The instruction about guessing is a bald-faced lie though, right? You should definitely guess (rather than leave blank) for questions 1-20.
I liked the names they chose, which are much less ham-handed than most US word problems aiming for diversity. Amrita, Jacob, Bertie, Tegwen, and Aimee are a nice little group!
Tatsu just did this through school. I'm not sure when the results will be back, or if he'll even tell me when he gets them, unless his score is high enough for him to want to boast about it.
"Amrita is baking a cake today. She bakes a cake every fifth day. How many days will it be before she next bakes a cake on a Thursday?"
Surely this depends on when you take the test?
4: Why would it make any difference if you took the test in the morning or afternoon?
1. Because of odds somehow? That doesn't seem like enough questions to be very sure you'd get any right
The questions on this paper challenge you to think, not to guess. You get more marks,
and more satisfaction, by doing one question carefully than by guessing lots of answers.
The UK IMC is about solving interesting problems, not about lucky guessing
I don't think it's telling you not to guess, if you don't have a clue or run out of time.
1,6: wouldn't it depend on the scoring and whether they disproportionately deduct for wrong answers?
7: How do you know? Did you get a piece of Amrita's cake?
No, it is telling you not to guess. A correct answer gains you points, and incorrect answer loses points, and nothing does nothing.
9: Five marks are awarded for each correct answer to Questions 1-15.
Six marks are awarded for each correct answer to Questions 16-25.
Each incorrect answer to Questions 16-20 loses 1 mark.
Each incorrect answer to Questions 21-25 loses 2 marks.
So guessing is definitely better than nothing on 1-15.
Never mind! I just re-read it with my math professor and figured it out.
Because my birthday was on a Wednesday.
I also liked the "dumb machine" explanation and think kids would respond better to it than the typical US version. I'm reminded of something Lisa Delpit cites in one of her books about a black teacher reminding her black students before a standardized test that they need to remember this is going to be graded by a bunch of little old white ladies in Tallahassee and the teacher already knows how much they know but they need to be able to explain it in ways the Tallahassee ladies will understand, which seemed like a useful remark on code-switching.
15: So you had to wait a day to get your birthday cake, because Amrita doesn't make any exceptions to her baking a cake every 5 days policy?
Ooh--the irregular polygon where you have to determine the value of one of the angles is unlike any geometry problem I had at that time. (Solved it by treating it like a vector problem.) Neat to see the small differences in style between countries.
Just barely got answers to all 25 in one hour. Don't know if they're right!
Didn't like the irregular polygon at all.
small differences in style
Also, vocabulary. "Trapezium" means trapezoid, I guess, based on problem 21. "Arrowhead" I assume means there is an interior angle greater than 180 degrees. I would have guessed that "kite" means a quadrilateral with one axis of symmetry, but then I think the problem doesn't work. Am I missing something, or what the hell is a kite?
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HASTY BAT SIGNAL
Any lawyers here? I need to know something about the application of the principle of joint enterprise in gang cases in the US and I need to know it quickly. Could anyone interested drop me a line at my real.name@liberal.newspaper.co.uk
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12 - on 1-20, right? And then not beyond that. What a weird way to set up incentives. I guess the idea is that everyone is guessing in 21-25.
OK, a kite is what I think it should be. I guess this thing is a trapezium.
21 -- i don't think there are any criminal lawyers here, but I can put you in touch with a very good one very quickly if you need it for a story
"Definitions for trapezoid and trapezium have caused controversy for more than two thousand years." Teach the controversy!!
20/23: I recall either being taught kites or helping younger students with them. My recollection (haven't checked it against the problem) is that a kite is a quadrilateral such that each side has an adjacent side of the same length, like a degenerate rhombus. I think that means they have two axes of symmetry, but that doesn't line up with what you have so I must be off.
Oh, I see--you could have a non-convex kite with only one axis.
