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The dataset generation failed because of a cast error
Error code: DatasetGenerationCastError
Exception: DatasetGenerationCastError
Message: An error occurred while generating the dataset
All the data files must have the same columns, but at some point there are 22 new columns ({'Solution_2_.28Simple.29', 'Solution_with_Explanations', 'Solution_where', 'Solution_4_Without_the_nasty_computations', 'Solution_.28No_Bash.29', 'Solution_3_.28Angle-Chasing.29', 'Solution_Number_Sense', 'Solution_3_.28INCORRECT.29', 'Solution_2_.28Taken_from_Twitch_Solves_ISL.29', 'Solution_3:_Graph_Theoretic', 'Problem_4', 'Solution_3_.28Less_technical_bary.29', 'Solution_with_Thought_Process', 'Solution_2_.28and_motivation.29', 'Solution_3_.28Not_a_formal_proof_but_understandable.29', 'Solution_5.28stupid_but_simple.29', 'Solution_4_.28Less_bashy_bary.29', 'Solution_3_.28without_induction.29', 'Solution_3_.28clear_Solution_2.29', 'Solution_6_.28Motivation_for_Solution.29', 'Solution_1:_Pigeonhole', 'Solution_2.2C_Barycentric_.28Modified_by_Evan_Chen.29'}) and 26 missing columns ({'Solution_2_2', 'Solution4', 'Solution_3_.28Visual.29', 'Solution_7_from_Brilliant_Wiki_.28Muirheads.29_.3D', 'Solution_2_.28duality_principle.29', 'Solution_7', 'Problem_2', 'Solution_2.28SFFT.29', 'Solution_2_.28Avoids_the_error_of_the_first.29', 'Solution_3_.28No_Miquel.27s_point.29', 'Problems', 'Solution_2_.28Sort_of_Root_Jumping.29', 'Solution_1_2', 'Solution_3_.28Complex_Bash.29', 'Solution1', 'Solution_2_.28Strong_Induction_and_Alternative_Case_Division.29', 'Solution_3:_The_only_one_that_actually_works', 'Solution_4:_This_works_as_well', 'Solution_2_.28elegant.29', 'Solution_1.1', 'Solution_3b', 'Solution_3_.28Trigonometry.29', 'Solution_8_.28fast_Titu.27s_Lemma_no_substitutions.29', 'Solution_1_.28Induction_and_Case_Division.29', 'Solution_1_.28Efficient.29', 'Solution_2_.28Three_perpendicular_bisectors.29'}).
This happened while the json dataset builder was generating data using
hf://datasets/autores/imo_hq/usajmo_hq.json (at revision 2e0d2a2e692e42e4ab668040208ca4d9e95fe337)
Please either edit the data files to have matching columns, or separate them into different configurations (see docs at https://hf.co/docs/hub/datasets-manual-configuration#multiple-configurations)
Traceback: Traceback (most recent call last):
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 2011, in _prepare_split_single
writer.write_table(table)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/arrow_writer.py", line 585, in write_table
pa_table = table_cast(pa_table, self._schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2302, in table_cast
return cast_table_to_schema(table, schema)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/table.py", line 2256, in cast_table_to_schema
raise CastError(
datasets.table.CastError: Couldn't cast
Solution_1: string
Solution_2_.28Simple.29: string
Solution_with_Explanations: string
Solution_where: string
Solution_4_Without_the_nasty_computations: string
Solution_2: string
Solution_.28No_Bash.29: string
Solution_Number_Sense: string
Solution_3_.28Angle-Chasing.29: string
Solution_3_.28INCORRECT.29: string
Solution_4: string
Solutions: string
Solution_2_.28Taken_from_Twitch_Solves_ISL.29: string
Solution_3:_Graph_Theoretic: string
Problem_4: string
Solution_3_.28Less_technical_bary.29: string
Solution_6: string
Solution_with_Thought_Process: string
Solution_2_.28and_motivation.29: string
Problem_5: string
Solution_3_.28Not_a_formal_proof_but_understandable.29: string
Solution_5.28stupid_but_simple.29: string
Solution: string
Solution_4_.28Less_bashy_bary.29: string
Problem: string
Solution_5: string
Solution_3_.28without_induction.29: string
