STEP II 2015 Solutions Thread

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Gawain
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#1
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STEP II 2015

Q1) team-f-maths (attached to first post)
Q2) Gawain
Q3) Gawain
Q4) team-f-maths (attached to first post)
Q5) Gawain
Q6)
Q7) Goose Lane
Q8) Gawain
Q9) Brammer ; Alt - mikelbird
Q10) mikelbird
Q11) mikelbird
Q12) brianeverit
Q13) brianeverit


Paper (Google Drive link courtesy of jneill)
Attached files
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Gawain
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Q5)

Base case:  \displaystyle S_1=\arctan(\frac12)

 \displaystyle \tan(S_1)=\frac12=\frac{1}{1+1}

Assume  \displaystyle \tan(S_k)=\frac{k}{k+1}

Then  \displaystyle \tan(S_{k+1})=\tan(S_k+\frac{1}{2(k+1)^2})

After simplifying with the tan addition formula we get:

 \displaystyle  \tan(S_{k+1})=\frac{(k+1)(2k^2+2k+1)}{(k+2)(2k^2+2k+1)}=\frac{k+1}{k+2}

If  \displaystyle \tan(a)=b and  \displaystyle b \geq 0 \; 0\leq a <\frac{\pi}{2}

then we can say

 \displaystyle a=\arctan(b).

Now since we have  \displaystyle \frac{n}{n+1} \geq 0 and  \displaystyle S_n < S_{\infty} = \frac{\pi}{4} < \frac{\pi}{2}

We can conclude that  \displaystyle S_n=\arctan(\frac{n}{n+1})

(induction is a safer bet here but I couldn't be bothered to type it up at the moment.)

ii)

 \displaystyle \tan(2 \alpha) = \frac{2\tan(\alpha)}{1-\tan^2(\alpha)} = \frac{4n^2}{4n^4-1}

After solving for tan we get  \displaystyle \tan(\alpha) = -2n^2  \; , \tan(\alpha)=\frac{1}{2n^2}

Therefore using our previous result and taking the limit to infinity we get:
 \displaystyle S_{\infty}=\sum_{n=1}^{\infty} \alpha_n = \arctan\left (1- \frac{1}{\infty} \right )=\frac{\pi}{4}
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Gawain
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Q8)
i)

 \displaystyle \frac{P-x_{1}}{P-x_{2}}=\frac{r_1}{r_2} \: \Rightarrow P=\frac{x_2r_1-x_1r_2}{r_1-r_2}

ii)

 \displaystyle Q=\frac{x_3r_1-x_1r_3}{r_1-r_3}\, , R=\frac{x_3r_2-x_2r_3}{r_2-r_3}

For PQR to be co-linear we need  \displaystyle P-Q=k(R-Q)

This is equivalent to  \displaystyle  aP+bQ+cR=0

with

 \displaystyle a+b+c=0

If we let:

 \displaystyle a=r_3\left ( r_1-r_2 \right )\, , b= -r_2\left ( r_1-r_3 \right )\, c=r_1\left ( r_2-r_3 \right )

Then we are done as a,b,c are scalar values.

iii)

For Q to be a midpoint we have the condition  \displaystyle P-R=2(P-Q)

Which is equivalent to:
 \displaystyle a=c

or

 \displaystyle \frac{2r_1r_3}{r_2}=r_1+r_3
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Gawain
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Q3)Extra triangles gained adding a rod length 8:
(8,7,1-6) = 6 Triangles
(8,6,2,5) = 4 Triangles
(8,5,3-4)= 2 Triangles

So  \displaystyle T_8-T_7=6+4+2 =12

Through a similar method  \displaystyle T_7-T_6=5+3+1=9

Therefore  \displaystyle T_8-T_6=21

Trivially:  \displaystyle T_{2m}-T_{2m-1}=m(m-1)

and

 \displaystyle T_{2m}-T{2m-2}=(2m-1)(m-1)

Induction:

Base: With 4 rods we can make 3 triangles, also  \displaystyle T_4=3 so our base case works.