But that would probably be an "arrowhead," so maybe convexity is necessary? I dunno.
HASTY BAT SIGNAL
At a glance, this looks like an anagram of my real name.
15: So you had to wait a day to get your birthday cake, because Amrita doesn't make any exceptions to her baking a cake every 5 days policy?
That's ridiculous. I have to wait until Amrita bakes one on a Wednesday. I'm expecting my cake March 2nd.
Is there a non-hasty version of the Bat Signal?
"Hey, Batman, this isn't urgent or anything, but somebody keeps taking my potato salad from the refrigerator in the break room. Could you look into that when you get a chance? Thanks!"
I violate heebie's anonymity by revealing that her real name is Tabitha S. Galnys. Or possibly Alia Stabshytgn. (In old country of course would be Alia Stabshytgnova.)
31 reminds me of that hilarious Christopher Walken Dead Zone trivial psychic sketch, " At lunch.. you're gonna treat yourself to a vanilla ice cream.. you're gonna eat it too fast.. you're gonna get an ice cream headache.. it's gonna hurt.. real bad... for eight, nine seconds.."
16: I wonder if that teacher's approach might backfire in terms of stereotype threat (if I'm remembering the term correctly).
35: I don't know! I'm probably not giving a good rendition of it, but I definitely taught test-taking skills as a form of code-switching when I helped the kids at church with ACT prep and so on and I think it's an important conversation to have, though maybe it backfires for that reason. This particular example seemed particularly empowering to me, but I'm not sure how it could best be addressed.
That seems like a stereotype threat they will encounter regardless, so perhaps repeated, explicit exposure helps?
Tracked down the answers; I messed up on #4 (the cake timing) by assuming that it had to be an answer that was independent of when today was and on #19 (list of integers with mean, median, mode) by forgetting the rules about medians in a set with an even number of elements.
In retrospect, none of my actual math classes involved nearly as much geometric construction or analysis as this test. Maybe the SAT-ish tests tended to? I'm not sure where else I would have picked it up.
So I'm almost as good at math as a pretty-good 16-year-old? At my age, I'll take that.
It took me a bit under an hour, although I was both distracted and had thought about some of the questions before. The coin one was tricky and fun; the last one was a bit of a pain since I couldn't remember how to derive the formula for 30/60/90 triangles and just guessed it (which happened to be right, but I got lucky there).
Nope, messed up that last one--no surprise since I was panicking to finish it quickly. But everything else was right, which I think gets me a score of one point less than Nathan. I'll take it.
Err, maybe a bit more, because he got no points for 4 (which was kind of silly about the date--that's a little meta). Whatever. I have an unfair advantage since I work for a company that works with high school math stuff so this is probably fresher in my mind than most.
Answers. Whoops, meant to post those in a timely manner.
38, 39 - if you're getting about 23 out of 25 right, you're as good at maths as a very good 16 year old. From memory, 62 gets you a gold medal this year, and 99 gets you through to the Olympiad round. Which is slightly lower I think than last year, so this year's was presumably a bit harder. The instructions are well written because the UKMT are sensible people, not government employees.
Kid C's a good mathematician (although not IMO standard, probably too old already, lol - the youngest boy in the current team of 6 has been in it since he was 14). Last year he got full marks in the intermediate challenge, this year (he's 15 now) he reckons he's got 113/135 (four questions wrong?).
He's done the senior challenge for the last couple of years - https://www.ukmt.org.uk/individual-competitions/senior-challenge/ if you want more - which potentially gets you through to the British Mathematical Olympiad rounds. He did pretty well in BMO1 - as he's younger there was a lower threshold to get through to BMO2. Waiting atm to hear the results for that, but he's crossing his fingers not to have got zero! (Which is the most common score.)
https://bmos.ukmt.org.uk/home/bmo.shtml If if you want to have a look at the BMO 1&2 papers. I can generally hope to do the first three questions on BMO1 and nothing on BMO2!
