| Chris
Tynan |
1) (iii) Show that: 12 -22 +32 -42 +(-1)n-1 n2 = (-1)n+1 (An2 +Bn) where A and B are constants to be determined. By removing the factor of (-1)n+1 and taking the sum of squares, I get the answer is: (-1)n+1 /6.n.(2n2 +3n), but I can't get it the form requested. This is driving me mad, it's probably something very simple but I just can't get it. 3) Quite a nice question, but not so nice to explain The Tour de Clochemerle is not yet as big as the Tour de France. This year there were five riders, Arouet, Barthes, Camus, Diderot and Eluard (I abbreviated them as A-E for ease), who took part in five stages. The winner of each stage got 5 points, the runner-up 4 and so on. The total number of points scored over the five stages was the riders score. Each rider obtained an overall different score and the riders finished the tour in alphabetical order with Arouet gaining a magnificent 24 points. Camus showed consistency by gaining the same position in four of the five stages and Eluard's rather dismal performance was relieved by a 3rd place in stage 4 and a 1st in the final stage. i) Show Eluard must have recieved 11 points overall ii)Find the scores obtained by Barthes, Camus and Diderot. iii)Where did Barthes come in the final stage? I have an answer to this but don't want to post it yet since I think this is a nice problem. I just want to check with others that my answer is correct. Chris |
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| Kerwin
Hui |
1(iii) One way to do this is to first find out what A,B should be (using the case n=1,2, for example), and proving this works is just a matter of computation. Q3 is a very nice problem, needs no more than about 5 minutes if one thinks clearly. Kerwin |
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| Chris
Tynan |
I've gotten A and B = 1/2, after this (in STEP) would induction be OK to use? Chris |
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| Kerwin
Hui |
Yes, the reason for using n=1,2 to find A,B is that you only need the induction step to complete the question. |
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| Philip
Ellison |
I did question 3 during preparation for my Cambridge interview (it's what my teacher calls "mathsless" maths questions... ones that require little mathematical knowledge, but can still be challenging). You should be able to check it yourself... just check that all the criteria are fulfilled by your answer! |
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| Alice
Thompson |
Simple cases of proof by induction are on the STEP II specification. Have you seen the new booklet of advanced problems at http://www.maths.cam.ac.uk/undergrad/advancedproblems/ |
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| Mark
Durkee |
I got B=15, C=13, D=12, with Barthes last in the final stage. |
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| Chris
Tynan |
q(6) from the same paper: Let . Show that is a root of and find the other roots in terms of . The number is a root of the quadratic equation where and are real. Find the numerical values of and . For part one, I thought that , where is an integer, would suffice. As for part two, I'm completely stuck of a nice solution. Thanks, Chris PS. Mark, I agree with you |
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| Alice
Thompson |
I get that: Which from my calculator = 1, but I'm having difficulty proving it. |
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| Chris
Tynan |
I perfectly agree with you Alice, however the next part of the question is to prove your above sum and the sine equivalent, so I imagine it can be answered in some other way. Note: A=-u-v and B = uv where u and v are the two roots, in case you didn't already know this. Chris |
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| Demetres
Christofides |
The second part is just a couple of lines long. Hint: A and B are real. What does this say you about the other root of the quadratic ? Demetres |
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| Mark
Durkee |
I've got number 6 out (I think) If you say (which was you had to show in the 1st part which Chris didn't post) then (as the 2nd root is the conjugate of the 1st - by looking at the roots on the Argand Diagram you can see what the conjugate is) But the sum of the seven roots found above =0 so Therefore, For , say Therefore, Multiplying out and simplifying the powers of that are > 6 gives: (as the sum of the roots is 0) The final part of the question was: Show that and find the value of I did this by comparing my answers for and with their values in sin and cos form (like Alice found). I got . The question asked you to show carefully how you obtained the sign of your answer (since it was obtained by square rooting something) - I did this graphically by showing that was more positive than is negative - does anyone know if this would be considered sufficient in a STEP exam? |
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| Chris
Tynan |
and are both greater than and is more than so it's positive... Thanks for the rest of the question Chris |
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| Chris
Tynan |
Q(7)(iii) A hole in the shape of the triangle ABC is cut in the top of a level table. A sphere of radius R rests in the hole. Find the height of the centre of the sphere above the level of the table-top, expressing your answer in terms of a,b,c,s and R, where s is the semiperimeter. In the two earlier parts I've proved Area ABC = rs and Heron's formula. I've just done this now by letting r as the radius of the incircle of ABC. Now, rs = sqrt s(s-a)(s-b)(s-c), r = sqrt s(s-a)(s-b)(s-c)/s and since R2 -r2 =h2 where h is the height, we get h = sqrt R2 -((s-a)(s-b)(s-c)/s). Is this right? Chris |
