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Preliminary mathematics marathon (3 Viewers)

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This marathon ain't so good cos liek most ppl haven't done as much as us, liek my school has only done basic algebra, and geometry.
 
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Liek the only people who have posted is me, you, 2010ers and ppl who have already done the HSC.
 

mitchy_boy

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New question?

LM and FG are two chords of the circle at right angles to each other, and they intersect at P. From P, a line perpendicular to GM is drawn, meeting it at S. SP is produced to meet LF at T.

a) Show that ∆ LTP is isosceles.
b) Show that T is the midpoint of LF.

 

Drongoski

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New question?

LM and FG are two chords of the circle at right angles to each other, and they intersect at P. From P, a line perpendicular to GM is drawn, meeting it at S. SP is produced to meet LF at T.

a) Show that ∆ LTP is isosceles.
b) Show that T is the midpoint of LF.


Let angle LFG = A
and angle FLM = B

,: A + B = 90 deg (since in triangle LPF angle LPF = 90 deg)

.: angle LMG = angle LFG = A (angles on same chord LG)
& angle FGM = angle FLM = B (angles on same chord MF)

.: angle SPM = B (angle PSM = 90 deg)

.: angle LPT = angle SPM = B (vert opp angles)

.: angle TLP = B = angle LPT in triangle LTP

.: triangle LTP is isosceles

Similarly in triangle GPS, angle GPS = A (angle PSG = 90 deg)
.: angle TPF = angle GPS = A (vert opp angles)
.: in triangle TFP, angle TPF = A = angle TFP
i.e TPF is an isosceles triangle
.: PT = TF
but TP = LT
.: PT = TF

i.e. T is mid-point of LF

QED


Edit

Didn't realise this is a marathon question; can someone pl post a question for me.
 
Last edited:

mitchy_boy

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Let angle LFG = A
and angle FLM = B

,: A + B = 90 deg (since in triangle LPF angle LPF = 90 deg)

.: angle LMG = angle LFG = A (angles on same chord LG)
& angle FGM = angle FLM = B (angles on same chord MF)

.: angle SPM = B since angle PSM = 90 deg

.: angle LPT = angle SPM = B (vert opp angles)

.: angle TLP = B = angle LPT in triangle LTP

.: trangle LTP is isosceles

Similarly in triangle GPS, angle GPS = A since angle PSG = 90 deg
.: angle TPF = angle GPS = A (vert opp angles)
.: in triangle TFP, angle TPF = A = angle TFP
i.e TPF is an isosceles triangle
.: PT = TF
but TP = LT
.: PT = TF

i.e. T is mid-point of LF

QED


Edit

Didn't realise this is a marathon question; can someone pl post a question for me.
Woah that was ext2 harder ext1 question. I don't have time to go through now but I think you might have hit it on the head. Nice work man.
 

Trebla

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Very tempting to do that one above using logarithmic differentiation....

Have you Prelimers done series? I have a nice series question here:



Otherwise, another one from differential calculus:

The tangent to the hyperbola xy = k at the point P(x1, y1) cuts the x-axis at X and the y-axis at Y. Show that the area of triangle OXY is constant.
 

hscishard

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Very tempting to do that one above using logarithmic differentiation....

Have you Prelimers done series? I have a nice series question here:



Otherwise, another one from differential calculus:

The tangent to the hyperbola xy = k at the point P(x1, y1) cuts the x-axis at X and the y-axis at Y. Show that the area of triangle OXY is constant.
I'll try the second one.
f'(x) = -k/x^2

Therefore tangent is

y-intercept:

x intercept


Area of triangle = 1/2AB
B is perp height.


k is a constant and points of P(x1,y1) are also constants.
Therefore area is constant?

Edit: Oh For the first one, I think S = 7/4 (calculator LOL)
 
Last edited:

nikkifc

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I'll try the second one.
f'(x) = -k/x^2

Therefore tangent is

y-intercept:

x intercept


Area of triangle = 1/2AB
B is perp height.


k is a constant and points of P(x1,y1) are also constants.
Therefore area is constant?
P is not a constant lol. You need to use the condition (ie. xy=k) since you are not using parametrics.

And Trebla, this is a typical 4U conics question isn't it? It's a bit mean asking it to the prelimers :p
 
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Otherwise, another one from differential calculus:

The tangent to the hyperbola xy = k at the point P(x1, y1) cuts the x-axis at X and the y-axis at Y. Show that the area of triangle OXY is constant.
I was thinking of like a logical approach...

A = 1/2 ab

height = y
length = x

xy=k

therefore A = 1/2 k

since k is constant, area is constant

amirite?
 

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