Carrotsticks' 2014 Extension 1 HSC Solutions (1 Viewer)

Thank you!

garfield qiu

New Member
Carrotsticks, I doubt your parametric solution for Q13 (c) (iii).
you said: This is the equation of a circle centred at the focus with radius a. We have the condition that
x =/= 0 because this can only occur if P lies precisely on the vertex of the parabola ie: t = 0 .
However, when this happens, then we cannot define Q because having a ratio of 0 :1 yields
an undefined point.

But please check out the definition of ratio, 0:1 means 0/1 which has meaning (P and Q have no distance); 1/0 is meaningless
Also, when you put t=0 back into (i), the coordinates of Q becomes (0,0) which is on the circle we got in (iii), and it completes the circle. That's why the question says it's a full circle, BOS is not that stupid as you think. So imo, you do need to separate the cases where x=0 or not.
When x=0, t=0, but at this moment t is not the slope of OQ because at this moment Q is on origin, then OQ doesn't have a slope, so (ii) has no problem.
Thus the whole question is good. It may become a reason of some people losing state rank no1

RealiseNothing

what is that?It is Cowpea
Carrotsticks, I doubt your parametric solution for Q13 (c) (iii).
you said: This is the equation of a circle centred at the focus with radius a. We have the condition that
x =/= 0 because this can only occur if P lies precisely on the vertex of the parabola ie: t = 0 .
However, when this happens, then we cannot define Q because having a ratio of 0 :1 yields
an undefined point.

But please check out the definition of ratio, 0:1 means 0/1 which has meaning (P and Q have no distance); 1/0 is meaningless
Also, when you put t=0 back into (i), the coordinates of Q becomes (0,0) which is on the circle we got in (iii), and it completes the circle. That's why the question says it's a full circle, BOS is not that stupid as you think. So imo, you do need to separate the cases where x=0 or not.
When x=0, t=0, but at this moment t is not the slope of OQ because at this moment Q is on origin, then OQ doesn't have a slope, so (ii) has no problem.
Thus the whole question is good. It may become a reason of some people losing state rank no1
The locus of Q is most definitely not a full circle.

The question also says that Q lies on a circle, this does not necessarily mean it maps out a full circle.

RealiseNothing

what is that?It is Cowpea
Also a cool way of doing the second part of the circle geo question:

We want to show that OP is a tangent to the circle through PQC. Let there be a point D that lies on both the arc PC and the line OP extended.

We know that angle PAO = angle APO by isosceles triangles, and angle APO = angle DPC by vertically opposite angles, and thus angle PAO = angle DPC.

However angle PAO = angle CQP from part (i). Thus angle DPC = angle CQP, and so CD=PC by equal angles subtend equal arcs.

If CD=PC then the point P and D coincide and so, other then P, the line OP does not pass through any points on the circle PQC. Hence OP is a tangent.

Carrotsticks

Retired
I am now looking a little more into the locus problem. Thanks to all who raised it.

RealiseNothing

what is that?It is Cowpea
Also a cool way of doing the second part of the circle geo question:

We want to show that OP is a tangent to the circle through PQC. Let there be a point D that lies on both the arc PC and the line OP extended.

We know that angle PAO = angle APO by isosceles triangles, and angle APO = angle DPC by vertically opposite angles, and thus angle PAO = angle DPC.

However angle PAO = angle CQP from part (i). Thus angle DPC = angle CQP, and so CD=PC by equal angles subtend equal arcs.

If CD=PC then the point P and D coincide and so, other then P, the line OP does not pass through any points on the circle PQC. Hence OP is a tangent.

Carrotsticks

Retired
Oh haha okay it didn't take long to see, just needed a minute of clear thinking to spot. I can confirm that my initial statement was erroneous, I will explain why below (it's been fixed now). Many thanks again to all who raised the problem.

When P slides along the parabola, Q slides along the arc of a circle and gets extremely close to (0,2a), but never reaches it.

Now our computations of the locus of Q to obtain the equation come from the assumption that x=/=0. HOWEVER, when we compute the case x=0 separately (I cannot believe that I did not do this, and that I disregarded it completely), it yields a point that 'fills the gap' and 'continues' the equation of the locus, thereby making the only 'hole' in the locus being the point (0,2a).

EDIT: 1:21AM grammar is not the strongest of grammars.

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RealiseNothing

what is that?It is Cowpea
The probability question part (ii):

$\bg_white Let the probability that player 1 eventually wins be A$

$\bg_white A+rA+q=1$

$\bg_white A=\frac{1-q}{1+r}$

$\bg_white A = \frac{p+r}{1+r}$ since $\bg_white p+r+q=1$

Carrotsticks

Retired
Carrotsticks, I doubt your parametric solution for Q13 (c) (iii).
you said: This is the equation of a circle centred at the focus with radius a. We have the condition that
x =/= 0 because this can only occur if P lies precisely on the vertex of the parabola ie: t = 0 .
However, when this happens, then we cannot define Q because having a ratio of 0 :1 yields
an undefined point.

