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− | ==Day 1==
| + | #redirect [[2020 USAMO Problems]] |
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− | ===Problem 1===
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− | Let <math>ABC</math> be a fixed acute triangle inscribed in a circle <math>\omega</math> with center <math>O</math>. A variable point <math>X</math> is chosen on minor arc <math>AB</math> of <math>\omega</math>, and segments <math>CX</math> and <math>AB</math> meet at <math>D</math>. Denote by <math>O_1</math> and <math>O_2</math> the circumcenters of triangles <math>ADX</math> and <math>BDX</math>, respectively. Determine all points <math>X</math> for which the area of triangle <math>OO_1O_2</math> is minimized.
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− | [[2020 USOMO Problems/Problem 1|Solution]] | |
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− | ===Problem 2===
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− | An empty <math>2020 \times 2020 \times 2020</math> cube is given, and a <math>2020 \times 2020</math> grid of square unit cells is drawn on each of its six faces. A beam is a <math>1 \times 1 \times 2020</math> rectangular prism. Several beams are placed inside the cube subject to the following conditions:
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− | <math>\bullet</math> The two <math>1 \times 1</math> faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are <math>3 \cdot 2020^2</math> possible positions for a beam.)
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− | <math>\bullet</math> No two beams have intersecting interiors.
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− | <math>\bullet</math> The interiors of each of the four <math>1 \times 2020</math> faces of each beam touch either a face
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− | of the cube or the interior of the face of another beam.
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− | What is the smallest positive number of beams that can be placed to satisfy these conditions?
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− | [[2020 USOMO Problems/Problem 2|Solution]]
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− | ===Problem 3===
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− | Let <math>p</math> be an odd prime. An integer <math>x</math> is called a <i>quadratic non-residue</i> if <math>p</math> does not divide <math>x - t^2</math> for any integer <math>t</math>.
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− | Denote by <math>A</math> the set of all integers <math>a</math> such that <math>1 \le a < p</math>, and both <math>a</math> and <math>4 - a</math> are quadratic non-residues. Calculate the remainder when the product of the elements of <math>A</math> is divided by <math>p</math>.
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− | [[2020 USOMO Problems/Problem 3|Solution]]
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− | ==Day 2==
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− | ===Problem 4===
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− | Suppose that <math>(a_1, b_1), (a_2, b_2), \ldots , (a_{100}, b_{100})</math> are distinct ordered pairs of nonnegative integers. Let <math>N</math> denote the number of pairs of integers <math>(i, j)</math> satisfying <math>1 \le i < j \le 100</math> and <math>|a_ib_j - a_j b_i|=1</math>. Determine the largest possible value of <math>N</math> over all possible choices of the <math>100</math> ordered pairs.
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− | [[2020 USOMO Problems/Problem 4|Solution]]
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− | ===Problem 5===
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− | A finite set <math>S</math> of points in the coordinate plane is called <i>overdetermined</i> if <math>|S| \ge 2</math> and there exists a nonzero polynomial <math>P(t)</math>, with real coefficients and of degree at most <math>|S| - 2</math>, satisfying <math>P(x) = y</math> for every point <math>(x, y) \in S</math>. For each integer <math>n \ge 2</math>, find the largest integer <math>k</math> (in terms of <math>n</math>) such that there exists a set of <math>n</math> distinct points that is not overdetermined, but has <math>k</math> overdetermined subsets.
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− | [[2020 USOMO Problems/Problem 5|Solution]]
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− | ===Problem 6===
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− | Let <math>n \ge 2</math> be an integer. Let <math>x_1 \ge x_2 \ge \cdots \ge x_n</math> and <math>y_1 \ge y_2 \ge \cdots \ge y_n</math> be <math>2n</math> real numbers such that
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− | <cmath>\begin{align*}
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− | 0 &= x_1 + x_2 + \cdots + x_n = y_1 + y_2 + \cdots + y_n\\
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− | \text{and }1 &= x_1^2+x_2^2+\cdots+x_n^2=y_1^2+y_2^2+\cdots+y_n^2.
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− | \end{align*}
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− | </cmath>
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− | Prove that
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− | <cmath>\sum_{i=1}^n(x_iy_i-x_iy_{n+1-i})\ge\frac{2}{\sqrt{n-1}}.</cmath>
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− | [[2020 USOMO Problems/Problem 6|Solution]]
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− | {{MAA Notice}}
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− | {{USAMO newbox|year= 2020 |before=[[2019 USAMO]]|after=[[2021 USAMO]]}}
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