You can let someone else take control of the OP whenever.

STEP I:

1: Solution by Unbounded

2: Solution by nuodai

3: Solution by nuodai

4: Solution by Unbounded

5: Solution by darkness9999

6: Solution by SimonM

7: Solution by SimonM

8: Solution by Glutamic Acid

9: Solution by Unbounded

10: Solution by SimonM

11: Solution by brianeverit

12: Solution by Farhan.Hanif93

13: Solution by Aurel-Aqua

14: Solution by Aurel-Aqua

STEP II:

1: Solution by DeanK22

2: Solution by DeanK22 and etothepiiplusone

3: Solution by SimonM

4: Solution by sonofdot

5: Solution by Daniel Freedman

6: Solution by Elongar

7: Solution by SimonM

8: Solution by Daniel Freedman

9: Solution by brianeverit

10: Solution by Farhan.Hanif93

11: Solution by Farhan.Hanif93

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by Farhan.Hanif93

STEP III:

1: Solution by Daniel Freedman

2: Solution by Daniel Freedman

3: Solution by Daniel Freedman

4: Solution by Dadeyemi

5: Solution by Elongar

6: Solution by Anonymous

7: Solution by tommm

8: Solution by DeanK22

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

Solutions written by TSR members:

1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007

STEP I:

1: Solution by Unbounded

2: Solution by nuodai

3: Solution by nuodai

4: Solution by Unbounded

5: Solution by darkness9999

6: Solution by SimonM

7: Solution by SimonM

8: Solution by Glutamic Acid

9: Solution by Unbounded

10: Solution by SimonM

11: Solution by brianeverit

12: Solution by Farhan.Hanif93

13: Solution by Aurel-Aqua

14: Solution by Aurel-Aqua

STEP II:

1: Solution by DeanK22

2: Solution by DeanK22 and etothepiiplusone

3: Solution by SimonM

4: Solution by sonofdot

5: Solution by Daniel Freedman

6: Solution by Elongar

7: Solution by SimonM

8: Solution by Daniel Freedman

9: Solution by brianeverit

10: Solution by Farhan.Hanif93

11: Solution by Farhan.Hanif93

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by Farhan.Hanif93

STEP III:

1: Solution by Daniel Freedman

2: Solution by Daniel Freedman

3: Solution by Daniel Freedman

4: Solution by Dadeyemi

5: Solution by Elongar

6: Solution by Anonymous

7: Solution by tommm

8: Solution by DeanK22

9: Solution by brianeverit

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by brianeverit

13: Solution by brianeverit

14: Solution by brianeverit

Solutions written by TSR members:

1987 - 1988 - 1989 - 1990 - 1991 - 1992 - 1993 - 1994 - 1995 - 1996 - 1997 - 1998 - 1999 - 2000 - 2001 - 2002 - 2003 - 2004 - 2005 - 2006 - 2007

(edited 6 years ago)

Scroll to see replies

STEP III 2006, Question 2

Spoiler

STEP III 2006, Question 3

Spoiler

(edited 12 years ago)

STEP III 2006, Question 1

Spoiler

STEP III 8

We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; $[br]\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}$

We will now prove property 1

We now prove property 2 is met by this operation

We prove property three is met by the above operation.

We now prove the final property;

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

As required

Spoiler

We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; $[br]\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}$

We will now prove property 1

Spoiler

We now prove property 2 is met by this operation

Spoiler

We prove property three is met by the above operation.

Spoiler

We now prove the final property;

Spoiler

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

Spoiler

As required

I/8:

(i) Let the base be OAC, so area = 1/2ca. Height is b, so the area is b*1/3*(1/2ca) = abc/6.

(ii) $\cos \theta$ is the angle between BC and AC. $BC = \left( \begin{array}{c} 0 \\ -b \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right)$

$AC = \left( \begin{array}{c} -a \\ 0 \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right)$

$\left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right) . \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right) = \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \cos \theta \Rightarrow \cos \theta = \dfrac{c^2}{\sqrt{(a^2 + c^2)(b^2 + c^2)}}$

Area of triangle = "$\dfrac{1}{2}ab \sin \theta$". Note that $\sin^2 \theta = 1 - \cos^2 \theta$, so $\sin^2 \theta = 1 - \dfrac{c^4}{(a^2 + c^2)(b^2 + c^2)} = \dfrac{a^2b^2 + a^2c^2 + b^2c^2}{(a^2 + c^2)(b^2 + c^2)}$

$\text{area} = \dfrac{1}{2} \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \dfrac{\sqrt{a^2b^2 + b^2c^2 +a^2c^2}}{\sqrt{(a^2 + c^2)(b^2 + c^2)}} = \dfrac{1}{2}\sqrt{a^2b^2 + b^2c^2 +a^2c^2}$

The area will clearly be equal to abc/6, only that the height of the new triangle is d.

