MathDB
2014 RMT + SMT Team Round - Rice + Stanford Math Tournament

Source:

January 25, 2022
algebrageometrynumber theoryRMTSMTcombinatorics

Problem Statement

p1. Given that the three points where the parabola y=bx22y = bx^2 -2 intersects the xx-axis and yy-axis form an equilateral triangle, compute bb.
p2. Compute the last digit of 2(3(4...2014)))2^{(3^{(4^{...2014)})})}
p3. A math tournament has a test which contains 1010 questions, each of which come from one of three different subjects. The subject of each question is chosen uniformly at random from the three subjects, and independently of the subjects of all the other questions. The test is unfair if any one subject appears at least 5 times. Compute the probability that the test is unfair.
p4. Let SnS_n be the sum Sn=1+11+111+1111+...+111...11S_n = 1 + 11 + 111 + 1111 + ... + 111... 11 where the last number 111...11111...11 has exactly nn 11’s. Find 102017/S2014\lfloor 10^{2017}/S_{2014} \rfloor.
p5. ABCABC is an equilateral triangle with side length 1212. Let OAO_A be the point inside ABCABC that is equidistant from BB and CC and is 3\sqrt3 units from AA. Define OBO_B and OCO_C symmetrically. Find the area of the intersection of triangles OABCO_ABC, AOBCAO_BC, and ABOCABO_C.
p6. A composition of a natural number nn is a way of writing it as a sum of natural numbers, such as 3=1+23 = 1 + 2. Let P(n)P(n) denote the sum over all compositions of nn of the number of terms in the composition. For example, the compositions of 33 are 33, 1+21+2, 2+12+1, and 1+1+11+1+1; the first has one term, the second and third have two each, and the last has 33 terms, so P(3)=1+2+2+3=8P(3) = 1 + 2 + 2 + 3 = 8. Compute P(9)P(9).
p7. Let ABCABC be a triangle with AB=7AB = 7, AC=8AC = 8, and BC=9BC = 9. Let the angle bisector of A A intersect BCBC at DD. Let EE be the foot of the perpendicular from CC to line ADAD. Let MM be the midpoint of BCBC. Find MEME.
p8. Call a function gg lower-approximating for ff on the interval [a,b][a, b] if for all x[a,b]x \in [a, b], f(x)g(x)f(x) \ge g(x). Find the maximum possible value of 12g(x)dx\int_1^2 g(x)dx where g(x)g(x) is a linear lower-approximating function for f(x)=xxf(x) = x^x on [1,2][1, 2].
p9. Determine the smallest positive integer xx such that 1.24x1.24x is the same number as the number obtained by taking the first (leftmost) digit of xx and moving it to be the last (rightmost) digit of xx.
p10. Let aa and bb be real numbers chosen uniformly and independently at random from the interval [10,10][-10, 10]. Find the probability that the polynomial x5+ax+bx^5+ax+b has exactly one real root (ignoring multiplicity).
p11. Let bb be a positive real number, and let ana_n be the sequence of real numbers defined by a1=a2=a3=1a_1 = a_2 = a_3 = 1, and an=an1+an2+ban3a_n = a_{n-1} + a_{n-2} + ba_{n-3} for all n>3n > 3. Find the smallest value of bb such that n=1an2n\sum_{n=1}^{\infty} \frac{\sqrt{a_n}}{2^n} diverges.
p12. Find the smallest LL such that (11a)b(112b)c(113c)aL\left( 1 - \frac{1}{a} \right)^b \left( 1 - \frac{1}{2b} \right)^c \left( 1 - \frac{1}{3c} \right)^a \le L for all real numbers a,ba, b, and cc greater than 11.
p13. Find the number of distinct ways in which 30(3030)30^{(30^{30})} can be written in the form a(bc)a^{(b^c)} , where a,ba, b, and cc are integers greater than 11.
p14. Convex quadrilateral ABCDABCD has sidelengths AB=7AB = 7, BC=9BC = 9, CD=15CD = 15. A circle with center II lies inside the quadrilateral, and is tangent to all four of its sides. Let MM and NN be the midpoints of ACAC and BDBD, respectively. It can be proven that II always lies on segment MNMN. If II is in fact the midpoint of MNMN, find the area of quadrilateral ABCDABCD.
p15. Marc has a bag containing 1010 balls, each with a different color. He draws out two balls uniformly at random and then paints the first ball he drew to match the color of the second ball. Then he places both balls back in the bag. He repeats until all the balls are the same color. Compute the expected number of times Marc has to perform this procedure before all the balls are the same color.
PS. You had better use hide for answers.