Difference between revisions of "AoPS Wiki:Competition ratings"
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* Problem 10 - 12: '''5''' | * Problem 10 - 12: '''5''' | ||
*: Let <math>z</math> be a complex number with <math>|z|=2014</math>. Let <math>P</math> be the polygon in the complex plane whose vertices are <math>z</math> and every <math>w</math> such that <math>\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}</math>. Then the area enclosed by <math>P</math> can be written in the form <math>n\sqrt{3}</math>, where <math>n</math> is an integer. Find the remainder when <math>n</math> is divided by <math>1000</math>. ([[2014 AIME II Problems/Problem 10|Solution]]) | *: Let <math>z</math> be a complex number with <math>|z|=2014</math>. Let <math>P</math> be the polygon in the complex plane whose vertices are <math>z</math> and every <math>w</math> such that <math>\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}</math>. Then the area enclosed by <math>P</math> can be written in the form <math>n\sqrt{3}</math>, where <math>n</math> is an integer. Find the remainder when <math>n</math> is divided by <math>1000</math>. ([[2014 AIME II Problems/Problem 10|Solution]]) | ||
− | * Problem | + | * Problem 13 - 15: '''6''' |
*: ''A right cone|right circular cone has a base with radius <math>600</math> and height <math> 200\sqrt{7}. </math> A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is <math>125</math>, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is <math>375\sqrt{2}.</math> Find the least distance that the fly could have crawled.'' ([[2004 AIME II Problems/Problem 11|Solution]]) | *: ''A right cone|right circular cone has a base with radius <math>600</math> and height <math> 200\sqrt{7}. </math> A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is <math>125</math>, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is <math>375\sqrt{2}.</math> Find the least distance that the fly could have crawled.'' ([[2004 AIME II Problems/Problem 11|Solution]]) | ||
− | |||
=== [[Indonesian MO]] === | === [[Indonesian MO]] === |
Revision as of 20:50, 7 March 2016
Shortcut: |
This page contains an approximate estimation of the difficulty level of various competitions. It is designed with the intention of introducing contests of similar difficulty levels (but possibly different styles of problems) that readers may like to try to gain more experience.
Each entry groups the problems into sets of similar difficulty levels and suggests an approximate difficulty rating, on a scale from 1 to 10 (from easiest to hardest). Note that many of these ratings are not directly comparable, because the actual competitions have many different rules; the ratings are generally synchronized with the amount of available time, etc. Also, due to variances within a contest, ranges shown may overlap. A sample problem is provided with each entry, with a link to a solution.
As you may have guessed with time many competitions got more challenging because many countries got more access to books targeted at olympiad preparation. But especially web site where one can discuss olympiad as our very own ML/AoPS! Thus when judging the difficulty level consider the last 10-15 years with more priority.
If you have some experience with mathematical competitions, we hope that you can help us make the difficulty rankings more accurate. Currently, the system is on a scale from 1 to 10 where 1 is the easiest level, e.g. early AMC problems and 10 is hardest level, e.g. China IMO Team Selection Test. When considering problem difficulty put more emphasis on problem-solving aspects and less so on technical skill requirements.[1]
Contents
- 1 Competitions
- 1.1 MOEMS
- 1.2 AMC 8
- 1.3 Mathcounts
- 1.4 AMC 10
- 1.5 Austrian MO
- 1.6 AMC 12
- 1.7 AIME
- 1.8 Indonesian MO
- 1.9 ARML
- 1.10 Central American Olympiad
- 1.11 Ibero American Olympiad
- 1.12 Canadian MO
- 1.13 Ibero American Olympiad
- 1.14 HMMT
- 1.15 APMO
- 1.16 Balkan MO
- 1.17 HMMT
- 1.18 China TST
- 1.19 Ibero American Olympiad
- 1.20 IMO
- 1.21 IMO Shortlist
- 1.22 JBMO
- 1.23 Putnam
- 1.24 USAMO
- 1.25 USAJMO
- 1.26 USAMTS
- 1.27 USA TST
- 2 Scale
- 3 See also
Competitions
MOEMS
- Division E: 1
- The whole number
is divisible by
.
