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The dataset generation failed
Error code: DatasetGenerationError
Exception: TypeError
Message: Couldn't cast array of type string to null
Traceback: Traceback (most recent call last):
File "/usr/local/lib/python3.14/site-packages/datasets/builder.py", line 1827, in _prepare_split_single
for key, table in generator:
^^^^^^^^^
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 613, in wrapped
for item in generator(*args, **kwargs):
~~~~~~~~~^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/packaged_modules/json/json.py", line 343, in _generate_tables
self._cast_table(pa_table, json_field_paths=json_field_paths),
~~~~~~~~~~~~~~~~^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/packaged_modules/json/json.py", line 132, in _cast_table
pa_table = table_cast(pa_table, self.info.features.arrow_schema)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2378, in table_cast
return cast_table_to_schema(table, schema)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2312, in cast_table_to_schema
cast_array_to_feature(
~~~~~~~~~~~~~~~~~~~~~^
table[name] if name in table_column_names else pa.array([None] * len(table), type=schema.field(name).type),
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
feature,
^^^^^^^^
)
^
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 1861, in wrapper
return pa.chunked_array([func(chunk, *args, **kwargs) for chunk in array.chunks])
~~~~^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2068, in cast_array_to_feature
_c(array.field(name) if name in array_fields else null_array, subfeature)
~~^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 1863, in wrapper
return func(array, *args, **kwargs)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2118, in cast_array_to_feature
casted_array_values = _c(array.values, feature.feature)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 1863, in wrapper
return func(array, *args, **kwargs)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2152, in cast_array_to_feature
return array_cast(
array,
...<2 lines>...
allow_decimal_to_str=allow_decimal_to_str,
)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 1863, in wrapper
return func(array, *args, **kwargs)
File "/usr/local/lib/python3.14/site-packages/datasets/table.py", line 2014, in array_cast
raise TypeError(f"Couldn't cast array of type {_short_str(array.type)} to {_short_str(pa_type)}")
TypeError: Couldn't cast array of type string to null
The above exception was the direct cause of the following exception:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1369, in compute_config_parquet_and_info_response
parquet_operations, partial, estimated_dataset_info = stream_convert_to_parquet(
~~~~~~~~~~~~~~~~~~~~~~~~~^
builder, max_dataset_size_bytes=max_dataset_size_bytes
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
)
^
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 948, in stream_convert_to_parquet
builder._prepare_split(split_generator=splits_generators[split], file_format="parquet")
~~~~~~~~~~~~~~~~~~~~~~^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
File "/usr/local/lib/python3.14/site-packages/datasets/builder.py", line 1694, in _prepare_split
for job_id, done, content in self._prepare_split_single(
~~~~~~~~~~~~~~~~~~~~~~~~~~^
gen_kwargs=gen_kwargs, job_id=job_id, **_prepare_split_args
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
):
^
File "/usr/local/lib/python3.14/site-packages/datasets/builder.py", line 1879, in _prepare_split_single
raise DatasetGenerationError("An error occurred while generating the dataset") from e
datasets.exceptions.DatasetGenerationError: An error occurred while generating the datasetNeed help to make the dataset viewer work? Make sure to review how to configure the dataset viewer, and open a discussion for direct support.
id string | title string | category string | subjects list | concepts list | source dict | difficulty dict | estimatedMinutes null | statement string | answerChoices list | answer dict | solutions list | images dict | rights dict |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
prob.aops-000000 | APMO 1989 Problem 1 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1989,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1989_APMO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7,
"note": null
} | null | Let \(x_1\), \(x_2\), \(\cdots\), \(x_n\) be positive real numbers, and let
S = x_1 + x_2 + \cdots + x_n.
Prove that
(1 + x_1)(1 + x_2) \cdots (1 + x_n) \leq 1 + S + \frac{S^2}{2!} + \frac{S^3}{3!} + \cdots + \frac{S^n}{n!} | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000001 | APMO 1989 Problem 2 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1989,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1989_APMO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.4,
"note": null
} | null | Prove that the equation 6(6a^2 + 3b^2 + c^2) = 5n^2 has no solutions in integers except \(a = b = c = n = 0\). | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000002 | APMO 1989 Problem 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1989,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1989_APMO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Let \(A_1\), \(A_2\), \(A_3\) be three points in the plane, and for convenience, let \(A_4= A_1\), \(A_5 = A_2\). For \(n = 1\), \(2\), and \(3\), suppose that \(B_n\) is the midpoint of \(A_n A_{n+1}\), and suppose that \(C_n\) is the midpoint of \(A_n B_n\). Suppose that \(A_n C_{n+1}\) and \(B_n A_{n+2}\) meet at \(... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000003 | APMO 1989 Problem 4 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1989,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/1989_APMO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.2,
"note": null
} | null | Let \(S\) be a set consisting of \(m\) pairs \((a,b)\) of positive integers with the property that \(1 \leq a < b \leq n\). Show that there are at least
4m \cdot \dfrac{(m - \dfrac{n^2}{4})}{3n}
triples \((a,b,c)\) such that \((a,b)\), \((a,c)\), and \((b,c)\) belong to \(S\). | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000004 | APMO 1989 Problem 5 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1989,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1989_APMO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.6,
"note": null
} | null | Determine all functions \(f\) from the reals to the reals for which
(1) \(f(x)\) is strictly increasing and (2) \(f(x) + g(x) = 2x\) for all real \(x\),
where \(g(x)\) is the composition inverse function to \(f(x)\). (Note: \(f\) and \(g\) are said to be composition inverses if \(f(g(x)) = x\) and \(g(f(x)) = x\) for... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000005 | APMO 1990 1 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1990,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1990_APMO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7,
"note": null
} | null | Let .I be the intersection of" AG and EF.
Let 6 = AI.IG - FI.IE. Then
AI:ADlz, IG:ADl6, FI:BCl4:IE. (1)
Fur"ther, applying the cosine rule to triangles ABD, ACD we get
i AB2 : BC2I4 + AD2 _ AD.BC.cos LBDA,
'''.. ACz : aC2l+ + AD2 + AD.BC.cos IBDA,
f ,.*j;-\, so ADz : (Or, + AC2 - BCr 12) 12.
-H"rr""
6 - (or, + ACz - zB... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000006 | APMO 1990 2 | combinatorics | [
"combinatorics"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1990,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1990_APMO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.4,
"note": null
} | null | Let a_1,a_2,\cdots,a_n be positive real numbers, and let S_k be the sum of products of a_1,a_2,\cdots,a_n taken k at a time.
Show that
S_k S_{n-k} \geq {n \choose k}^2 a_1 a_2 \cdots a_n. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000006.s0",
"kind": "proof-sketch",
"text": "Let, \n\\begin{align*}\nS_k & = \\sum_{i=1}^{n\\choose k}t_i \\\\\nS_{n-k} & = \\left(\\prod_{m=1}^{n}a_m \\right)\\left(\\sum_{i=1}^{n\\choose k}\\dfrac{1}{t_i}\\right)\n\\end{align*}\nThus, \n\\begin{align*}\nS_k S_{n-k} &=\\left(\\sum_{i=... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000007 | APMO 1990 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1990,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1990_APMO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Consider all the triangles ABC which have a fixed base AB and whose altitude from C is a constant h. For which of these triangles is the product of its altitudes a maximum? | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000008 | APMO 1990 5 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1990,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1990_APMO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.6,
"note": null
} | null | Not a perfect solution, but I think it works.
\\
The construction would obviously depend on the parity of n, let's denote m to be \(ceiling((n-1)/2)\), so either \(n=2*m+2\) or \(n=2*m+1\).
\\
By the condition, we know that \(m>=2\), then we construct some rectangles with side lengths \(m\) and \(\sqrt{m^2-1}\).