$\dfrac{1}{6}abc = \dfrac{1}{6}(a^2b^2 + b^2c^2 +a^2c^2)^{1/2}d \Rightarrow a^2b^2c^2 = d^2(a^2b^2 + b^2c^2 +a^2c^2)$

$\Rightarrow \dfrac{1}{d^2} = \dfrac{a^2b^2 + b^2c^2 +a^2c^2}{a^2b^2c^2} \Rightarrow \dfrac{1}{d^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}$

(i) Let the base be OAC, so area = 1/2ca. Height is b, so the area is b*1/3*(1/2ca) = abc/6.

(ii) $\cos \theta$ is the angle between BC and AC. $BC = \left( \begin{array}{c} 0 \\ -b \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right)$

$AC = \left( \begin{array}{c} -a \\ 0 \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ c \end{array} \right) = \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right)$

$\left( \begin{array}{c} 0 \\ -b \\ c \end{array} \right) . \left( \begin{array}{c} -a \\ 0 \\ c \end{array} \right) = \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \cos \theta \Rightarrow \cos \theta = \dfrac{c^2}{\sqrt{(a^2 + c^2)(b^2 + c^2)}}$

Area of triangle = "$\dfrac{1}{2}ab \sin \theta$". Note that $\sin^2 \theta = 1 - \cos^2 \theta$, so $\sin^2 \theta = 1 - \dfrac{c^4}{(a^2 + c^2)(b^2 + c^2)} = \dfrac{a^2b^2 + a^2c^2 + b^2c^2}{(a^2 + c^2)(b^2 + c^2)}$

$\text{area} = \dfrac{1}{2} \sqrt{a^2 + c^2} \sqrt{b^2 + c^2} \dfrac{\sqrt{a^2b^2 + b^2c^2 +a^2c^2}}{\sqrt{(a^2 + c^2)(b^2 + c^2)}} = \dfrac{1}{2}\sqrt{a^2b^2 + b^2c^2 +a^2c^2}$

The area will clearly be equal to abc/6, only that the height of the new triangle is d.

$\dfrac{1}{6}abc = \dfrac{1}{6}(a^2b^2 + b^2c^2 +a^2c^2)^{1/2}d \Rightarrow a^2b^2c^2 = d^2(a^2b^2 + b^2c^2 +a^2c^2)$

$\Rightarrow \dfrac{1}{d^2} = \dfrac{a^2b^2 + b^2c^2 +a^2c^2}{a^2b^2c^2} \Rightarrow \dfrac{1}{d^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}$

DeanK22

STEP III 8

We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; $[br]\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}$

We will now prove property 1

We now prove property 2 is met by this operation

We prove property three is met by the above operation.

We now prove the final property;

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

As required

Spoiler

We introduce some notation and an operation we wish to show is equivalent to the operation delta. We define the operation upon a polynomial f(x) as follows; $[br]\displaystyle \frac{d}{dx} f(x) = \lim_{dx \to 0} \frac{f(x+dx) - f(x)}{dx}$

We will now prove property 1

Spoiler

We now prove property 2 is met by this operation

Spoiler

We prove property three is met by the above operation.

Spoiler

We now prove the final property;

Spoiler

We now need to show that delta (the operation given in the question) returns nx^{n-1} after being applied to x^n.

Spoiler

As required

I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that $\Delta(x^n) = nx^{n-1}$, using the four rules. Then let $\displaystyle h(x) = \sum_{k=0}^n a_k x^k$, and apply the operation to this function, making use of rules ii) and iii) to show that $\displaystyle \Delta(h(x)) = \sum_{k=0}^n k a_k x^{k-1}$.

