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Really!!
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#121
Report Thread starter 11 years ago
#121
(Original post by DFranklin)
I know I said I wouldn't reply again, but I was bored.

I looked up the statement of Cevi's theorem as I didn't really understand the one you posted. Anyhow, it took me about half an hour to prove it using vectors, and I didn't use isometry, barycentric coordinates, or more than a few lines of algebra.

Seeing as you don't seem to know that much about STEP you might want to be careful with the eye-rolling.

I said this reminded me of a STEP question, but I hadn't actually done the question, and I specifically avoided looking at it before working out my own proof (so you wouldn't say I'd got a hint from it or something stupid).

But having done my own proof, I took the time to look at the STEP question (it is from this year: STEP II, 2007, Q8), and it seems to me a pretty much exact match for part (a), and although it doesn't ask for part (b), you can just use the workings from the solution to part (a) to prove the converse with very little trouble.

Note that STEP II actually was quite easy (for a STEP paper) this year; there are people on this forum who answered 8 questions for STEP II, which implies they would have taken around 20 minutes to do this problem. If you think it would take days, maybe you wouldn't find STEP quite as easy as you think...
lol...I've just finished to solve it and the proof was very long. You need barycenter and some bit of isometry and perhaps a tiny bit of algebra. Vectors alone aren't enough I can tell you that now. Maybe you didn't prove it properly. I checked the mark scheme (a guided mark scheme) and you needed barycenter and a bit of isometry plus the rest are just algebra manipulations. At first, I felt that vectors were enough but after consulting the guided mark scheme, I realised that it required the above (quoted).
I could pretty much do everything in a short amount of time expect that lone question. Plus I repeat, the STEPS papers I had done aren't that difficult (and I did plenty of them). The most recent STEP III paper I did dates from 2002 (of course I did all the 90's ones as well plus 05 and 06(i'm not sure)). They were 14 questions and I managed to do 10 in 2 hours and 1/2. Although I must admit that the last probability question took quite some time (about 25 min) . First question (still 2002) find the area of the region between the x-axis and the function y = lnx/x for 1<=x<=a. I'm quite surprised you call that difficult. Plus there was this question where they asked to solve the df equation y^4(dy/dx)^4= (y^2-1)^2. are you suggesting that this is extremely difficult?!. But that wasn't the only paper I did. By far the only Difficult STEP III paper (I had done*) was the 1998's one where two of the questions were really tough but I managed to do 10 in 2h50min. If you want I'll scan these STEP III papers and post it here (or maybe ask my friend for the link) so that you can see for yourself. You've actually made me say things I didn't want to say. I can tell a lot of people will feel offended and frustrated. I hope you're proud of yourself :mad:. So amigo there you have it. I'm not going to say this again.

Out of all of the problems I had done from my teacher's book, Q1 is by far the only one I respect. A simple vector manipulation isn't enough to prove this theorem and that is a fact. Maybe I exagerated when I said days but it takes a lot of time. As a matter of fact, I've never seen this kind of problem in a STEP III paper not even II. Unless of course you prove me wrong. Why don't you try Q2? :p:.


PS: I'm not your equal because you're far above me (since you're a graduate and I respect you for that). So stop pestering me :mad:. Now can you please stop arguying with me?. I gave my personal opinion based on what I had done. If you agree then fine.

* All these past papers exams were done during my spare time and I timed myself.
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#122
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#122
(Original post by coffeym)
I think this impression is put across because of his use of English...although I would tend to agree .

Franckii05, where are you from (originally)?*

OMG...relax friend. I told you these questions were translated from Greek to english (Q1 and 2) actually by my teacher. well didn't I?!
Q3,4,5 were Translated from French to English. So sorry for the akward wording but I tried to be as simplistic and realistic as possible. I guess I made a mess out of it...:confused:. But like I said before the questions themselves lack consistency.


*Where am I from? hmm....I'm an alien from Maaaars :rolleyes:.
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henryt
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#123
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#123
I'm officially leaving this thread forever. It's been... erm... time-consuming! Ciao!!
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#124
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#124
(Original post by Dystopia)
Question 1 is a fairly elementary result in geometry...

It's been a while since I've studied this kind of stuff, but I don't think the condition for concurrence given is actually sufficient; it implies either that the lines meet or that they are parallel. Admittedly, it would be sufficient if the lines weren't extended, but if they weren't then would be no need to use vectors.


Nonsense. You make it sound far more difficult than it really is. I'm not very good at geometry or vectors, but can produce a proof of Ceva's theorem covering all possible cases without difficulty.

Proof using barycentrics

Say that \frac{A'B}{A'C} = \frac{n}{m}, \; \frac{B'C}{B'A} = \frac{l}{n}, \; \frac{C'A}{C'B} = \frac{m}{l} then the point G\left(\frac{l}{l + m + n}, \frac{m}{l + m + n}, \frac{n}{l + m + n}\right) lies on AA', BB' and CC': G = \textbf{a} + \frac{n + m}{l + m + n}(\textbf{a'} - \textbf{a}) and so on. If l + m + n = 0 then the lines are parallel.
*

It seems to me that the primary source of difficulty is the vague way in which the questions are written and the 'unusual' terms used.

ok if that's opinion then fine. I'm not going to argue with you. Yes the English in these question (Q1 and 2) seems akward because it was originally written in other language (Greek). I guess they were poorly translated . But I think that Q3,4 and 5 are perfectly fine.


*I'm afraid your proof isn't valid in this case. There is more to it than you actually think.
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Dystopia
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#125
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#125
(Original post by Franckii05)
*I'm afraid your proof isn't valid in this case. There is more to it than you actually think.
How is my proof not valid? The sign conventions take care of the cases when the lines BC, CA, AB are extended.
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#126
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#126
I don't think we'll ever come to an agreement...aaand I'm giving up.
Can we close this thread now?!!

PS: I know where I stand though....
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#127
Report Thread starter 11 years ago
#127
(Original post by Dystopia)
How is my proof not valid? The sign conventions take care of the cases when the lines BC, CA, AB are extended.
Sorry I meant your method of using the barycenters isn't quite right. You need to cover all the possibilities.

Don't fuss this thread is officially closed and there is no need in wasting your time doing Q1 again...I'll probably post the actual solution (according to the guided mark scheme) on the Maths' forum or something.


Well bye
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