Tried doing the last question on the most recent BMO 2 since I've been doing some abstract algebra in my free time and it's devolved into a nasty case analysis that probably wouldn't be necessary if I could grok what's really going on. I couldn't imagine doing these under pressure, which is why I never took the Putnam exam.
43. Yeah, for 2015 on BMO1, I *miiight* be able to do 3 or 6 if I sat a while. On BMO2, all is lost. Perhaps I could make something out of 2, but I'm not betting anything without amazing odds.
I hate the putnam! Or rather, I feel like the putnam is emblematic of all the extra math scaffolding and support systems that I never knew existed.
One of my prouder moments this past year was solving a Putnam problem posed to me at my neighborhood bar after having downed a bottle of wine. I just got 21 out of 25 on this test. I obviously need to drink more.
I'm really rusty at quick arithmetic and I made a few careless logical errors. I did spot myself the trapezium/arrowhead nonsense. The geometry came easier than I expected it to.
46: Every time I look at one of those exams I feel like I would spend most of the time deciphering the questions and notation. The ones I can actually understand often turn out to be kind of fun with elegant solutions (that I would in no way be able to come up with in the time limit).
45: I think I found a pretty simple and elegant solution to 2015 BMO1 question 6 (the one about charming numbers), but I'm somewhat convinced I screwed it up since I don't need any of the powers of 5 to ever be greater than 1.
46: I took a one credit Putnam prep class in college, and yes, it revealed that yes, everyone else at my school had a much better math education than I did and that everything would be an uphill battle.
I didn't know that people studied for the Putnam until I was well-into grad school.
46, 47: I don't see solutions posted for sufficiently old Putnams. Let's pick a fun problem from 1988 and solve it in these little boxes.
All I remember about series convergence is the integral test, so I say either A-3 or B-4. For A-3, series expansion of the sine, so the quantity in parens can be evaluated, series less the first term over the series, does that go anywhere?
A-2 doesn't seem that hard. So we need to find a g and an interval such that f'*g' = f*g' + f'*g, where f(x)=e^(x^2). f' = e^(x^2)*2x, so
f'*g' = f*g' + f'*g
e^(x^2)*2x*g' = e^(x^2)*g' + e^(x^2)*2x*g
exponentials are never zero, so
2x*g' = g' + 2x*g
g'(2x-1) = 2x*g
g'/g = 2x/(2x-1) (x != 1)
So suppose that g is something like e^(h(x)). Then g' = e^(h(x))*h'(x), which would mean g'/g = h'(x). So this reduces to finding a function that has the derivative 2x/(2x-1) on some interval that doesn't include 1. That should be an easy integration, not that I quite remember how to do that.
Those are easier reading than I'd remembered. But I can't be sucked in. I mustn't.
A-1 is easy, right? It's twice the area of a trapezoid with bases 2 and 4 and height 1? Or did I fuck up the inequalities?
Looking at these makes me feel dumb and out of practice (and like I don't have time to wrestle with them right now -- I'm still looking forward to having time to go through the problems in the OP). That said, it's nice when they look manageable:
A-1 is easy, right? It's twice the area of a trapezoid with bases 2 and 4 and height 1?
That's what it looks like to me.
Also, for A-4. it looks like you can avoid having adjacent points of the same color by filling the space using consecutive diagonal lines of color. E.G, for a 6x6 grid
ABCABC
BCABCA
CABCAB
ABCABC
BCABCA
CABCAB
Nick
For the first part of A-4, if you tile the plane with equilateral triangles side length 1 and then try to paint every vertex, there is only one way to do so up to symmetry (each triangle must have unique colorings on each corner). If we then try to overlay a second tiling, the vertices must each fall within one of the first set of triangles, and thus will be within 1 inch, and thus be uncolorable.
I don't understand a thing about math, but somehow I knew Eggplant was going to get sucked in.