2006 I Question 13

Part (i)

The number of diamonds in one kilogram has a Poisson distribution of $D \sim \text{Poisson}(1)$. We find that per 100 grams, $D \sim \text{Poisson}(\lambda = 0.1)$. For 100T grams, $D_{T} \sim \text{Poisson}(\lambda = 0.1T)$. $T$ is the distribution of scores of the die.

Drawing 1, we have $P(T = 1 \cap D_{1}=0) = \frac{1}{6}(e^{-0.1})$.

Similarly, $\displaystyle P((T=1 \cap D_{1})\cup(T=2\cap D_{2}=0)\cup\ldots\cup(T=6\cap D_{6}=0)) = \sum_{t=1}^6 \frac{1}{6}e^{-0.1t}=$

$\displaystyle = \frac{1}{6}e^{-0.1}\sum_{t=1}^6 e^{(-0.1)(t-1)} = \frac{e^{-0.1}}{6}\frac{1-e^{-0.6}}{1-e^{-0.1}}$ (recognising a finite geometric sum).

Part (i) - expectation

$\displaystyle E[D_T] = \sum_{t=1}^6 \frac{1}{6}E[D_t] = \frac{1}{6}\sum_{t=1}^6 0.1t = \frac{1}{6}(0.1+0.2+0.3+0.4+0.5+0.6) = \frac{2.1}{6} = \frac{0.7}{2} = 0.35$

Part (ii)

This time, $T$ is a geometric distribution with $p = \frac{5}{6}$. Thus, $P(T=t)=\frac{1}{6}\left(\frac{5}{6}\right)^{t-1}$.

This time, our probability is:

$\displaystyle P((T=1 \cap D_{1}=0)\cup(T=2\cap D_{2}=0)\cup\ldots) = \sum_{t=1}^{\infty}P(T=t\cap D_{t}=0) =$

$\displaystyle = \sum_{t=1}^{\infty} \left(\frac{5}{6}\right)^{t-1}\frac{1}{6}e^{-0.1t} = \frac{e^{-0.1}}{6}\sum_{t=1}^{\infty} \left(e^{-0.1}\frac{5}{6}\right)^{t-1} =$

$\displaystyle = \frac{e^{-0.1}}{6}\frac{1}{1-\frac{5}{6}e^{-0.1}} = \frac{e^{-0.1}}{6-5e^{-0.1}}$, as required.

Part (ii) - expectation

$\displaystyle E[D_T] = \sum_{t=1}^\infty E[D_t]P(T=t) = \sum_{t=1}^\infty 0.1t\frac{1}{6}\left(\frac{5}{6}\right)^{t-1} =\frac{1}{60}\sum_{t=1}^\infty t\left(\frac{5}{6}\right)^{t-1}$.

Consider: $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} ((1-x)^{-1}) = (1-x)^{-2} = \frac{\mathrm{d}}{\mathrm{d}x} \sum_{t=0}^\infty x^t = \sum_{t=1}^\infty tx^{t-1}$.

Thus, $\displaystyle E[D_T] = \frac{1}{60}\left(1-\frac{5}{6}\right)^{-2} = \frac{36}{60} = 0.6$.

Part (i)

The number of diamonds in one kilogram has a Poisson distribution of $D \sim \text{Poisson}(1)$. We find that per 100 grams, $D \sim \text{Poisson}(\lambda = 0.1)$. For 100T grams, $D_{T} \sim \text{Poisson}(\lambda = 0.1T)$. $T$ is the distribution of scores of the die.

Drawing 1, we have $P(T = 1 \cap D_{1}=0) = \frac{1}{6}(e^{-0.1})$.

Similarly, $\displaystyle P((T=1 \cap D_{1})\cup(T=2\cap D_{2}=0)\cup\ldots\cup(T=6\cap D_{6}=0)) = \sum_{t=1}^6 \frac{1}{6}e^{-0.1t}=$

$\displaystyle = \frac{1}{6}e^{-0.1}\sum_{t=1}^6 e^{(-0.1)(t-1)} = \frac{e^{-0.1}}{6}\frac{1-e^{-0.6}}{1-e^{-0.1}}$ (recognising a finite geometric sum).

