gamma+x+50=180, thus, gamma=130-x For the left side, we know that beta=x+30 (from 20+gamma+beta=180), and from that we derive that alpha=x-10 (solve alpha+beta+40=180 for alpha) If you've followed me so far, you can solve it yourself, the result is below in another spoiler if you wanna find it out yourself. + Show Spoiler [Result] +Now we combine the equations for alpha with what we know about the small triangle, that alpha+x+70=180, and get x-10+x+70=180, or 2x=120, thus x=60. Not too hard, but it took me half an hour to find an easy way to describe it. Hope you enjoyed, I wasted my 3000th post for this
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On February 07 2009 04:21 Cpt Obvious wrote:
^Wrong.
I dunno man
I'm too tired now to follow your solution even if it's simple basic maths, I will however do so in the morning;I really think my solution is good. I can't find what angle you said doesn't fit.
I'm preparing for architecture college, so I drawed the triangle using perfectly measured angles, as I'm used to doing that lately, and if I manually measure X, it's 80 degrees as well. The initial drawing presented in the op is flawed, the actual angles there are not what they should be (for example the angle in A is waay larger than 20 degrees)
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I did not measure the angle, I calculated it. As you said, there is only one correct solution, which I provided.
I really hope you look into my solution and try to understand it, I think it is correct. All I can say for sure is that your solution gives me an error with a previously calculated angle. I made a drawing to show where alpha, beta, and gamma are, as well as show what the other angles are and how I calculated them.
In the image you can see that if I set x=80, then alpha=70, thus the sum of all angles in the small triangle is 80+70+70=220, whereas it should be 180. I conclude your solution is wrong. That, or ALL my calculations are wrong, which I allow myself to doubt strongly.
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Obvious i think you have wrong :O
Wait ... we get another result with the isocele triangles but both methods seem legit T-T
We need to check something ...
Two different solutions maybe ? oo
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Would you be so kind as to elaborate where exactly my fault lies, then, Sir?
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On February 07 2009 04:21 Cpt Obvious wrote:^Wrong. If you calculate all the angles but the 4 which depend on x (set x=80), the bottom angle of the small triangle comes out as 60 degrees, whereas it should give you 70 degrees (using symmetry and the sum of a triangle's angles being 180 degrees) Solution below, using only middle-school geometry and simple equations: + Show Spoiler [Full solution] +We calculate all angles possible like I said before, and get 4 unknown angles, all of which depend on what x is. From there, we calculate x so that it solves all the equations at once. The unknown angles are x, the left angle of the small triangle (let's call it alpha), and the left and right angles missing from the top triangle (beta and gamma). We know that gamma+x+50=180, thus, gamma=130-x For the left side, we know that beta=x+30 (from 20+gamma+beta=180), and from that we derive that alpha=x-10 (solve alpha+beta+40=180 for alpha) If you've followed me so far, you can solve it yourself, the result is below in another spoiler if you wanna find it out yourself. + Show Spoiler [Result] +Now we combine the equations for alpha with what we know about the small triangle, that alpha+x+70=180, and get x-10+x+70=180, or 2x=120, thus x=60. Not too hard, but it took me half an hour to find an easy way to describe it. Hope you enjoyed, I wasted my 3000th post for this
+ Show Spoiler +
Not sure how you got alpha = x - 10
I tried the same method
alpha + beta + 40 = 180
alpha + beta = 140
alpha + 30 + x = 140
alpha = 110 - x
did i mess up?!
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On February 07 2009 08:02 Cpt Obvious wrote:
Would you be so kind as to elaborate where exactly my fault lies, then, Sir?
Well you are right but there are two different solutions i guess oO
Rbzzzz recalculating stuff ~_^
B3tty seems righ. Your beta is wrong. ( but it works  ? )
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On February 07 2009 07:57 Cpt Obvious wrote:I did not measure the angle, I calculated it. As you said, there is only one correct solution, which I provided. I really hope you look into my solution and try to understand it, I think it is correct. All I can say for sure is that your solution gives me an error with a previously calculated angle. I made a drawing to show where alpha, beta, and gamma are, as well as show what the other angles are and how I calculated them. In the image you can see that if I set x=80, then alpha=70, thus the sum of all angles in the small triangle is 80+70+70=220, whereas it should be 180. I conclude your solution is wrong. That, or ALL my calculations are wrong, which I allow myself to doubt strongly.
how did you get x-10? Did you mean 180-((x+30)+40)?
