(another) past ruse Q .. hmm (2 Viewers)

[mreditor16]

anyway, anyone have a nice / quick / neat way of doing part vii? the ruse solutions doesn't have an answer for this Q unfortunately.

so I was wondering if anyone has a nice solution for it?

thanks

here is the Q

 

[mreditor16]

anyone???

 

[mreditor16]

.... :O

 

[mreditor16]

bumpity bump

 

[Blueblade11]

I can't be bothered to write anything out or find B, C or D, but I'm assuming you can do the following:

- Prove AB = BC = CD = AD
- Prove AB is perpendicular to BC or CD.
- Therefore, square. (equal sides, right angle.)

 

[Kurosaki]

poor mreditor LOL

 

[Kurosaki]

It looks like a lot of tedious algebra crunching =( and coordinate geometry to prove one of the properties Blueblade11 said. Can't seen any super elegant way at the moment.

 

[Immortality]

I remember this question, reminded me of 2010's integration spam

 

[mreditor16]

I remember this question, reminded me of 2010's integration spam

thanks for the valuable contribution.

 

[Immortality]

anyway, anyone have a nice / quick / neat way of doing part vii? the ruse solutions doesn't have an answer for this Q unfortunately.

so I was wondering if anyone has a nice solution for it?

thanks

here is the Q

Basically what Ruse did was a shorter variation of another question that asked you to prove it was a parallelogram/rhombus before that (I think it was an Independent Paper Question can't remember).

But essentially if i remember quickly, the fastest way to prove this was to prove rhombus (diagonals perpendicular to each other) followed by proving 1 pair of adjacent sides was perpendicular too, thus making a rhombus with one right angle a square.

 

[IR]

Why not use complex numbers?

 

[mreditor16]

Basically what Ruse did was a shorter variation of another question that asked you to prove it was a parallelogram/rhombus before that (I think it was an Independent Paper Question can't remember).

But essentially if i remember quickly, the fastest way to prove this was to prove rhombus (diagonals perpendicular to each other) followed by proving 1 pair of adjacent sides was perpendicular too, thus making a rhombus with one right angle a square.

sounds quicker than what the others suggested - so i'll try that...

Why not use complex numbers?

ceebs, besides I'd have to set it up myself.

------------------

thanks everyone for the help

 

[Tugga]

anyway, anyone have a nice / quick / neat way of doing part vii? the ruse solutions doesn't have an answer for this Q unfortunately.

so I was wondering if anyone has a nice solution for it?

thanks

here is the Q

The tangents are perpendicular to each other (this is apparent when using point gradient to find their equations). Squares have diagonals that are perpendicular and bisect (I think), so you only need to prove that the midpoint is the same?

Also, this midpoint has to be the point P.

 

[Kurosaki]

The tangents are perpendicular to each other (this is apparent when using point gradient to find their equations). Squares have diagonals that are perpendicular and bisect (I think), so you only need to prove that the midpoint is the same?

Also, this midpoint has to be the point P.

that only proves it's a rhombus though I think.

 

[Tugga]

that only proves it's a rhombus though I think.

Oh hmm, yeah true! My bad, it would probably be easiest to find the adjacent sides equal then (or that the distance from P is the same if that makes it easier algebra-wise)

 

[IR]

Editor what paper was that? Which JRAHS Paper?

 

[mreditor16]

Editor what paper was that? Which JRAHS Paper?

James Ruse 2011 4U Trial

 

[IR]

James Ruse 2011 4U Trial

Thanks.

 

[Blueblade11]

I just figured an easier way to prove, for anyone interested:

-Prove AC = BD (ie diagonals equal length)
-Prove m(AC) * m(BD) = -1 (ie perpindicular.)

Thus, Square.

...Well, if anyone's still wondering.

 

[mreditor16]

I just figured an easier way to prove, for anyone interested:

-Prove AC = BD (ie diagonals equal length)
-Prove m(AC) * m(BD) = -1 (ie perpindicular.)

Thus, Square.

...Well, if anyone's still wondering.

hmmm

I don't think equal and perpendicular diameters is enough to prove it is a square.

I think you also have to prove that the diagonals bisect each other as well.