We can... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000009 | APMO 1991 2 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1991,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1991_APMO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.4,
"note": null
} | null | Suppose there are 997 points given in a plane. If every two points are joined by a line segment with its midpoint colored in red, show that there are at least 1991 red points in the plane. Can you find a special case with exactly 1991 red points? | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000009.s0",
"kind": "proof-sketch",
"text": "For the number of red points to be minimum, the number of red points that coincide must be maximum. Therefore, set all points as (0, 0), (0, 1), \\ldots, (0, 996). Notice that possible red points will be (0, 0.5), (0, 1), (0, 1.5), \\ldots, ... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000010 | APMO 1991 Problem 3 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1991,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1991_APMO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Let \(a_1\), \(a_2\), \(\cdots\), \(a_n\), \(b_1\), \(b_2\), \(\cdots\), \(b_n\) be positive real numbers such that \(a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n\). Show that
\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2} | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000010.s0",
"kind": "proof-sketch",
"text": "By Cauchy-Schwarz, \\(\\left(\\sum \\dfrac{a_i^2}{a_i+b_i}\\right)\\left(\\sum a_i+b_i\\right)\\geq \\left(\\sum a_i\\right) ^2\\). Since \\(\\sum a_i=\\sum b_i\\), we get\\(\\sum \\dfrac{a_i^2}{a_i+b_i}\\geq \\dfrac{\\left(\\sum a_i\\right)... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000011 | APMO 1991 Problem 5 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1991,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1991_APMO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.6,
"note": null
} | null | Given are two tangent circles and a point \(P\) on their common tangent perpendicular to the lines joining their centres. Construct with ruler and compass all the circles that are tangent to these two circles and pass through the point \(P\). | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000012 | APMO 1997 Problem 2 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1997,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1997_APMO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.4,
"note": null
} | null | Find an integer in the range 100 \leq n \leq 1997 such that \frac{2^n+2}{n} is an integer. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000012.s0",
"kind": "proof-sketch",
"text": "We proceed by contradiction. Assume there exists an odd integer n > 1 such that\nn | (2n + 2).\nSince n is odd, gcd(n, 2) = 1. We can factor out a 2 from the expression and divide\nit out without altering the divisibility condition:\nn | 2(2... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000013 | APMO 1998 Problem 2 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1998,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1998_APMO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.4,
"note": null
} | null | Show that for any positive integers \(a\) and \(b\), \((36a + b)(36b + a)\) cannot be a power of \(2\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000013.s0",
"kind": "proof-sketch",
"text": "First, assume that \\((36a + b)(36b + a)\\) is a power of \\(2\\). \nLet \\(36a + b = 2^m = p\\) and \\(36b + a = 2^n = q\\).\nThen \\frac{1}{36} < \\frac{p}{q} = \\frac{2^m}{2^n} = 2^{m-n} < 36\n-6 < m - n < 6\n\nConsider \\(4^m - 4^n = p^2... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000014 | APMO 1998 Problem 3 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1998,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1998_APMO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Let \(a, b, c\) be positive real numbers. Prove that \((1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a}) \geq 2(1+\frac{a+b+c}{\sqrt[\leftroot{-2}\uproot{2}3]{abc}})\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000014.s0",
"kind": "proof-sketch",
"text": "WLOG, assume \\(abc = 1\\). Then\n(a+b+c)(ab+ac+bc-2)-abc \\geq 2\n(a+b+c)(ab+ac+bc-2) \\geq 3\n\nThen by AM-GM, \n\\frac{a+b+c}{3} \\geq \\sqrt[\\leftroot{-2}\\uproot{2}3]{abc}\na+b+c \\geq 3\nand\nab+ac+bc \\geq 3\nso \nab+ac+bc-2 \\geq 1\... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000015 | APMO 1998 Problem 4 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1998,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/1998_APMO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.2,
"note": null
} | null | (Răzvan Gelca)
Let \(ABC\) be a triangle and \(D\) the foot of the altitude from \(A\). Let \(E\) and \(F\) be on a line through \(D\) such that \(AE\) is perpendicular to \(BE\), \(AF\) is perpendicular to \(CF\), and \(E\) and \(F\) are different from \(D\). Let \(M\) and \(N\) be the midpoints of the line segments... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000015.s0",
"kind": "proof-sketch",
"text": "We use directed angles mod \\(\\pi\\).\n\nsize(200);\ndefaultpen(1);\n\npair B=(0,0), A=(1,3), C=(4,0), X=(4,4);\npair D=foot(A,B,C);\npath O1=circumcircle(A,B,D), O2=circumcircle(A,C,D);\npair E=IntersectionPoint(O1,D--X,1), F=IntersectionP... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000016 | APMO 1999 Problem 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 1999,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1999_APMO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Let \(\Gamma_1\) and \(\Gamma_2\) be two circles intersecting at \(P\) and \(Q\). The common tangent, closer to \(P\), of \(\Gamma_1\) and \(\Gamma_2\) touches \(\Gamma_1\) at \(A\) and \(\Gamma_2\) at \(B\). The tangent of \(\Gamma_1\) at \(P\) meets \(\Gamma_2\) at \(C\), which is different from \(P\), and the extens... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000017 | APMO 2001 2 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2001,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/2001_APMO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.4,
"note": null
} | null | Find the largest positive integer \(N\) so that the number of integers in the set \(\{1,2,\dots,N\}\) which are divisible by 3 is equal to the number of integers which are divisible by 5 or 7 (or both). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000017.s0",
"kind": "proof-sketch",
"text": "The equation\n\\left\\lfloor\\frac{N}{3}\\right\\rfloor=\\left\\lfloor\\frac{N}{5}\\right\\rfloor+ \\left\\lfloor\\frac{N}{7}\\right\\rfloor-\\left\\lfloor\\frac{N}{35}\\right\\rfloor\nis equivalent to the problem statement. The period (of r... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000018 | APMO 2001 Problem 1 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2001,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/2001_APMO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7,
"note": null
} | null | For any positive integer \(n\), let \(S(n)\) be the sum of digits in the decimal representation of \(n\). Any positive integer obtained by removing one or more digits from the right end of the decimal representation of \(n\) is called a stump of \(n\). Let \(T(n)\) be the sum of all stumps of \(n\). Prove that \(n = S(... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000019 | APMO 2013 Problem 5 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2013,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/2013_APMO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.6,
"note": null
} | null | Let \(ABCD\) be a quadrilateral inscribed in a circle \(\omega\), and let \(P\) be a point on the extension of \(AC\) such that \(PB\) and \(PD\) are tangent to \(\omega\). The tangent at \(C\) intersects \(PD\) at \(Q\) and the line \(AD\) at \(R\). Let \(E\) be the second point of intersection between \(AQ\) and \(\o... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000019.s0",
"kind": "proof-sketch",
"text": "Solution 1 \n\nLet \\(R'=AD \\cap BE\\). Note that the tangents at \\(B\\) and \\(D\\) concur on \\(AC\\) at \\(P\\), so \\(ABCD\\) is harmonic, hence the tangents at \\(C\\) and \\(A\\) concur on \\(BD\\) at \\(X\\), say.\n\nNow apply Pasca... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000020 | APMO 2016 Problem 1 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2016,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/2016_APMO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7,
"note": null
} | null | We say that a triangle \(ABC\) is great if the following holds: for any point \(D\) on the side \(BC\), if \(P\) and \(Q\) are the feet of the perpendiculars from \(D\) to the lines \(AB\) and \(AC\), respectively, then the reflection of \(D\) in the line \(PQ\) lies on the circumcircle of the triangle \(ABC\). Prove t... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000021 | APMO 2016 Problem 2 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2016,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/2016_APMO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.4,
"note": null
} | null | A positive integer is called fancy if it can be expressed in the form2^{a_1}+2^{a_2}+ \cdots+ 2^{a_{100}},where \(a_1,a_2, \cdots, a_{100}\) are non-negative integers that are not necessarily distinct. Find the smallest positive integer \(n\) such that no multiple of \(n\) is a fancy number. | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000022 | APMO 2016 Problem 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2016,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/2016_APMO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Let \(AB\) and \(AC\) be two distinct rays not lying on the same line, and let \(\omega\) be a circle with center \(O\) that is tangent to ray \(AC\) at \(E\) and ray \(AB\) at \(F\). Let \(R\) be a point on segment \(EF\). The line through \(O\) parallel to \(EF\) intersects line \(AB\) at \(P\). Let \(N\) be the inte... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000023 | APMO 2016 Problem 4 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2016,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/2016_APMO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.2,
"note": null
} | null | The country Dreamland consists of \(2016\) cities. The airline Starways wants to establish some one-way flights between pairs of cities in such a way that each city has exactly one flight out of it. Find the smallest positive integer \(k\) such that no matter how Starways establishes its flights, the cities can always ... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000024 | APMO 2016 Problem 5 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2016,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/2016_APMO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.6,
"note": null
} | null | Find all functions \(f: \mathbb{R}^+ \to \mathbb{R}^+\) such that
(z + 1)f(x + y) = f(xf(z) + y) + f(yf(z) + x),for all positive real numbers \(x, y, z\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000024.s0",
"kind": "proof-sketch",
"text": "We claim that \\(f(x)=x\\) is the only solution. It is easy to check that it works. Now, we will break things down in several claims. Let \\(P(x.y,z)\\) be the assertion to the Functional Equation.\n\nClaim 1: \\(f\\) is injective.\n\nProof:... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000025 | APMO 2023 1 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2023,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/2023_APMO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7,
"note": null
} | null | Let \(n\ge 5\) be an integer. Consider \(n\) squares with side lengths \(1, 2, \dots , n\),
respectively. The squares are arranged in the plane with their sides parallel to the \(x\) and
\(y\) axes. Suppose that no two squares touch, except possibly at their vertices.
Show that it is possible to arrange these squares i... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000025.s0",
"kind": "proof-sketch",
"text": "https://youtu.be/xkIm0k1FE-8",
"hints": [],
"images": []
}
] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000026 | APMO 2023 2 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2023,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/2023_APMO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.4,
"note": null
} | null | Find all integers \(n\) satisfying \(n \ge 2\) and \(\frac{\sigma(n)}{p(n)-1}=n\), in which \(\sigma(n)\) denotes the sum of all positive divisors of \(n\), and \(p(n)\) denotes the largest prime divisor of \(n\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000026.s0",
"kind": "proof-sketch",
"text": "Answer: n=6.\n\nRearrange the equation to the form p(n)-1=\\frac{\\sigma(n)}{n}, then do case work on the values of p(n). \n\nAs p(n) grows large, we can see that the right hand side becomes always smaller than the left hand side. We can use... | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000027 | APMO 2023 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "APMO",
"year": 2023,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/2023_APMO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Let ABCD be a parallelogram. Let W, X, Y , and Z be points on sides AB, BC, CD, and
DA, respectively, such that the incenters of triangles AWZ, BXW, CYX and DZY form a
parallelogram. Prove that WXYZ is a parallelogram. | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "apmo",
"holder": "Asian Pacific Mathematics Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found."
