Royal Statistical Society Christmas Quiz.
December 21, 2018 4:38 AM   Subscribe

"Solving the RSS’s fiendishly tricky festive quiz will require general knowledge, logic and lateral thinking." [Spoilers for the quiz] Metfites everywhere are invited to join me in trying to tackle this notoriously difficult quiz. I've given my own handful of answers so far below the fold.

1a) Twas the night before Christmas and all through the house not a creature was stirring not even a mouse.

2a) Art Garfunkel.
(Simon & Garfunkel first worked together as Tom & Jerry. Garfunkel is the younger of the two men. Military rank seems to be the theme here, and he had an early solo B-side called "Private World".)

2h) Sacha Baron Cohen's Ali G.

4a) Motorhead is the odd one out. Their most famous song is Ace of Spades, but all the others are dealing queens.
(KC & The Sunshine Band: Queen of Clubs / Tom Odell: Queen of Diamonds / Styx: Queen of Spades / Dave Edmunds: Queen of Hearts.)

That's all I've got so far - and all I'm likely to get on my own too.
posted by Paul Slade (26 comments total)

This post was deleted for the following reason: at the original poster's request, in deference to the wishes of the quiz organizers. For people seeking other quiz fun, this year's King William's College General Knowledge quiz thread is live. -- LobsterMitten



 
Q6. The top right image is from Five O'Clock Tea by Mary Cassatt.
posted by snakeling at 5:13 AM on December 21, 2018


Oh wow...this quiz is making me feel...very not smart. I'll keep mulling it over, but...wow..
posted by sharp pointy objects at 6:20 AM on December 21, 2018


2d. Anna Gunn
posted by umber vowel at 6:31 AM on December 21, 2018


Q2b is surely Alec Guinness.
posted by ricochet biscuit at 6:57 AM on December 21, 2018


2b. Sir Alec Guinness
posted by Four Ds at 6:58 AM on December 21, 2018


2f Al Gore?
posted by Four Ds at 7:38 AM on December 21, 2018


I saw this on my RSS feed.
posted by Mr.Encyclopedia at 7:51 AM on December 21, 2018


2g is Amy Grant, who was born in Augusta, GA.
posted by coppermoss at 7:55 AM on December 21, 2018


Ooh, I know #8! Releases of R, of which there were four in 2018, get their names from Peanuts strips. Someone has compiled a list of them, and the 2018 releases with comic strips corresponding to those dates are "Joy in Playing" (3.5.0), "Feather Spray" (3.5.1), and "Eggshell Igloo" (3.5.2).

There were actually four releases in 2018 but the first one, 3.4.4, seems to be left out. Eggshell Igloo was just released yesterday.
posted by madcaptenor at 9:00 AM on December 21, 2018 [1 favorite]


12) No. (It's a simple logic problem. She's the head of the math department. She does not need my help to solve a math problem.)
posted by Garm at 9:32 AM on December 21, 2018 [1 favorite]


2e: Antoine Griezmann
- Jersey #7 for both French national team and Atlético Madrid
- As part of the French national team at the World Cup: won the World Cup (gold), the Silver Boot (2nd best scorer), and the Bronze Ball (3rd best player).

Common theme to #2 seems to be all their names start with A.
posted by Upton O'Good at 9:32 AM on December 21, 2018


There are ten different letters in #10 and they anagram to the word "hyperbolic". Not sure how to use that though.
posted by madcaptenor at 9:40 AM on December 21, 2018


Common theme to #2 seems to be all their names start with A.

All with the initials AG, in fact:

a) Art Garfunkel
b) Alec Guinness
c) ?????????????
d) Anna Gunn
e) Antoine Griezmann
f) Al Gore
g) Amy Grant
h) Ali G.

How this fits in with the question itself being headed "Bilko", I don't know. My earlier idea about the theme being military rank was evidently wrong, though.
posted by Paul Slade at 9:50 AM on December 21, 2018


Bilko = Phil Silvers
Silver = Ag
posted by Four Ds at 9:55 AM on December 21, 2018 [4 favorites]


Oh, and on 2' s followup question, both Garfunkel and Grant were on The Animal's Christmas.
posted by Four Ds at 10:28 AM on December 21, 2018


Four Ds, I was working my way around to the periodic table Ag and how that could involve fraud. when i scrolled down and saw your comment RE: Bilko. You were so many steps ahead of me!!
posted by Megafly at 11:43 AM on December 21, 2018


For Q5, I have an inkling it might be basketball as I can find NBA all stars that match up with a lot of the initials and in approximately the right order but not all.
posted by Four Ds at 1:40 PM on December 21, 2018


Q1, point 5 is from A Christmas Carol. "I am as light as a feather, I am as happy as an angel, I am as merry as a schoolboy. I am as giddy as a drunken man. A merry Christmas to everybody. A happy New Year to all the world. Hallo here. Whoop. Hallo."
posted by coppermoss at 6:53 PM on December 21, 2018 [2 favorites]


Assuming the 10 letters in Q10's first part each correspond to a digit, I brute-forced it with a Python script and got (COPIER, BICYCLE, PYRRHIC, HORROR, EERIER) for values H=8, Y=1, P=7, E=4, R=9, B=3, O=0, L=5, I=6, C=2.
posted by valrus at 7:46 PM on December 21, 2018 [2 favorites]


Ah - using the values I determined previously, Q10's title, "BICHHOO (YO!)," translates to "3628800 (10!)" (10! = 10 factorial = 10 × 9 × ... × 2 × 1 = 3628800) which is how many possibilities there were for the values of the 10 letters.
posted by valrus at 8:17 PM on December 21, 2018


Q11 - P.I.G. stands for "Primes In German." The first 25 primes are all numbers under 100, whose names in German are determined by the ones digit, e.g. 29 = "neunundzwanzig" = "nine and twenty."