Part (i) - expectation

$\displaystyle E[D_T] = \sum_{t=1}^6 \frac{1}{6}E[D_t] = \frac{1}{6}\sum_{t=1}^6 0.1t = \frac{1}{6}(0.1+0.2+0.3+0.4+0.5+0.6) = \frac{2.1}{6} = \frac{0.7}{2} = 0.35$

Part (ii)

This time, $T$ is a geometric distribution with $p = \frac{5}{6}$. Thus, $P(T=t)=\frac{1}{6}\left(\frac{5}{6}\right)^{t-1}$.

This time, our probability is:

$\displaystyle P((T=1 \cap D_{1}=0)\cup(T=2\cap D_{2}=0)\cup\ldots) = \sum_{t=1}^{\infty}P(T=t\cap D_{t}=0) =$

$\displaystyle = \sum_{t=1}^{\infty} \left(\frac{5}{6}\right)^{t-1}\frac{1}{6}e^{-0.1t} = \frac{e^{-0.1}}{6}\sum_{t=1}^{\infty} \left(e^{-0.1}\frac{5}{6}\right)^{t-1} =$

$\displaystyle = \frac{e^{-0.1}}{6}\frac{1}{1-\frac{5}{6}e^{-0.1}} = \frac{e^{-0.1}}{6-5e^{-0.1}}$, as required.

Part (ii) - expectation

$\displaystyle E[D_T] = \sum_{t=1}^\infty E[D_t]P(T=t) = \sum_{t=1}^\infty 0.1t\frac{1}{6}\left(\frac{5}{6}\right)^{t-1} =\frac{1}{60}\sum_{t=1}^\infty t\left(\frac{5}{6}\right)^{t-1}$.

Consider: $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} ((1-x)^{-1}) = (1-x)^{-2} = \frac{\mathrm{d}}{\mathrm{d}x} \sum_{t=0}^\infty x^t = \sum_{t=1}^\infty tx^{t-1}$.

Thus, $\displaystyle E[D_T] = \frac{1}{60}\left(1-\frac{5}{6}\right)^{-2} = \frac{36}{60} = 0.6$.

2006 I Question 14

Either I missed something crucial, or this question is incredibly easy. I'll write up the incredibly-easy interpretation of this question:

Part (i)

This is a geometric progression, where $\displaystyle P(\text{Red on rth attempt}) = pq^{r-1}$, where $\displaystyle p=\frac{1}{n}$, $\displaystyle q = \frac{n-1}{n}$, so $\displaystyle P(\text{Red on rth attempt}) = \left(\frac{n-1}{n}\right)^{r-1}\frac{1}{n} = \frac{(n-1)^{r-1}}{n^r}$.

Differentiating, we get: $\displaystyle \frac{\mathrm{d}P}{\mathrm{d}n} = \frac{n^r(r-1)(n-1)^{r-2} - rn^{r-1}(n-1)^{r-1}}{n^{2r}}$. Equating to zero, $n^r(r-1)(n-1)^{r-2} = rn^{r-1}(n-1)^{r-1} \iff n(r-1) = r(n-1) \iff n(r-1-r) = -r \iff -n=-r \iff n = r$. We know that this is the maximum as r tends to infinity, P tends to zero.

Part (ii)

$P(\text{Red on 1st}) = \frac{1}{n}$

$P(\text{Red on 2nd}) = \frac{n-1}{n}\cdot\frac{1}{n-1} = \frac{1}{n}$

$P(\text{Red on rth}) = \frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdots\frac{n-(r-2)}{n-(r-3)}\cdot\frac{n-(r-1)}{n-(r-2)}\cdot\frac{1}{n-(r-1)}=\frac{1}{n}$. Clearly maximum at n = 1.

Did I make a mistake here? It just looks too simple for a STEP question, I'm afraid.

Edit: see GHOSH-5's post here: http://www.thestudentroom.co.uk/showpost.php?p=19496483&postcount=41

Either I missed something crucial, or this question is incredibly easy. I'll write up the incredibly-easy interpretation of this question:

Part (i)

This is a geometric progression, where $\displaystyle P(\text{Red on rth attempt}) = pq^{r-1}$, where $\displaystyle p=\frac{1}{n}$, $\displaystyle q = \frac{n-1}{n}$, so $\displaystyle P(\text{Red on rth attempt}) = \left(\frac{n-1}{n}\right)^{r-1}\frac{1}{n} = \frac{(n-1)^{r-1}}{n^r}$.