} |
prob.aops-000028 | Canadian MO 1969 Problem 1 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | Show that if \(a_1/b_1=a_2/b_2=a_3/b_3\) and \(p_1,p_2,p_3\) are not all zero, then \(\left(\frac{a_1}{b_1} \right)^n=\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}\) for every positive integer \(n.\) | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000028.s0",
"kind": "proof-sketch",
"text": "Instead of proving the two expressions equal, we prove that their difference equals zero.\n\nSubtracting the LHS from the RHS,\n\\(0=\\frac{p_1a_1^n+p_2a_2^n+p_3a_3^n}{p_1b_1^n+p_2b_2^n+p_3b_3^n}-\\frac{a_1^n}{b_1^n}.\\)\n\nFinding a common ... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000029 | Canadian MO 1969 Problem 10 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 10,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_10",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Let \(ABC\) be the right-angled isosceles triangle whose equal sides have length 1. \(P\) is a point on the hypotenuse, and the feet of the perpendiculars from \(P\) to the other sides are \(Q\) and \(R\). Consider the areas of the triangles \(APQ\) and \(PBR\), and the area of the rectangle \(QCRP\). Prove that regard... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000029.s0",
"kind": "proof-sketch",
"text": "Let \\(AQ=x.\\) Because triangles \\(APQ\\) and \\(BPR\\) both contain a right angle and a \\(45^\\circ\\) angle, they are isosceles right triangles. Hence, \\(PQ=RC=x\\) and \\(QC=PR=BR=1-x.\\)\n\nNow let's consider when \\(\\frac13 <x<\\f... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000030 | Canadian MO 1969 Problem 2 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | Determine which of the two numbers \sqrt{c+1}-\sqrt{c}, \sqrt{c}-\sqrt{c-1} is greater for any c\ge 1. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000030.s0",
"kind": "proof-sketch",
"text": "Multiplying and dividing \\sqrt{c+1}-\\sqrt c by its conjugate,\n\n\\sqrt{c+1}-\\sqrt c=\\frac{(\\sqrt{c+1})^2-(\\sqrt c)^2}{\\sqrt{c+1}+\\sqrt{c}}=\\frac1{\\sqrt{c+1}+\\sqrt{c}}.\n\nSimilarly, \\sqrt c-\\sqrt{c-1}=\\frac{1}{\\sqrt c+\\sqrt{... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000031 | Canadian MO 1969 Problem 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | Let \(c\) be the length of the hypotenuse of a right triangle whose two other sides have lengths \(a\) and \(b\). Prove that \(a+b\le c\sqrt{2}\). When does the equality hold? | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000031.s0",
"kind": "proof-sketch",
"text": "We square both sides of the equation to get a^2+b^2+2ab \\leq 2c^2 . Since , by Pythagorean Theorem, a^2 + b^2 = c^2, allowing us to reduce the original inequality to 2ab \\leq c^2. By AM-GM, 2ab \\leq a^2 + b^2 = c^2, which leads to 2ab ... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000032 | Canadian MO 1969 Problem 4 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.5,
"note": null
} | null | Let \(ABC\) be an equilateral triangle, and \(P\) be an arbitrary point within the triangle. Perpendiculars \(PD,PE,PF\) are drawn to the three sides of the triangle. Show that, no matter where \(P\) is chosen, \(\frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000032.s0",
"kind": "proof-sketch",
"text": "Let a side of the triangle be \\(s\\) and let \\([ABC]\\) denote the area of \\(ABC.\\) Note that because \\(2[\\triangle ABC]=2[\\triangle APB]+2[\\triangle APC]+2[\\triangle BPC],\\) \\(\\frac{\\sqrt3}{2}s^2=s(PD+PE+PF).\\) Dividing both s... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000033 | Canadian MO 1969 Problem 5 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Let \(ABC\) be a triangle with sides of length \(a\), \(b\) and \(c\). Let the bisector of the \(\angle C\) cut \(AB\) at \(D\). Prove that the length of \(CD\) is \(\frac{2ab\cos \frac{C}{2}}{a+b}.\) | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000033.s0",
"kind": "proof-sketch",
"text": "Let \\(CD=d.\\) Note that \\([\\triangle ABC]=[\\triangle ACD]+[\\triangle BCD].\\) This can be rewritten as \\(\\frac12 a b \\sin C=\\frac12 a d \\sin \\frac C2 +\\frac12 b d \\sin \\frac C2 .\\)\n\nBecause \\(\\sin C=2\\sin \\frac C2 \\cos... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000034 | Canadian MO 1969 Problem 6 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 6,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_6",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Find the sum of \(1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!\), where $ n!=n(n-1)(n-2)\cdots2\cdot1$. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000034.s0",
"kind": "proof-sketch",
"text": "Note that for any positive integer $ n,$ $ n\\cdot n!+(n-1)\\cdot(n-1)!=(n^2+n-1)(n-1)!=(n+1)!-(n-1)!.$\nHence, pairing terms in the series will telescope most of the terms.\n\nIf $ n$ is odd, $ (n+1)!-(n-1)!+(n-1)!-(n-3)!\\cdots -2!+2!-0!.... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000035 | Canadian MO 1969 Problem 7 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 7,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_7",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Show that there are no integers \(a,b,c\) for which \(a^2+b^2-8c=6\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000035.s0",
"kind": "proof-sketch",
"text": "Note that all perfect squares are equivalent to \\(0,1,4\\pmod8.\\) Hence, we have \\(a^2+b^2\\equiv 6\\pmod8.\\) It's impossible to obtain a sum of \\(6\\) with two of \\(0,1,4,\\) so our proof is complete.",
"hints": [],
"images"... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000036 | Canadian MO 1969 Problem 8 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 8,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_8",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Let \(f\) be a function with the following properties:
1) \(f(n)\) is defined for every positive integer \(n\);
2) \(f(n)\) is an integer;
3) \(f(2)=2\);
4) \(f(mn)=f(m)f(n)\) for all \(m\) and \(n\);
5) \(f(m)>f(n)\) whenever \(m>n\).
Prove that \(f(n)=n\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000036.s0",
"kind": "proof-sketch",
"text": "We can manipulate the 4th equation into \\frac{f(mn)}{f(n)}=f(n). Then, we substitute 2 as m and n. This allows us to calculate that f(1)=1. Plugging in 2 for the original condition allows us to get f(4)=4. \nApplying the 5th rule yields f(2... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000037 | Canadian MO 1969 Problem 9 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1969,
"number": 9,
"url": "https://artofproblemsolving.com/wiki/index.php/1969_Canadian_MO_Problems/Problem_9",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Show that for any quadrilateral inscribed in a circle of radius \(1,\) the length of the shortest side is less than or equal to \(\sqrt{2}\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000037.s0",
"kind": "proof-sketch",
"text": "Let a,b,c,d be the edge-lengths and e,f be the lengths of the diagonals of the quadrilateral. By Ptolemy's Theorem, ab+cd = ef. However, each diagonal is a chord of the circle and so must be shorter than the diameter: e,f \\le 2 and thus ab+... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000038 | Canadian MO 1970 Problem 1 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | Find all number triples $ (x,y,z)$ such that when any of these numbers is added to the product of the other two, the result is \(2\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000038.s0",
"kind": "proof-sketch",
"text": "We have:\n\\(\\begin{matrix} x + yz &=& 2\\\\\ny + xz &=& 2\\\\\nz + yx &=& 2 \\end{matrix}\\)\n\nFrom the first equation minus the second, we get :\n\n \\begin{matrix}x-y + yz - xz &=& 0 \\\\(1-z)(x-y) &=& 0\\end{matrix} \n\nSo either \\(z=... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000039 | Canadian MO 1970 Problem 10 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 10,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_10",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Given the polynomial \(f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n\) with integer coefficients \(a_1,a_2,\ldots,a_n\), and given also that there exist four distinct integers \(a, b, c\) and \(d\) such that \(f(a)=f(b)=f(c)=f(d)=5\), show that there is no integer \(k\) such that \(f(k)=8\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000039.s0",
"kind": "proof-sketch",
"text": "Set g(x) = f(x) − 5. Since a, b, c, d are all roots of g(x), we must have\n\ng(x) = (x − a) (x − b) (x − c) (x − d) h(x)\n\nfor some h(x) ∈ Z[x]. \n\nLet k be an integer such that f(k) = 8, giving g(k) = f(k) − 5 = 3. Using the factorization... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000040 | Canadian MO 1970 Problem 2 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | Given a triangle \(ABC\) with angle \(A\) obtuse and with altitudes of length \(h\) and \(k\) as shown in the diagram, prove that \(a+h\ge b+k\). Find under what conditions \(a+h=b+k\).