Solution_3_.28clear_Solution_2.29: string
Solution_6_.28Motivation_for_Solution.29: string
Solution_1:_Pigeonhole: string
Problem_6: string
Solution_2.2C_Barycentric_.28Modified_by_Evan_Chen.29: string
Solution_3: string
to
{'Solution_1': Value(dtype='string', id=None), 'Solution_2_2': Value(dtype='string', id=None), 'Solution4': Value(dtype='string', id=None), 'Solution_3_.28Visual.29': Value(dtype='string', id=None), 'Solution_2': Value(dtype='string', id=None), 'Solution_7_from_Brilliant_Wiki_.28Muirheads.29_.3D': Value(dtype='string', id=None), 'Solution_2_.28duality_principle.29': Value(dtype='string', id=None), 'Solution_4': Value(dtype='string', id=None), 'Solution_7': Value(dtype='string', id=None), 'Problem_2': Value(dtype='string', id=None), 'Solutions': Value(dtype='string', id=None), 'Solution_2.28SFFT.29': Value(dtype='string', id=None), 'Solution_2_.28Avoids_the_error_of_the_first.29': Value(dtype='string', id=None), 'Solution_6': Value(dtype='string', id=None), 'Solution_3_.28No_Miquel.27s_point.29': Value(dtype='string', id=None), 'Problems': Value(dtype='string', id=None), 'Solution_2_.28Sort_of_Root_Jumping.29': Value(dtype='string', id=None), 'Solution_1_2': Value(dtype='string', id=None), 'Solution_3_.28Complex_Bash.29': Value(dtype='string', id=None), 'Problem_5': Value(dtype='string', id=None), 'Solution1': Value(dtype='string', id=None), 'Solution_3': Value(dtype='string', id=None), 'Solution_2_.28Strong_Induction_and_Alternative_Case_Division.29': Value(dtype='string', id=None), 'Solution': Value(dtype='string', id=None), 'Solution_3:_The_only_one_that_actually_works': Value(dtype='string', id=None), 'Solution_4:_This_works_as_well': Value(dtype='string', id=None), 'Problem': Value(dtype='string', id=None), 'Solution_2_.28elegant.29': Value(dtype='string', id=None), 'Solution_1.1': Value(dtype='string', id=None), 'Solution_5': Value(dtype='string', id=None), 'Solution_3b': Value(dtype='string', id=None), 'Solution_3_.28Trigonometry.29': Value(dtype='string', id=None), 'Solution_8_.28fast_Titu.27s_Lemma_no_substitutions.29': Value(dtype='string', id=None), 'Solution_1_.28Induction_and_Case_Division.29': Value(dtype='string', id=None), 'Solution_1_.28Efficient.29': Value(dtype='string', id=None), 'Problem_6': Value(dtype='string', id=None), 'Solution_2_.28Three_perpendicular_bisectors.29': Value(dtype='string', id=None)}
because column names don't match
During handling of the above exception, another exception occurred:
Traceback (most recent call last):
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 1321, in compute_config_parquet_and_info_response
parquet_operations = convert_to_parquet(builder)
File "/src/services/worker/src/worker/job_runners/config/parquet_and_info.py", line 935, in convert_to_parquet
builder.download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1027, in download_and_prepare
self._download_and_prepare(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1122, in _download_and_prepare
self._prepare_split(split_generator, **prepare_split_kwargs)
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 1882, in _prepare_split
for job_id, done, content in self._prepare_split_single(
File "/src/services/worker/.venv/lib/python3.9/site-packages/datasets/builder.py", line 2013, in _prepare_split_single
raise DatasetGenerationCastError.from_cast_error(
datasets.exceptions.DatasetGenerationCastError: An error occurred while generating the dataset
