HSC 2014 MX2 Marathon (archive) (1 Viewer)

dunjaaa · Jul 27, 2014

Re: HSC 2014 4U Marathon

Screen shot 2014-07-27 at 12.44.44 AM.png

An alternative solution that avoids substitution.

Sy123 · Jul 28, 2014

Re: HSC 2014 4U Marathon

$\\ $If$ \ x,y,z \ $are non-negative real numbers, show that$ \\ \\ (x^2+1)(y^2+1)(z^2+1) \geq (x+y+z- xyz)^2$

mathsbrain · Jul 28, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $If$ \ x,y,z \ $are non-negative real numbers, show that$ \\ \\ (x^2+1)(y^2+1)(z^2+1) \geq (x+y+z- xyz)^2$

$$Given$\ x,y,z \geq 0\ ,\\ $R.T.P.$\ (x^2+1)(y^2+1)(z^2+1)-(x+y+z- xyz)^2 \geq 0\\ $LHS$=x^2y^2+x^2z^2+y^2z^2-2xy-2xz-2yz+2x^2yz+2xy^2z+2xyz^2+1\\ $But$\ (xy+xz+yz)^2=(x^2y^2+x^2z^2+y^2z^2)+2x^2yz+2xy^2z+2xyz^2\\ \therefore \ $LHS$=(xy+xz+yz)^2-2(xy+xz+yz)+1=(xy+xz+yz-1)^2\geq0\\ $with equality iff$\ xy+xz+yz=1$

mathsbrain · Jul 28, 2014

Re: HSC 2014 4U Marathon

Without letting z=x+iy, explain why the locus of $arg(z-3-4i)-arg(z-2+2i)=\frac{\pi}{3}$ is the major arc of a circle. Find the centre of this circle without letting z=x+iy.

Sy123 · Jul 29, 2014

Re: HSC 2014 4U Marathon

mathsbrain said:
Without letting z=x+iy, explain why the locus of $arg(3+4i)-arg(2-2i)=\frac{\pi}{3}$ is the major arc of a circle. Find the centre of this circle without letting z=x+iy.

Do you mean

$arg(z-(3+4i))-arg(z-(2-2i))=\frac{\pi}{3}$

?

---------

$\\ $Let$ \ a = \sqrt[n]{n} \\ \\ $Consider$ \ \underbrace{a^{a^{\dots ^{a}}}}_{n} \ \ $and$ \ n \\ \\ $Which is bigger?$$

mathsbrain · Jul 29, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
Do you mean

$arg(z-(3+4i))-arg(z-(2-2i))=\frac{\pi}{3}$

?

Yes thats exactly what i mean, thanks and i have edited the question now.

aDimitri · Jul 30, 2014

Re: HSC 2014 4U Marathon

mathsbrain said:
Without letting z=x+iy, explain why the locus of $arg(z-3-4i)-arg(z-2+2i)=\frac{\pi}{3}$ is the major arc of a circle. Find the centre of this circle without letting z=x+iy.

ok i'm not 100% sure what constitutes letting z = x+iy
Am i allowed to simultaneously solve

$arg(z-3-4i)-arg(z-2+2i)=\frac{2\pi}{3} \\ |z-3-4i| = |z-2+2i|$

Thus finding the equation of the circle?

gahyunkk · Jul 31, 2014

Re: HSC 2014 4U Marathon

Hey guys how do i do this

Draw diagram of
Argz[z-(root3+i)] =pie/6

mathsbrain · Jul 31, 2014

Re: HSC 2014 4U Marathon

aDimitri said:
ok i'm not 100% sure what constitutes letting z = x+iy
Am i allowed to simultaneously solve

$arg(z-3-4i)-arg(z-2+2i)=\frac{2\pi}{3} \\ |z-3-4i| = |z-2+2i|$

Thus finding the equation of the circle?

We're looking for a geometrical approach without using algebra. So no solving equations(at least not simultaneous equations) if that makes sense

mathsbrain · Jul 31, 2014

Re: HSC 2014 4U Marathon

gahyunkk said:
Hey guys how do i do this

Draw diagram of
Argz[z-(root3+i)] =pie/6

Go http://community.boredofstudies.org/14/mathematics-extension-2/243515/complex-locus-help.html

Sy123 · Aug 10, 2014

Re: HSC 2014 4U Marathon

$\\ $Given$ \ \sec x + \tan x = A \ $for some non-zero real$ \ A \ $find$ \ \csc x + \cot x \ $in terms of$ \ A$

dunjaaa · Aug 11, 2014

Re: HSC 2014 4U Marathon

sec(x)+tan(x)=A
1/(1-sin(x)) = A
sin(x) = 1 - 1/A
cos^2(x) = 1 - (1-(1/A))^2 = (2A-1)/A^2
cos(x) = ±[sqrt(2A-1)]/A
1/(1 - cos(x)) = A/(A±sqrt(2A-1))
Therefore, cosec(x) + cot(x) = A/(A±sqrt(2A-1)) where A>= 1/2