Trigonometry also solves the question. Call BC=a. Label the inside triangle DEF (with D lying on AB and F lying on AC). Use sine rule on triangle BFC to get BF in terms of a. Use sine rule on triangle BDC to get BD in terms of a. Use cosine rule on triangle BDF to get DF in terms of a. Use cosine rule again on BDF to get the required angle. The answer comes out once again to be
+ Show Spoiler +
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That's where I didn't follow his solution as well, his alpha. I think its actually alpha = 110 - x as b3tty wrote, but if hes getting x-10 from somewhere, it must be somewhere else (because both work if x=60 as he concludes)
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+ Show Spoiler +
the way i did it was split x into two parts
i got x = 80
can provide a diagram if anyone wants
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On February 07 2009 08:26 geno wrote:
That's where I didn't follow his solution as well, his alpha. I think its actually alpha = 110 - x as b3tty wrote, but if hes getting x-10 from somewhere, it must be somewhere else (because both work if x=60 as he concludes)
+ Show Spoiler +
I think where he messed up was
instead of making -> alpha + beta + 40 = 180
he made it -> alpha + beta = 40
substituting beta as 30+x
so he would've gotten
alpha = 10 - x
which is incorrect, so i think his solution is wrong, x = 80 is the only solution.
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HAHAHA
110-x is the same as x-10 because x=60, lol.
I somehow calculated alpha differently, and ended up with x-10 instead of 110-x, which is the same thing, which is also why my solution is still correct.
edit: also x=80 can easily be proven to be a wrong solution. If we set x=80, then alpha has to be 30, thus beta becomes 120, and gamma has to be 40, in order for all the triangle-sums to work.
But then, on the right side, you have gamma+x+50=180, which is not true because gamma+x+50=40+80+50=170!
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no.+ Show Spoiler + Alpha is 30. Beta becomes 110, gamma becomes 50, all is right with the world.
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klizzer, everything you said assumes my equations for alpha, beta and gamma are correct. Then please calculate the sum of the small triangle for me, if you'd be so kind.
x+alpha+70 gives 220 for x=80, where it should be 180. All is not right with the world.
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Oh, no, I used alpha=110-x, not alpha=x-10, which is wrong, I should have clarified that.
You can't say alpha=x-10 and justify it "because x=60" plainly because you used that expression for alpha to obtain x, didn't you?
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Another attempt using linear algebra:
We got four variables; x,alpha,beta and gamma, and four equations:
(1) 20+beta+gamma=180
(2) gamma=130-x
(3) alpha=110-x
(4) alpha+beta+40=180
we now want to eliminate alpha, beta and gamma consecutively until we get an equation for x that doesn't eliminate x in the process.
first i use (2) in (1) to get: 20+beta+130-x=180 -> beta=30+x
then I use that in (4) and substitute alpha with (3) to get: (110-x)+(30+x)+40=180 -> x free variable
Whatever I do, unless I somehow magically find the calculation which got me alpha=x-10, there is no way you can calculate x from this set of equations. It is indeterminate (or whatever the proper term for that is). There are 4 equations for 4 variables, but this system has infinite solutions as it is. The only thing bothering me about this is that x is definitely determined by construction. There can be only one solution, but both 80 and 60 solve all my equations. Sorry if it took me so long to acknowledge that.
edit: also yes, alpha is 110-x for now. I KNOW I came to that result not by guessing, I knew I calculated it SOMEHOW, I just can't seem to recall that trick.
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the way i did it was split x into two parts
i got x = 80
can provide a diagram if anyone wants
YES PLEASE!
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On February 07 2009 09:41 Cpt Obvious wrote:
then I use that in (4) and substitute alpha with (3) to get: (110-x)+(30+x)+40=180 -> x free variable
Hi, I said this
:p
algebra > geometry imo
+ Show Spoiler +
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Ok thanks, I think I understood that. That's trigonometry though, not algebra. Algebra is what I tried to do
Still can't grasp why it's impossible to solve with my path, but hey, that's math.
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On February 07 2009 23:12 Cpt Obvious wrote:Ok thanks, I think I understood that. That's trigonometry though, not algebra. Algebra is what I tried to do Still can't grasp why it's impossible to solve with my path, but hey, that's math.
No it's lack of understanding of maths 
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+ Show Spoiler [Easier Solution] +found an easier way to solve it then the way i suggested before i think this works.. if not, point out my errors, cheers! extend BC out connect the top line of the smaller triangle to this extended line you have now created a new isosceles triangle you know the large angle of this triangle is 120 degrees you can solve for the two smaller angles because they are equal. X = 80 + Show Spoiler [Diagram] +
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How do you know that the new triangle is isoceles?
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Since your triangle looks isosceles, it probably isn't. As I've mentioned before, that drawing, the one u based your solution upon, is incorrect. The angles aren't really what it is written on them. So if your triangle looks isosceles on the wrong drawing, it most probably has way disproportionate angles on an accurate drawing, therefore it isn't isosceles
..unless you have an explanation to why beta = alfa, I haven't bothered to verify what I wrote above
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I also don't get how you know that the other angle is 100 degrees. Can you elaborate on that?
