Fall+Interschool+2011

Interschool answers!! And by the way, feel free to check anyone's answers, especially if you were thinking of working on the problem originally.

The solutions are posted! Scoring results will probably be announced at the first competition. []

Here's the test itself, if you want to check out the questions again: []

Then, f(x) = 2(x - a)(x - b)(x - c)(x - d)(x - e). [The uber-useful root notation!] Multiplying this out we get f(x) = 2x^5 + 2*( -a - b - c - d - e )*x^4 + 2*( a*b + a*c + a*d + ... + d*e )*x^3 + ... + 2(-a*b*c*d*e). Now that it's back in a nice polynomial, we know the coefficients of this and the one in the problem (2x^5 - 4x^4 + + x^3 - ... - 5) must be the same! So 2(-a - b - c - d - e) = -4, 2(ab + ac + ... + de) = 1, etc. Simplifying, we get a+b+c+d+e = (Sum of roots) = 2 ab + ac + ... + de = (Sum of all possible products of 2 roots) = 1/2 Now, as it seems to turn out, these two equations are all we need to find the answer! Using a magic little equation that is true no matter what a, b, c, d, and e are: ( a + b + c + d + e ) ^ 2 = (a^2 + b^2 + c^2 + d^2 + e^2) + 2*( ab + ac + ad + ... + de ) 2 ^ 2 = (Sum of squares of roots) + 2*1/2 __Sum of squares of roots = 3__ -Matt || I am so lost.... -Max and Sebi || A=3 - 2root3 and B= 3 + 2root3. Since w ehave the absolute value of the integral, we get 36 regardless of which combination we use. :) ||  || So, every element in M2011 is 1/sqrt(2011).  Multiplying that by itself using matrix multiplication, the new matrix (M2011)^2 should end up having every element be a 1, and doing this again, (M2011)^4 should be full of 2011's.  The sum of the elements of a 2011 x 2011 matrix full of 2011's will thus be __2011^3__.  -Matt ||   || Thanks to Mr. Dubard for really helping out! :) || So the right hand side is just 2012/2(X1 + X2010) using Carl Friedich Gauss's very useful sum of sequence formula. If you factor the left in a similar way you get factors of X1+X2012 in each term. So diving both sides by X1+X2010 yields... X1-X2+X3-X4+X5-....-X2012=1006 but since it's an arithmetic progression, X1-X2 = X3=X4 and so on. So 1006(-common difference) = 1006. So the common difference= -1 That means the equation is true for any numbers starting at 2012 or bigger and decreasing by 1. so X1-X2010 = 2011. :) ||  || I Christmas(1225) ||   || - || Bradley or Matt, looking for a number || Missing an l. "noel" is definitely the nicest answer... yet, the packet says it's supposed to be a number somehow and I'm not sure how to make that a number! Only thing I could think of is some sort of "keycode number" that the keyboard sends to the computer when you press the letter L - Javascript gives it a 76 || Well done! XD -Matt Had the same question in my old logic book!- Bradley || 1979 the smashing pumpkins and 21 guns by green day??? So the answer is 2000. || 2000!!! || up to 8 || Bradley Al Alexa, Lisette ||   || Maybe 1st Noel has to do with it, or the Tip-Taps, which made a song on youtube called First Noel 1123... || -Ashley || -Ashley Then please put your name in the "people working on it" column and write something like "do not work on it, I got it". Otherwise, we will not know that you are working on it! -Matt Sorry :'( !! i got the answer before this answer thing was here (and also before it was actually here). I'll remember next time! -Ashley || Book Title: The Meanings of It All: Thoughts of a Citizen-Scientist ||  || Book Title:Real and Complex Analysis ||   || [Which is kind of tough to do on 2-D paper, so I first made a B vs. D graph and labeled each of the more significant-looking points (B,D) with the range of values of P that were bigger than the sum of B and D] 0 <= P,B,D <=1, so all the possible combinations of values these fellows can pick make a 1x1x1 cube with one vertex at (0,0,0) and its opposite at (1,1,1). Now, the only points (P,B,D) where P > B + D form a nice little tetrahedron with vertices at (0,0,0), (1,0,0), (1,1,0), and (1,0,1)! So, the probability that P > B + D will be the ratio (volume of the *tetrahedron of fulfillment*) : (volume of the *cube of possibility*) = 1/6 : 1 = __1/6__ -Matt || Not really sure if this is a valid way of going about doing this, but it made sense to me - Stat people catch any problems with this? - Matt || Tempted to say 1/3 myself, but here's my chain of logic: Given Events: -Hannah chooses A -Host opens B, and there is a goat. Assumptions: -Choice of door to put car behind is random (A, B, or C) -Choice of door for the host to slip and open is random. [at least, I assume] Possible scenarios: 1) Car is under A, host randomly opened B 2) Car is under C, host randomly opened B [Any other scenarios, i.e. "Car is under B, host randomly opened A", were eliminated by the given events] Probabilities of each scenario having occured in the first place: 1) (Chance of car being under A) * (Chance of host tripping and opening B) = 1/3 * 1/3 = 1/9 2) (Chance of car being under C) * (Chance of host tripping and opening B) = 1/3 * 1/3 = 1/9 Since we know these are the only possibilities that result in what ocurred (the Host opening B and there being a goat), having observed this event I think we can make them account for 100% while keeping the ratio of 1) : 2) the same by scaling them up to 1) 1/2 2) 1/2 And thus the answer would be .50 -Matt ||  || Using //pure// Nash equilibrium analysis (using a web applet! =D) we get Z,C, but the problem directly hints that it takes "mixed strategies", so the simulation is probably off... ||  || 2nd statement: Aaron has the conjugate of Rodrigo's root 3rd statement: Pratik either has 1 or -1 4th statement: The real parts of Rodrigo and Aaron's roots are both 2, and we get that the inflection point is either 1 or some fraction 5th statement: It must be 1 || Various steps in between, these are just the most information we can eventually deduce from each statement || You Tube link: [] ||  ||
 * Number || Solution || Correct Answer || Pts won || Solver || Explanation || Thoughts? ||
 * 1a || 8 || 8 || 1 || Matt ||  ||   ||
 * 1b || 3 || 3 || 1 || Tafur or Matt || Let's call the 5 roots a, b, c, d, and e.
 * 1c || 0 || 0 || 1 || Tafur+Matt || Subtract 2010 from all elements and the determinant will then be 0; add 1 to all //these// elements and the determinant is still 0; same when you add 1 yet again, so I'm guessing the trend continues as you keep adding 1's all the way up to the matrix in the problem -Matt ||  ||
 * 1d || 36 || 6 || 0 || Morin || I got A=3 + 2root3, B=3 - 2root, or
 * 1e ||  || 1 || 0 ||   ||   ||   ||
 * 2a || 2011^3 || 2011 || 0 (dispute?) || Matt || The crazy exp(~!@$#&) thing all ends up working out to 1, since e^(2PI)=1 according to an __//awesome//__ result of complex numbers and the whole thing in the exp works out to be (e^(2PI))^(-jk) = 1.
 * 2b || 17/190 or 33/190? || 33/190 || 2? || Matt+Clara or Matt || First answer involves the sum of only powers of 2 that differ by 3 powers; second takes into account powers of 2 that differ by 9 and 15 powers (which should still count) -Matt ||  ||
 * 2c || 72 || 72 || 2 || Matt || Every term n+3 works out to be (term n)*-1/8, and adding up all terms to infinity we get that x2 = -1/2. Then, after moving numbers around and applying the infinite geometric series formula a bunch of times, I got 72 as the final answer -Matt ||  ||
 * 2d || 2011 || 2011 || 2 || Morin & Dubard
 * 2e || 125 || 125!! || 2 || Diego ||  ||   ||
 * 3a || 3 + (cant figure out the second song!) || 8 (3 Doors Down + The Click 5) || 0 ||  || Precal & others ||   ||
 * 3b || 7 || 7 || up to 3 ||  || Precal & others || Hey! -Sebi ||
 * 3c || 247 || 247 || up to 3 ||  || Precal & others ||   ||
 * 3d || 8 || 8 || up to 3 ||  || Precal & others ||   ||
 * 3e || 80 || 80 || up to 3 ||  || Preal & others ||   ||
 * 4a || http://www.filedropper.com/interschool-question4 ||  || - || Matt || !!!!!!!! ||   ||
 * 4b || noel or 76
 * 4c || 1123 ||  || - || Maty ||   ||   ||
 * 4d || bit.ly ||  || - || Carlos+Jack Kim ||   ||   ||
 * 4e || Both
 * 5a || Please Excuse My Dear Aunt Sally || Parentheses, exponents, multiplication, division, addition, subtraction || 1? || Tafur ||  || or Parenthesis, "Exponent, Multipication, Devision, Addition, Subtraction" but that's probably too obvious.
