# HSC 2017-2019 MX2 Marathon ADVANCED (1 Viewer)

#### Kingom

##### Member
Re: HSC 2017 MX2 Marathon ADVANCED

A square pyramid has height exactly half it's base side length, a. Find the volume of the pyramid.
It's just $\bg_white \frac { { a }^{ 3 } }{ 6 }$.
Consider a cube with side length a. If you cut along its diagonals u get 6 identical pyramids with height half the base.

#### dan964

##### MOD
Moderator
Re: HSC 2017 MX2 Marathon ADVANCED

why can the standard V=1/3Ah be using used
where A is the area of the base = a^2
and h = a/2

#### Kingom

##### Member
Re: HSC 2017 MX2 Marathon ADVANCED

why can the standard V=1/3Ah be using used
where A is the area of the base = a^2
and h = a/2
ewwww formula pleb jkjk

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon ADVANCED

Find the maximal possible angle that a circle centred at the origin can meet the ellipse x^2/a^2 + y^2/b^2 = 1.

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon ADVANCED

(The above Q isn't too hard, just something to get this thread rolling again .)

#### dan964

##### MOD
Moderator
Re: HSC 2017-2018 MX2 Marathon ADVANCED

Update 2018
This marathon will continue into 2018.

#### Paradoxica

##### -insert title here-
Re: HSC 2017 MX2 Marathon ADVANCED

Find the maximal possible angle that a circle centred at the origin can meet the ellipse x^2/a^2 + y^2/b^2 = 1.
WLOG, assuming a=1, the maximum angle is 2cot⁻¹(b)

If this is correct, I will post the working out later, I'm a bit all over the place atm.

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon ADVANCED

WLOG, assuming a=1, the maximum angle is 2cot⁻¹(b)

If this is correct, I will post the working out later, I'm a bit all over the place atm.
Honestly can't remember, but will check this afternoon. (It's just a simple calculation of course.)

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon ADVANCED

Also, a new problem in the meantime of a different flavour:

An ellipse is inscribed in the triangle ABC, with points of contact X,Y,Z on the triangle sides opposite A,B,C respectively.

Prove that AX,BY,CZ are concurrent. (That is, prove that there exists a point that all three of these lines pass through.)

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon ADVANCED

WLOG, assuming a=1, the maximum angle is 2cot⁻¹(b)

If this is correct, I will post the working out later, I'm a bit all over the place atm.
I don't think that is quite correct, but it is close. (Perhaps an erroneous trig manipulation?)

As a test, if $\bg_white b\approx 1$, then we expect the maximal angle of contact to be close to $\bg_white 0$, because the circle and the ellipse are basically identical. Also, the maximal angle should never exceed $\bg_white \pi/2$.

#### Paradoxica

##### -insert title here-
Re: HSC 2017 MX2 Marathon ADVANCED

I don't think that is quite correct, but it is close. (Perhaps an erroneous trig manipulation?)

As a test, if $\bg_white b\approx 1$, then we expect the maximal angle of contact to be close to $\bg_white 0$, because the circle and the ellipse are basically identical. Also, the maximal angle should never exceed $\bg_white \pi/2$.
what do you mean by that? is the circle not allowed to have arbitrary radius?

also I know what my mistake was, inverse trig simplification is a pain

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#### Paradoxica

##### -insert title here-
Re: HSC 2017 MX2 Marathon ADVANCED

I don't think that is quite correct, but it is close. (Perhaps an erroneous trig manipulation?)

As a test, if $\bg_white b\approx 1$, then we expect the maximal angle of contact to be close to $\bg_white 0$, because the circle and the ellipse are basically identical. Also, the maximal angle should never exceed $\bg_white \pi/2$.
So I believe the answer is 2tan⁻¹(b) - π/2 or something like that.

Probably should have an absolute around that since we're dealing with absolute angle measures.

#### seanieg89

##### Well-Known Member
Re: HSC 2017-2018 MX2 Marathon ADVANCED

My meaning was that "angle of contact" means "acute angle of contact", otherwise the two circles that meet the ellipse tangentially give you the maximal angle (pi) and the problem is boring.

Your new answer seems correct with absolute values.

My answer (including the a-dependence) was the equivalent expression

$\bg_white \tan^{-1}\left(\frac{a^2-b^2}{2ab}\right)\quad \quad (a>b).$

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon ADVANCED

Also, a new problem in the meantime of a different flavour:

An ellipse is inscribed in the triangle ABC, with points of contact X,Y,Z on the triangle sides opposite A,B,C respectively.

Prove that AX,BY,CZ are concurrent. (That is, prove that there exists a point that all three of these lines pass through.)
Bump to move the unanswered question to the front. This one is a bit harder (in a certain sense) than the previous problem, unless there is a shortcut other than my intended method. Good to see some thread activity!

#### Paradoxica

##### -insert title here-
Re: HSC 2017 MX2 Marathon ADVANCED

Bump to move the unanswered question to the front. This one is a bit harder (in a certain sense) than the previous problem, unless there is a shortcut other than my intended method. Good to see some thread activity!
Affine Transformation Bash

#### seanieg89

##### Well-Known Member
Re: HSC 2017 MX2 Marathon ADVANCED

Affine Transformation Bash
Haha not even bashy, once you spot you can parallel project / affine transform the circle to an ellipse, there are no calculations to do, you just need to prove that the cevians meeting the incircle contact points are concurrent.

This is within the scope of HSC geometry. Alternatively, at this stage the concurrence is trivial from Ceva's theorem, which isn't hard to prove itself.

#### Sy123

##### This too shall pass
Re: HSC 2017-2018 MX2 Marathon ADVANCED

$\bg_white \\ Let \ u_n \ be a sequence defined by, \ u_0 = u_1 = u_2 = 1 \ and, \\ u_n u_{n+3} - u_{n+1} u_{n+2} = n! \\\\ Show that \ u_n \ is an integer for all n \geq 0$