draw((0,0)--(5,0)--(16/5,12/5)--cycle,dot);
draw((2.5,0)--(2.5,7.5/4)--(5,0)--cycle,black);
MP("C",(0,0),SW);MP("D",(16/5,12/5),N);M... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000040.s0",
"kind": "proof-sketch",
"text": "We first realize \\frac{a}{b}=\\frac{k}{h} and a \\> b and k \\> h. Let \n \\frac{a}{b} = n. We want to show nb+h \\geq nh+b. We can simplify this to (n-1)(b-h) \\geq 0. \\Delta ABE is a right triangle with hypotenuse b and h as a leg. The... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000041 | Canadian MO 1970 Problem 3 | arithmetic | [
"arithmetic"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | A set of balls is given. Each ball is coloured red or blue, and there is at least one of each colour. Each ball weighs either \(1\) pound or \(2\) pounds, and there is at least one of each weight. Prove that there are two balls having different weights and different colours. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000041.s0",
"kind": "proof-sketch",
"text": "Let \\(R1, R2, B1, B2\\) each be the 4 types of balls such that the number is the weight and the letter be the color of the ball.\n\nWe can prove by contradiction.\n\nAssume \\(R1\\) exists, then, \\(B2\\) can't exist. We know there has to b... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000042 | Canadian MO 1970 Problem 4 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.5,
"note": null
} | null | a) Find all positive integers with initial digit \(6\) such that the integer formed by deleting \(6\) is \(1/25\) of the original integer.
b) Show that there is no integer such that the deletion of the first digit produces a result that is \(1/35\) of the original integer. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000042.s0",
"kind": "proof-sketch",
"text": "a)\nLet the integer have n digits. Then the integer can be represented in the form of \\(6 *10^{n-1} + x\\) where x is a non-negative integer less than \\(10^{n-1}\\). The problem is asking for integers such as \\(6 *10^{n-1} + x = 25x\\). S... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000043 | Canadian MO 1970 Problem 5 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | A quadrilateral has one vertex on each side of a square of side-length 1. Show that the lengths \(a\), \(b\), \(c\) and \(d\) of the sides of the quadrilateral satisfy the inequalities \(2\le a^2+b^2+c^2+d^2\le 4.\) | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000043.s0",
"kind": "proof-sketch",
"text": "Let the quadrilateral be \\(ABCD\\). Suppose \\(A\\) is a distance \\(w\\), \\(1-w\\) from the two nearest vertices of the square. Define \\(x\\), \\(y\\), \\(z\\) similarly. Then the sum of the squares of the sides of the quadrilateral is \... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000044 | Canadian MO 1970 Problem 6 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 6,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_6",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Given three non-collinear points \(A,B,C,\) construct a circle with centre \(C\) such that the tangents from \(A\) and \(B\) are parallel. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000044.s0",
"kind": "proof-sketch",
"text": "Construct segment \\(AB\\). Find the midpoint of \\(AB\\) (denote the midpoint \\(M\\)), by constructing its perpendicular bisector, and the point where the perpendicular bisector and \\(AB\\) meet is the midpoint of \\(AB\\). Join \\(M\\) t... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000045 | Canadian MO 1970 Problem 7 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 7,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_7",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Show that from any five integers, not necessarily distinct, one can always choose three of these integers whose sum is divisible by \(3\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000045.s0",
"kind": "proof-sketch",
"text": "Claim 1: if \\(a\\), \\(b\\), and \\(c\\) have equal modularity, then \\((a+b+c)\\) is divisible by \\(3\\)\n\nProof of Claim 1: Let \\(a\\), \\(b\\), and \\(c\\) be three integers with equal modularity.\n\nThat is, \\(a \\equiv b\\;(mod\\;3... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000046 | Canadian MO 1970 Problem 8 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 8,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_8",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Consider all line segments of length 4 with one end-point on the line y=x and the other end-point on the line y=2x. Find the equation of the locus of the midpoints of these line segments. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000046.s0",
"kind": "proof-sketch",
"text": "Point on \\(y=x\\) line: \\((a,a)\\)\n\nPoint on \\(y=2x\\) line: \\((b,2b)\\)\n\nThen,\n\n\\((b-a)^2+(2b-a)^2=4^2\\)\n\n\\(5b^2-6ab+2a^2-16=0\\)\n\nUsing the quadratic equation,\n\n\\(b=\\frac{3a \\pm \\sqrt{80 - a^2}}{5}\\)\n\nMidpoint \\(... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000047 | Canadian MO 1970 Problem 9 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1970,
"number": 9,
"url": "https://artofproblemsolving.com/wiki/index.php/1970_Canadian_MO_Problems/Problem_9",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Let \(f(n)\) be the sum of the first \(n\) terms of the sequence
0, 1,1, 2,2, 3,3, 4,4, 5,5, 6,6, \ldots\, .
a) Give a formula for \(f(n)\).
b) Prove that \(f(s+t)-f(s-t)=st\) where \(s\) and \(t\) are positive integers and \(s>t\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000047.s0",
"kind": "proof-sketch",
"text": "Part a):\n\n\\(f(n)=\\begin{cases} 2\\sum_{k=1}^{\\frac{n-1}{2}}k,\\; & n\\;is\\;odd \\\\ 2\\sum_{k=1}^{\\frac{n}{2}}k-\\frac{n}{2},\\; & n\\;is\\;even\\end{cases}\\)\n\n\\(f(n)=\\begin{cases} \\left( \\frac{n-1}{2} \\right)\\left( \\frac{n-... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000048 | Canadian MO 1971 Problem 1 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | \(DEB\) is a chord of a circle such that \(DE=3\) and \(EB=5 .\) Let \(O\) be the center of the circle. Join \(OE\) and extend \(OE\) to cut the circle at \(C.\) Given \(EC=1,\) find the radius of the circle.
Image:CanadianMO_1971-1.jpg | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000048.s0",
"kind": "proof-sketch",
"text": "First, extend \\(CO\\) to meet the circle at \\(P.\\) Let the radius be \\(r.\\) Applying power of a point,\n\\((EP)(CE)=(BE)(ED)\\) and \\(2r-1=15.\\) Hence, \\(r=8.\\)",
"hints": [],
"images": []
}
] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000049 | Canadian MO 1971 Problem 10 | problem-solving-techniques | [
"problem-solving-techniques"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 10,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_10",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Suppose that \(n\) people each know exactly one piece of information, and all \(n\) pieces are different. Every time person \(A\) phones person \(B\), \(A\) tells \(B\) everything that \(A\) knows, while \(B\) tells \(A\) nothing. What is the minimum number of phone calls between pairs of people needed for everyone to ... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000049.s0",
"kind": "proof-sketch",
"text": "If all the n people know one distinct thing each, we can, without loss of generality, label them \\(P_1, P_2, \\dots, P_n\\). One way that is always possible is as follows:\n1. Starting at \\(P_1\\), each \\(P_k\\) calls \\(P_{k+1}\\), givin... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000050 | Canadian MO 1971 Problem 2 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | Let \(x\) and \(y\) be positive real numbers such that \(x+y=1\). Show that \(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\ge 9\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000050.s0",
"kind": "proof-sketch",
"text": "Solution 1\n\\left(1+\\frac{1}{x}\\right)\\left(1+\\frac{1}{y}\\right) \\ge 9\n1 + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{xy} \\ge 9 \n\\frac{xy+x+y+1}{xy}\\ge 9\n\\frac{xy+2}{xy}\\ge 9\n 1 + \\frac{2}{xy}\\ge 9\n\\frac{2}{xy} \\ge 8\n1\\ge... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000051 | Canadian MO 1971 Problem 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | \(ABCD\) is a quadrilateral with \(AD=BC\). If \(\angle ADC\) is greater than \(\angle BCD\), prove that \(AC>BD\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000051.s0",
"kind": "proof-sketch",
"text": "Consider triangles \\(ADC\\) and \\(BDC\\). These triangles share two of the same side lengths. Thus, by the Hinge Theorem, since we are given that \\(\\angle ADC \\ge \\angle BCD\\), we have \\(AC>BD\\).",
"hints": [],
"image... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000052 | Canadian MO 1971 Problem 4 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.5,
"note": null
} | null | Determine all real numbers \(a\) such that the two polynomials \(x^2+ax+1\) and \(x^2+x+a\) have at least one root in common. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000052.s0",
"kind": "proof-sketch",
"text": "Solution 1 \n\nLet this root be \\(r\\). Then we have\n\n$\\begin{matrix} r^2 + ar + 1 &=& r^2 + r + a\\\\\nar + 1 &=& r + a\\\\\n(a-1)r &=& (a-1)\\end{matrix} $\n\nNow, if $a = 1 $, then we're done, since this satisfies the problem's condi... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000053 | Canadian MO 1971 Problem 5 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Let \(p(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x+a_0\), where the coefficients $ a_i$ are integers. If \(p(0)\) and \(p(1)\) are both odd, show that \(p(x)\) has no integral roots. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000053.s0",
"kind": "proof-sketch",
"text": "Inputting \\(0\\) and \\(1\\) into \\(p(x)\\), we obtain\n\n\\(p(0)=a_0\\)\n\nand\n\n\\(p(1)=a_0+a_1+a_2+\\cdots+a_n\\)\n\nThe problem statement tells us that these are both odd. We will keep this in mind as we begin our proof by contradicti... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000054 | Canadian MO 1971 Problem 6 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 6,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_6",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Show that, for all integers \(n\), \(n^2+2n+12\) is not a multiple of \(121\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000054.s0",
"kind": "proof-sketch",
"text": "Solution 1 \n\nNotice \\(n^{2} + 2n + 12 = (n+1)^{2} + 11\\). For this expression to be equal to a multiple of 121, \\((n+1)^{2} + 11\\) would have to equal a number in the form \\(121x\\). Now we have the equation \\((n+1)^{2} + 11 = 121x\... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000055 | Canadian MO 1971 Problem 7 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 7,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_7",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Let \(n\) be a five digit number (whose first digit is non-zero) and let \(m\) be the four digit number formed from n by removing its middle digit. Determine all \(n\) such that \(n/m\) is an integer. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000055.s0",
"kind": "proof-sketch",
"text": "Let \\(n=10000a+1000b+100c+10d+e\\) and \\(m=1000a+100b+10d+e\\), where \\(a\\), \\(b\\), \\(c\\), \\(d\\), and \\(e\\) are base-10 digits and \\(a\\neq 0\\). If \\(n/m\\) is an integer, then \\(m|n\\), or\n\n1000a+100b+10d+e|10000a+1000b+10... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000056 | Canadian MO 1971 Problem 8 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 8,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_8",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | A regular pentagon is inscribed in a circle of radius \(r\). \(P\) is any point inside the pentagon. Perpendiculars are dropped from \(P\) to the sides, or the sides produced, of the pentagon.