 * 5b || Simon's Favorite Factoring Trick || Simon's Favorite Factoring Trick || 1 || Tafur+Morin+Max Grad+Matt || Says that xy - x - y + 1 can be factored into (x-1)(y-1), which is a useful / semi-well-known fact featured in the artofproblemsolving.com forums || You stole my question! :(
 * 5c || Don't Forget To Carry The 1 || Yup || 1 || Matt || This took days... but it was worth it ||  ||
 * 5d || Complete The Square || Yup || 1 || Morin ||  ||   ||
 * 5e || Solve the [General] Cubic || Square the Circle || their judgment || Matt || You can't solve the general cubic ax^3 + bx^2 + cx + d, as far as I know -Matt ||  ||
 * 6a || F. Scott Fitzgerald || Yup || 1 || Ashley || The Great Gatsby-it's the letter on the keyboard next to the letter written ||  ||
 * 6b ||  || Barbara Kingsolver (used hexadecimal) ||   ||   ||   ||   ||
 * 6c || Richard Phillips Feynman || Correct || 1 || Ashley || This is the ISBN number for his book
 * 6d ||  || Walter Rudin [aaargh! shifted by 1] || 0 ||   ||   ||   ||
 * 6e || Walter Rudin??? || C.S. Lewis || 0 || Ashley || There are 486 pages in it and it was published on 05/01/1986
 * 7a ||  || Most: 502631 ||   ||   ||   ||   ||
 * 7b ||  || Least: 0 ||   ||   ||   ||   ||
 * 7c || //**Note to people taking part in this problem: if you are going to vote for us on the site, keep doing it until you get a prime number of votes.**// || Most prime: 502631 ||  ||   || Get a program / team of people to make our number of votes prime just seconds before they check the results, and while we're at it add 1 to everybody else's so that theirs will be even and therefore not prime! :D -Matt ||   ||
 * 7d ||  || Median: 4 ||   ||   ||   ||   ||
 * 7e || Our number: 19? || Mean: 15500 ||  ||   ||   ||   ||
 * 8a ||  || Anagram || 0 ||   ||   ||   ||
 * 8b || Hardy-Ramanujan number || Hardy and Ramanujan || 1 || Bradley || First number that can be expressed as a sum of 2 cubes in 2 ways ||  ||
 * 8c || Akshay Venkatesh (there may be others, but accoding to wikipedia his did this at princeton, age 20) || Correct || 1 || Tafur ||  ||   ||
 * 8d ||  || Kolmogorov ||   ||   ||   ||   ||
 * 8e ||  || Ramanujan ||   ||   ||   ||   ||
 * 9a || Princeton || Correct || 1.5 || Alexa K. || Google Images ||  ||
 * 9b || The Shim - Sutcliffe Integral House (designed for homeowner James Stewart) || Correct || 1.5 || Alexa K. || Google Images ||  ||
 * 9c || Wolfram Research || Correct || 1.5 || Alexa K. || Google Images ||  ||
 * 9d || Casio Headquarters || Correct || 1.5 || Alexa K. || Google Images ||  ||
 * 9e || New York City??? || Correct || 1.5 || Ashley || Based off of Belvedere Castle in Central Park || Confirmed - Alexa K and Lisette (Belvedere Castle) ||
 * 10a || 1 || Correct! f(x) = 0 is the only one || 1.5 || Matt ||  ||   ||
 * 10b || .17 || Correct || 1.5 || Matt || Make a 3D graph with P-, B-, and D-axes to represent the combination of values (P,B,D) that Patrick, Bishoy, and Dhyan respectively can draw.