All the data files must have the same columns, but at some point there are 22 new columns ({'Solution_2_.28Simple.29', 'Solution_with_Explanations', 'Solution_where', 'Solution_4_Without_the_nasty_computations', 'Solution_.28No_Bash.29', 'Solution_3_.28Angle-Chasing.29', 'Solution_Number_Sense', 'Solution_3_.28INCORRECT.29', 'Solution_2_.28Taken_from_Twitch_Solves_ISL.29', 'Solution_3:_Graph_Theoretic', 'Problem_4', 'Solution_3_.28Less_technical_bary.29', 'Solution_with_Thought_Process', 'Solution_2_.28and_motivation.29', 'Solution_3_.28Not_a_formal_proof_but_understandable.29', 'Solution_5.28stupid_but_simple.29', 'Solution_4_.28Less_bashy_bary.29', 'Solution_3_.28without_induction.29', 'Solution_3_.28clear_Solution_2.29', 'Solution_6_.28Motivation_for_Solution.29', 'Solution_1:_Pigeonhole', 'Solution_2.2C_Barycentric_.28Modified_by_Evan_Chen.29'}) and 26 missing columns ({'Solution_2_2', 'Solution4', 'Solution_3_.28Visual.29', 'Solution_7_from_Brilliant_Wiki_.28Muirheads.29_.3D', 'Solution_2_.28duality_principle.29', 'Solution_7', 'Problem_2', 'Solution_2.28SFFT.29', 'Solution_2_.28Avoids_the_error_of_the_first.29', 'Solution_3_.28No_Miquel.27s_point.29', 'Problems', 'Solution_2_.28Sort_of_Root_Jumping.29', 'Solution_1_2', 'Solution_3_.28Complex_Bash.29', 'Solution1', 'Solution_2_.28Strong_Induction_and_Alternative_Case_Division.29', 'Solution_3:_The_only_one_that_actually_works', 'Solution_4:_This_works_as_well', 'Solution_2_.28elegant.29', 'Solution_1.1', 'Solution_3b', 'Solution_3_.28Trigonometry.29', 'Solution_8_.28fast_Titu.27s_Lemma_no_substitutions.29', 'Solution_1_.28Induction_and_Case_Division.29', 'Solution_1_.28Efficient.29', 'Solution_2_.28Three_perpendicular_bisectors.29'}).
This happened while the json dataset builder was generating data using
hf://datasets/autores/imo_hq/usajmo_hq.json (at revision 2e0d2a2e692e42e4ab668040208ca4d9e95fe337)
Please either edit the data files to have matching columns, or separate them into different configurations (see docs at https://hf.co/docs/hub/datasets-manual-configuration#multiple-configurations)Need help to make the dataset viewer work? Make sure to review how to configure the dataset viewer, and open a discussion for direct support.
Solution_1_2 null | Solution_6 string | Solution_2_2 null | Solution_1 string | Problem_5 null | Solution_1_.28Efficient.29 null | Solution_2_.28Avoids_the_error_of_the_first.29 null | Solution_2_.28Strong_Induction_and_Alternative_Case_Division.29 null | Problem_6 null | Solution_1.1 null | Solution1 null | Solution_2_.28Sort_of_Root_Jumping.29 null | Solution_2_.28elegant.29 null | Solution_3_.28No_Miquel.27s_point.29 null | Solution4 null | Solution_2_.28Three_perpendicular_bisectors.29 null | Solution_7_from_Brilliant_Wiki_.28Muirheads.29_.3D null | Solution_3:_The_only_one_that_actually_works null | Solution_8_.28fast_Titu.27s_Lemma_no_substitutions.29 null | Solution_3_.28Complex_Bash.29 null | Solution_4:_This_works_as_well null | Solution_3_.28Visual.29 null | Solution_3 string | Problems null | Solutions null | Solution_1_.28Induction_and_Case_Division.29 null | Solution_5 string | Problem_2 null | Solution_2_.28duality_principle.29 null | Solution_4 string | Solution_7 string | Problem string | Solution string | Solution_2 string | Solution_2.28SFFT.29 null | Solution_3b null | Solution_3_.28Trigonometry.29 null |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
null | To understand why its irreducible, let's take a closer look at the fraction itself. If we were to separate both fractions, we end up with $\frac{21n}{14n}$ + $\frac{4}{3}$.
Simplifying the fraction, we end up with. $\frac{3}{2}$. Now combining these fractions with addition as shown before in the problem, we end up w... | null | Denoting the greatest common divisor of $a, b$ as $(a,b)$, we use the Euclidean algorithm:
Their greatest common divisor is 1, so $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D. | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Proof by contradiction:
Assume that $\dfrac{14n+3}{21n+4}$ is a reducible fraction.
