Geoff Smith

UKMT volunteer

[Geoff Smith @GeoffBath]

@SupportRequests #0310630823

[Geoff Smith @GeoffBath]

@ProfSmudge 1/n -1/(n+1) is a decreasing function of n.

[Geoff Smith @GeoffBath]

@PStilance @Cshearer41 Swap x and y to be consistent with Piotr's notation.

[Geoff Smith @GeoffBath]

@PStilance @Cshearer41 You don't need to determine x and y. The intersecting chords theorem (twice) and subtraction give x-y =2 and substituting back y^2+8y-4 =0. Now (x+y+6)^2 = 4(y^2+8y+16)= 80. The area is therefore 10π.

[Geoff Smith @GeoffBath]

@HoneywillTim Here is a fast argument. Consider any bead which has a predecessor. There are 5 beads different to the predecessor and 4 the same, so the probability that the beads are diffeent is 5/9. Generalizes of course. @Ridermeister

[Geoff Smith @GeoffBath]

@HoneywillTim This seems closely related to the expected waiting time until HT compared to e.w.t. for HH when tossing a fair coin. There are neat martingale arguments for those different answers.

[Geoff Smith @GeoffBath]

@HoneywillTim Not the obvious answer, unless your notion of obvious is carefully chosen. Try 2 reds and 2 blues by hand!

[Geoff Smith @GeoffBath]

@sonukg4india Typo. 5/2.

[Geoff Smith @GeoffBath]

@sonukg4india An enlargement with scale factor -3/2 from the coloured circles contact point carries the red circle to the yellow circle. So an enlargement with scale factor 3/2 from the far left point carries the red circle to the plain circle. So the answer is 5.

[Geoff Smith @GeoffBath]

@sonukg4india The cutting if circles in half is just done to confuse 😀

[Geoff Smith @GeoffBath]

@blatherwick_sam .. and use the spirit in other ways. y = ln x? Then e^y = x. so y' e^y = 1 so y' = 1/x.

[Geoff Smith @GeoffBath]

@blatherwick_sam Wonderful method. Use it to derive the quotient rule, or to avoid the quot. rule when differentiating x^{-n}, or tan x thought of as sin x / cos x, or cot, sec, cosec etc.

[Geoff Smith @GeoffBath]

@Pouriaaa @janemerrick23 Many congratulations: richly deserved.

[Geoff Smith @GeoffBath]

@UKMathsTrust This is a very demanding exam. Fee waivers have been awarded to high scorers in BMO1 who are eligible to represent the UK in international competition. Anyone else considering entry should look at past papers first, to make sure that they will enjoy the experience.

[Geoff Smith @GeoffBath]

This World Cup has disrupted the Premier League for far too long. Looking forward to football between well coached teams, mediated by officials who are mostly more competent than those recruited by FIFA.

[Geoff Smith @GeoffBath]

@Statsyman @Ridermeister @goyalam @UKMathsTrust Correction. The final "Qualification" should read "Participation".

[Geoff Smith @GeoffBath]

@Statsyman @Ridermeister @goyalam @UKMathsTrust Matters are in hand to eliminate the possibility of double dipping competitions in China. As for certificates: they are already discriminated. Fee waiver entrants get a Qualification certificate, whereas normal fee-paying students get a Qualification certificate.

[Geoff Smith @GeoffBath]

@ProfDPrabhat The office think that the problem is fixed.

[Geoff Smith @GeoffBath]

@ProfDPrabhat I have had an assuranc that the UKMT Office has identified the problem and is working on it. Try again tomorrow. I hope that you enjoy it. I am just finishing Mathematical Olympiad Primer II. Hope to publish in 2023.

[Geoff Smith @GeoffBath]

@ProfDPrabhat Very annoying. As you may know, there were governance issues with the Trust recently. That is now resolved, but the administration is currently done by a skeleton staff. Sorry. I will chase them.