Hard Circle geometry question (1 Viewer)

Unravel

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As expected, on that page there is nothing to support your claim that if "Each of the corresponding vertices are the midpoints of the exterior vertices" then the figures are similar

If you disagree, provide your full solution
 

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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
 

Paradoxica

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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
No, 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.
 

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No, 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.
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
 
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Paradoxica

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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
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.
 

Unravel

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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.
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).

I mean, i suppose i could do what you did and whenever im presented with a problem i could just say something along the lines of " oh by the theory of addition this is true. I have no need to justify myself, im sure this theory and the problem is linked somehow, its up to you to make the links haha i totally solved the problem".

So, explain your steps in a clear and logical manner.
 

Paradoxica

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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.
 

Unravel

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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.
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.

Dont feel too bad about not being able to do it though, even some ruse and sydney boys students in my class couldnt do it


The problem still remains open, if you want solutions provided by the question maker, you can pm
 
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leehuan

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Judging 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.
 

Unravel

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Judging 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.
Ah, sorry about that, that was my mistake. All euclidean geometry is welcome and fine. No vectors though
 

si2136

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why are all the maths kids so egotistical lel
Not 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.
 

KingOfActing

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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 not 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

edit2: nevermind -eternal sigh-
 
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Unravel

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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
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?
 

KingOfActing

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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?
AY, sorry. Typo'd.

I rewrote it now and noticed I made a mistake in one step, so it doesn't lead to that conclusion. Oopsies.
 

Paradoxica

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Not 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.
I never said I would get 99.95

And FYI I am aware of the propensity of me not getting even 70 tyvm.

I have realistic expectations.
 

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