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| Kerwin
Hui |
That is correct, assuming R> r. Out of curiousity, how did you prove Heron's? Kerwin |
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| Chris
Tynan |
Well, it must be said that the question offered a fair bit of guidance... Using Area = 1/2bcsin[a] we get A2 =1/16(4b2 c2 -(2bccos[a])2 ) Using 2bccos[a]=b2 +c2 -a2 and a bit of factorising gives us A2 =1/16(a2 -(b-c)2 )((b+c)2 -a2 ) and it follows that A = sqrt s(s-a)(s-b)(s-c) The question made the whole deduction quite easy... Chris |
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| Mark
Durkee |
I thought that this formula was Hero's formula, not Heron's??? Did you get anywhere with Q5 on this paper? 5. The famous film star Birkhoff Maclane is sunning herself by the side of her enormous circular swimming pool, centre O at a point A in its circumference. She wants a drink from a small jug of iced tea diametrically opposite point B. She has 3 choices: a) To swim directly to B b) To choose x with 0 < x < pi to run around the pool to a point X with angle AOX = x and then swim directly from X to B c) To run around the pool from A to B She can run k times as fast as she can swim and wants to get there as fast as possible. Explain which of (a), (b) and (c) she should take for each value of k. Is there one choice that she will never take? Writing an expression for time in terms of x (treating (a) as x=0 and (c) as x=pi) and differentiating showed me that there are no minimums in this range - does this mean that the time can never be minimised by taking option (c) and that (a) or (c) must always be the answer - if so this seems to me to be a rather surprising result. (It is obvious that (a) is best for k < = 1, and (c) is best as k tends to infinity, and if (b) is never correct then is is not that hard to find the ranges where each of the others are best) |
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| Colin
Prue |
if you think about the geometry of the problem it can be seen that (b) would NEVER be a solution. that is, while swimming you are always heading towards the drink. while running you are only heading directly towards teh drink at the limit as you approach it, but your path does become more direct the further around the circle you go. therefore, as you run around the circle it becomes INCREASINGLY beneficial to keep on running, but NO MORE beneficial to swim. therefore if it was worth running to begin with it is worth running at any point after this decision. this geometrical argument does not, however, help in nailing down a range, only in disregarding (b) |
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| Colin
Prue |
whoops! silly statement, of course it is now possible to specify the range!: k < pi/2 we swim, k > pi/2 we run! is it not an option to swim to X and then run the rest of the way, as this could be beneficial for some k...!! |
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| Colin
Prue |
no it couldn't - you would never combine the two as that would require you to do the slow action (swimming) non directly. have i read this correctly? is A diametrically opposite to B? |
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| Chris
Tynan |
I have said, WLOG, the diameter is 2m and she swims at 1ms-1 . So if she swims it will take her 2s and if she runs it will take her pi/k seconds. From this we see Colin's 3:13 statement is correct concerning k. Now it's left to rule out (b). Say we pick a point X with < AOX=t then |XB|2 =2-2cos(180-t) by cosine rule, so |XB|=2cos[t/2]. Using this, it takes t/k secs. to run to X and then 2cos[t/2] to swim to B, so the total time is: t/k + 2cos[t/2] which seems to hold no use whatsoever and so I agree with Colin's earlier post about (b). I can see that, since you maintain a constant running speed, if you run at the start, then swim later, there is little point, because whatever the value of k, you could either run the remaining distance faster than you can swim directly, or the converse. But I'm still not totally certain about the argument... Chris |
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| Chris
Tynan |
By the way Mark, Heron's/Hero's formula are the same thing: http://mathworld.wolfram.com/HeronsFormula.html Chris |
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| Kerwin
Hui |
The second derivative is always negative, so it cannot be a minimum. Kerwin |
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| Demetres
Christofides |
The correct name as translated from Greek should be Heron. Demetres |
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| Mark
Durkee |
Thanks to all (Colin in particular - it was an argument like yours that I needed to convince myself that what I had derived by differentiating could be correct) Mark ps I suppose that Heron's name got translated incorrectly because someone thought that Hero was a better name for a famous mathematician than Heron! |
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| Chris
Tynan |
Thanks Kerwin, that seemed to be quite involved just to rule out (b). Is there not a simpler method? |
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| Kerwin
Hui |
That is only 2 lines of working at most: You know that ![]() So for (case (b)):![]() Kerwin |
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| Chris
Tynan |
I was also referring to the preceding working to work out the distance of XB and the total time, etc. But I agree it's not difficult at all in hindsight. Chris |