Sy123 · Aug 11, 2014

Re: HSC 2014 4U Marathon

$\\ $Explain why there is only 1 circle that will pass through 3 non-collinear points$$

aDimitri · Aug 11, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Explain why there is only 1 circle that will pass through 3 non-collinear points$$

is it sufficient to consider the construction of perpendicular bisectors of each of the sides of the triangle formed by the 3 points, and noting that for any 3 points, the 3 perpendicular bisectors all meet at one location, meaning there is only one possible center for a circle touching these three points.
could prove this with co-ordinate geometry but i cbb

Sy123 · Aug 11, 2014

Re: HSC 2014 4U Marathon

aDimitri said:
is it sufficient to consider the construction of perpendicular bisectors of each of the sides of the triangle formed by the 3 points, and noting that for any 3 points, the 3 perpendicular bisectors all meet at one location, meaning there is only one possible center for a circle touching these three points.
could prove this with co-ordinate geometry but i cbb

What you are talking about is the orthocenter, but the orthocenter is not the centroid of a triangle and is not equidistant to all 3 points.

Here is a hint:

The general equation of a circle in the co-ordinate plane is: $(x-h)^2 + (x-k)^2 = r^2$

Axio · Aug 11, 2014

Re: HSC 2014 4U Marathon

glittergal96 · Aug 11, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
What you are talking about is the orthocenter, but the orthocenter is not the centroid of a triangle and is not equidistant to all 3 points.

Here is a hint:

The general equation of a circle in the co-ordinate plane is: $(x-h)^2 + (x-k)^2 = r^2$

I think Dimitri is correct. The orthocenter is the intersection of the perpendicular cevians, the intersection of perpendicular bisectors is the circumcenter...which is what your question is asking for.

Think about it this way. Lying on a perpendicular bisector of a side means you are equidistant from a certain two of the triangles vertices. So lying on all three perpendicular bisectors means you are equidistant from all three.

(And the centroid is something different altogether. That is the meeting point of the bisecting cevians.)

aDimitri · Aug 11, 2014

Re: HSC 2014 4U Marathon

glittergal96 said:
I think Dimitri is correct. The orthocenter is the intersection of the perpendicular cevians, the intersection of perpendicular bisectors is the circumcenter...which is what your question is asking for.

Think about it this way. Lying on a perpendicular bisector of a side means you are equidistant from a certain two of the triangles vertices. So lying on all three perpendicular bisectors means you are equidistant from all three.

(And the centroid is something different altogether. That is the meeting point of the bisecting cevians.)

Yes this is exactly what i meant.
Don't the perpendicular bisectors meet at the circumcenter?

Stygian · Aug 11, 2014

Re: HSC 2014 4U Marathon

aDimitri said:
is it sufficient to consider the construction of perpendicular bisectors of each of the sides of the triangle formed by the 3 points, and noting that for any 3 points, the 3 perpendicular bisectors all meet at one location, meaning there is only one possible center for a circle touching these three points.
could prove this with co-ordinate geometry but i cbb

your proof is more elegant and yes the the perpendicular bisectors do meet at the circumcentre

(interesting to note is that this occurs irrespective of whether this circumcentre is external to the triangle or not)

In terms of what Sy said, the orthocentre is slightly different, it being the intersection of the altitudes of a triangle

Sy123 · Aug 11, 2014

Re: HSC 2014 4U Marathon

glittergal96 said:
I think Dimitri is correct. The orthocenter is the intersection of the perpendicular cevians, the intersection of perpendicular bisectors is the circumcenter...which is what your question is asking for.

Think about it this way. Lying on a perpendicular bisector of a side means you are equidistant from a certain two of the triangles vertices. So lying on all three perpendicular bisectors means you are equidistant from all three.

(And the centroid is something different altogether. That is the meeting point of the bisecting cevians.)

aDimitri said:
Yes this is exactly what i meant.
Don't the perpendicular bisectors meet at the circumcenter?

Stygian said:
your proof is more elegant and yes the the perpendicular bisectors do meet at the circumcentre

(interesting to note is that this occurs irrespective of whether this circumcentre is external to the triangle or not)

In terms of what Sy said, the orthocentre is slightly different, it being the intersection of the altitudes of a triangle

Ah I see, ok, my bad

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