Assumption:  \displaystyle T_{2k}=\frac{k(k-1)(4k+1)}{6}

Induction:  \displaystyle T_{2(k+1)}=(2k+1)(k) + T_{2k}

Simplifying we get: \displaystyle T_{2k}=\frac{(k+1((k+1)-1)(4(k+1)+1)}{6}

Using the second difference formula we get: \displaystyle T_{2m-1}=\frac{m(m-1)(4m-5)}{6}
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Gawain
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Q2)
i)

 \displaystyle \frac{x}{\sin(\alpha)}=\frac{AB}{sin(3 \alpha)}\, \Rightarrow AB=\frac{x\sin(3 \alpha)}{\sin(\alpha )}

 \displaystyle \frac{x}{\sin(\alpha)}=\frac{AB}{sin(3 \alpha)}\, \Rightarrow AB=\frac{x\sin(3 \alpha)}{\sin(\alpha )}

So

 \displaystyle AB=\frac{x\left ( \sin\left ( \alpha \cos\left ( 2\alpha)  + \cos\left ( \alpha \sin\left ( 2\alpha  \right ) \right ) }{\sin(\alpha )}=x\left ( 4\cos^2(\alpha ) -1 \right )=x\left ( 3-4\sin^2(\alpha ) \right )

ii)

 \displaystyle DE=\frac{AB}{2}-x\cos(2\alpha )=x(3-4\sin^2(\alpha) -2 +4\sin^2{\alpha})=\frac{x}{2}

iii)

Consider the triangle CFG where G is the foot of the perpendicular from F to CE.

Also let angle FCG be  \displaystyle \theta

Obviously DE=FG which implies that  \displaystyle \sin(\theta)=\frac{CF}{FG}=\frac12

As \theta = \frac{\pi}{6} , this tells us that BCF is

 \displaystyle \frac{2\pi}{3} - 2\alpha

But then ACF is  \displaystyle \pi - 3\alpha +2\alpha - \frac{2\pi}{3}=\frac{\pi}{3} - \alpha

Since the ratio of BCF:ACF is 2:1 then FC trisects ACB.
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programmer<3
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(Original post by Gawain)
Q8)
i)

 \displaystyle \frac{P-x_{1}}{P-x_{2}}=\frac{r_1}{r_2} \: \Rightarrow P=\frac{x_2r_1-x_1r_2}{r_1-r_2}

ii)

 \displaystyle Q=\frac{x_3r_1-x_1r_3}{r_1-r_3}\, , R=\frac{x_3r_2-x_2r_3}{r_2-r_3}

For PQR to be co-linear we need  \displaystyle P-Q=k(R-Q)

This is equivalent to  \displaystyle  aP+bQ+cR=0

with

 \displaystyle a+b+c=0

If we let:

 \displaystyle a=r_3\left ( r_1-r_2 \right )\, , b= -r_2\left ( r_1-r_3 \right )\, c=r_1\left ( r_2-r_3 \right )

Then we are done as a,b,c are scalar values.

iii)

For Q to be a midpoint we have the condition  \displaystyle P-R=2(P-Q)

Which is equivalent to:
 \displaystyle a=c

or

 \displaystyle \frac{2r_1r_3}{r_2}=r_1+r_3
How tough was part 2 compared to part 1. Part 3 simply follows from part 2, doesnt it?
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programmer<3
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#7
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Question 6 solution:
i) sec^2 (pi/4 - x/2) = 1/cos^2(pi/4 - x/2) = 2/2cos^2(pi/4 - x/2) = 2/(1+cos(pi/2-x)) = 2/(1+sin x)

ii) take y = pi - x.
=> dy = -dx. Substituting and noticing that sin x = sin(pi - x) gives that integral( xf(sin(x) ) = integral( (pi-x)f(sin x)).
We use this fact in evaluating I = integral(x/(1+sin x)). We get 2I = integral(pi/(1+sin x)). Hence I = pi/2.integral(1/(1+sin x)).

iii) let (1+sinx)^2 = t for simplicity of writing this answer.
now J = integral ( (2x^3 - 3pix^2) / t )
Let J1 = integral(2x^3/t) and J2 = integral(-3pix^2/t).
so J = J1 + J2.