a) Prove that the sum of the lengths of these perpendiculars is constant.
b) Express this constant in terms of the radius \(r... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000056.s0",
"kind": "proof-sketch",
"text": "Let the pentagon be \\(ABCDE\\), and the perpendiculars from \\(P\\) to sides \\(AB\\), \\(BC\\), \\(CD\\), \\(DE\\), and \\(EA\\) have lengths \\(p_1\\), \\(p_2\\), \\(p_3\\), \\(p_4\\), and \\(p_5\\), respectively.\n\nSince \\(P\\) is insi... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000057 | Canadian MO 1971 Problem 9 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1971,
"number": 9,
"url": "https://artofproblemsolving.com/wiki/index.php/1971_Canadian_MO_Problems/Problem_9",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Two flag poles of height \(h\) and \(k\) are situated \(2a\) units apart on a level surface. Find the set of all points on the surface which are so situated that the angles of elevation of the tops of the poles are equal. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000057.s0",
"kind": "proof-sketch",
"text": "Let the top of the flagpole of height h be denoted by H and its intersection with the surface be denoted by I. Similarly, let the top of the flagpole of length k be denoted by K and its intersection with the surface be denoted by J. Finally,... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000058 | Canadian MO 1973 Problem 1 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1973,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1973_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | \text{(i)} Solve the simultaneous inequalities, x<\frac{1}{4x} and x<0; i.e. find a single inequality equivalent to the two simultaneous inequalities.
\text{(ii)} What is the greatest integer that satisfies both inequalities 4x+13 < 0 and x^{2}+3x > 16.
\text{(iii)} Give a rational number between 11/24 and 6/13.
\te... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000058.s0",
"kind": "proof-sketch",
"text": "\\text{(i)} <center>x<\\frac{1}{4x} \\Rightarrow 4x^2 > 1 \\Rightarrow x^2> \\frac{1}{4}. \nSince from the second inequality x<0, our solution is x<-1/2.\n\n\\text{(ii)} From 4x+13 < 0, we get that 4x+13 < 0 \\Rightarrow x < -\\frac{13}{4} \... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000059 | Canadian MO 1973 Problem 2 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1973,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1973_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | Find all real numbers that satisfy the equation |x+3|-|x-1|=x+1. (Note: |a| = a if a\ge 0; |a|=-a if a<0.) | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000059.s0",
"kind": "proof-sketch",
"text": "We can break this up into cases based upon if x+3 and x-1 are positive or negative.\n\n\\text{Case 1}: x+3 > 0, x - 1 > 0\n\nIn this case x > 1. Then we have x+3-x+1 = x+1 \\implies x = 3. \n\n\\text{Case 2}: x+3 > 0, x - 1 \\leq 0\n\nIn thi... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000060 | Canadian MO 1973 Problem 3 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1973,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1973_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | Prove that if p and p+2 are prime integers greater than 3, then 6 is a factor of p+1. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000060.s0",
"kind": "proof-sketch",
"text": "Prime numbers greater than 3 are odd. Thus, if p and p+2 are prime integers greater than 3, then they are odd, and p+1 is a multiple of 2. Also, consider each group of three consecutive integers. One has remainder 1 after division upon 3, on... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000061 | Canadian MO 1973 Problem 4 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1973,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/1973_Canadian_MO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.5,
"note": null
} | null | size(200);
pair A=dir(120), B=dir(80);
for(int k=0;k<9;++k)
{
pair C=dir(120-(40)*(k+2));
D(A--B);
A=B;B=C;
}
for(int k=0;k<3;++k)
{
pair A1=dir(120-(40)*(3*k));
pair B1=dir(120-(40)*(3*k+2));
pair C1=dir(120-(40)*(3*k+3));
D(A1--B1);
D(A1--C1);
}
for(int k=0;k<9;++k)
{
pair A=dir(120+(40)*(k));
MP("P_{"+string(k)+"}",... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000062 | Canadian MO 1973 Problem 5 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1973,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1973_Canadian_MO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | For every positive integer n, let h(n) = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}.
For example, h(1) = 1, h(2) = 1+\frac{1}{2}, h(3) = 1+\frac{1}{2}+\frac{1}{3}.
Prove that n+h(1)+h(2)+h(3)+\cdots+h(n-1) = nh(n)\qquad for n=2,3,4,\ldots | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000063 | Canadian MO 1973 Problem 6 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1973,
"number": 6,
"url": "https://artofproblemsolving.com/wiki/index.php/1973_Canadian_MO_Problems/Problem_6",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | If A and B are fixed points on a given circle not collinear with center O of the circle, and if XY is a variable diameter, find the locus of P (the intersection of the line through A and X and the line through B and Y). | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000064 | Canadian MO 1973 Problem 7 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1973,
"number": 7,
"url": "https://artofproblemsolving.com/wiki/index.php/1973_Canadian_MO_Problems/Problem_7",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | \text{(i)} Observe that
\frac{1}{1}= \frac{1}{2}+ \frac{1}{2}; \quad \frac{1}{2}= \frac{1}{3}+ \frac{1}{6};\quad \frac{1}{3}= \frac{1}{4}+\frac{1}{12};\quad...
State a general law suggested by these examples, and prove it.
\text{(ii)} Prove that for any integer n greater than 1 there exist positive integers i and j su... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000064.s0",
"kind": "proof-sketch",
"text": "\\text{i}: We see that \\frac{1}{n(n+1)}+\\frac{1}{n+1} = \\frac{1}{n}. By simply evaluating the left hand side using common denominators, we see that both sides are equal. So therefore we have proved the law.\n\n\\text{ii}: Using partial fr... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000065 | Canadian MO 1975 Problem 1 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1975,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1975_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | Simplify
\left(\frac{1\cdot2\cdot4+2\cdot4\cdot8+\cdots+n\cdot2n\cdot4n}{1\cdot3\cdot9+2\cdot6\cdot18+\cdots+n\cdot3n\cdot9n}\right)^{1/3}. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000065.s0",
"kind": "proof-sketch",
"text": "\\left(\\frac{1\\cdot2\\cdot4+2\\cdot4\\cdot8+\\cdots+n\\cdot2n\\cdot4n}{1\\cdot3\\cdot9+2\\cdot6\\cdot18+\\cdots+n\\cdot3n\\cdot9n}\\right)^{1/3}\n\n=\\left(\\frac{2^3\\cdot1+2^3\\cdot2^3+2^3\\cdot3^3+2^3\\cdot4^3+\\cdots+2^3n^3}{3^3+3^3\\c... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000066 | Canadian MO 1975 Problem 2 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1975,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1975_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | A sequence of numbers a_1, a_2, a_3, \dots satisfies
(i) a_1 = \frac{1}{2}
(ii) a_1+a_2+\cdots+a_n=n^2a_n\quad(n\ge1).