The only "Z" is 2, "zwei." No other primes end in 2. The only "F" is 5, "fünf," similarly the only prime ending in 5.

After the 25th prime, you pass 100 and the story changes: all numbers 100-199 start with "einhundert" (or maybe just "hundert"), numbers 200-299 start with "zweihundert", etc. So the composition of the next 25 elements is 21 "E"s and 4 "Z"s.

The fifth element in the sequence is indeed also the shortest: the fifth prime, 11, is "elf" in German, which also fits the theme.
posted by valrus at 10:07 PM on December 21, 2018 [2 favorites]


Q12: ratio between the longest and shortest segments is sqrt(21) (and yes, the information provided is sufficient.)

I didn't end up using the number of threads, but mostly just properties of parallelograms. In particular:
  • if you have a quadrilateral whose opposite sides are the same length, it must be a parallelogram and the opposite sides must be parallel
  • if you connect the opposite corners of a parallelogram, two resulting lines will bisect each other
Consider a parallelogram with sides of length A and B and diagonals C and D. If we call T the angle between A and C, then by the law of cosines we have B2 = A2 + C2 - 2AC cos(T) for the triangle bounded by the far side and the diagonal, and (D/2)2 = A2 + (C/22 - 2 A (C/2) cos(T) for the triangle bounded by and the two half-diagonals. The first equation gives 2AC cos(T) = A2 + C2 - B2, and substituting into the second equation we get D2/4 = A2 + C2/4 - 1/2(A2 + C2 -B2) or after some manipulation D2 = 2A2 + 2B2 - C2.

The above equation will hold for any parallelogram, and we have five of them in the figure, giving five constraints:
  1. red and yellow border, blue and red diagonals: (blue)2 = 2(yellow)2 + (red)2
  2. red and green border, orange and red diagonals: (orange)2 = 2(green)2 + (red)2
  3. red and magenta border, blue and orange diagonals: (blue)2 = 2(red)2 + 2(magenta)2 - (orange)2
  4. green and magenta border, blue and yellow diagonals: (blue)2 = 2(green)2 + 2(magenta)2 - (yellow)2
  5. all-magenta border, magneta and orange diagonals: (orange)2 = 3(magenta)2
For the sixth constraint, we look at the two nested isosceles triangles (red and orange). Both share a yellow base. Consider a line bisecting the angle of one of them. This line will also a) bisect the other triangle, b) bisect the yellow base, c) form a right angle with the yellow line it bisects, and d) the purple segment will lie along it! Now, as the yellow base is parallel to the yellow segment meeting the purple line, and the line containing the purple segment intersects the yellow base at a right angle, it will also intersect the other yellow piece at a right angle, and more specifically that yellow segment and the purple segment will form a right angle. In this special case, the law of cosines reduces to the Pythagorean theorem, and we have our sixth constraint: (green)2 = (purple)2 + (yellow)2

At this point, we can divide all six constraint equations by (purple)2 and do some variable substitutions: A = (green)2/(purple)2, B = (yellow)2/(purple)2, C = (magenta)2/(purple)2, D = (red)2/(purple)2, E = (orange)2/(purple)2, F = (blue)2/(purple)2. This gives us six linear equations with 6 unknowns:
| 0  2  0  1  0 -1 | |A|     |0|
| 2  0  0  1 -1  0 | |B|     |0|
| 0  0  2  2 -1 -1 | |C|  =  |0|
| 0  0  3  0 -1  0 | |D|     |0|
| 2 -1  2  0  0 -1 | |E|     |0|
| 1 -1  0  0  0  0 | |F|     |1|
and we can solve for A-F: A=4, B=3, C=7, D=13, E=21, F=19. Since none of these are less than 1, purple is the smallest, orange is the largest, and (orange)2/(purple)2 = 21, or the ratio between largest and smallest is sqrt(21).
posted by Upton O'Good at 11:27 PM on December 21, 2018 [5 favorites]


Hi puzzle fans,
Glad to see you're enjoying the challenge!
A quick request - since this thread has recently appeared on Google, I was wondering if budding solvers might consider ROT13-encoding any solutions / discoveries (e.g. using https://cryptii.com/pipes/rot13) so that they are obscured to Google's search indexing. This will avoid accidentally spoiling the quizzical fun for those who are trying to crack the puzzles themselves!
Doing ROT13 encoding or decoding is literally a 5-second copy/paste via the above link - e.g. "Glad to see you're enjoying the challenge!" becomes "Tynq gb frr lbh'er rawblvat gur punyyratr!" (and vice versa).
Wishing you all a very Merry Christmas - and happy solving!
T
posted by SpoilerAlert at 3:57 AM on December 22


[Hi SpoilerAlert, welcome to Metafilter! I've added an explicit spoiler warning above-the-fold, so hopefully folks will have fair warning if they click through to the thread. We typically ask people not to use ROT-13 except for very brief things, since a whole thread of ROT-13 becomes impossible to follow as a conversation.]
posted by LobsterMitten (staff) at 7:40 AM on December 22


Q14. HEX 19 [6 points]
A:
    __    __ 
   /██\__/  \
   \██/  \__/
 __/  \__/  \
/  \__/▓▓\__/
\__/  \▓▓/  \
   \__/  \__/
█ = start
▓ = highest chance after 19 moves, ~ 6.64% chance

Nathaniel will land on one of the 9 shown ~40.76% of the time.
posted by Pig Tail Orchestra at 10:37 AM on December 22


(No surprise that the hex farthest to the left has the lowest probability, with less than 0.0009% chance.)
posted by Pig Tail Orchestra at 10:43 AM on December 22


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