leaves a remainder of
when divided by
or
. What is the smallest value that
can be? (Solution)
- The whole number
- Division M: 1
- The value of a two-digit number is
times more than the sum of its digits. The units digit is 1 more than twice the tens digit. Find the two-digit number. (Solution)
- The value of a two-digit number is
AMC 8
- Problem 1 - Problem 12: 1
- What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock? (Solution)
- Problem 13 - Problem 25: 1.5
- A fifth number,
, is added to the set
to make the mean of the set of five numbers equal to its median. What is the number of possible values of
? (Solution)
- A fifth number,
Mathcounts
- Countdown: 0.5 (School, Chapter), 1 (State, National)
- Sprint: 1-1.5 (school), 1.5-2 (Chapter),2-2.5 (State), 2-3 (National)
- Target: 1.5 (school), 2 (Chapter), 2.5 (State), 3 (National)
AMC 10
- Problem 1 - 5: 1
- The larger of two consecutive odd integers is three times the smaller. What is their sum? (Solution)
- Problem 6 - 20: 2
- How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression? (Solution)
- Problem 21 - 25: 3
- Mr. Jones has eight children of different ages. On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" she exclaims. "That number is evenly divisible by the age of each of us kids!" "That's right," replies Mr. Jones, "and the last two digits just happen to be my age." Which of the following is not the age of one of Mr. Jones's children? (Solution)
Austrian MO
- Regional Competition for Advanced Students, Problems 1-4: 2
- Federal Competition for Advanced Students, Part 1. Problems 1-4: 3
- Federal Competition for Advanced Students, Part 2, Problems 1-6: 4
AMC 12
- Problem 1-10: 2
- A solid box is
cm by
cm by
cm. A new solid is formed by removing a cube
cm on a side from each corner of this box. What percent of the original volume is removed? (Solution)
- A solid box is
- Problem 11-20: 3
- An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point? (Solution)
- Problem 21-25: 4
- Functions
and
are quadratic,
, and the graph of
contains the vertex of the graph of
. The four
-intercepts on the two graphs have
-coordinates
,
,
, and
, in increasing order, and
. The value of
is
, where
,
, and
are positive integers, and
is not divisible by the square of any prime. What is
? (Solution)
- Functions
AIME
- Problem 1 - 5: 3
- Starting at
an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let
be the probability that the object reaches
in six or fewer steps. Given that
can be written in the form
where
and
are relatively prime positive integers, find
(Solution)
- Starting at
- Problem 6 - 10: 4
- Problem 10 - 12: 5
- Let
be a complex number with
. Let
be the polygon in the complex plane whose vertices are
and every
such that
. Then the area enclosed by
can be written in the form
, where
is an integer. Find the remainder when
is divided by
. (Solution)
- Let
- Problem 13 - 15: 6
- A right cone|right circular cone has a base with radius
and height
A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is
, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is
Find the least distance that the fly could have crawled. (Solution)
- A right cone|right circular cone has a base with radius
Indonesian MO
- Problem 1/5: 3.5
- In a drawer, there are at most
balls, some of them are white, the rest are blue, which are randomly distributed. If two balls were taken at the same time, then the probability that the balls are both blue or both white is
. Determine the maximum amount of white balls in the drawer, such that the probability statement is true? <url>viewtopic.php?t=294065 (Solution)</url>
- In a drawer, there are at most
- Problem 2/6: 4.5
- Find the lowest possible values from the function
for any real numbers .<url>viewtopic.php?t=294067 (Solution)</url>
- Problem 3/7: 5
- A pair of integers
is called good if
- A pair of integers
Given 2 positive integers which are relatively prime, prove that there exists a good pair
with
and
, but
and
. <url>viewtopic.php?t=294068 (Solution)</url>
- Problem 4/8: 6
- Given an acute triangle
. The incircle of triangle
touches
respectively at
. The angle bisector of
cuts
and
respectively at
and
. Suppose
is one of the altitudes of triangle
, and
be the midpoint of
.
- Given an acute triangle
(a) Prove that and
are perpendicular with the angle bisector of
.
(b) Show that is a cyclic quadrilateral. <url>viewtopic.php?t=294069 (Solution)</url>
ARML
- Individuals, Problem 6 and 8: 4
- Individuals, Problem 10: 6
- Team/power, Problem 1-5: 3.5
- Team/power, Problem 6-10: 5\
Central American Olympiad
- Problem 1: 4
- Find all three-digit numbers
(with
) such that
is a divisor of 26. (<url>viewtopic.php?p=903856#903856 Solution</url>)
- Find all three-digit numbers
- Problem 2,4,5: 5-6
- Show that the equation
has no integer solutions. (<url>viewtopic.php?p=291301#291301 Solution</url>)
- Show that the equation
- Problem 3/6: 6.5
- Let
be a convex quadrilateral.