 * 10c || 0.3 or 0.5 || 0.5 - not the Monty Hall problem, they say || 0 or 1.5 || Bradley or Matt || Can't change her choice. Chose from 3 doors, prob. remains 1/3 - Bradley
 * 10d || 6.5 Units^2 || 5.5 || 0 || Bradley || Separated into triangles. ||  ||
 * 10e || 1 || Correct || 1.5 || Matt || Graphing the real part of the numbers as x-coordinates and the imaginary parts as y-coordinates, using the fact that squaring a complex number (or point, in this set-up) will square its distance to the origin and double its angle away from the horizontal, and the formula Area = 1/2 a*b*sin(x), we get that z must have a distance away from the origin (or |z|) of 1. (very, very hard if you ) ||  ||
 * 11a || solved ||  ||   || Tafur ||   ||   ||
 * 11b || solved ||  ||   || Tafur ||   ||   ||
 * 11c || solved ||  ||   || Tafur ||   ||   ||
 * 11d || solved ||  ||   || I saw someone else had it... ||   ||   ||
 * 11e || solved ||  ||   || Tafur ||   ||   ||
 * 12a || .22? || 10.85 || 0 || Matt... just putting a guess out there ||  || Erp.. sorry, thought it was supposed to be a number between 0 and 1, not 0 and 100 :0 Sorry everyone! (and multiplied by 100, we still lost...)- Matt ||
 * 12b || .12? || 8.157 || 0 || Matt (see above) ||  ||   ||
 * 12c || .22 || 6.67 || 0 || Matt (see above) ||  ||   ||
 * 12d || .01? || 21.11 || 0 || Matt (see above) ||  ||   ||
 * 12e || 0 || 16.67 || 0 || Matt (see above) ||  ||   ||
 * 13a || 0 || 0.01 - must be greater than 0... || 0 || Matt || I wrote an amusing self-dialogue to explain how a "perfectly rationalizable" person would approach this (and, an infinite number of hours later, end up with 0), but it somehow got erased from the page =( ||  ||
 * 13b || 1/2 || Correct || 1.5 || Matt || P1 has an expected payoff of 5/2 by choosing L and just 2 for choosing R - then B has a 50-50 chance of getting 1 or 0, so the expected would be 1/2 ||  ||
 * 13c || d,b || Correct || 1.5 || Tafur || In a Nash equilibrium, neither player benefits from changing their strategy. || I agree -Matt ||
 * 13d || (w,d)? [or (z,c)?] || (c,z) || 0 or 1.5 || Matt || Row W has the highest average return for Player 1, and Column D has the highest average return for Player 2, so if each assumes the other is picking a row/column at random, they'd best be going for W and D respectively.
