Elegance Revisited (1 Viewer)

[OLDMAN]

We had a thread last year on elegance -Spice Girl and ND were the main correspondents.

To do well in Ext. 2, students need to develop their EQ elegance quotient.

Here are three problems to practise on, the first two could easily be done by a Year 11 doing the preliminary, the third - well, lets just say it could happily be embedded in a Question 8 Ext2. But all three share things in common (indeed most maths problems) , multiple approaches -choose the most elegant.


1) P lies on 8y = 15x. Q lies on 10y = 3x and the midpoint of PQ is (8,6). Find distance PQ.

2) A line through the origin divides the parallelogram with vertices (10,45), (10,114), (28,153), (28,84) into two congruent pieces. Find its slope.

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.

 

[ND]

Am i missing something or is 3) just ...<R(a, a), Q(a, 0)>?
<ok i should probably post up my soln:
when R(a,a) and Q(a, 0), PQ+RQ+PR=2RQ, which is the minimum cos in a triangle the sum of two sides is greater than or euqal to the third: RP+PQ>=RQ.>

edit: Hidden.

 

[Affinity]

nope

 

[Affinity]

(a,a) and (a,0) doesn't work..Suppose P is a point very close to (a,0)...

 

[ND]

Ok is it R((a+b)/2, (a+b)/2) Q(a+b, 0)?

edit: duh that's obviosuly wrong.

edit2: my last guess (before i go out) is R((a+b)/2, (a+b)/2) Q(a, 0). Heh.

 

[Affinity]

nope

 

[santazlilhelpa]

Q (a,o)
R [(a+b)/2, (a+b)/2]

???
just browsing through

 

[:: ck ::]

for a)
i got the answer... itll be a bit annoying to type on comp... the distance should be 36 and 36/49 units

 

[OLDMAN]

__________________________________________________
for a)
i got the answer... itll be a bit annoying to type on comp... the distance should be 36 and 36/49 units -ryan.cck
_________________________________________________

your answer's too large. Would be interested in a brief description of your approach though.

 

[CM_Tutor]

Originally posted by :: ryan.cck ::
for a)
i got the answer... itll be a bit annoying to type on comp... the distance should be 36 and 36/49 units

That's way too big, IMO. Look at a diagram. Both lines are only in quadrants 1 and 3, and for (8, 6) to be the midpoint, P and Q must both be in the first quadrant. Your answer has both P and Q more than 15 units from (8, 6), but the only points on the lines that far from (8, 6) can't possibly have (8, 6) as a midpoint. (BTW, FWIW I got
PQ = 60 / 7 units in a back-of-an-envelope calculation.)

Edit: Oldman and I were clearly typing at the same time...

 

[:: ck ::]

oh CRAP I FORGOT TO SQUARE ROOT coz i was doin usin calculator and a small post it note.. lol

i got coordinates for P = 4/4/7, 8/4/7

Q = 11/3/7, 3/3/7

use distance formula u get 8/4/7 = 60/7 same as CM_Tutor

scanning it now

 

[CM_Tutor]

OK, so you've now got the correct answer (which is good), but was the approach as elegant as it might have been? For example, did you really need to find the coordinates of P and Q? I only actually found the coordinate of P...

 

[nike33]

hehe i did it using vectors, ill post 2morrow when i get on main comp with scanner

 

[:: ck ::]

rarr..vectors .. hardcore ><"

heres my unelegant attempt at 1...

 

[CM_Tutor]

:: ryan.cck ::, your method can certainly be shortened, as it's better to avoid 4 variables. Here is an abbreviated version of my approach...

P lies on 8y = 15x, so let P be at (x₀, y₀ = 15x₀ / 8), and similarly Q (on 10y = 3x) is at (x₁, y₁ = 3x₁ / 10)

Now, (8, 6) is the midpoint of PQ, so examining x co-ordinates gives x₀ + x₁ = 2 * 8 = 16 ===> x₁ = 16 - x₀ ___(*)

Y co-ordinates tell us that 5 * 15x₀ + 4 * 3x₁ = 12 * 40, which, using (*), gives x₀ = 32 / 7, and so y₀ = 60 / 7

Now, PQ² = (x₀ - x₁)² + (y₀ - y₁)² = (2x₀ - 16)² + (2y₀ - 12)² = 4(x₀ - 8)² + 9[(5x₀ / 4) - 4]², again using (*)

Simplifying this, with x₀ = 32 / 7 and noting PQ > 0, gives PQ = 60 / 7 units

(On reflection, since we know that P is (32 / 7, 60 / 7), it would be quicker to find the distance from P to (8, 6), and then double it.)

Note: I'm pretty sure there is a better way to do this, perhaps using the fact that OQRP is a parallelogram, where
R is (16, 12) ...

 

[OLDMAN]

Short, shorter, shortest. Maths' such a perverse world .

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). The distance of P to (8,6) is
sqrt((24/7)^2+(18/7)^2) = 30/7. Hence PQ = 60/7.

 

[:: ck ::]

ahh crap that seems much more easier.. lol

 

[nike33]

ahh nice soln err i ment oldman lol

 

[:: ck ::]

umm for 2.. is the answer 5/4/19 ?

i dont wanna embarass myself like last time.. lolz...

 

[CM_Tutor]

Originally posted by :: ryan.cck ::
umm for 2.. is the answer 5/4/19 ?

Yes, I got slope = 99 / 19

PS: To All, Oldman's solution to 1 applies the intercept theorems to the parallelogram OQRP that I mentioned above - It's a good example of how the intercept theorems can make your life easier...