, and
,
,
and
are points on
,
,
and
respectively, such that
. If
,
, show that
(<url>viewtopic.php?p=828841#p828841 Solution</url>
- Let
Ibero American Olympiad
- Problem 1/4: 5.5
- Problem 2/5: 6.5
- Problem 3/6: 7.5
Canadian MO
- Problem 1: 5
- Problem 2: 6
- Problem 3: 6.5
- Problem 4: 7-7.5
- Problem 5: 7.5-8
Ibero American Olympiad
- Problem 1/4: 5.5
- Problem 2/5: 6.5
- Problem 3/6: 7.5
HMMT
February Contest
- Individual Round, Problem 1-5: 5
- Individual Round, Problem 6-10: 6
- Team Round: 7.5
- HMIC: 8
APMO
- Problem 1: 6
- Problem 2: 7
- Problem 3: 7
- Problem 4: 7.5
- Problem 5: 8
Balkan MO
- Problem 1: 6
- Solve the equation
in positive integers.
- Solve the equation
- Problem 2: 6.5
- Let
be a line parallel to the side
of a triangle
, with
on the side
and
on the side
. The lines
and
meet at point
. The circumcircles of triangles
and
meet at two distinct points
and
. Prove that
.
- Let
- Problem 3: 7.5
- A
rectangle is partitioned into unit squares. The centers of all the unit squares, except for the four corner squares and eight squares sharing a common side with one of them, are coloured red. Is it possible to label these red centres
in such way that the following to conditions are both fulfilled
- A
the distances
are all equal to
the closed broken line
has a centre of symmetry?
- Problem 4: 8
- Denote by
the set of all positive integers. Find all functions
such that
- Denote by
for all
. '
HMMT
February Contest
- Individual Round, Problem 1-5: 5
- Individual Round, Problem 6-10: 6
- Team Round: 7.5
- HMIC: 8
V
China TST
- Problem 1/4: 7
- Given an integer
prove that there exist odd integers
and a positive integer
such that
- Given an integer
- Problem 2/5: 8.5
- Given a positive integer
and real numbers
such that
prove that for any positive real number
- Given a positive integer
- Problem 3/6: 10
- Let
be an integer and let
be non-negative real numbers. Define
for
. Prove that
- Let
Ibero American Olympiad
- Problem 1/4: 5.5
- Problem 2/5: 6.5
- Problem 3/6: 7.5
IMO
- Problem 1/4: 6.5
- Find all functions
(so that
is a function from the positive real numbers) such that
- Find all functions
![$\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}$](http://latex.artofproblemsolving.com/9/a/b/9abfd05b9258aa6b964529540de233fce9a2edd6.png)
for all positive real numbers satisfying
(Solution)
- Problem 2/5: 7.5-8
- Let
be a polynomial of degree
with integer coefficients, and let
be a positive integer. Consider the polynomial
, where
occurs
times. Prove that there are at most
integers
such that
. (Solution)
- Let
- Problem 3/6: 9.5
- Assign to each side
of a convex polygon
the maximum area of a triangle that has
as a side and is contained in
. Show that the sum of the areas assigned to the sides of
is at least twice the area of
. (<url>viewtopic.php?p=572824#572824 Solution</url>)
- Assign to each side
IMO Shortlist
- Problem 1-2: 5.5-7
- Problem 3-4: 7-8
- Problem 5+: 8-10
JBMO
- Problem 1: 4
- Find all real numbers
such that
- Find all real numbers
- Problem 2: 5
- Let
be a convex quadrilateral with
,
and
. The diagonals intersect at point
. Determine the measure of
.
- Let
- Problem 3: 5
- Find all prime numbers
, such that
.
- Find all prime numbers
- Problem 4: 6
- A
table is divided into
white unit square cells. Two cells are called neighbors if they share a common side. A move consists in choosing a cell and changing the colors of neighbors from white to black or from black to white. After exactly
moves all the
cells were black. Find all possible values of
.
