Circles (1 Viewer)

[lil tiger]

a lil question

Lines are drawn on a clock face, joining the numberals 12 with 7 and 3 with 10. Show that these lines are perpendicular

 

[Giant Lobster]

heh... join the lines to the centre of the clock. ull know every interval subtends 30 degrees at the centre. work it out from there

 

[~*HSC 4 life*~]

thats an odd question!

 

[lil tiger]

i've tried connecting the lines to the centre but i can't establish a relationship to find the angles which they intersect (which is suppose to be 90)

 

[Grey Council]

hmm, that IS rather strange. try this:

put both lines onto a set of co-ordinate axis. ie, draw it up using that x-y axis graph.

oh shit. this is strange. You can't get the equation of either line, can you?

...

wtf, still can't think of a solution.

let me mull it over, i'll see if i can post up the solution tommorow.

 

[lil tiger]

Hehe, the strangeness is what strikes me. Guess my lateral thinking ability this is mediocre. I tried to draw a diagram and attach it, but it was too big(i did it in paint).

 

[lil tiger]

Actually there's another two i don't get.

1. Prove that if two tangents are parallel then the line joining their points of contact is a diameter
2. If a parallelogram is circumscirbed about a circle (ie. each side is a tangent) then by listing equal segments or otherwise show that the parallelogram is a rhombus.

 

[Grey Council]

ok, this is one dodgy answer, but here goes:
draw up a circle
mark it like a clock
draw lines 12-6, 12-7, 3-9, 3-10, 6-7, 9-10
see how 12-6 and 3-9 are diameteres of a semi-circle? so 6-7-12 is a 90 degree angle, and so is 9-10-3
now note that 6-12-7 = 9-3-10 (obviously, its a darn clock)
also note that 6-12 and 9-3 meet at 90 degrees. ie perpendicular.
so now notice that 9-3-10 has been rotated by 90 degrees.
therefore 3-10 has been rotated 90 degrees about 0 (centre of clock)
therefore 12-7 and 3-10 are perpendicular.

yes, i know, that is the dodgiest proof ever. But its the first one that came to mind after I drew it up. sorry, will do one more polished in the morning. Its 10:30

 

[Grey Council]

oh the first one is easy.

there is a rule that you are allowed to state in the HSC that says that perpendiculars to tangents to a circle pass through the centre. Therefore any two parallel tangents' perpendiculars will meet at the centre, forming a straight line.

 

[Grey Council]

second one also seems easy. here goes:

use the fact that any two tangents drawn to a circle from one external point makes both tangents equal from point of contact to point. gee, how vague. Hope you see what i mean.
and you know they are parallel. therefore opposite sides are parallel and equal. go figure it out, very easy

 

[lil tiger]

good effort and yes its a lil dodgy, especially at the end where is says therefore 3-10 has been rotated 90 degrees about 0 (centre of clock)
therefore 12-7 and 3-10 are perpendicular.
but yeah its a dodgy question in the first place

 

[lil tiger]

Originally posted by GuardiaN
oh the first one is easy.

there is a rule that you are allowed to state in the HSC that says that perpendiculars to tangents to a circle pass through the centre. Therefore any two parallel tangents' perpendiculars will meet at the centre, forming a straight line.

can you just say this? i understand but i don't think its clear enuf

 

[Grey Council]

your allowed to say it. why not? its a rule that you are allowed to state. Just word it differntly if you want, but thats the gist of it.

 

[Grey Council]

sorry, as i said, VERY dodgy proof. and the point you picked up is the one i was fearing. Its called fudging by some. teeheehee, but technically its sound. Just how do you state it in a way that sounds pro? hmph

and thats an interesting question. ^__^

 

[lil tiger]

Originally posted by GuardiaN
second one also seems easy. here goes:

use the fact that any two tangents drawn to a circle from one external point makes both tangents equal from point of contact to point. gee, how vague. Hope you see what i mean.
and you know they are parallel. therefore opposite sides are parallel and equal. go figure it out, very easy

erm, yes and no, i need to prove one pair of sides equal but i can't

 

[lil tiger]

Originally posted by GuardiaN
your allowed to say it. why not? its a rule that you are allowed to state. Just word it differntly if you want, but thats the gist of it.

don't you have to mention the fact that the two perpendiculars to form a straight line, they must be co-interior. Then you have to mention that they are co-interior by mentioning that the angles they make(to the perpendicular) sum to 180. then you say what you said earlier? wrong?

 

[Grey Council]

mmm

why can't you just say that perpendiculars from two lines meet at one point. Both lines are parallel. Hence perpendiculars form straight line.

Or is there a logic error in that? :S

 

[lil tiger]

Its probably right. Just one of the things i want to make sure

 

[Xayma]

GuardiaN you could say that since the gradients are equal ie (m₁=m₂) and the gradients of a perpendicular line to the other line is m₃m₄=-1 the perpendicular lines will have equal gradient so when they do touch they will form an angle of 180 degrees (as shown by the tan theta formula for an angle between two lines)

 

[Grey Council]

hrmph. In circle geo I usually say:
"becuase it has to be"
lol, my teacher wrote that in my report. He commended my logic but I need to think analytically instead of saying:
"Becuase it has to be"

lol.

But do what Xayma said. I think i'll save myself any more embarrasment.