Circle Geo. (1 Viewer)

[CrashOveride]

http://img2.exs.cx/img2/9578/q53.jpg

Thanks.

 

[Jase]

argh..got too messy. ill try again on a fresh sheet.
Where did you get this question from?

 

[ngai]

yeah where did u get this from?
q53...so its question 53 out of a whole heap of questions?

oh btw, have u got ruse 4u trial 1998?

 

[BlackJack]

I have a non-standard solution, I'll post it up in a sec.

EDIT: Solution, in detail/step by step.
Consider a circle of the above description, and extend AB to become the x-axis of the cartesian plane, with the center of the circle as the origin, O. (Therefore the circle is described by x²+y²=r².)

Therefore, Q and R can be described using cartesian coordinates. Construct triangle OQR, and 2 lines connecting Q and R perpendicularly to the x-axis, mark these crossings on the x-axis S & T, respectively.

Note QPS and PRT form triangles with:
PQ = root(2) QS (QS is the y-value of Q in the plane)
PR = root(2) RT. (RT is the y-value of R in the plane)

Also, noting ^BOR is the same angle as the angle btwn OQ and the y-axis, we can express QS and RT in terms of the said angle, say θ

Then QS=r cosθ and RT = r sinθ

Therefore, 2*PQ² + 2*PR²
= 2*( 2*QS² + 2*RT²)
= 4* ( r² (cos²θ + sin²θ ))
= (2r)²
= d² = AB². QED

 

[Rorix]

i recognise this question, i did it with demos in yr10 ngai >_>

 

[CrashOveride]

For some reason the server decided to give it that filename, it was originally "q2.jpg".
It is from my 4unit textbook, from the harder circles exercise question 2 from about 30 or so.

Thanks for that black jack, i also managed to come up wiht a more "standard" solution ...about one page long i can scan it up if anybody is interested?

Ngai: Was that Ruse trial question directed at me? If so, yes i have a collection of JR trials.

 

[tempco]

Yep I'd appreciate it if you posted up the standard solution

 

[Rorix]

(this question was asked in a ruse trial)

 

[CrashOveride]

Also, noting ^BOR is the same angle as the angle btwn OQ and the y-axis, we can express QS and RT in terms of the said angle, say θQUOTE]

How do you show these angles to be equal?

 

[mojako]

Yep I'd appreciate it if you posted up the standard solution

i also managed to come up wiht a more "standard" solution ...about one page long i can scan it up if anybody is interested?

post it!!

 

[CrashOveride]

post it!!

Here it is.


Also:

Also, noting ^BOR is the same angle as the angle btwn OQ and the y-axis, we can express QS and RT in terms of the said angle, say θ

How do you show these angles to be equal?

 

[tempco]

Clicked on the attachment and this popped up:

"The image “http://www.boredofstudies.org/community/attachment.php?attachmentid=5604” cannot be displayed, because it contains errors."

Thanks anyway.

 

[mojako]

when i clicked it it just displayed a small box with "X"
no error message

 

[CrashOveride]

Works here:

http://img32.exs.cx/img32/1277/circleQ2.jpg