-insert title here-
- Jun 19, 2014
- Outside reality
No, that is not what I meant.I've interpreted your claim as: "Suppose we have two similar quadrilaterals, ABCD and EFGH. Call the midpoint of AE as X, and CG as Y. Also, label the midpoint of CD as P and FH as Q. Then, XYPQ is similar to both these quadrilaterals." If this is what you mean, there are counterexamples to this claim, so let me know if you meant something else
Lol, even now you still make claims without any proof (you are still yet to prove your claim), instead of just making vague references to spiral similarityNo, that is not what I meant.
The quadrilaterals MUST be oriented similarly.
i.e. if the vertices A, B, C, D progresses clockwise, then E, F, G, H must also progress clockwise (for your example above)
Your counterexample only works when the figures are not oriented similarly.
In this case, the Kites are oriented similarly, so there is no problem.
You present this as a general question without specifying HSC tools only. So I took advantage of the matter of the fact.Lol, even now you still make claims without any proof (you are still yet to prove your claim), instead of just making vague references to spiral similarity
To make yourself clear, I suggest you prove your claims via using hsc tools
Lol, you lack understanding. The whole point of my last post wasnt because i specifically wanted a hsc proof for this question, i merely suggested that you present your "solution" in a clear and logical manner (presenting a solution using hsc tools was just one way i suggested in order to achieve this).You present this as a general question without specifying HSC tools only. So I took advantage of the matter of the fact.
Spiral Similarity can be proven using vectors, complex numbers, or projective geometry.
If you wanted a proof that a HSC student could come up with, then I suggest you specify that condition.
don't go brandishing your ego on "HSC" level proofs if you never specified that.
Yeah dw, im also used to people who cant solve a problem and try to cover it up by making a load of excuses such as claiming to merely be "leaving a hole" when they are actually stuck but need to maintain their ego.get used to it sweetie, that's the mathematical culture of BOS, we tend to leave holes that can be filled by the reader, just some of us take it further than others.
Ah, sorry about that, that was my mistake. All euclidean geometry is welcome and fine. No vectors thoughJudging by the presence of the midpoint and etc. I'd have been inclined to combine a few vectorial methods in this question.
When you say only circle geometry do you mean only "Euclidean geometry"? Because circle geometry feels as though (whilst probably pointless) things like alternate angles are denied.
Or are you saying it can be done using ONLY those theorems and no need for the 2U bunch.
Sounds different to what i was expecting, but it seems like legit progress so can you upload your solution? But what did you mean by T?I think I got some of it? If it's right I'll type it all up but it was a long trek
I ended up with proving where n is half CD and m is half AB
edit: in case I'm not here later, my solution was to label the midpoint of CD as N and the midpoint of AB as M. Then call angle XCN alpha and YDN beta. Use the cosine rule on XCN, YDN, XBM, YBM, and then through lots of messy algebra end up with the result I got above
I never said I would get 99.95Not all, but I know many who are. It's the people who think Mathematics can get them 99.95 ATAR with only studying Maths and failing all of the other subjects because Maths is life.