HSC 2013 MX2 Marathon (archive) (1 Viewer)

iBibah · Jan 10, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Ok first we want to know what this looks like, it is an argand diagram with complex numbers $z$ and $w$ .

Now since it is an equilateral triangle, $z=w$ due to all sides being the same length. Hence $z^2+w^2=z^2+z^2=2z^2$

Now let $z=x+iy$ , then by substitution we get:

$2z^2 = 2(x+iy)^2 = 2(x^2-y^2+2xyi)$

So we end up with:

$z^2 + w^2 = 2z^2 = 2(x^2-y^2+2xyi)$

A really big thanks to your contributions for helping me and other students for our upcoming half yearlies!

I'm really looking forward to future questions that you have to offer!

hahaha

iBibah · Jan 10, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
Ok first we want to know what this looks like, it is an argand diagram with complex numbers $z$ and $w$ .

Now since it is an equilateral triangle, $z=w$ due to all sides being the same length. Hence $z^2+w^2=z^2+z^2=2z^2$

Now let $z=x+iy$ , then by substitution we get:

$2z^2 = 2(x+iy)^2 = 2(x^2-y^2+2xyi)$

So we end up with:

$z^2 + w^2 = 2z^2 = 2(x^2-y^2+2xyi)$

A really big thanks to your contributions for helping me and other students for our upcoming half yearlies!

I'm really looking forward to future questions that you have to offer!

hahaha

twinklegal19 · Jan 10, 2013

Re: HSC 2013 4U Marathon

A nice circle geo question

$\\$Two circles with centres O and P and radii r and s (where r $<$ s) respectively touch externally at T. ABC and ADE are common tangents to the two circles.\\\\a) Show that A, O, T and P are collinear\\b) Show that AO= $\frac{r(r+s)}{s-r}$

twinklegal19 · Jan 10, 2013

Re: HSC 2013 4U Marathon

A nice circle geo question

$\\$Two circles with centres O and P and radii r and s (where r $<$ s) respectively touch externally at T. ABC and ADE are common tangents to the two circles.\\\\a) Show that A, O, T and P are collinear\\b) Show that AO = $\frac{r(r+s)}{s-r}$

GoldyOrNugget · Jan 10, 2013

Re: HSC 2013 4U Marathon

Omed62 said:
Thanks RealiseNothing, your one of the best guys out their, so nice and keen on helping students.

I hope god be with you always... lol could you please answer the probability questions... please

no need, they're both correct

cutemouse · Jan 11, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
A nice circle geo question

$\\$Two circles with centres O and P and radii r and s (where r $<$ s) respectively touch externally at T. ABC and ADE are common tangents to the two circles.\\\\a) Show that A, O, T and P are collinear\\b) Show that AO = $\frac{r(r+s)}{s-r}$

Is this from Cambridge?

twinklegal19 · Jan 11, 2013

Re: HSC 2013 4U Marathon

cutemouse said:
Is this from Cambridge?

dunno, got it from a maths revision seminar lol

(wouldn't be surprised though)

RealiseNothing · Jan 11, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
A nice circle geo question

$\\$Two circles with centres O and P and radii r and s (where r $<$ s) respectively touch externally at T. ABC and ADE are common tangents to the two circles.\\\\a) Show that A, O, T and P are collinear\\b) Show that AO= $\frac{r(r+s)}{s-r}$

I'll post my solution if no one answers it, cbf'd right now.

iBibah · Jan 11, 2013

RealiseNothing said:
I'll post my solution if no one answers it, cbf'd right now.

Haha it took you a while to explain it today, let alone writing/typing it.

Sy123 · Jan 11, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
A nice circle geo question

$\\$Two circles with centres O and P and radii r and s (where r $<$ s) respectively touch externally at T. ABC and ADE are common tangents to the two circles.\\\\a) Show that A, O, T and P are collinear\\b) Show that AO= $\frac{r(r+s)}{s-r}$

Woah, I can't picture anything with that information heh.

Any chance of a diagram?

RealiseNothing · Jan 11, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Woah, I can't picture anything with that information heh.

Any chance of a diagram?

The question came without a diagram iirc, but if some one can be bothered posting one up then I guess so.

SpiralFlex · Jan 11, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
A nice circle geo question

$\\$Two circles with centres O and P and radii r and s (where r $<$ s) respectively touch externally at T. Let A be an external point. ABC and ADE are common tangents touching the circles P and O respectively.\\\\a) Show that A, O, T and P are collinear\\b) Show that AO = $\frac{r(r+s)}{s-r}$

Twinkle's question rephrased for Sy.

iBibah · Jan 11, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Woah, I can't picture anything with that information heh.

Any chance of a diagram?