But please check out the definition of ratio, 0:1 means 0/1 which has meaning (P and Q have no distance); 1/0 is meaningless
Also, when you put t=0 back into (i), the coordinates of Q becomes (0,0) which is on the circle we got in (iii), and it completes the circle. That's why the question says it's a full circle, BOS is not that stupid as you think. So imo, you do need to separate the cases where x=0 or not.
When x=0, t=0, but at this moment t is not the slope of OQ because at this moment Q is on origin, then OQ doesn't have a slope, so (ii) has no problem.
Thus the whole question is good. It may become a reason of some people losing state rank no1
Thank you for pointing out this erroneous statement. You are one of two students who have pointed this out. I am quite certain that the reason why I had made that mistake in the first place was because I had confused a ratio of 0:1 with the ratio 1:0 amongst the rush, without thinking about the bigger picture overall pointing out that the origin is most certainly a part of the locus equation.

Some things to note about your response:

1. I'm not too sure how I feel about your comment "BOS is not that stupid as you think", I would appreciate you not putting words in my mouth, thanks.

2. The equation of the locus is most certainly NOT a 'full circle'. It is excluding the point (0,2a) for reasons I have explained below.

3. I can see that you are talking about (ii) in a rather defensive manner, and I am not quite sure why you are doing so since nobody in the entire thread (or forum section for that matter) has had any problems with it.

Carrotsticks

Retired
The probability question part (ii):

$\bg_white Let the probability that player 1 eventually wins be A$

$\bg_white A+rA+q=1$

$\bg_white A=\frac{1-q}{1+r}$

$\bg_white A = \frac{p+r}{1+r}$ since $\bg_white p+r+q=1$
Can you elaborate on your first line?

RealiseNothing

what is that?It is Cowpea
Can you elaborate on your first line?
If person 1 has a probability of A of eventually winning, we note that person 2 must also have a probability of A of eventually winning given that person 1 does not win on his first turn.

So we arrive at 3 outcomes:

Person 1 eventually wins - A

Person 2 eventually wins after person 1 does not win/lose on his first turn - rA

Person 2 wins due to person 1 losing on the first turn = q

This covers all outcomes and hence $\bg_white A+rA+q=1$

Carrotsticks

Retired
That works very nicely, I did not see it myself.

trapizi

╰(゜Д゜)╯
the probability question part (ii):

$\bg_white let the probability that player 1 eventually wins be a$

$\bg_white a+ra+q=1$

$\bg_white a=\frac{1-q}{1+r}$

$\bg_white a = \frac{p+r}{1+r}$ since $\bg_white p+r+q=1$
omg this is exactly what i did

kimsungho

l0l
hey carrot, would you need to show that there were 2 solutions not just one for the ski jump question to find max d? because i showed the pi/8 and justified with 2nd derivative test, but i didnt include the 2nd possible solution. will the marker take marks off for this?

mreditor16

Well-Known Member
hey carrot, would you need to show that there were 2 solutions not just one for the ski jump question to find max d? because i showed the pi/8 and justified with 2nd derivative test, but i didnt include the 2nd possible solution. will the marker take marks off for this?
Best to confirm with Carret, but that sounds fine to me

mreditor16

Well-Known Member
arghh so I showed that there were two stat points as 22.5 degrees (pi/8) and -67.5 degrees. I showed that the 22.5 degrees was a maxima by testing points around it, but just inspection of the other one in terms of the derivative told me that it was a minima so I didn't bother proving it (I wrote that it was a stationary point though).

Carrot do you think thats okay :/
it'll be fine, because the question specified that

As a result, you shouldn't have considered -67.5 to begin with.

Chris100

Member
Yes, you should have calculated 2 different solutions but then you should have mentioned that because one of them isn't in the domain given by the question, only pi/8 can be considered
then u confirm that its a maximum value with a table or 2nd derivative or whatever

Samir97

Member
Carrotsticks, for the parametric circle q, i didnt know how to use the previous part to find the curcle equation but I could immediately see that the centre was going to be S(0,a) so I just squared the X-coordinate of Q, subtracted a from the y co-ordinate (gave me [y-a]) and i squared that, then added both squares and the final term cancelled out to just a^2. Is that considered fudging? And would I still obtain full marks?

Gumball

New Member
Carrotsticks, for the parametric circle q, i didnt know how to use the previous part to find the curcle equation but I could immediately see that the centre was going to be S(0,a) so I just squared the X-coordinate of Q, subtracted a from the y co-ordinate (gave me [y-a]) and i squared that, then added both squares and the final term cancelled out to just a^2. Is that considered fudging? And would I still obtain full marks?
even though im not carret i believe it was fudging as there was a legit way to figure it out lol. you may be lucky to get even 1 but definitely not more than that. remember the test wasn't very hard so marking will be

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