Now, lets solve J1.
J1 = integral(2x^3/t) ....................(1)
=> J1 = integral( 2(pi-x)^3/t ).......(2)
Expand the cubic term and add (1) and (2).
you will get 2J1 = integral( 2(pi^3 - 3pi^2x + 3pix^2)/t ).
So J1 = integral( (pi^3 - 3pi^2x + 3pix^2)/t ).

Now consider J = J1+J2.The 3pix^2 term gets cancelled.
so J = integral( (pi^3 - 3pi^x)/t ).
Integral 3pi^2x/t = 3pi^3/2 integral((1/t))
So J = -pi^3/2 integral( 1/t ).
=> J = pi^3.integral(sec^4z/4) for z = pi/4-x/2.

sec^4z can broken into sec^2z(1+tan^2z) and integrated to get tanz + tan^3z/3.
replace z and put the limits. you will get the answer.

The final expression is (pi^3)/4.(tan z - tan^3 z/3) for z = pi/4-x/2. thus, putting limits 0 and pi gives -2pi^3/3.
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DomStaff
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#8
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(Original post by Gawain)
Q5)

Base case:  \displaystyle S_1=\arctan(\frac12)

 \displaystyle \tan(S_1)=\frac12=\frac{1}{1+1}

Assume  \displaystyle \tan(S_k)=\frac{k}{k+1}

Then  \displaystyle \tan(S_{k+1})=\tan(S_k+\frac{1}{2(k+1)^2})

After simplifying with the tan addition formula we get:

 \displaystyle  \tan(S_{k+1})=\frac{(k+1)(2k^2+2k+1)}{(k+2)(2k^2+2k+1)}=\frac{k+1}{k+2}

If  \displaystyle \tan(a)=b and  \displaystyle b \geq 0 \; 0\leq a &lt;\frac{\pi}{2}

then we can say

 \displaystyle a=\arctan(b).

Now since we have  \displaystyle \frac{n}{n+1} \geq 0 and  \displaystyle S_n &lt; S_{\infty} = \frac{\pi}{4} &lt; \frac{\pi}{2}

We can conclude that  \displaystyle S_n=\arctan(\frac{n}{n+1})

(induction is a safer bet here but I couldn't be bothered to type it up at the moment.)

ii)

 \displaystyle \tan(2 \alpha) = \frac{2\tan(\alpha)}{1-\tan^2(\alpha)} = \frac{4n^2}{4n^4-1}

After solving for tan we get  \displaystyle \tan(\alpha) = -2n^2  \; , \tan(\alpha)=\frac{1}{2n^2}

Therefore using our previous result and taking the limit to infinity we get:
 \displaystyle S_{\infty}=\sum_{n=1}^{\infty} \alpha_n = \arctan\left (1- \frac{1}{\infty} \right )=\frac{\pi}{4}
It is probably slightly easier to notice:

 \dfrac{4n^2}{4n^4 -1} = \dfrac{ \dfrac{1}{2n^2} + \dfrac{1}{2n^2}}{1-(\dfrac{1}{2n^2})^2}

From which the conclusion follows immediately.
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DomStaff
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You might as well look for Physicsmaths's post in the STEP thread and link that for the final part of Q6.
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lllllllllll
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For Q11 you had to do CLM and conservation of energy. Then you had to show the something about the second collision. The last part you had to show whether theta reached a particular value. That's all I remember. I think we'll be needing a paper to confirm most of the questions
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username1432489
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(Original post by Gawain)
STEP II 2015

Q1) team-f-maths (attached to first post)
Q2) Gawain
Q3) Gawain
Q4) team-f-maths (attached to first post)
Q5) Gawain
Q6)
Q7)
Q8) Gawain
Q9)
Q10)
Q11)
Q12)


I'll upload Q2 once I've managed to upload my diagram.
The last part of Q4 isn't right here. Obviously ignore the vertical lines, and translate it (0 pi) The graph doesn't look like an arctan(x) graph for (iii). Name:  Screen Shot 2015-06-18 at 16.22.58.png
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username1162972
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Someone gona do Q10? I can't remember it. I did not have time to finish it in the exam. I wrote down a load of crap hoping it was right last minute.