Determine the value of a_n\quad(n\ge1). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000066.s0",
"kind": "proof-sketch",
"text": "I claim a_n=\\frac{1}{(n)(n+1)} with a proof by induction. First we can use partial fraction decomposition to rewrite \\frac{1}{(n)(n+1)} as \\frac{A}{n}+\\frac{B}{n+1}. We have \\frac{An+A+Bn}{(n)(n+1)}=\\frac{1}{(n)(n+1)}. We can set coeff... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000067 | Canadian MO 1975 Problem 3 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1975,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1975_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | For each real number r, [r] denotes the largest integer less than or equal to r, e.g., [6] = 6, [\pi] = 3, [-1.5] = -2. Indicate on the (x,y)-plane the set of all points (x,y) for which [x]^2+[y]^2 = 4. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000067.s0",
"kind": "proof-sketch",
"text": "Both [x] and [y] are integers. The only pairs of integers whose squares sum to 4 by testing are just (0,2) and (2,0) and (0,-2) and (-2,0). These are the floors, so the solutions are the set of four unit squares on the (x,y) plane whose bott... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000068 | Canadian MO 1975 Problem 4 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1975,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/1975_Canadian_MO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.5,
"note": null
} | null | For a positive number such as 3.27, 3 is referred to as the integral part of the number and .27 as the decimal part. Find a positive number such that its decimal part, its integral part, and the number itself form a geometric progression. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000068.s0",
"kind": "proof-sketch",
"text": "Let a be the integer part and b be the decimal part, thus, we have the G.P.\n\n(b,a,a+b)\n\nSo,\n\nb(a+b)=a^2\n\nab+b^2=a^2\n\nThere we have a quadratic equation. We must isolate a or b.\n\na^2-ab-b^2=0\n\nb=\\frac{-a\\pm a\\sqrt{5}}{2}\n\nI... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000069 | Canadian MO 1975 Problem 5 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1975,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1975_Canadian_MO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | A, B, C, D are four "consecutive" points on the circumference of a circle and P, Q, R, S are points on the circumference which are respectively the midpoints of the arcs AB, BC, CD, DA. Prove that PR is perpendicular to QS. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000069.s0",
"kind": "proof-sketch",
"text": "Define T=PR\\cap QS. Then,\n\\angle PTQ=\\frac{1}{2}\\left(\\widehat{PB}+\\widehat{BQ}+\\widehat{RD}+\\widehat{DS}\\right)=\\frac{1}{4}\\left(\\widehat{AB}+\\widehat{BC}+\\widehat{CD}+\\widehat{DA}\\right)=\\frac{1}{4}\\cdot360^\\circ=90^\\c... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000070 | Canadian MO 1975 Problem 6 | combinatorics | [
"combinatorics"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1975,
"number": 6,
"url": "https://artofproblemsolving.com/wiki/index.php/1975_Canadian_MO_Problems/Problem_6",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | (i)15 chairs are equally place around a circular table on which are name cards for 15 quests. The guests fail to notice these cards until after they have sat down, and it turns out that no one is sitting in the correct seat. Prove that the table can be rotated so that at least two of the guests are simultaneously corre... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000070.s0",
"kind": "proof-sketch",
"text": "(i)\nNotice that for each person there will only be one configuration(achieved through rotation) for which the name card matches the guest, therefore, for the 15 possible configurations, there will be 15 correct placed cards. But since one o... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000071 | Canadian MO 1975 Problem 7 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1975,
"number": 7,
"url": "https://artofproblemsolving.com/wiki/index.php/1975_Canadian_MO_Problems/Problem_7",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | A function f(x) is \textit{periodic} if there is a positive integer such that f(x+p) = f(x) for all x. For example, \sin x is periodic with period 2\pi. Is the function \sin(x^2) periodic? Prove your assertion. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000071.s0",
"kind": "proof-sketch",
"text": "To prove that \\sin(x^2) is periodic, we need to check if there exists a positive number p such that \\sin((x + p)^2) = \\sin(x^2) for all x. Using the trigonometric property \\sin(a) = \\sin(b) \\iff a - b = 2n\\pi (where n \\in \\mathbb{Z}... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000072 | Canadian MO 1975 Problem 8 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1975,
"number": 8,
"url": "https://artofproblemsolving.com/wiki/index.php/1975_Canadian_MO_Problems/Problem_8",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Let k be a positive integer. Find all polynomials
P(x) = a_0+a_1x+\cdots+a_nx^n
where the a_i are real, which satisfy the equation
P(P(x)) = \{P(x)\}^k. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000072.s0",
"kind": "proof-sketch",
"text": "Let f(n) be the degree of polynomial n. We begin by noting that f(P(x)) = k. This is because the degree of the LHS is f(P(x))^{f(P(x))} and the RHS is f(P(x))^k. Now we split P(x) into two cases.\n\nIn the first case, P(x) is a constant. Thi... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000073 | Canadian MO 1976 1 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1976,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1976_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | Given four weights in geometric progression and an equal arm balance, show how to find the heaviest weight using the balance only twice. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000073.s0",
"kind": "proof-sketch",
"text": "Note that for any 4 weights a,ar,ar^2,ar^3, the heaviest weight, ar^3, always dominates on the balance when these 4 weights are measured with two weights on both sides. I like to tackle these problems by putting some numbers in.Let the 4 wei... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000074 | Canadian MO 1977 Problem 1 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1977,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1977_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | If \(f(x)=x^2+x,\) prove that the equation \(4f(a)=f(b)\) has no solutions in positive integers \(a\) and \(b.\) | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000074.s0",
"kind": "proof-sketch",
"text": "Directly plugging \\(a\\) and \\(b\\) into the function, \\(4a^2+4a=b^2+b.\\) We now have a quadratic in \\(a.\\)\n\nApplying the quadratic formula, $a=\\frac{-1\\pm \\sqrt{b^2+b+1}}{2}. $\n\nIn order for both \\(a\\) and \\(b\\) to be integ... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000075 | Canadian MO 1977 Problem 2 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1977,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1977_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | Let \(O\) be the center of a circle and \(A\) be a fixed interior point of the circle different from \(O.\) Determine all points \(P\) on the circumference of the circle such that the angle \(OPA\) is a maximum.
Image:CanadianMO-1977-2.jpg | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000075.s0",
"kind": "proof-sketch",
"text": "If \\(AB\\) is the chord perpendicular to \\(OX\\) through point \\(P\\), then extend \\(AO\\) to meet the circle at point \\(C\\). It is now evident that \\(O\\) is the midpoint of \\(AC\\), \\(X\\) is the midpoint of \\(AB\\), and hence \\... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000076 | Canadian MO 1977 Problem 3 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1977,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1977_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | N is an integer whose representation in base b is 777. Find the smallest positive integer b for which N is the fourth power of an integer. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000076.s0",
"kind": "proof-sketch",
"text": "Rewriting N in base 10, N=7(b^2+b+1)=a^4 for some integer a. Because 7\\mid a^4 and 7 is prime, a^4 \\ge 7^4. Since we want to minimize b, we check to see if a=7 works.\n\nWhen a=7, b^2+b+1=7^3. Solving this quadratic, b = 18 .\n\nCategory:I... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000077 | Canadian MO 1977 Problem 4 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1977,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/1977_Canadian_MO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.5,
"note": null
} | null | Let
p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0
and
q(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots +b_1x+b_0
be two polynomials with integer coefficients. Suppose that all of the coefficients of the product \(p(x)\cdot q(x)\)
are even, but not all of them are divisible by 4. Show that one of \(p(x)\) and \(q(x)\) has all even co... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000077.s0",
"kind": "proof-sketch",
"text": "suppose p(x) and q(x) have all even coefficient then as p(x).q(x) involves all the ai(x).bj(x).. all the coefficient of p(x).q(x) must be divisible by 4. which is certainly not the case, hence p(x) or q(x) has atleast one odd coefficient. \n... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000078 | Canadian MO 1977 Problem 5 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1977,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1977_Canadian_MO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | A right circular cone of base radius \(1\) cm and slant height of \(3\) cm is given. \(P\) is a point on the circumference of
the base and the shortest path from \(P\) around the cone is drawn (see diagram). What is the minimum distance from
the vertex \(V\) to this path?
path p1=yscale(.25)*arc((0,0),1,0,180);
path... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000079 | Canadian MO 1977 Problem 6 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1977,
"number": 6,
"url": "https://artofproblemsolving.com/wiki/index.php/1977_Canadian_MO_Problems/Problem_6",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | Let \(0<u<1\) and define
u_1=1+u\quad ,\quad u_2=\frac{1}{u_1}+u\quad \ldots\quad u_{n+1}=\frac{1}{u_n}+u\quad ,\quad n\ge 1
Show that $u_n>1 $ for all values of \(n=1,2,3\ldots\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000079.s0",
"kind": "proof-sketch",
"text": "Prove by induction that \\(\\forall i>1\\) \\(1<u_i<\\frac{1}{1-u}\\)\n\n\\(u_1=1+k\\) that \\(1<u_1\\) and \\(1+u<\\frac{1}{1-u} \\to 1-u^2<1 \\to -u^2<0\\).\n\nBy induction if \\(u_i<\\frac{1}{1-u}\\) then \\(u_{i+1}=\\frac{1}{u_i}+u>(1-u)... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000080 | Canadian MO 1977 Problem 7 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1977,
"number": 7,
"url": "https://artofproblemsolving.com/wiki/index.php/1977_Canadian_MO_Problems/Problem_7",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 8.1,
"note": null
} | null | A rectangular city is exactly \(m\) blocks long and \(n\) blocks wide (see diagram).