Differentiating, we get: $\displaystyle \frac{\mathrm{d}P}{\mathrm{d}n} = \frac{n^r(r-1)(n-1)^{r-2} - rn^{r-1}(n-1)^{r-1}}{n^{2r}}$. Equating to zero, $n^r(r-1)(n-1)^{r-2} = rn^{r-1}(n-1)^{r-1} \iff n(r-1) = r(n-1) \iff n(r-1-r) = -r \iff -n=-r \iff n = r$. We know that this is the maximum as r tends to infinity, P tends to zero.

Part (ii)

$P(\text{Red on 1st}) = \frac{1}{n}$

$P(\text{Red on 2nd}) = \frac{n-1}{n}\cdot\frac{1}{n-1} = \frac{1}{n}$

$P(\text{Red on rth}) = \frac{n-1}{n}\cdot\frac{n-2}{n-1}\cdots\frac{n-(r-2)}{n-(r-3)}\cdot\frac{n-(r-1)}{n-(r-2)}\cdot\frac{1}{n-(r-1)}=\frac{1}{n}$. Clearly maximum at n = 1.

Did I make a mistake here? It just looks too simple for a STEP question, I'm afraid.

Edit: see GHOSH-5's post here: http://www.thestudentroom.co.uk/showpost.php?p=19496483&postcount=41

Daniel Freedman

I'm pretty sure you don't need to prove that each of the four rules works for differentiation. You just need to prove, by induction as you have, that $\Delta(x^n) = nx^{n-1}$, using the four rules. Then let $\displaystyle h(x) = \sum_{k=0}^n a_k x^k$, and apply the operation to this function, making use of rules ii) and iii) to show that $\displaystyle \Delta(h(x)) = \sum_{k=0}^n k a_k x^{k-1}$.

They've said "given that the operator $\Delta$ has these properties; show that...".

Dean has said "A particular operator $\delta$ has those properties, and for $\delta$, we have ...."

But he hasn't shown that $\delta$ is the only possible operator with the properties.

This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised).

DFranklin

Actually, it's worse than that.

They've said "given that the operator $\Delta$ has these properties; show that...".

Dean has said "A particular operator $\delta$ has those properties, and for $\delta$, we have ...."

But he hasn't shown that $\delta$ is the only possible operator with the properties.

This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised).

They've said "given that the operator $\Delta$ has these properties; show that...".

Dean has said "A particular operator $\delta$ has those properties, and for $\delta$, we have ...."

But he hasn't shown that $\delta$ is the only possible operator with the properties.

This is one of those unfortunate cases where what he's done makes a reasonable amount of sense, but very definitely doesn't answer the actual question. I think you'd lose a ton of marks for this (even if the Examiner sympathised).

Thanks for pointing that out - certainly a mistake that will not be made this Summer (hopefully). Unfortunately I do not really know how to prove the Uniquness of this operator - could you please shed some light on to how you do that? Thanks.

Well, that's effectively what the question is asking you to show. I would:

(1) Prove by induction that $\Delta x^n = n x^{n-1}$ (using (i) and (iv)).

(2) Then use (iii) to prove $\Delta a_n x^n = n a_n x^{n-1}$

(3) Then use (ii) to prove $\Delta \sum_0^N a_n x^n = \sum_0^N n a_n x^{n-1}$.

[Which effectively proves uniqueness of $\Delta$ for polynomials].

Then simply observe this is the same as the derivative in the case of polynomials.

(1) Prove by induction that $\Delta x^n = n x^{n-1}$ (using (i) and (iv)).

(2) Then use (iii) to prove $\Delta a_n x^n = n a_n x^{n-1}$

(3) Then use (ii) to prove $\Delta \sum_0^N a_n x^n = \sum_0^N n a_n x^{n-1}$.

[Which effectively proves uniqueness of $\Delta$ for polynomials].

Then simply observe this is the same as the derivative in the case of polynomials.

I have different answers to STEP I Q 14 Im not to sure who is wrong...

Part (i)

Part (ii)

Part (i)

Spoiler

Part (ii)

Spoiler

STEP I: Q5

Part (i)

Part (ii)

STEP III 2006, Q5

We wish to show that $\alpha, \beta, \gamma$ form an equilateral iff

$\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = 0$

Multiplying by two and factorising, we wish to show that,

$\displaystyle (\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 0$

holds for all equilateral triangles.