- A
Putnam
- Problem A/B,1-2: 7
- Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola
and both branches of the hyperbola
(A set
in the plane is called convex if for any two points in
the line segment connecting them is contained in
) (<url>viewtopic.php?p=978383#p978383 Solution</url>)
- Find the least possible area of a convex set in the plane that intersects both branches of the hyperbola
- Problem A/B,3-4: 8
- Let
be an
matrix all of whose entries are
and whose rows are mutually orthogonal. Suppose
has an
submatrix whose entries are all
Show that
. (<url>viewtopic.php?p=383280#383280 Solution</url>)
- Let
- Problem A/B,5-6: 9
- For any
, define the set
. Show that there are no three positive reals
such that
. (<url>viewtopic.php?t=127810 Solution</url>)
- For any
USAMO
- Problem 1/4: 7
- Let
be a convex polygon with
sides,
. Any set of
diagonals of
that do not intersect in the interior of the polygon determine a triangulation of
into
triangles. If
is regular and there is a triangulation of
consisting of only isosceles triangles, find all the possible values of
. (Solution)
- Let
- Problem 2/5: 8
- Three nonnegative real numbers
,
,
are written on a blackboard. These numbers have the property that there exist integers
,
,
, not all zero, satisfying
. We are permitted to perform the following operation: find two numbers
,
on the blackboard with
, then erase
and write
in its place. Prove that after a finite number of such operations, we can end up with at least one
on the blackboard. (Solution)
- Three nonnegative real numbers
- Problem 3/6: 9
- Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree
with real coefficients is the average of two monic polynomials of degree
with
real roots. (Solution)
- Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree
USAJMO
- Problem 1/4: 6
- Problem 2/5: 7
- Problem 3/6: 8
USAMTS
USAMTS generally has a different feel to it than olympiads, and is mainly for proofwriting practice instead of olympiad practice depending on how one takes the test. USAMTS allows an entire month to solve problems, with internet resources and books being allowed. However, the ultimate gap is that it permits computer programs to be used, and that Problem 1 is not a proof problem. However, it can still be roughly put to this rating scale:
- Problem 1-2: 4
- Find three isosceles triangles, no two of which are congruent, with integer sides, such that each triangle’s area is numerically equal to 6 times its perimeter. (Solution)
- Problem 3-5: 6
- Call a positive real number groovy if it can be written in the form
for some positive integer
. Show that if
is groovy, then for any positive integer
, the number
is groovy as well. (Solution)
- Call a positive real number groovy if it can be written in the form
USA TST
(seems to vary more than other contests; estimates based on 08 and 09)
- Problem 1/4/7: 7
- Problem 2/5/8: 8
- Problem 3/6/9: 9.5
Scale
[1] All levels estimated and refer to averages. The following is a rough standard based on the USA tier system AMC 8 – AMC 10 – AMC 12 – AIME – USAMO/USAJMO, representing Middle School – Junior High – High School – Challenging High School – Olympiad levels. Other contests can be interpolated against this.
- Problems strictly for beginners, on the easiest elementary school or middle school levels. Examples would be MOEMS, easy Mathcounts questions, #1-20 on AMC 8s, #1-5 AMC 10s, and others that involve standard techniques introduced up to the middle school level
- For motivated beginners, harder questions from the previous categories (#21-25 on AMC 8, Challenging Mathcounts questions, #5-20 on AMC 10, #5-10 on AMC 12, the easiest AIME questions, etc).
- For those not too familiar with standard techniques, MathCounts National, #21-25 on AMC 10, #11-20ish on AMC 12, #1-5 on AIMEs, etc.
- Intermediate-leveled problem solvers, the most difficult questions on AMC 12s (#21-25s), more difficult AIME-styled questions #6-10
- Difficult AIME problems (#10-13), others, simple proof-based problems (JBMO etc)
- High-leveled AIME-styled questions, not requiring proofs (#12-15). Introductory-leveled Olympiad-level questions (#1,4s).
- Intermediate-leveled Olympiad-level questions, #1,4s that require more technical knowledge than new students to Olympiad-type questions have, easier #2,5s, etc.
- High-level difficult Olympiad-level questions, eg #2,5s on difficult Olympiad contest and easier #3,6s, etc.
- Difficult Olympiad-level questions, eg #3,6s on difficult Olympiad contests.
- Problems occasionally even unsuitable for normal grade school level competitions due to being exceedingly tedious/long/difficult (eg very few students are capable of solving, even on a worldwide basis), or involving techniques beyond high school level mathematics.