A woman lives on the southwest corner of the city and works in the northeast corner. She walks
to work each day but, on any given trip, she makes sure that her path does not include any intersection
twice. Show that the number \(f(m,n... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000080.s0",
"kind": "proof-sketch",
"text": "The woman's path divides the grid into two regions, one containing the northwest corner, and one containing the southeast corner. Shade all the cells in the region containing the northwest corner. Thus every path corresponds to a shading of ... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000081 | Canadian MO 1980 Problem 1 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1980,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/1980_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | If \(a679b\) is the decimal expansion of a number in base \(10\), such that it is divisible by \(72\), determine \(a,b\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000081.s0",
"kind": "proof-sketch",
"text": "Write \\(a679b\\) as \\(a679b = 10000a + 6790 + b = 72n\\), for some integer \\(n\\). Since \\(72 = 8 \\times 9\\), we look at this equation \\(\\mod 8\\), then \\(\\mod 9\\). \n\nLooking at the equation \\(\\mod 8\\):\n$\\begin{matrix} 1000... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000082 | Canadian MO 1980 Problem 2 | uncategorized | [] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1980,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/1980_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | The numbers from \(1\) to \(50\) are printed on cards. The cards are shuffled and then laid out face up in \(5\) rows of \(10\) cards each. The cards in each row are rearranged to make them increase from left to right. The cards in each column are then rearranged to make them increase from top to bottom. In the final a... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000083 | Canadian MO 1980 Problem 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1980,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/1980_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | A triangle is given with one constant angle and a constant inradius. Explicitly define all triangles for which the minimum perimeter is achieved. | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000084 | Canadian MO 1980 Problem 4 | probability | [
"probability"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1980,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/1980_Canadian_MO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.5,
"note": null
} | null | A gambling student tosses a fair coin. She gains \(1\) point for each head that turns up, and gains \(2\) points for each tail that turns up. Prove that the probability of the student scoring exactly \(n\) points is \(\boxed{\frac{1}{3}\cdot\left(2+\left(-\frac{1}{2}\right)^{n}\right)}\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000084.s0",
"kind": "proof-sketch",
"text": "Yes\nProof:\nEach row has 10 numbers.\nFor the 1rst row let's call them a_1,a_2\\dots,a_{10}\nFor the 2nd row let's call them b_1,b_2\\dots,b_{10}\nFor the 3rd row let's call them c_1,c_2\\dots,c_{10}\nFor the 4th row let's call them d_1,d_2... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000085 | Canadian MO 1980 Problem 5 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 1980,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/1980_Canadian_MO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | A parallelepiped has the property that all cross sections, which are parallel to any fixed face \(F\), have the same perimeter as \(F\). Determine whether or not any other poyhedron has this property | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000086 | Canadian MO 2005 Problem 1 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2005,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/2005_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | Consider an equilateral triangle of side length \(n\), which is divided into unit triangles, as shown. Let \(f(n)\) be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in our path share a common edge and the path never travels up (from a lower r... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000086.s0",
"kind": "proof-sketch",
"text": "Define a last triangle of a row as the triangle in the row that the path visits last. From the last triangle in row \\(k\\), the path must move down and then directly across to the last triangle in row \\(k+1\\). Therefore, there is exactl... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000087 | Canadian MO 2005 Problem 2 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2005,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/2005_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | Let \((a,b,c)\) be a Pythagorean triple, i.e., a triplet of positive integers with \({a}^2+{b}^2={c}^2\).
Prove that \((c/a + c/b)^2 > 8\).
Prove that there does not exist any integer \(n\) for which we can find a Pythagorean triple \((a,b,c)\) satisfying \((c/a + c/b)^2 = n\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000087.s0",
"kind": "proof-sketch",
"text": "We have \n\n\\(\\left(\\frac ca + \\frac cb\\right)^2 = \\frac{c^2}{a^2} + 2\\frac{c^2}{ab} + \\frac{c^2}{b^2} = \\frac{a^2 + b^2}{a^2} + 2\\frac{a^2 + b^2}{ab} + \\frac{a^2+b^2}{b^2} = 2 + \\left(\\frac{a^2}{b^2} + \\frac{b^2}{a^2}\\right) ... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000088 | Canadian MO 2005 Problem 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2005,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/2005_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | Let \(S\) be a set of \(n\ge 3\) points in the interior of a circle.
Show that there are three distinct points \(a,b,c\in S\) and three distinct points \(A,B,C\) on the circle such that \(a\) is (strictly) closer to \(A\) than any other point in \(S\), \(b\) is closer to \(B\) than any other point in \(S\) and \(c\)... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000088.s0",
"kind": "proof-sketch",
"text": "(a) Let \\(H\\) be the convex hull of \\(S\\), and choose any three vertices \\(V_1, V_2, V_3\\) of \\(H\\).\n\nNow, draw a line \\(l_1\\) through \\(V_1\\) such that all points of \\(S\\) lie on one side of \\(l_1\\). (This is possible beca... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000089 | Canadian MO 2005 Problem 4 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2005,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/2005_Canadian_MO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.5,
"note": null
} | null | Let \(ABC\) be a triangle with circumradius \(R\), perimeter \(P\) and area \(K\). Determine the maximum value of \(KP/R^3\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000089.s0",
"kind": "proof-sketch",
"text": "Use the formula \\(K=\\dfrac{abc}{4R}\\) to get \\(KP/R^3=\\dfrac{abc(a+b+c)}{4R^4}\\). Then use the extended sine law to get something in terms of sines, and use AM-GM and Jensen's to finish. (Jensen's is used for \\(\\sin A+\\sin B+\\sin... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000090 | Canadian MO 2005 Problem 5 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2005,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/2005_Canadian_MO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | Let's say that an ordered triple of positive integers \((a,b,c)\) is \(n\)-powerful if \(a \le b \le c\), \(\gcd(a,b,c) = 1\), and \(a^n + b^n + c^n\) is divisible by \(a+b+c\). For example, \((1,2,2)\) is 5-powerful.
Determine all ordered triples (if any) which are \(n\)-powerful for all \(n \ge 1\).
Determine all... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000090.s0",
"kind": "proof-sketch",
"text": "Partial Solution:\n\nConsider \\(P(x)=(x-a)(x-b)(x-c)\\).\nLet \\(S_k= a^k+b^k+c^k\\).\n\nAccording to Newton’s Sum:\n\n\\(S_{k+3}-(a+b+c)S_{k+2}+(ab+bc+ca)S_{k+1}-(abc)S_k=0\\). \nSo clearly if \\(a+b+c \\vert S_k, S_{k+1},\\) then \\(a+b+c... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000091 | Canadian MO 2006 Problem 1 | combinatorics | [
"combinatorics"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2006,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/2006_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | Let \(f(n,k)\) be the number of ways distributing \(k\) candies to \(n\) children so that each child receives at most two candies. For example, \(f(3,7)=0\), \(f(3,6)=1\), and \(f(3,4)=6\). Evaluate \(f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000091.s0",
"kind": "proof-sketch",
"text": "\\(x_1 + \\cdots + x_{2006} = k\\)\n\n\\(x_i \\in \\{0,1,2\\}\\)\n\n\\(\\therefore\\) generating function for one student is \\(f(x) = (x^0 + x^1 + x^2)\\)\n\ngenerating function for \\(n\\) many students is \\(f(x)^n = (x^0 + x^1 + x^2)^n\\... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000092 | Canadian MO 2006 Problem 2 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2006,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/2006_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | Let \(ABC\) be an acute angled triangle. Inscribe a rectangle \(DEFG\) in this triangle so that \(D\) is on \(AB\), \(E\) on \(AC\), and \(F\) and \(G\) on \(BC\). Describe the locus of the intersections of the diagonals of all possible rectangles \(DEFG\). | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000092.s0",
"kind": "proof-sketch",
"text": "The locus is the line segment which joins the midpoint of side \\(BC\\) to the midpoint of the altitude to side \\(BC\\) of the triangle.\n\nLet \\(r = \\frac{AD}{AB}\\) and let \\(H\\) be the foot of the altitude from \\(A\\) to \\(BC\\). ... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000093 | Canadian MO 2006 Problem 3 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2006,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/2006_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | In a rectangular array of nonnegative real numbers with m rows and n columns, each row and each
column contains at least one positive element. Moreover, if a row and a column intersect in a positive
element, then the sums of their elements are the same. Prove that m = n. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000093.s0",
"kind": "proof-sketch",
"text": "Assume that m does not equal n. WLOG, m < n. Now we proceed by strong induction. We will assume that the statement is true if a < m and b < n. The base case, where either m or n is 1, is clearly true. Consider the bipartite graph between the... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000094 | Canadian MO 2006 Problem 4 | combinatorics | [
"combinatorics"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2006,
"number": 4,
"url": "https://artofproblemsolving.com/wiki/index.php/2006_Canadian_MO_Problems/Problem_4",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.5,
"note": null
} | null | Consider a round robin tournament with \(2n+1\) teams, where two teams play exactly one match and there are no ties. We say that the teams \(X\), \(Y\), and \(Z\) form a cycle triplet if \(X\) beats \(Y\), \(Y\) beats \(Z\), and \(Z\) beats \(X\).
(a) Find the minimum number of cycle triplets possible.
(b) Find the m... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000094.s0",
"kind": "proof-sketch",
"text": "700px",
"hints": [],
"images": []
}
] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000095 | Canadian MO 2006 Problem 5 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2006,
"number": 5,
"url": "https://artofproblemsolving.com/wiki/index.php/2006_Canadian_MO_Problems/Problem_5",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.8,
"note": null
} | null | The vertices of a right triangle \(ABC\) inscribed in a circle divide the circumference into three arcs.
