HSC 2016 MX2 Marathon (archive) (1 Viewer)

leehuan · Jan 7, 2016

Re: HSC 2016 4U Marathon

hedgehog_7 said:
Given P ( a cos theta , b sin theta ) and Q ( a cos phi , b sin phi ) lie on the ellipse x^2/a^2 + y^2/b^2 = 1

IF PQ subtends a right angle at (a,0) show that ( tan theta/2 ) X ( tan phi/2 ) = -b^2/a^2

How does b (sin theta - sin phi ) / a (cos theta - cos phi) = b/a * ( 2 sin ( (theta - phi) /2 ) ) ) * cos ( theta + phi /2 ) ) / - 2 sin ( (theta - phi) /2 ) ) sin (theta + phi /2 )

You already got an answer from Paradoxica when you accidentally posted this in the wrong thread. It's a simple sum to product identity.
The 4U course expects you to know how to prove these identities!

Paradoxica · Jan 10, 2016

Re: HSC 2016 4U Marathon

$$\noindent \mathbf{NEW QUESTION}$ \\ $Two adjacent triangles share a common side, $X$. One triangle has side lengths $4,\;8$ and the other triangle has side lengths $5,\;15$. If $X$ is an integer, what is the value of $X$?$

anjumiyake · Jan 10, 2016

Re: HSC 2016 4U Marathon

By the triangle inequality, we have the following:

(1) 4 + 8 > X
(2) X + 5 > 15

The only integer satisfying those conditions is X = 11.

Edit: Strange question to be putting in this thread

InteGrand · Jan 10, 2016

Re: HSC 2016 4U Marathon

$\noindent \textbf{NEW QUESTION}$

$\noindent (i) Let $a>1$. Find the value of $a$ for which the graph of $y=a^x$ is tangent to the graph of $y=x$.$

$\noindent (ii) Hence, deduce the set of all values of $a>1$ for which the equation $a^x = x$ has a real solution.$

Paradoxica · Jan 10, 2016

Re: HSC 2016 4U Marathon

$\noindent A particular tetrahedron has integer edge lengths of either 5 or 7 units.$\\$What is the smallest possible volume of this object?$

Drsoccerball · Jan 10, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
$\noindent \textbf{NEW QUESTION}$

$\noindent (i) Let $a>1$. Find the value of $a$ for which the graph of $y=a^x$ is tangent to the graph of $y=x$.$

$\noindent (ii) Hence, deduce the set of all values of $a>1$ for which the equation $a^x = x$ has a real solution.$

Should I be "values of x" ? or is it a because I have an equation for a but it involves heeps of logs.

InteGrand · Jan 10, 2016

Re: HSC 2016 4U Marathon

Drsoccerball said:
Should I be "values of x" ? or is it a because I have an equation for a but it involves heeps of logs.

$\noindent Values of $a$. E.g. If $a= 10$, the equation becomes $10^x = x$, which has no real solutions.$

Paradoxica · Jan 10, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
$\noindent Values of $a$. E.g. If $a= 10$, the equation becomes $10^x = x$, which has no real solutions.$

$$\noindent I got $x= e^{\frac{1}{e}}$ through experimentation. The equation $a^x = x$ has a real solution when $a = e^{\frac{1}{e}}$ and for all $0<a \le 1$. There are two solutions for all $1<a<e^{\frac{1}{e}} \\ $Now somebody please prove this, because right now, I can't see a means of doing so. Perhaps we should consider the curve $y=x^x \dots$

InteGrand · Jan 10, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
$\noindent I got $x= e^{\frac{1}{e}}$ through experimentation. The equation $a^x = x$ has a real solution when $a = e^{\frac{1}{e}}$ and for all $0<a \le 1$. There are two solutions for all $1<a<e^{\frac{1}{e}} \\ $Now somebody please prove this.$

Correct! Here's a hint that makes it easier:

$Let $a = \mathrm{e}^\lambda$ and solve for $\lambda$ first instead.$

KX · Jan 10, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
$\noindent \textbf{NEW QUESTION}$

$\noindent (i) Let $a>1$. Find the value of $a$ for which the graph of $y=a^x$ is tangent to the graph of $y=x$.$