57: Ah, I was misreading the question (which I thought was likely).
Wait, I misread the question. I missed the "exactly".
I HAVE WORK I NEED TO BE DOING PEOPLE.
Thinking about regular tilings won't answer A-4, I'm pretty sure. For two colors, consider a black point and the circle centered on it of radius one. I don't see how to proceed for three colors though, and I don't think 9 is solvable.
I think A-1 is the interior of the triangle (2,1, -2,1, 0,-1).
Oh, hey, Nick, have you heard of this Brazilian vinyl project?
http://www.thevinylfactory.com/vinyl-factory-releases/inside-the-worlds-biggest-record-collection-an-interview-with-zero-freitas/
50
I think that's correct, though I can't figure out how to state a formal proof. My informal proof is that you can cover the integers by addition using a basis set, as long as you have no consecutive pairs of integers in said basis set with ratio greater than 2. The powers of two form just such a basis set and you can clearly create every integer by summing unique powers of two. So, starting with 3 (and taking 1 and 2 as given), you can always multiply by 5/3 (1.67) or 9/5 (1.8) to get the next member of the basis set and they will always be closer together than 2x. And more strongly, you never need more than one power of 5 in your basis set.
Oh, hey, Nick, have you heard of this Brazilian vinyl project?
No I hadn't; that's wild. Thanks for the link.
62.2:
The solution has to be symmetric with respect to both the x and y axes.
Any fully colored solution will be a superposition of correctly colored lattices, and any rotation around a vertex yields a lattice that can be colored at most two ways. Seems like some set of rotations should result in a contradiction.
Vertices in straight lines necessarily rotate through the three colors (re green blue red green blue), so consider a 3-4-5 triangle...
Crap, that still isn't sufficient. Uh, consider a 3-3-2 triangle...
A thousand times 61. Math people are like vampires; throw numbers at us and we are lost in this useless world.
I'm immensely sad about Eco's passing. I loved A Name of the Rose and Focault's Pendelum.
A thousand times 61.
61,000! That was way easier than these other problems.
Anyway, all my flailing adds up to a proof by contradiction. And let's agree to pretend I said 3-3-1 on the first go. The other part seems relatively easy: tile the plane with a parallelogram made from 8 differently colored triangles with suitably chosen open boundaries.
64: Sorry F, I somehow completely missed your post last night. That's really good intuition; I just cranked out something algebraically kind of mechanically.
So here's my idea: Any positive integer can be represented in ternary. The terms of a ternary representation are almost charming numbers, except some might have a coefficient of 2. We can get rid of these fairly easily; note that if n >= 1,
2*3^n = 6*3^{n-1} = 5*3^{n-1}+3^{n-1}
(We can't do this for 2*3^0, but 2 is conveniently also included as a charming number.) So we can use this property as an operation lets us replace a multiple of 2 with two charming numbers. If we apply this process starting at the largest ternary digit and working towards the lowest, we know that the 5*3^{n-1} won't cause any issues as there are no multiples of 5 in the part of the ternary representation we haven't worked on yet; alas, there might be a multiple of 3^{n-1} depending upon what the (n-1)th digit of the ternary representation is. So, three cases:
0*3^{n-1}: No problem.
2*3^{n-1}: (doing this out of order since it makes the next case easier) Applying the operation above, we have
2*3^{n}+2*3^{n-1} -> 5*3^{n-1}+3*3^{n-1} = 3^n+5*3^{n-1}
This is really convenient since we just removed two 2's and it doesn't require any iteration.
1*3^{n-1}: Again, applying the above operation,
2*3^{n}+3^{n-1} -> 5*3^{n-1}+2*3^{n-1}
So in this case we've moved the 2 down a power. This doesn't reduce the number of 2's, but it still makes it easier: if we iterate this, we keep on pushing that 2 down until we come across another one and either we do the 2*3^{n-1} case above, or we push it down to the 2*3^0 base case.