• Want to help us with this year's BoS Trials?
    Let us know before 30 June. See this thread for details
  • Looking for HSC notes and resources?
    Check out our Notes & Resources page

Merry Christmas from Math Man (1 Viewer)

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
I drew my diagram differently, on the first quadrant. Oh and by the way math man. What program did you use to draw that diagram. I want to know. :)
 
Last edited:

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
hmm..well since the question said "if they are cyclic" it works out to be 90...if you think otherwise show me your working
 

IamBread

Member
Joined
Oct 24, 2011
Messages
757
Location
UNSW
Gender
Male
HSC
2011
Working is up there ^

Sorry about them being pictures, unfortunately it seems to be the best I can do :/
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
where did you get angle DBC = pi/8 ...btw your approach to this question is not the best...the question said "if they are cyclic..." meaning you use circle geo to do it
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
i saw it...dont worry...well the question only works because i said they are cyclic...even though they may not be cyclic since the question told you to assume they are you use that fact to prove it
 

IamBread

Member
Joined
Oct 24, 2011
Messages
757
Location
UNSW
Gender
Male
HSC
2011
Triangle ODB is isosceles, with COB = pi/2, therefore OBC = pi/4. Line OA bisects angle COB, so angle AOB = pi/8. OBAC is a rhombus so diagonals cut at pi/2, using angle sum of a triangle you get angle OBC = 3pi/8.
OBC = OBD + DBC,
DBC = 3pi/8 - pi/4 = pi/8

This may not be the best method, though it does show it can't be done.
 

IamBread

Member
Joined
Oct 24, 2011
Messages
757
Location
UNSW
Gender
Male
HSC
2011
Ah yes it did, I misread the question, I thought it asked to show that they were cyclic, then prove that thing. Seems like an odd question though.
 

seanieg89

Well-Known Member
Joined
Aug 8, 2006
Messages
2,662
Gender
Male
HSC
2007
Just seems a bit silly to assume something like those points being concyclic. From that assumption we can also prove things like:

-any two real numbers are equal
-triangles have angle sum 1,000,000
-any two functions are the same
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Of course MathMan, it's excellent that you wish to develop your student's thinking abilities with such questions.

However, it would be nice if you verified that the question works, just to avoid future confusion and shenanigans like this =)

But please, do post more problems! We would love to attempt them!
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
well here is another complex number/circle geo question i wrote...dw this one makes no silly assumptions

circlecomplex.png
 

lydia1992

scientificfield
Joined
Oct 7, 2011
Messages
52
Location
australia
Gender
Female
HSC
2012
Uni Grad
2016
:santa:MERRY CHRISTMAS :D
wish you all a good year full of successes
may God bless you
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
well here is another complex number/circle geo question i wrote...dw this one makes no silly assumptions

View attachment 23992
I just managed to 'prove' that Arg(z) + Arg(w) = 90 degrees. I have probably made an arithmetic error somewhere to achieve something silly like this, so I think this implies bed time.

Will be continuing this tomorrow morning.
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
I just managed to 'prove' that Arg(z) + Arg(w) = 90 degrees. I have probably made an arithmetic error somewhere to achieve something silly like this, so I think this implies bed time.

Will be continuing this tomorrow morning.
yeh that is silly..but the question only took me 10 secs to do so i dont think it is bed time just yet..yet again i did make the question :p
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top