$\noindent (ii) Hence, deduce the set of all values of $a>1$ for which the equation $a^x = x$ has a real solution.$

i) a=e
$$ii) logging both sides, lna= (lnx)/x. The quantity lnx/x achieves a max of 1/e, at x=e, and range is all real values$ \ <1/e. \ $Thus, there exists a real solution as long as 'a' satisfies $0<lna<1/e \ or \ lna<0, \ or \ 1< a< e^{1/e} \ and \ 0<a<=1$

Paradoxica · Jan 10, 2016

Re: HSC 2016 4U Marathon

KX said:
i) a=e
$$ii) logging both sides, lna= (lnx)/x. The quantity lnx/x achieves a max of 1/e, at x=e, and range is all real values$ \ <1/e. \ $Thus, there exists a real solution as long as 'a' satisfies $0<lna<1/e \ or \ lna<0, \ or \ 1< a< e^{1/e} \ and \ 0<a<=1$

This doesn't justify anything.

KX · Jan 10, 2016

Re: HSC 2016 4U Marathon

More working shown:

$i)$on the exponential curve,$ \frac{dy}{dx}=1 $ at $ x=-\frac{ln(lna)}{lna}. $For the curve to be tangent, at this point P on the curve the y coordinate must equal the x coordinate. $ a^{\frac{-ln(lna)}{lna}}=\frac{-ln(lna)}{lna}. \ $After logging both sides and manipulating the expression, we reach: $ ln(-ln(lna))=0. \therefore -ln(lna)=1. \therefore lna=\frac{1}{e}. \ \Rightarrow a= e^{\frac{1}{e}} \\ ii) $If$ a > e^{\frac{1}{e}}, $then the exponential will be too steep to intersect y=x, since for this value of a the curve is tangent to the line. Therefore, for the curve to intersect with the straight line, $1<a\leq e^{\frac{1}{e}}$

InteGrand · Jan 10, 2016

Re: HSC 2016 4U Marathon

KX said:
More working shown:

$i)$on the exponential curve,$ \frac{dy}{dx}=1 $ at $ x=-\frac{ln(lna)}{lna}. $For the curve to be tangent, at this point P on the curve the y coordinate must equal the x coordinate. $ a^{\frac{-ln(lna)}{lna}}=\frac{-ln(lna)}{lna}. \ $After logging both sides and manipulating the expression, we reach: $ ln(-ln(lna))=0. \therefore -ln(lna)=1. \therefore lna=\frac{1}{e}. \ \Rightarrow a= e^{\frac{1}{e}} \\ ii) $If$ a > e^{\frac{1}{e}}, $then the exponential will be too steep to intersect y=x, since for this value of a the curve is tangent to the line. Therefore, for the curve to intersect with the straight line, $1<a\leq e^{\frac{1}{e}}$

$\noindent Well done!$

$\noindent Here's the method for (i) with $a=\mathrm{e}^\lambda$.$

$\noindent Our curves are $y=\mathrm{e}^{\lambda x}$ and $y=x$. Let their point of osculation be $P\left(t,t \right)$, which clearly lies on the line $y=x$. Since it also lies on the curve $y=\mathrm{e}^{\lambda x}$, we must have $t=\mathrm{e}^{\lambda t}\quad (1)$.$

$\noindent Since the curves have equal slope here (as they are tangent), we must also have $1=\lambda \mathrm{e}^{\lambda t}\quad (2)$ (equating slopes at the point $P$). Substituting from (1) into (2), we have $1= \lambda t $ (as $\mathrm{e}^{\lambda t} = t$ by (1)). So back in (1), we have $t=\mathrm{e}^1 = \mathrm{e}$.$

$So $\lambda t = 1 \Rightarrow \lambda = \frac{1}{t}=\frac{1}{\mathrm{e}}$. So we have $a=\mathrm{e}^{\lambda}=\mathrm{e} ^{\frac{1}{\mathrm{e}}}$.$

InteGrand · Jan 14, 2016

Re: HSC 2016 4U Marathon

$\noindent Let $R$ be a real number and $n$ be a positive integer.$

$\noindent (i) Suppose $R<0$, so $-R>0$. For any integer $k$, let $\omega_k := \sqrt [n]{-R}\,\mathrm{c\textit{i}s}\left(\frac{(2k+1)\pi}{n} \right)$, so that $\omega_k$ is an $n^{\text{th}}$ root of $R$. Show that $\omega_k$ and $\omega_{(n-1)-k}$ are complex conjugates.$

