# STEP 2005 Solutions Thread

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1: Solution by nuodai

2: Solution by Unbounded

3: Solution by Unbounded

4: Solution by Unbounded

5: Solution by Unbounded

6: Solution by Unbounded

7: Solution by Unbounded

8: Solution by Unbounded

9: Solution by nuodai

10: Solution by Sk1lz

11: Solution by nuodai

12: Solution by darkness9999

13: Solution by brianeverit

14: Solution Farhan.Hanif93

1: Solution by SimonM

2: Solution by Daniel Freedman

3: Solution by SimonM

4: Solution by Glutamic Acid

5: Solution by SimonM

6: Solution by Daniel Freedman

7: Solution by SimonM

8: Solution by sonofdot

9: Solution by Farhan.Hanif93

10: Solution by Glutamic Acid

11: Solution by Farhan.Hanif93

12: Solution by nuodai

13: Solution by brianeverit

14: Solution by brianeverit

1: Solution by DeanK22

2: Solution by Adje

3: Solution by SimonM

4: Solution by SimonM

5: Solution by Daniel Freedman

6: Solution by sonofdot

7: Solution by sonofdot

8: Solution by SimonM

9: Solution by sonofdot

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by sweeneyrod

13: Solution by brianeverit

14: Solution by sonofdot

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 nuodai

2: Solution by Unbounded

3: Solution by Unbounded

4: Solution by Unbounded

5: Solution by Unbounded

6: Solution by Unbounded

7: Solution by Unbounded

8: Solution by Unbounded

9: Solution by nuodai

10: Solution by Sk1lz

11: Solution by nuodai

12: Solution by darkness9999

13: Solution by brianeverit

14: Solution Farhan.Hanif93

**STEP II:**1: Solution by SimonM

2: Solution by Daniel Freedman

3: Solution by SimonM

4: Solution by Glutamic Acid

5: Solution by SimonM

6: Solution by Daniel Freedman

7: Solution by SimonM

8: Solution by sonofdot

9: Solution by Farhan.Hanif93

10: Solution by Glutamic Acid

11: Solution by Farhan.Hanif93

12: Solution by nuodai

13: Solution by brianeverit

14: Solution by brianeverit

**STEP III:**1: Solution by DeanK22

2: Solution by Adje

3: Solution by SimonM

4: Solution by SimonM

5: Solution by Daniel Freedman

6: Solution by sonofdot

7: Solution by sonofdot

8: Solution by SimonM

9: Solution by sonofdot

10: Solution by brianeverit

11: Solution by brianeverit

12: Solution by sweeneyrod

13: Solution by brianeverit

14: Solution by sonofdot

**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

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#4

I'm not going to type it up, but STEP III Question 7 is one of my favourite STEP questions

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#5

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#9

(Original post by

What happens when n = 2 in part ii)?

**Daniel Freedman**)What happens when n = 2 in part ii)?

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#10

**STEP I: Question 12**

Part 1

Part 2

Spoiler:

Group 1:

Group 2:

Show

Group 1:

Group 2:

Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.

`\implies \lambda p=\frac{2}{5}\times({\lambda p +(1-\lambda)q -\frac{1}{4}[\lambda p +(1-\lambda)q] + \frac{1}{4}[\lambda\times(1-p) + (1- \lambda)\times(1-q)])`

\Rightarrow 10\lambda p=4\lambda p + 4(1-\lambda)q - \lambda p -(1-\lambda)q +\lambda(1-p) + (1-\lambda)(1-q)

\Rightarrow 10\lambda p= 4\lambda p + 4q - 4\lambda q - \lambda p - q + \lambda q + \lambda - \lambda p + 1 - q - \lambda - \lambda q

\Rightarrow 10\lambda p = 2\lambda p + 2q - 2\lambda q + 1

\Rightarrow \boxed{\lambda= \frac{4q + 2}{5+6p + 4q}}

\Rightarrow 10\lambda p=4\lambda p + 4(1-\lambda)q - \lambda p -(1-\lambda)q +\lambda(1-p) + (1-\lambda)(1-q)

\Rightarrow 10\lambda p= 4\lambda p + 4q - 4\lambda q - \lambda p - q + \lambda q + \lambda - \lambda p + 1 - q - \lambda - \lambda q

\Rightarrow 10\lambda p = 2\lambda p + 2q - 2\lambda q + 1

\Rightarrow \boxed{\lambda= \frac{4q + 2}{5+6p + 4q}}

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#16

__STEP III 2005 Question 6__
Spoiler:

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Spoiler:

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Let

By comparing coefficients, and

So other roots are satisfied by:

Using the quadratic formula:

This gives roots and , where

By comparing coefficients, and

So other roots are satisfied by:

Using the quadratic formula:

This gives roots and , where

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#19

(Original post by

**GHOSH-5**)__Question 2, STEP I, 2005__The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.

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#20

(Original post by

For this part, is it not far simpler to note that as the gradient of PR is and the gradient of SQ is , and as it's clear that and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.

**Hashshashin**)For this part, is it not far simpler to note that as the gradient of PR is and the gradient of SQ is , and as it's clear that and therefore PR and SQ are parallel?

The same method works to show QR and PS are parallel, and as PR and PS are definition perpendicular, as are QR and QS, therefore PSQR is a rectangle.

The saves a fair bit of algebra, and as far as I can see there's no flawed reasoning.

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