 * 13e ||  || 2.5 ||   ||   ||   ||   ||
 * 14a || 120.25 || 120 - to the nearest integer, they say, but they might take it || 0 or 1.5 || Tafur ||  || I agree, Ashley (I got the same thing) ||
 * 14b || Joey Votto (I think...This could be wrong) || Miguel Cabrera || 0 || Ashley || I made a chart of all the players from the 2011 regular season and did the math. It's too long to put here but if you want to see it, just ask me. ||  ||
 * 14c ||  || Edmonton Oilers ||   ||   ||   ||   ||
 * 14d || 70% (17/29) || Correct || 1.5 || Tafur ||  ||   ||
 * 14e ||  || 61% ||   ||   ||   ||   ||
 * 15 ||  ||   ||   ||   ||   ||   ||
 * 16a || 6 || 7 || 0 || Matt || Wikipedia says the size of a graph is the number of vertices, which makes sense || Curses!!! -Matt ||
 * 16b || 2021055 edges || Correct || 1 || Bradley || 2(n-1)/2 || Nice, just like the diagonals formula for polygons except the sides count too! =) assuming you made a typo in the explanation, did you mean n*(n-1)/2? (which would give the answer you have... just making sure) -Matt ||
 * 16c || a is 2 || Correct || 1 || Tafur ||  ||   ||
 * 16d || 0 || Correct || 1 || Matt || Thank you Wikipedia! || does anyone have a TI-89? It would help with messy derivatives. Thanks! ||
 * 16e || c || Correct || 1 || Ashley || The empty set is both open and closed || for some reason I feel like this is a trick question tho....??? ||
 * 17a || 60 || Correct || 1 || Ashley or Matt || 5*4*3*2*1 ways to arrange letters F, A1, A2, M, T; then divide by two to account for the fact that you can switch A1 and A2 without changing the word ||  ||
 * 17b || 970 || Correct || 1 || Ashley ||  ||   ||
 * 17c || 9 || Correct || 1 || Ashley ||  ||   ||
 * 17d || 0 || Correct || 1 ||  ||   ||   ||
 * 17e || 2011 || Correct || 1 || Tafur ||  || Yay! Great job, everybody! ||
 * 18a || 1337 || Correct || 1 || Jack + Harrison || "Leet Speak" ||  ||
 * 18b || 668 || Correct || 1 || Matt ||  ||   ||
 * 18c || 2380 || Correct || 1 || Matt || 17 Choose 4. Not really sure why it works yet, the web says so - if anyone can explain it to me, I am all ears =) ||  ||
 * 18d ||  || 748 ||   ||   ||   ||   ||
 * 18e ||  || 17 ||   ||   ||   ||   ||
 * 19a || 1 || Correct || 2.5 || Matt || 1st statement: Rodrigo must have a complex root (x-intercept)
 * 19b || 5*10^5 || 12580 or 490 (dispute?) || 0? || Matt || They don't say it has to be an //arithmetic// sequence - the sequence 1, 0, 0, 0, 0, 0, 0, ..., 1/2 also satisfies the sum, so the answer is then [10^6 * 1/2] ||  ||
 * 19c ||  || Sylvester's 4-Point Thm ||   ||   ||   ||   ||
 * 20a || 17 || 16 || 0 || Dr. Stavisky ||  ||   ||
 * 20b || Three wooden benches || Playpen balls || 0 || Matt || http://collegiateway.org/colleges/mit/ ||  ||
 * 20c || RRT || Correct || 1 || Ashley || There is some song about this called "Complementary in Kinetic Typography"
 * 20d ||  || Maseeh Hall ||   ||   ||   ||   ||
 * 20e ||  || ADP ||   ||   ||   ||   ||
 * 21a ||  ||   ||   ||   ||   ||   ||
 * 21b ||  ||   ||   ||   ||   ||   ||
 * 21c ||  ||   ||   ||   ||   ||   ||
 * 21d ||  ||   ||   ||   ||   ||   ||
 * 21e ||  ||   ||   ||   ||   ||   ||
 * 22 || POLYNOMIAL || Correct! || Up to 6 || Geo Team || Each row and column contains one math-related word; crossing out all the letters contained in these words leaves just the word "POLYNOMIAL" ||  ||
 * 23 || SQUID INK || Correct! || Up to 6 || Matt + Geo Team || Googled the first line decoded, "REPORT + GRACES = SNORED", and found the exact same question on a puzzle website (along with the solution)! Answer supposedly has to do with there being 325 ways to superimpose two different letters, and then each superimposed letter-thing can count for a digit in base 325 (where AB=0, AC=1, BC=25, and YZ = 324). Then the final answer is a 4-digit number in base 325, and the letters that make this up can be rearranged to spell "SQUID INK" ||  ||
 * 24 ||  ||   ||   ||   ||   ||   ||
 * 25 ||  ||   ||   ||   ||   ||   ||
 * ||  || Total || Around 78!!! Great work, everyone! ||   ||   ||   ||