$\noindent (ii) Suppose $R>0$. For any integer $k$, let $\omega_k := \sqrt [n]{R}\,\mathrm{c\textit{i}s}\left(\frac{2k\pi}{n} \right)$, so that $\omega_k$ is an $n^{\text{th}}$ root of $R$. Show that $\omega_k$ and $\omega_{n-k}$ are complex conjugates.$

seanieg89 · Jan 14, 2016

Re: HSC 2016 4U Marathon

Perhaps more suitable for the advanced thread, but there are enough unanswered questions there for now:

A friend of yours (who knows very little about maths) is building a calculator and has programmed in the following features:
-It can display 20 digits on it's screen (and a decimal point between any two consecutive digits.)
-It can store some large but finite amount of rational numbers in memory.
-It can add/subtract/multiply/divide two rational numbers to give a third number. It can either get these source numbers from you typing them in or from memory. Similarly it can either output the result to the display or it can store the result in another memory slot.
-It can compare two rational numbers and decide which one is larger.

* (The word number in this sense means a rational number even if I don't alway write it.)

Given that his calculator takes 1 unit of time to do an elementary operation between two rational numbers, 1 unit of time to compare two numbers, and 1 unit of time to write a rational number to a vacant memory slot (assume that it takes 0 time to read/delete from a memory slot and to read from / write to the display), advise him on how to program in a square root function, and do this as efficiently as you can.

I.e. Write an algorithm for calculating the n-th root of an arbitrary positive rational number x in terms of the elementary operations and perhaps using memory. Prove that your algorithm will give you an arbitrarily good approximation to the exact n-th root of x, and include a upper bound for how many units of time are required to obtain an answer that is correct rounded to its first 20 digits.

You can also continue this exercise by thinking about how you could efficiently compute logarithms, exponentials, trignonometric functions, the values of special constants etc.

This question is quite open-ended. It is easy to come up with algorithms that "work", but the point of the exercise is to be creative and choose efficient ways (as small an upper bound for computation time as possible).

InteGrand · Jan 14, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
Perhaps more suitable for the advanced thread, but there is enough unanswered questions there for now:

A friend of yours (who knows very little about maths) is building a calculator and has programmed in the following features:
-It can display 20 digits on it's screen (and a decimal point between any two consecutive digits.)
-It can store some large but finite amount of rational numbers in memory.
-It can add/subtract/multiply/divide two rational numbers to give a third number. It can either get these source numbers from you typing them in or from memory. Similarly it can either output the result to the display or it can store the result in another memory slot.
-It can compare two rational numbers and decide which one is larger.

* (The word number in this sense means a rational number even if I don't alway write it.)

Given that his calculator takes 1 unit of time to do an elementary operation between two rational numbers, 1 unit of time to compare two numbers, and 1 unit of time to write a rational number to a vacant memory slot (assume that it takes 0 time to read/delete from a memory slot and to read from / write to the display), advise him on how to program in a square root function, and do this as efficiently as you can.

I.e. Write an algorithm for calculating the n-th root of an arbitrary positive rational number x in terms of the elementary operations and perhaps using memory. Prove that your algorithm will give you an arbitrarily good approximation to the exact n-th root of x, and include a upper bound for how many units of time are required to obtain an answer that is correct rounded to its first 20 digits.

You can also continue this exercise by thinking about how you could efficiently compute logarithms, exponentials, trignonometric functions, the values of special constants etc.

This question is quite open-ended. It is easy to come up with algorithms that "work", but the point of the exercise is to be creative and choose efficient ways (as small an upper bound for computation time as possible).

Is this an unsolved problem (i.e. there may be faster ways than the currently fastest known way?)?

seanieg89 · Jan 14, 2016

Re: HSC 2016 4U Marathon

InteGrand said:
Is this an unsolved problem (i.e. there may be faster ways than the currently fastest known way?)?

Well my definition of algorithm time etc was very simplistic. There might be a provably optimal solution to the problem I posed, but I am certainly not asking anyone to obtain or prove an optimal algorithm. Just hoping the competitive aspect of "beating someone elses algorithm" is a motivating aspect haha.

In general I don't know if there is such an algorithm for square roots that is clearly better than all the rest (lots are GOOD, but more or less equal), and not much time is devoted to proving optimality or anything, because square roots aren't a "difficult" thing to compute. Vs say integrals over multi-dimensional domains, or numerical solutions to PDEs.

