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2021 MMATHS Mathathon Rounds 4-5 Math Majors of America Tournament for HS

Source:

August 10, 2023
MMATHSalgebrageometrycombinatoricsnumber theory

Problem Statement

Round 4
p10. How many divisors of 101110^{11} have at least half as many divisors that 101110^{11} has?
p11. Let f(x,y)=xy+yxf(x, y) = \frac{x}{y}+\frac{y}{x} and g(x,y)=xyyxg(x, y) = \frac{x}{y}-\frac{y}{x} . Then, if f(f(...f(f(2021fsf(f(1,2),g(2,1)),2),2)...,2),2)\underbrace{f(f(... f(f(}_{2021 fs} f(f(1, 2), g(2,1)), 2), 2)... , 2), 2) can be expressed in the form a+bca + \frac{b}{c}, where aa, bb,cc are nonnegative integers such that b<cb < c and gcd(b,c)=1gcd(b,c) = 1, find a+b+(log2(log2c)a + b + \lceil (\log_2 (\log_2 c)\rceil
p12. Let ABCABC be an equilateral triangle, and letDEF DEF be an equilateral triangle such that DD, EE, and FF lie on ABAB, BCBC, and CACA, respectively. Suppose that ADAD and BDBD are positive integers, and that [DEF][ABC]=97196\frac{[DEF]}{[ABC]}=\frac{97}{196}. The circumcircle of triangle DEFDEF meets ABAB, BCBC, and CACA again at GG, HH, and II, respectively. Find the side length of an equilateral triangle that has the same area as the hexagon with vertices D,E,F,G,HD, E, F, G, H, and II.
Round 5
p13. Point XX is on line segment ABAB such that AX=25AX = \frac25 and XB=52XB = \frac52. Circle Ω\Omega has diameter ABAB and circle ω\omega has diameter XBXB. A ray perpendicular to ABAB begins at XX and intersects Ω\Omega at a point YY. Let ZZ be a point on ω\omega such that YZX=90o\angle YZX = 90^o. If the area of triangle XYZXYZ can be expressed as ab\frac{a}{b} for positive integers a,ba, b with gcd(a,b)=1gcd(a, b) = 1, find a+ba + b.
p14. Andrew, Ben, and Clayton are discussing four different songs; for each song, each person either likes or dislikes that song, and each person likes at least one song and dislikes at least one song. As it turns out, Andrew and Ben don't like any of the same songs, but Clayton likes at least one song that Andrew likes and at least one song that Ben likes! How many possible ways could this have happened?
p15. Let triangle ABCABC with circumcircle Ω\Omega satisfy AB=39AB = 39, BC=40BC = 40, and CA=25CA = 25. Let PP be a point on arc BCBC not containing AA, and let QQ and RR be the reflections of PP in ABAB and ACAC, respectively. Let AQAQ and ARAR meet Ω\Omega again at SS and TT, respectively. Given that the reflection of QRQR over BCBC is tangent to Ω\Omega , STST can be expressed as ab\frac{a}{b} for positive integers a,ba, b with gcd(a,b)=1gcd(a,b)= 1. Find a+ba + b.
PS. You should use hide for answers. Rounds 1-3 have been posted [url=https://artofproblemsolving.com/community/c4h3131401p28368159]here and 6-7 [url=https://artofproblemsolving.com/community/c4h3131434p28368604]here ,Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.