The right angle is at \(A\), so that the opposite arc \(BC\) is a semicircle while arc \(AB\) and arc \(AC\) are
supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is the
mid... | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000095.s0",
"kind": "proof-sketch",
"text": "Let the intersection of the tangents at \\(D\\) and \\(E\\), \\(E\\) and \\(F\\), \\(F\\) and \\(D\\) be labeled \\(Z, X,Y\\), respectively.\n\nIt is a well-known fact that in a right triangle \\(PQR\\) with \\(M\\) the midpoint of hypotenus... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000096 | Canadian MO 2013 1 | algebra | [
"algebra"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2013,
"number": 1,
"url": "https://artofproblemsolving.com/wiki/index.php/2013_Canadian_MO_Problems/Problem_1",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.5,
"note": null
} | null | Determine all polynomials \(P(x)\) with real coefficients such that
(x+1)P(x-1)-(x-1)P(x)
is a constant polynomial. | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000096.s0",
"kind": "proof-sketch",
"text": "Let \\(F(x)=(x+1)P(x-1)-(x-1)P(x)\\)\n\n\\(P(x)=\\sum_{i=0}^{n}c_ix^i\\)\n\n\\(F(x)=(x+1)\\sum_{i=0}^{n}(x-1)^ic_i-(x-1)\\sum_{i=0}^{n}c_ix^i\\)\n\n\\(F(x)=\\sum_{i=0}^{n}x(x-1)^ic_i+\\sum_{i=0}^{n}(x-1)^ic_i-\\sum_{i=0}^{n}c_ix^{i+1}+\\sum_... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000097 | Canadian MO 2013 2 | number-theory | [
"number-theory"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2013,
"number": 2,
"url": "https://artofproblemsolving.com/wiki/index.php/2013_Canadian_MO_Problems/Problem_2",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 6.8,
"note": null
} | null | The sequence \(a_1, a_2, \dots, a_n\) consists of the numbers \(1, 2, \dots, n\) in some order. For which positive integers \(n\) is it possible that the \(n+1\) numbers \(0, a_1, a_1+a_2, a_1+a_2+a_3,\dots, a_1 + a_2 +\cdots + a_n\) all have different remainders when divided by \(n + 1\)? | [] | {
"letter": null,
"value": null
} | [
{
"id": "sol.aops-000097.s0",
"kind": "proof-sketch",
"text": "First, we show that this is impossible for even numbers.\nLet n be an even number. Consider the first sum: 0. This obviously leaves a remainder of 0 when divided by \\(n+1\\). Then, consider the final sum. This sum will be equal to the sum o... | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
prob.aops-000098 | Canadian MO 2013 3 | geometry | [
"geometry"
] | [] | {
"dataset": "aops",
"collection": "Canadian MO",
"year": 2013,
"number": 3,
"url": "https://artofproblemsolving.com/wiki/index.php/2013_Canadian_MO_Problems/Problem_3",
"tags": []
} | {
"course": null,
"source": null,
"mohs": 7.1,
"note": null
} | null | Let \(G\) be the centroid of a right-angled triangle \(ABC\) with \(\angle BCA = 90^\circ\). Let \(P\) be the point on ray \(AG\) such that \(\angle CPA = \angle CAB\), and let \(Q\) be the point on ray \(BG\) such that \(\angle CQB = \angle ABC\). Prove that the circumcircles of triangles \(AQG\) and \(BPG\) meet at a... | [] | {
"letter": null,
"value": null
} | [] | {
"statement": [],
"solutions": []
} | {
"part": "cmo",
"holder": "Canadian Mathematical Olympiad organizers",
"distribution": null,
"use": "notice-only",
"notice": "Copyright the Canadian Mathematical Olympiad organizers. No open license was found."
} |
End of preview.
Arquimedes
1,038,963 competition and benchmark math problems in one schema.
Every record keeps its text. rights names the holder and the license
the distributor attached. distribution is null when no open license
was found. This stamp does not create a new grant.
Code for the tools that built the corpus is MIT. The Mohs value is a calibration produced for this project, not a measurement from the source. Project page: https://github.com/richer-richard/arquimedes
Rows with rights.use of notice-only are the contest problems this
project collected itself (core and aops). Rows with noncommercial
are CEMC, CC BY-NC 4.0. Every other row is passed through from a dataset
already published under Apache 2.0 or MIT, and the underlying holder is
named on the row.
Drop notice-only and noncommercial to leave those 11,748 rows out.
Nothing is dropped in this copy.
| part | problems are described as | distribution | use |
|---|---|---|---|
cemc |
Copyright the University of Waterloo. CEMC materials are CC BY-NC 4.0. Do not charge a fee for access to past contests. | CC-BY-NC-4.0 | noncommercial |
maa |
Copyright the Mathematical Association of America. After the contest period, booklet notices allow educational copies that are not for profit or commercial advantage and that keep this notice. Wording differs by year; some AMC 8 booklets allow paper copies only. | none | notice-only |
hmmt |
Copyright the Harvard-MIT Mathematics Tournament. Past problems are published by HMMT. No open license was found. | none | notice-only |
imo |
Copyright the IMO organizers. No open license was found. | none | notice-only |
imo-shortlist |
Copyright the IMO shortlist organizers. No open license was found. | none | notice-only |
apmo |
Copyright the Asian Pacific Mathematics Olympiad organizers. No open license was found. | none | notice-only |
cmo |
Copyright the Canadian Mathematical Olympiad organizers. No open license was found. | none | notice-only |
math |
MIT, copyright 2021 Dan Hendrycks, via the MATH-lighteval mirror. The MIT grant is the dataset author's. The problems are contest problems. | MIT | pass-through |
olympiadbench |
Hugging Face card Apache 2.0. The OpenBMB code repository is MIT and licenses the code, not the problem text. Problems were OCR'd from official olympiad PDFs. | Apache-2.0 | pass-through |
omnimath |
Hugging Face card tagged Apache 2.0. The GitHub repository has no LICENSE file. Problems were taken from official contest sites and AoPS. | Apache-2.0 | pass-through |
deepmath |
MIT, DeepMath-103K. Questions were drawn from Apache 2.0 collections that include third-party text. The three traces on each row are DeepSeek-R1 generations. DeepSeek licenses the R1 code and weights under MIT. | MIT | pass-through |
numina/amc_aime |
Released by Project Numina in NuminaMath 1.5 under Apache 2.0. The problems are copyright the Mathematical Association of America. MAA booklet notices allow educational, non-commercial copies that keep the notice. | Apache-2.0 | pass-through |
numina/aops_forum |
Released by Project Numina under Apache 2.0, from the AoPS forum. AoPS terms allow use of site materials for information purposes only and do not grant the public a license to the posts. | Apache-2.0 | pass-through |
numina/olympiads |
Released by Project Numina under Apache 2.0. The problems are olympiad problems. The Apache tag is Numina's, not a grant from the contest organizers. | Apache-2.0 | pass-through |
numina/olympiads_ref |
Released by Project Numina under Apache 2.0. The problems are olympiad problems. The Apache tag is Numina's, not a grant from the contest organizers. | Apache-2.0 | pass-through |
numina/cn_k12 |
Released by Project Numina under Apache 2.0, from Chinese K-12 exams. The Apache tag is Numina's, not a grant from the exam authors. | Apache-2.0 | pass-through |
numina/cn_contest |
Released by Project Numina under Apache 2.0, from Chinese contests. The Apache tag is Numina's, not a grant from the contest organizers. | Apache-2.0 | pass-through |
numina/orca_math |
Released by Project Numina under Apache 2.0 from the orca_math source. A separate Orca-Math license file was not re-checked for this release. | Apache-2.0 | pass-through |
numina/synthetic_math |
Synthetic problems released by Project Numina under Apache 2.0. | Apache-2.0 | pass-through |
numina/metamath |
Included in NuminaMath 1.5 under Apache 2.0. MetaMathQA's own Hugging Face card is MIT. | Apache-2.0 | pass-through |
numina/inequalities |
Curated by Project Numina and released under Apache 2.0. | Apache-2.0 | pass-through |
numina/number_theory |
Curated by Project Numina and released under Apache 2.0. | Apache-2.0 | pass-through |
bigmath/big_math |
Source label big_math on Big-Math-RL-Verified, whose dataset card is Apache 2.0. This is the label stored on the row. | Apache-2.0 | pass-through |
bigmath/olympiads |
Included in Big-Math under Apache 2.0. The problems are olympiad problems. The Apache tag is SynthLabsAI's, not a grant from the contest organizers. | Apache-2.0 | pass-through |
bigmath/harp |
Included in Big-Math under Apache 2.0 from the harp source. A separate HARP license was not found for this release. | Apache-2.0 | pass-through |
bigmath/cn_k12 |
Included in Big-Math under Apache 2.0 from Chinese K-12 exams. The Apache tag is SynthLabsAI's, not a grant from the exam authors. | Apache-2.0 | pass-through |
bigmath/aops_forum |
Included in Big-Math under Apache 2.0, from the AoPS forum. AoPS terms allow use of site materials for information purposes only. | Apache-2.0 | pass-through |
bigmath/orca_math |
Included in Big-Math under Apache 2.0 from the orca_math source. A separate Orca-Math license file was not re-checked for this release. | Apache-2.0 | pass-through |
bigmath/gsm8k |
Included in Big-Math under Apache 2.0. GSM8K's own Hugging Face card is MIT. | Apache-2.0 | pass-through |
bigmath/openmath |
Included in Big-Math under Apache 2.0 from the openmath source. A separate OpenMath license was not found for this release. | Apache-2.0 | pass-through |
bigmath/omnimath |
Included in Big-Math under Apache 2.0 from Omni-MATH. The Omni-MATH Hugging Face card is tagged Apache 2.0 and its GitHub repository has no LICENSE file. | Apache-2.0 | pass-through |
bigmath/amc_aime |
Included in Big-Math under Apache 2.0. The problems are copyright the Mathematical Association of America. MAA booklet notices allow educational, non-commercial copies that keep the notice. | Apache-2.0 | pass-through |
Full wording also lives in rights.json.
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