Note that $\alpha - \beta, \beta - \gamma, \gamma - \alpha$ are the sides of the triangle. If they are shown on an Argand diagram, the angle between any two consecutive sides is $\frac{\pi}{3}$. Note also that the lengths of the sides are all of equal length, so the magnitudes of the complex numbers that represent them will also be equal.

Using this, we reconstruct the equation:

$\displaystyle \left(re^{i \theta}\right)^2 + \left(re^{i (\theta +\frac{\pi}{3})}\right)^2 + \left(re^{i (\theta +\frac{2\pi}{3})}\right)^2 = 0$

or, more simply,

$\displaystyle r^2 \left( e^{2 i \theta} + e^{2 i (\theta +\frac{\pi}{3})} + e^{2 i (\theta +\frac{2 \pi}{3})} \right) = 0$

$\displaystyle r^2 e^{2 i \theta} \left( 1 + e^{i \frac{\pi}{3}} + e^{i \frac{2 \pi}{3}} \right) = 0$

The bracket evaluates to 0, so we are done.

For the next part, we take the roots of the cubic to be $\alpha, \beta, \gamma$, and expand, yielding Vieta's formula:

$\displaystyle z^3 - (\alpha + \beta + \gamma) z^2 + (\alpha \beta + \beta \gamma + \gamma \alpha) z - \alpha \beta \gamma = 0$

Algebra shows that:

$\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = (\alpha + \beta + \gamma)^2 - 3 (\alpha \beta + \beta \gamma + \gamma \alpha) = a^2 - 3b = 0$

and the result follows immediately.

Now we write,

$p = Ae^{i \phi}$

$q = B + Ci$

The transformation $z = p w + q$ has the effect of rotating our equilateral triangle by the angle $\phi$, magnifying it by a factor of $A$ and translating it through the vector represented by $q$ (would I need to show this?). Under these transformations, the equilateral triangle formed by the roots of the initial cubic remains an equilateral triangle.

We wish to show that $\alpha, \beta, \gamma$ form an equilateral iff

$\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = 0$

Multiplying by two and factorising, we wish to show that,

$\displaystyle (\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 0$

holds for all equilateral triangles.

Note that $\alpha - \beta, \beta - \gamma, \gamma - \alpha$ are the sides of the triangle. If they are shown on an Argand diagram, the angle between any two consecutive sides is $\frac{\pi}{3}$. Note also that the lengths of the sides are all of equal length, so the magnitudes of the complex numbers that represent them will also be equal.

Using this, we reconstruct the equation:

$\displaystyle \left(re^{i \theta}\right)^2 + \left(re^{i (\theta +\frac{\pi}{3})}\right)^2 + \left(re^{i (\theta +\frac{2\pi}{3})}\right)^2 = 0$

or, more simply,

$\displaystyle r^2 \left( e^{2 i \theta} + e^{2 i (\theta +\frac{\pi}{3})} + e^{2 i (\theta +\frac{2 \pi}{3})} \right) = 0$

$\displaystyle r^2 e^{2 i \theta} \left( 1 + e^{i \frac{\pi}{3}} + e^{i \frac{2 \pi}{3}} \right) = 0$

The bracket evaluates to 0, so we are done.

For the next part, we take the roots of the cubic to be $\alpha, \beta, \gamma$, and expand, yielding Vieta's formula:

$\displaystyle z^3 - (\alpha + \beta + \gamma) z^2 + (\alpha \beta + \beta \gamma + \gamma \alpha) z - \alpha \beta \gamma = 0$

Algebra shows that:

$\displaystyle \alpha^2 + \beta^2 + \gamma^2 - \alpha \beta - \beta \gamma - \gamma \alpha = (\alpha + \beta + \gamma)^2 - 3 (\alpha \beta + \beta \gamma + \gamma \alpha) = a^2 - 3b = 0$

and the result follows immediately.

Now we write,

$p = Ae^{i \phi}$

$q = B + Ci$

The transformation $z = p w + q$ has the effect of rotating our equilateral triangle by the angle $\phi$, magnifying it by a factor of $A$ and translating it through the vector represented by $q$ (would I need to show this?). Under these transformations, the equilateral triangle formed by the roots of the initial cubic remains an equilateral triangle.

