HSC 2016 MX1 Marathon (archive) (3 Viewers)

seanieg89 · Jun 28, 2016

Re: HSC 2016 3U Marathon

Also you have cut off the text so idk, but the question itself might even be asking you to state the vertices in the order of correspondence.

davidgoes4wce · Jun 28, 2016

Re: HSC 2016 3U Marathon

seanieg89 said:
I don't know how strict the HSC is on such things, but it is generally a bad idea to state congruences and similarities without listing the vertices in order of their correspondence (which is not the case with your statement, the angles are not in the same order).

Even in the best-case scenario that they don't mark you down for such things, I think it is still a much better idea to state things in order in the first place, so then you can directly read off the resulting statements about side lengths and angles from your similarity/congruence statement without having to refer back to the diagram to check that angles "match up".

What do you mean by 'order'? I am a person that likes to follow the order as well.

My reason for going AD (was to go against the opposite direction of the double arrow) , was that I thought BC (like AD, going against the opposite direction of the double arrow )

My thinking is another possible solution would be:

$\bigtriangleup ABD \equiv \bigtriangleup DCB$ ?

davidgoes4wce · Jun 28, 2016

Re: HSC 2016 3U Marathon

Question:

$Prove that each pair of triangles is congruent, giving reasons. Write the vertices in matching order.$

dan964 · Jun 28, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Question:

$Prove that each pair of triangles is congruent, giving reasons. Write the vertices in matching order.$

is there more information, or a diagram/context?

seanieg89 · Jun 28, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
What do you mean by 'order'? I am a person that likes to follow the order as well.

My reason for going AD (was to go against the opposite direction of the double arrow) , was that I thought BC (like AD, going against the opposite direction of the double arrow )

My thinking is another possible solution would be:

$\bigtriangleup ABD \equiv \bigtriangleup DCB$ ?

By order, I mean the "matching order" referred to in the question.

Congruence of polygons means you can rotate, translate, and flip one till it is "on top" of the other, and in doing so you get a pairing between vertices of the first polygon and the vertices of the second one. This pairing need not be unique, but you should state congruences and similarities so that the order of vertices are in correspondence, which is not the case in the two statements you made.

(Similarity is the same, but includes the operation of scaling).

In particular, note that if triangles ABC and XYZ are similar, then the angles A=X, B=Y, C=Z.

davidgoes4wce · Jun 28, 2016

Re: HSC 2016 3U Marathon

dan964 said:
is there more information, or a diagram/context?

This is the original question:

davidgoes4wce · Jun 28, 2016

Re: HSC 2016 3U Marathon

seanieg89 said:
By order, I mean the "matching order" referred to in the question.

Congruence of polygons means you can rotate, translate, and flip one till it is "on top" of the other, and in doing so you get a pairing between vertices of the first polygon and the vertices of the second one. This pairing need not be unique, but you should state congruences and similarities so that the order of vertices are in correspondence, which is not the case in the two statements you made.

(Similarity is the same, but includes the operation of scaling).

OK you cleared it up I redid Q5 (b) and listed down the matching pair.

davidgoes4wce · Jul 2, 2016

Re: HSC 2016 3U Marathon

When describing congruent triangles in a Quadrilateral:

Is it ok to say

$\angle ABD= \angle BDC ?$ instead of what the book has as : $\angle ABD=\angle CDB ?$

InteGrand · Jul 2, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
When describing congruent triangles in a Quadrilateral:

Is it ok to say

$\angle ABD= \angle BDC ?$ instead of what the book has as : $\angle ABD=\angle CDB ?$

It probably is OK, but the book's way is probably better because it lists the vertices in matching order (like matching with the corresponding vertices in the other congruent triangle). Always writing these things in matching order will also help you keep track of which sides and angles etc. in the congruent triangle are corresponding ones, which can help as you won't need to look carefully at the diagram to see this, you can just read it off your arrangement of the vertices' letters.

But I don't think you'd lose a mark in the HSC if you wrote the angles your way.

davidgoes4wce · Jul 2, 2016

Re: HSC 2016 3U Marathon

^^^ Thanks I want to develop good habits for Circle Geometry as I'll be honest I never touched that stuff in high school and university.

davidgoes4wce · Jul 2, 2016

Re: HSC 2016 3U Marathon

Is it ok to say for the reasoning for

$$ BC = AD (Given) instead of saying BC=AD (opposite sides of a parallelogram are equal) $ ?$

InteGrand · Jul 2, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
Is it ok to say for the reasoning for

$$ BC = AD (Given) instead of saying BC=AD (opposite sides of a parallelogram are equal) $ ?$

You should probably say that opposite sides of a parallelogram/rectangle are equal.

InteGrand · Jul 2, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
Simplify n(n-1 1)+n(n-1 2)+...+n(n-1 n-2)

Are those meant to be binomial coefficients?

leehuan · Jul 3, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
Simplify n(n-1 1)+n(n-1 2)+...+n(n-1 n-2)

Weird question that does not have a truly neat solution

$(1+x)^{n-1}=\binom{n-1}{0}+\binom{n-1}{1}x+\binom{n-1}{2}x^2+\dots+\binom{n-1}{n-2}x^{n-2}+\binom{n-1}{n-1}x^{n-1}$

$\text{Put }x=1\\ 2^{n-1} = \binom{n-1}{0}+\binom{n-1}{1}+\binom{n-1}{2}+\dots+\binom{n-1}{n-2}+\binom{n-1}{n-1}$

$\text{Rearranging}\\ 2^{n-1}-2 = \binom{n-1}{1}+\binom{n-1}{2}+\dots+\binom{n-1}{n-2}$

$\text{Then just smack }n\text{ in front of everything...}$

Paradoxica · Jul 4, 2016

Re: HSC 2016 3U Marathon

$\int_{0}^{64} \sqrt{1 + \frac{4}{\sqrt[3]{x^2}}} \,\,$d$x$

jathu123 · Jul 4, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
$\int_{0}^{64} \sqrt{1 + \frac{4}{\sqrt[3]{x^2}}} \,\,$d$x$

I have a feeling I did it the long way

Paradoxica · Jul 4, 2016

Re: HSC 2016 3U Marathon

integral95 · Jul 4, 2016

Re: HSC 2016 3U Marathon

jathu123 said:
View attachment 33316

I have a feeling I did it the long way

ya and also this is 3U, which means you normally wouldn't need to use substitution if you're not asked to

jathu123 · Jul 4, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
What is the derivative of x^2/3

ohh ye my bad thanks. Is this better?

$\int_{0}^{64}\sqrt{1+\frac{4}{x^{2/3}}}$ d$x \\ \int_{0}^{64}\frac{\sqrt{x^{2/3}+4}}{x^{1/3}} $ d$x\\ $ let $ u=x^{2/3}, $ d$u=\frac{2}{3x^{1/3}}$ d$x \\\therefore $ d$x= \frac{3}{2}x^{1/3}$ d$u \\ \int_{0}^{16}\frac{3}{2}\sqrt{u+4}$ d$u \\\\ = \left [(u+4)^{3/2} \right ] $from 0 to 16$ \\ =20^{3/2}-4^{3/2} = 40\sqrt{5}-8$

leehuan · Jul 4, 2016

Re: HSC 2016 3U Marathon

integral95 said:
ya and also this is 3U, which means you normally wouldn't need to use substitution if you're not asked to

It's a marathon. Whilst in the HSC you're given the substitution, anything appears to be fine here.

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