Elegance Revisited (1 Viewer)

[:: ck ::]

ahah yay... i dunno if some of my working is necessary or is as elegant.. but here goes ... [ignore the stuff in the square]

i kinda had a half guess as to how to make two congruent figures in the first place tho...

editz : oh crap 3 looks hard...

 

[CM_Tutor]

I did question 2 by letting the line in question be y = mx, and noting that it meets one side at (10, 10m) and the other at (28, 28m). To cut the parallelogram into two congruent parts, the distance from (10, 45) to (10, 10m) must be the same as the distance from (28, 28m) to (28, 153).

So, 10m - 45 = 153 - 28m, and hence m = 99 / 19

 

[:: ck ::]

woah... outclassed... haha ... thats way more elegant ... nice way to think of the problem~!

 

[OLDMAN]

Sorry guys, can't get classier than CM_tutor's.

 

[:: ck ::]

post up some more problems! >.<

 

[DcM]

Originally posted by OLDMAN
Line through (8,6) parallel to OQ is 3x - 10y +36 = 0. This meets OP at (16/7, 30/7) which should also be the midpoint of OP.
Thus P is (32/7,60/7).

y does the parallel line cut thru OP at midpt?

 

[CM_Tutor]

Originally posted by DcM
y does the parallel line cut thru OP at midpt?

Intercept theorems applied to parallelogram OQRP, where R is (16, 12).

Take the point (8, 6) as X, and let the line through X parallel to OQ meet OP at Y.

By intercept theorems, PY / YO = PX / XQ.

But X is the midpoint of PQ, and so PX = XQ, and thus PY = YO. Thus, Y is the midpoint of OP.

 

[nike33]

"3) Let P be the point (a,b) with 0 < b < a. Find Q on the x-axis and R on y=x, so that PQ+QR+RP is minimized."


my attempt at finding Q.. got to

Q( b[sqr(2) - 1] + a, 0 ) ill attempt R tommorow..is this right???

 

[:: ck ::]

working

arghh ill scan wot i got... its prolli totally wrong ... =.="

 

[nike33]

if someone can confirm its right ill post working for Q and R

 

[OLDMAN]

Originally posted by :: ryan.cck ::
post up some more problems! >.<

In a similar vein to the other questions : a car travels at 2/3 km/min due east. A circular storm, w/ radius 51, starts with its center 110 kms due north of the car and travels southeast at 1/sqrt(2) km/min. The car enters the circle of the storm at time t1 and leaves at t2 (mins). Find (t1+t2)/2.

___________________________________________________
nike33:
Q( b[sqr(2) - 1] + a, 0 ) ill attempt R tommorow..is this right???
___________________________________________________

Not right.

 

[:: ck ::]

umm is the distance supposed to be

<ans : sqrt[(b-a)^2 + (a+b)^2] >

ans hidden : quote to c

 

[OLDMAN]

Originally posted by :: ryan.cck ::
umm is the distance supposed to be

<ans : sqrt[(b-a)^2 + (a+b)^2] >

ans hidden : quote to c

Yes ryan.cck. Show us how, so we can see if its got class.


*** also note, I've edited the storm problem***

 

[SeDaTeD]

For the storm question,
let the intial coordinates of the car and cantre of the storm be C(0,0) and S(0,110) respectively.
As parametrics of time, the coordinates in terms of t will be: C(2t/3,0) and S(t/2, 110 - t/2) since the storm is moving SE at 1/sqrt(2) km/min.
Now to form a quadratic equation in t where CS = 51.
(too messy to have a big square root so i'll start off already squared)
(2t/3 - t/2)^2 + (110 - t/2)^2 = 51^2
t^2/36 + 12100 - 110t + t^2/4 = 2601
5t^2/18 -110t + 9499 = 0
t1 and t2 will be the roots of this equation.
Since we are only looking for (t1 + t2)/2, we can use -b/2a
= 110/2(5/18) = 198 min
If you want to find the times when it enters or exits, solve the quadratic.

 

[CM_Tutor]

SeDaTeD, two points - 1 major, 1 minor

1. (Minor) - you really should define S as the position of the centre of the storm.

2. (Major, but not hard to fix) - If S is initially (0, 110), then shouldn't the coordinates of S at time t be
(t / 2, 110 - t / 2), in which case

CS = 51
(2t/3 - t/2)^2 + (100 - t/2)^2 = 51^2
t^2/36 + 10000 - 100t + t^2/4 = 2601
5t^2/18 -100t + 7399 = 0
t1 and t2 will be the roots of this equation.
Since we are only looking for (t1 + t2)/2, we can use -b/2a
= 100/2(5/18) = 180 min

is wrong, and the answer is actually 198 min.

 

[SeDaTeD]

Woops, i didn't notice that. Yes you are correct, i accidently put 100 in. Don't know what I was doing. .

Cheers.

 

[:: ck ::]

aight here it goes... i got a hint from a family member on how to do this tho.. a very BIG hint.. so yeh hah... newaiz heres wot i got for 3

 

[nike33]

so what did you get for Q and R ?? (cant access it )

 

[:: ck ::]

theres no need to find Q and R <-- rar stupid

umm how come u cant access???

if u want ill upload it and give u the link...


the working is pretty wonky tho... all over the place.. hah.. if u dun understand i can try clarify

http://www.members.iinet.com.au/~rchiu/bos/q3.jpg


edit : BAH CRAP i feel like an idiot i only did half the question... T_T

 

[CM_Tutor]

Originally posted by :: ryan.cck ::
theres no need to find Q and R

I like the method, and you have certainly shown that the minimum distance is sqrt[2(a² + b²)], but there is a need to find Q and R to answer Oldman's question, as it was

Originally posted by OLDMAN
3) Let P be the point (a,b) with 0 < b < a. Find Q on the x-axis and R on y=x, so that PQ+QR+RP is minimized.

and not "Find the minimum possible value of PQ + QR + RP".

Edit: On reflection, it's unfortunate that the question wasn't just to find the minimum, as it's nice that it can be done without actucally finding the points Q and R. Oh, well...