RE; Noudai

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.

opting for the use of x as the distance from the side of the nearest corner of the barn and a diagram saves you considering the cases and gives the answer far quicker.

- MAT 1996-2006 Solution Thread
- Oxford pat
- Your favourite childhood yt channels?
- A Level chemistry question
- How should I go about preparing for the STEP Maths exam?
- OCR S1 Combinations and Permutations Exam paper question.
- What are the best places online for maths revision and why?
- STEP Maths I, II, III 1993 Solutions
- Maths Problem Solving
- Can someone please explain factoring quadratic equations?
- SQA Nat 5 Mathematics - Paper 2 - 3rd May 2024 [Exam Chat]
- STEP Maths I,II,III 1987 Solutions
- How should I prepare for the STEP?
- STEP maths I, II, III 1990 solutions
- Mathematics Admissions Test (MAT)
- Hard A* GCSE Question Challenge - Perfect for Edexcel Calculator Paper Revision
- STEP 2016 Solutions
- Biology maths
- AS Maths Exponential Question
- Quotes-Who like quotes?

Latest

Trending

Last reply 3 weeks ago

Edexcel A Level Mathematics Paper 3 (9MA0 03) - 20th June 2024 [Exam Chat]Maths Exams

386

1104

Last reply 1 month ago

Edexcel GCSE Mathematics Paper 1 Higher (1MA1 1H) - 16th May 2024 [Exam Chat]Maths Exams

419

1460

Last reply 1 month ago

AQA Level 2 Further Maths 2024 Paper 2 (8365/2) - 19th June [Exam Chat]Maths Exams

94

461

Last reply 1 month ago

Edexcel A-level Further Mathematics Paper 1 (9FM0 01) - 22nd May 2024 [Exam Chat]Maths Exams

180

615

Last reply 1 month ago

AQA A-level Mathematics Paper 3 (7357/3) - 20th June 2024 [Exam Chat]Maths Exams

140

627

Last reply 1 month ago

Edexcel A Level Mathematics Paper 2 (9MA0 02) - 11th June 2024 [Exam Chat]Maths Exams

498

1842

Last reply 1 month ago

AQA A-level Mathematics Paper 2 (7357/2) - 11th June 2024 [Exam Chat]Maths Exams

171

545

Last reply 3 months ago

AQA GCSE Mathematics Paper 1 Foundation (8300/1F) - 16th May 2024 [Exam Chat]Maths Exams

45

63

Last reply 3 months ago

Edexcel GCSE Mathematics Paper 3 Higher (1MA1 3H) - 10th June 2024 [Exam Chat]Maths Exams

265

1273

Last reply 3 months ago

Edexcel GCSE Statistics Paper 2 Higher Tier (1ST0 2H) - 17th June 2024 [Exam Chat]Maths Exams

23

30

Last reply 3 months ago

Edexcel GCSE Mathematics Paper 1 Foundation (1MA1 1F) - 16th May 2024 [Exam Chat]Maths Exams

42

47

Last reply 3 months ago

AQA GCSE Mathematics Paper 2 Higher (8300/2H) - 3rd June 2024 [Exam Chat]Maths Exams

136

414

Last reply 3 months ago

OCR GCSE Mathematics Paper 4 Higher (J560/04) - 16th May 2024 [Exam Chat]Maths Exams

87

197

Last reply 3 months ago

Edexcel A-level Further Mathematics Paper 2 (9FM0 02) - 3rd June 2024 [Exam Chat]Maths Exams

132

297

Last reply 3 months ago

Edexcel A Level Further Mathematics Paper 3D (9FM0 3D) - 21st June 2024 [Exam Chat]Maths Exams

53

121

Last reply 3 months ago

AQA Level 2 Further Maths 2024 Paper 1 (8365/1) - 11th June [Exam Chat]Maths Exams

168

709

Last reply 3 months ago

OCR A Level Mathematics A Paper 2 (H240/02) - 11th June 2024 [Exam Chat]Maths Exams

109

222

Last reply 3 months ago

Edexcel A Level Mathematics Paper 1 (9MA0 01) - 4th June 2024 [Exam Chat]Maths Exams

553

1452

Last reply 3 months ago

Edexcel GCSE Mathematics Paper 2 Higher (1MA1 2H) - 3rd June 2024 [Exam Chat]Maths Exams

298

1257