This solution looks elegant if it is correct, but I am not sure of that.
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the other angle is 100 because 60 + 20 = 80
100 degrees is missing from the straight line
and you guys are right, i have no way of proving that the new triangle is isoceles, this solution is wrong.
gonna draw up my original one later on today
and see if i messed up there again.
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Wow im actually glad this thread interested some people 1200 or so visits is good for a mathematics related one thank you guys, im still stucked in sigma_x's continuation, so im not writing my solution yet (but i guess its still fun for those still interested). I just want to add that the "pure algebra" doesnt work (cpt obbious and others first attempt) because it relies only in the a+b+c=180 formula and keeps away other data (namely linear data ac=ab and others).
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i used a protractor and determined that the angle is 80 degrees.
however, i have no solid way of proving it without using trigonometric calculations (sin, cos, tan)
is it possible to solve it using only angles?
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On February 08 2009 10:20 b3tty wrote:
i used a protractor and determined that the angle is 80 degrees.
however, i have no solid way of proving it without using trigonometric calculations (sin, cos, tan)
is it possible to solve it using only angles?
I dunno I tried but my linear algebra sucks
I have 4 equations that I can't solve simultaneously
Although I do know I'm on the right track because I keep proving that my equations are correct -_-;;
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On February 08 2009 10:20 b3tty wrote:
i used a protractor and determined that the angle is 80 degrees.
however, i have no solid way of proving it without using trigonometric calculations (sin, cos, tan)
is it possible to solve it using only angles?
Yes. Ill post the solution later as i said (in fact my point here is finding another solutions check the spoiler in the op if you want dome hint full solution is not up yet)
On February 08 2009 11:26 koreakool wrote:
I dunno I tried but my linear algebra sucks
I have 4 equations that I can't solve simultaneously
Although I do know I'm on the right track because I keep proving that my equations are correct -_-;;
You need more geometry than algebra for this, algebra skips some data.
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On February 08 2009 11:47 malongo wrote:Show nested quote +On February 08 2009 10:20 b3tty wrote:
i used a protractor and determined that the angle is 80 degrees.
however, i have no solid way of proving it without using trigonometric calculations (sin, cos, tan)
is it possible to solve it using only angles? Yes. Ill post the solution later as i said (in fact my point here is finding another solutions check the spoiler in the op if you want dome hint full solution is not up yet) Show nested quote +On February 08 2009 11:26 koreakool wrote:
I dunno I tried but my linear algebra sucks
I have 4 equations that I can't solve simultaneously
Although I do know I'm on the right track because I keep proving that my equations are correct -_-;; You need more geometry than algebra for this, algebra skips some data.
My solution uses pure algebra and trigonometric rules
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On February 08 2009 12:01 ydg wrote:Show nested quote +On February 08 2009 11:47 malongo wrote:On February 08 2009 10:20 b3tty wrote:
i used a protractor and determined that the angle is 80 degrees.
however, i have no solid way of proving it without using trigonometric calculations (sin, cos, tan)
is it possible to solve it using only angles? Yes. Ill post the solution later as i said (in fact my point here is finding another solutions check the spoiler in the op if you want dome hint full solution is not up yet) On February 08 2009 11:26 koreakool wrote:
I dunno I tried but my linear algebra sucks
I have 4 equations that I can't solve simultaneously
Although I do know I'm on the right track because I keep proving that my equations are correct -_-;; You need more geometry than algebra for this, algebra skips some data. My solution uses pure algebra and trigonometric rules
Omg i didnt saw that.... hey cheater X=79.crap!!! ahah i guess its more messy by pure geometry, but thats what i wanted an alternative solution thanX a lot im so bad at trigonometry.
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Well the 79.9986342 degrees is just from rounding odd numbers during the calculation, and since you only have whole degrees in every other angle of the triangles, it's pretty safe to assume that x is a whole number too.
@b3tty: I think I've proven that the algebraic attempt using only angles doesn't give a determinate solution. This is because I am missing one bit of information to include into my system of equations, and it ends up being indeterminate. All it does is prove that 180=180, basically.
I will not conclude that it is impossible, but I don't see where that additional information should come from but trigonometry. I did use AB=AC by calculating the angles complementing the 50 and 60 degree angles in the corners, so that can't be it.
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Regarding OP's solution:
+ Show Spoiler +How do you get that triangle IFD is isosceles in the final step of your solution?
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I refuse to give up.
However, I have midterms this week, so I think I'm gonna have to hold off on finding a solution until this week is over.
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I will SO solve this bitch
even if it takes me all week
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Well I actually did before, but I admit, it's really sketchy
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