SpiralFlex · Jan 11, 2013

Re: HSC 2013 4U Marathon

Guys just a tip, if a question is too easy or you have seen it before don't answer it, give others a go first then guide them if necessary. A few people have PMed me saying they are scared of this thread. We want more people.

Sy123 · Jan 11, 2013

Re: HSC 2013 4U Marathon

SpiralFlex said:
Twinkle's question rephrased for Sy.

Ok thats good, thank you.

GoldyOrNugget · Jan 11, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
A nice circle geo question

$\\$Two circles with centres O and P and radii r and s (where r $<$ s) respectively touch externally at T. ABC and ADE are common tangents to the two circles.\\\\a) Show that A, O, T and P are collinear\\b) Show that AO= $\frac{r(r+s)}{s-r}$

a)
∠ABO = ∠ACP = ∠ADO = ∠AEP = 90deg (angle of radius and tangent = 90 deg)
therefore OB || PC (corresponding angles)
therefore ∠AOB = ∠APC (corresponding angles), let this be θ
∠ABP = 180 - ∠AOB = 180 - θ (angles on a straight line are supplementary)
AOT = ∠AOB + ∠BOP = θ + (180 - θ) = 180deg therefore A,O,T are collinear

construct OT, TP
the angles of the tangent to both circles through T, and OT and TP are 90 deg (angle of radius and tangent = 90 deg)
so O,T,P are collinear

if A is collinear to O,T, and also P is collinear to O,T, then A,O,T,P are collinear

b) too tired to work it out, but im pretty sure just use pythagoras and that weird circle geometry rule involving tangents and chords where everything gets squared

twinklegal19 · Jan 11, 2013

Re: HSC 2013 4U Marathon

GoldyOrNugget said:
a)
∠ABO = ∠ACP = ∠ADO = ∠AEP = 90deg (angle of radius and tangent = 90 deg)
therefore OB || PC (corresponding angles)
therefore ∠AOB = ∠APC (corresponding angles), let this be θ
∠ABP = 180 - ∠AOB = 180 - θ (angles on a straight line are supplementary)
AOT = ∠AOB + ∠BOP = θ + (180 - θ) = 180deg therefore A,O,T are collinear

construct OT, TP
the angles of the tangent to both circles through T, and OT and TP are 90 deg (angle of radius and tangent = 90 deg)
so O,T,P are collinear

if A is collinear to O,T, and also P is collinear to O,T, then A,O,T,P are collinear

b) too tired to work it out, but im pretty sure just use pythagoras and that weird circle geometry rule involving tangents and chords where everything gets squared

you just assumed AOP to be a straight line which is what you are trying to prove

GoldyOrNugget · Jan 11, 2013

Re: HSC 2013 4U Marathon

twinklegal19 said:
you just assumed AOP to be a straight line which is what you are trying to prove

derp. it's too late at night for this shit haha. someone else do it

Sy123 · Jan 11, 2013

Re: HSC 2013 4U Marathon

Ok, for the first one, I don't know if my proof is allowed, but:

Consider the triangles ABO and ACP
Now, since radius perpendicular to tangent,

$\angle ABO = \angle ACP \ \ \ ($radius perpendicular to common tangents$)$

$\angle BAO = \angle CAP \ \ \ ($common angle$)$

$\angle BOA = \angle CPA \ \ \ ($angle sum of triangle$)$

$\therefore \ \ \triangle ABO ||| \triangle ACP \ \ \ $(equiangular$)$

Now, since both triangles have common vertice A, and ABC is collinear, therefore it must come about that AOP is collinear

But due to the theorem of intersecting circles thing (not sure the proper name), we already know that OTP is collinear

Hence it must follow that A, O, T, P is collinear.

EDIT: I can make it more rigorous by applying the same arguments for the triangles opposite the ones that I proved it for.
==========

Before I continue, I must know whether this proof is correct.

Thanks

iBibah · Jan 11, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Ok, for the first one, I don't know if my proof is allowed, but:

Consider the triangles ABO and ACP
Now, since radius perpendicular to tangent,

$\angle ABO = \angle ACP \ \ \ ($radius perpendicular to common tangents$)$

$\angle BAO = \angle CAP \ \ \ ($common angle$)$

$\angle BOA = \angle CPA \ \ \ ($angle sum of triangle$)$

$\therefore \ \ \triangle ABO ||| \triangle ACP \ \ \ $(equiangular$)$

Now, since both triangles have common vertice A, and ABC is collinear, therefore it must come about that AOP is collinear

But due to the theorem of intersecting circles thing (not sure the proper name), we already know that OTP is collinear

Hence it must follow that A, O, T, P is collinear.

EDIT: I can make it more rigorous by applying the same arguments for the triangles opposite the ones that I proved it for.
==========

Before I continue, I must know whether this proof is correct.

Thanks

Your second line (angle BAO = angle CAP) assumes that A, O, T and P are collinear.

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