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1999 DMM Tiebreaker Round - Duke Math Meet

Source:

February 16, 2024
algebrageometrynumber theoryDMM

Problem Statement

p1A. Compute 1+123+133+143+153+...1 + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \frac{1}{5^3} + ... 1−123+133−143+153−...1 - \frac{1}{2^3} + \frac{1}{3^3} - \frac{1}{4^3} + \frac{1}{5^3} - ...
p1B. Real values aa and bb satisfy ab=1ab = 1, and both numbers have decimal expansions which repeat every five digits: a=0.(a1)(a2)(a3)(a4)(a5)(a1)(a2)(a3)(a4)(a5)... a = 0.(a_1)(a_2)(a_3)(a_4)(a_5)(a_1)(a_2)(a_3)(a_4)(a_5)... and b=1.(b1)(b2)(b3)(b4)(b5)(b1)(b2)(b3)(b4)(b5)... b = 1.(b_1)(b_2)(b_3)(b_4)(b_5)(b_1)(b_2)(b_3)(b_4)(b_5)... If a5=1a_5 = 1, find b5b_5.
p2. P(x)=x4−3x3+4x2−9x+5P(x) = x^4 - 3x^3 + 4x^2 - 9x + 5. Q(x)Q(x) is a 33rd-degree polynomial whose graph intersects the graph of P(x)P(x) at x=1x = 1, 22, 55, and 1010. Compute Q(0)Q(0).
p3. Distinct real values x1x_1, x2x_2, x3x_3, x4x_4 all satisfy ∣∣x−3∣−5∣=1.34953 ||x - 3| - 5| = 1.34953. Find x1+x2+x3+x4x_1 + x_2 + x_3 + x_4.
p4. Triangle ABCABC has sides AB=8AB = 8, BC=10BC = 10, and CA=11CA = 11. Let LL be the locus of points in the interior of triangle ABCABC which are within one unit of either AA, BB, or CC. Find the area of LL.
p5. Triangles ABCABC and ADEADE are equilateral, and ADAD is an altitude of ABCABC. The area of the intersection of these triangles is 33. Find the area of the larger triangle ABCABC.
PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here.
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