Difference between revisions of "2020 INMO Problems"
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<math>\emph{Proposed by Prithwijit De}</math> | <math>\emph{Proposed by Prithwijit De}</math> | ||
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+ | [[2020 INMO Problems/Problem 1|Solution]] | ||
== Problem 2 == | == Problem 2 == | ||
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<math>\emph{Proposed by C.R. Pranesacher}</math> | <math>\emph{Proposed by C.R. Pranesacher}</math> | ||
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+ | [[2020 INMO Problems/Problem 2|Solution]] | ||
== Problem 3 == | == Problem 3 == | ||
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<math>\emph{Proposed by Sutanay Bhattacharya}</math> | <math>\emph{Proposed by Sutanay Bhattacharya}</math> | ||
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+ | [[2020 INMO Problems/Problem 3|Solution]] | ||
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== Problem 4 == | == Problem 4 == | ||
Let <math>n \geqslant 2</math> be an integer and let <math>1<a_1 \le a_2 \le \dots \le a_n</math> be <math>n</math> real numbers such that <math>a_1+a_2+\dots+a_n=2n</math>. Prove that<cmath>a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n.</cmath> | Let <math>n \geqslant 2</math> be an integer and let <math>1<a_1 \le a_2 \le \dots \le a_n</math> be <math>n</math> real numbers such that <math>a_1+a_2+\dots+a_n=2n</math>. Prove that<cmath>a_1a_2\dots a_{n-1}+a_1a_2\dots a_{n-2}+\dots+a_1a_2+a_1+2 \leqslant a_1a_2\dots a_n.</cmath> | ||
+ | [[2020 INMO Problems/Problem 4|Solution]] | ||
== Problem 5 == | == Problem 5 == | ||
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(c) Determine whether <math>5</math> is frameable. | (c) Determine whether <math>5</math> is frameable. | ||
+ | [[2020 INMO Problems/Problem 5|Solution]] | ||
== Problem 6 == | == Problem 6 == | ||
A stromino is a <math>3 \times 1</math> rectangle. Show that a <math>5 \times 5</math> board divided into twenty-five <math>1 \times 1</math> squares cannot be covered by <math>16</math> strominos such that each stromino covers exactly three squares of the board, and every square is covered by one or two strominos. (A stromino can be placed either horizontally or vertically on the board.) | A stromino is a <math>3 \times 1</math> rectangle. Show that a <math>5 \times 5</math> board divided into twenty-five <math>1 \times 1</math> squares cannot be covered by <math>16</math> strominos such that each stromino covers exactly three squares of the board, and every square is covered by one or two strominos. (A stromino can be placed either horizontally or vertically on the board.) | ||
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+ | [[2020 INMO Problems/Problem 6|Solution]] |
Latest revision as of 04:43, 25 September 2023
Problem 1
Let and be two circles of unequal radii, with centres and respectively, intersecting in two distinct points and . Assume that the centre of each circle is outside the other circle. The tangent to at intersects again in , different from ; the tangent to at intersects again at , different from . The bisectors of and meet and again in and , respectively. Let and be the circumcentres of triangles and , respectively. Prove that is the perpendicular bisector of the line segment .
Problem 2
Suppose is a polynomial with real coefficients, satisfying the condition , for every real . Prove that can be expressed in the formfor some real numbers and non-negative integer .
Problem 3
Let be a subset of . Suppose there is a positive integer such that for any integer , one can find positive integers so that and all the digits in the decimal(Base 10) representations of (expressed without leading zeros) are in . Find the smallest possible value of .
Problem 4
Let be an integer and let be real numbers such that . Prove that
Problem 5
Infinitely many equidistant parallel lines are drawn in the plane. A positive integer is called frameable if it is possible to draw a regular polygon with sides all whose vertices lie on these lines, and no line contains more than one vertex of the polygon.
(a) Show that are frameable.
(b) Show that any integer is not frameable.
(c) Determine whether is frameable.
Problem 6
A stromino is a rectangle. Show that a board divided into twenty-five squares cannot be covered by strominos such that each stromino covers exactly three squares of the board, and every square is covered by one or two strominos. (A stromino can be placed either horizontally or vertically on the board.)