Paradoxica · Jan 14, 2016

Re: HSC 2016 4U Marathon

seanieg89 said:
Well my definition of algorithm time etc was very simplistic. There might be a provably optimal solution to the problem I posed, but I am certainly not asking anyone to obtain or prove an optimal algorithm. Just hoping the competitive aspect of "beating someone elses algorithm" is a motivating aspect haha.

In general I don't know if there is such an algorithm for square roots that is clearly better than all the rest (lots are GOOD, but more or less equal), and not much time is devoted to proving optimality or anything, because square roots aren't a "difficult" thing to compute. Vs say integrals over multi-dimensional domains, or numerical solutions to PDEs.

if we have a fast algorithm for logarithms, we have a fast algorithm for square roots, and any roots.

...I just noticed algorithm and logarithm are anagrams. O_O

seanieg89 · Jan 14, 2016

Re: HSC 2016 4U Marathon

Paradoxica said:
if we have a fast algorithm for logarithms, we have a fast algorithm for square roots, and any roots.

...I just noticed algorithm and logarithm are anagrams. O_O

Sure, but "fast" is relative. It is not immediately clear whether the time taken to exponentiate and take the logarithm of a quantity will be short enough to offset the time saved by translating an "n-th root problem" to a trivial one-step "division by n problem".

The idea is to explore these possibilities with concrete maths

.

KingOfActing · Jan 15, 2016

Re: HSC 2016 4U Marathon

$The first method that comes to mind is Newton's Method for approximating the roots of an function. In this case, our function is $ f(x) = x^2 - a $ where $a$ is the number we want to calculate the root of. Newton's method is as follows. Start with some guess $x_0$ and then define the following relation: $ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$, which in our case becomes: $ x_{n+1} = x_n - \frac{x_n^2-a}{2x_n} = \frac{x_n^2+a}{2x_n}$. To prove that this will converge towards our desired value, take the limit as n approaches infinity: Let $x = \lim_{n \to \infty}x_n$, then $ x = \frac{x^2+a}{2x} \Longleftrightarrow x = \pm\sqrt{a} \\$Assuming we make an appropriate first guess, the function will always remain positive, hence converging towards the principal square root.$

$The operations to complete one step of the recursion would be, assuming $a$ is stored in memory slot 0 and the previous $x_n$ is stored in memory slot 1: $\\\\$Copy slot 1 to 2, copy slot 1 to 3, multiply slot 1 and 2, store in slot 1, add from slot 0 and 1, store in slot 1, divide slot 1 by the number 2, store in slot 1, divide slot 1 by slot 3, store in slot 1$\\\\$This would take 10 time units, but can be arbitrarily repetead to achieve any precision required.$

$Precision checking, however, will also take up a few time units in every round. In order to ensure at least 20 digits of precision, we must make sure $ x_{n+1} $ and $x_{n}$ agree up to 20 digits. This can be done by copying the previous guess, and at the end of the recursion checking whether $ \frac{|x_{n+1}-x_n|}{x_{n+1}} \leq 10^{-20} $ (note that absolute value can be replaced by checking which of the two is greater, and subtracting the smaller from the greater). Unfortunately, adding this step to every recursion adds 8 time units/recursion. It is left to the friend to decide how often they would like to check in the recursion, if at all.$

$There are a few more hurdles that need to be dealt with. The original number must be greater than 0 before beginning the recursion. An additional check for the number equalling 0 should result in a display of 0. Finally, the original guess must be made to be appropriate. A possible guess could be $\frac{a}{2}$. This is only detrimental in the case $a < \sqrt2$ and so can be probablistically argued to be a good option. Ultimately, any positive guess will eventually converge to 20 significant figures. If access to the number in its mantissa-exponent form is available, a very good starting guess would be keep the mantissa the same, and halve the exponent.$

After typing all that I now realise that it said n-th root not square root, but the idea would be the same, still using the Newtonian method just with the function $x^n - a$ . As for an upper bound on time units, I'm way too tired but a very weak one would be 20*time units/recursion. I also treated the numbers as having to be stored somewhere before any operations being done to them (You can't just multiply a number by itself, for example. You have to store that number in two different slots (register addresses), multiply them, then store the result somewhere in some slot.

I don't know if any of what I wrote makes any sense but I am too tired to read over it again. Yolo.

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