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username1162972
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#13
Here is the solution to 6.
Name:  ImageUploadedByStudent Room1434653046.825182.jpg
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Did not do the first parts as they were trivial.


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Kolasinac138
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How much would that last part be out of , d you think?
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raff97
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#15
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(Original post by Chazatthekeys)
The last part of Q4 isn't right here. Obviously ignore the vertical lines, and translate it (0 pi) The graph doesn't look like an arctan(x) graph for (iii). Name:  Screen Shot 2015-06-18 at 16.22.58.png
Views: 1272
Size:  10.7 KB
Actually the graph looks exactly like that for x<-1, translated up pi for -1<x<1, and translated up 2pi for x>1. So it all joins up at -1 and 1.

Its h(x) = arctan(x/(1-x^2)) + npi, where n varies depending on where you are
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username1162972
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#16
(Original post by raff97)
Actually the graph looks exactly like that for x<-1, translated up pi for -1<x<1, and translated up 2pi for x>1. So it all joins up at -1 and 1.

Its h(x) = arctan(x/(1-x^2)) + npi, where n varies depending on where you are
Weird. Guess I got that wrong


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raff97
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#17
(Original post by physicsmaths)
Weird. Guess I got that wrong


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I really want DFranklin and shamika come in and look at our solutions for STEP I and II. Have we still not got our hands on a paper? They should definitely be here for III tomorrow morning
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charles98
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#18
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#18
Can someone please do Q9?
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username2037821
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Nitpick: In the answer to Q4(ii), arctan(1/2) isn't pi/4
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username2037821
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7.(i) It is clear that there are many options, simplest is perhaps choosing a circle with centre on the x-axis that intersects at (0,a) and (0,-a). Our circle has equation  (x-c)^2 + y^2 = r^2 Subbing in for the points of intersection, we get c = \sqrt{r^2 - a^2} .

If  r &lt; a, the circle of radius r cannot intersect C at diametrically opposed points, since any two points are at most
 2r &lt; 2a apart. And if  r = a, if there were two such points of intersection, the circles would need to overlap, which doesn't satisfy the exactly two condition.


Name:  IMAG0333.jpg
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(ii)We can see from the diagram that we have two isoceles triangles, each defined by the centre of D and the points of intersection on the circles C1 and C2. Let us call the height of these triangles z_1 and z_2 respectively. We have {z_i}^2 = r^2 - {a_i}^2. Now consider the triangle defined by the centre of D, which we will give coordinates of (p,q), and the points   (d,0)  and   (-d,0) . Adding a perpendicular from the centre of D to the x-axis, pythagoras gives us

{z_1}^2 = (d+p)^2 + q^2 = d^2 + p^2 +2dp + q^2  (1)
and
{z_2}^2 = (d-p)^2 + q^2 = d^2 + p^2 - 2dp + q^2  (2)

Subtracting (2) from (1), we obtain:

{z_1}^2 - {z_2}^2 = 4dp

Substituting for z_i in terms of r and  a_i and rearranging gives the required result for p.

Now to find q, we add the equations (1) and (2), and substitute for p, resulting in
q^2 = r^2 - \frac{({a_1}^2 + {a_2}^2)}{2} - d^2 - \frac{({a_2}^2 - {a_1}^2)}{16d^2} . Note that q could be either positive or negative, since there are 2 possible circles D reflected on the x-axis.

Clearly  q^2\geq 0, so r^2 \geq \frac{({a_1}^2 + {a_2}^2)}{2} + d^2 + \frac{({a_2}^2 - {a_1}^2)}{16d^2}, and a little more rearranging gives the final result.
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