If a certain fraction $\dfrac{a}{b}$ is reducible, then the fraction $\dfrac{2a}{3b}$ is reducible, too.
In this case, $\dfrac{2a}{3b} = \dfrac{42n+8}{42n+9}$.
This fraction consists of two consecutives numbers, which never share any fa... | null | null | null | By Bezout's Lemma, $3 \cdot (14n+3) - 2 \cdot (21n + 4) = 1$, so the GCD of the numerator and denominator is $1$ and the fraction is irreducible. | null | null | We notice that:
So it follows that $7n+1$ and $14n+3$ must be coprime for every natural number $n$ for the fraction to be irreducible. Now the problem simplifies to proving $\frac{7n+1}{14n+3}$ irreducible. We re-write this fraction as:
Since the denominator $2(7n+1) + 1$ differs from a multiple of the numerator $7n+1$... | Let $\gcd (21n+4,14n+3)=a.$ So for some co-prime positive integers $x,y$ we have Multiplying $(1)$ by $2$ and $(2)$ by $3$ and then subtracting $(1)$ from $(2)$ we get We must have $a=1$, since $a$ is an positive integer. Thus, $\gcd(21n+4,14n+3)=1$ which means the fraction is irreducible, as needed. | Prove that the fraction $\frac{21n+4}{14n+3}$ is irreducible for every natural number $n$. | null | Proof by contradiction:
Assume that $\dfrac{14n+3}{21n+4}$ is a reducible fraction where $p$ is the greatest common factor of $14n+3$ and $21n + 4$. Thus,
Subtracting the second equation from the first equation we get $1\equiv 0\pmod{p}$ which is clearly absurd.
Hence $\frac{21n+4}{14n+3}$ is irreducible. Q.E.D. | null | null | null |
null | null | null | Let $N = 100a + 10b+c$ for some digits $a,b,$ and $c$. Then for some $m$. We also have $m=a^2+b^2+c^2$. Substituting this into the first equation and simplification, we get For an integer divisible by $11$, the the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by $11$. ... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Note that there are only 900 three-digit numbers. Of these, only 81 are divisible by 11. This brings to mind the great strategy of listing and bashing. Being organized while listing will get you the answer. There is plenty of time on the IMO, and listing is highly unlikely to miss cases. We end up getting $\boxed{N=550... | null | null | null | null | null | null | https://artofproblemsolving.com/community/c6h54826p22231968
~franzliszt | null | Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\dfrac{N}{11}$ is equal to the sum of the squares of the digits of $N$. | null | Define a ten to be all ten positive integers which begin with a fixed tens digit.
We can make a systematic approach to this:
By inspection, $\dfrac{N}{11}$ must be between 10 and 90 inclusive. That gives us 8 tens to check, and 90 as well.
For a given ten, the sum of the squares of the digits of $N$ increases faster th... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | (Hungary)
Solve the system of equations:
where $a$ and $b$ are constants. Give the conditions that $a$ and $b$ must satisfy so that $x, y, z$ (the solutions of the system) are distinct positive numbers. | Note that $x^2 + y^2 = (x+y)^2 - 2xy = (x+y)^2 - 2z^2$, so the first two equations become
We note that $(x+y)^2 - z^2 = \Big[ (x+y)+z \Big]\Big[ (x+y)-z\Big]$, so if $a$ equals 0, then $b$ must also equal 0. We then have $x+y = -z$; $xy = (x+y)^2$. This gives us $x^2 + xy + y^2 = 0$. Mutiplying both sides by $(x-y)$... | null | null | null | null |
null | null | null | As the new number starts with a $6$ and the old number is $1/4$ of the new number, the old number must start with a $1$.
As the new number now starts with $61$, the old number must start with $\lfloor 61/4\rfloor = 15$.
We continue in this way until the process terminates with the new number $615\,384$ and the old numb... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Find the smallest natural number $n$ which has the following properties:
(a) Its decimal representation has 6 as the last digit.
(b) If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number $n$. | null | Let the original number = $10n + 6$, where $n$ is a 5 digit number.
Then we have $4(10n + 6) = 600000 + n$.
=> $40n + 24 = 600000 + n$.
=> $39n = 599976$.
=> $n = 15384$.
=> The original number = $\boxed{153\,846}$. | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Find all real roots of the equation
where $p$ is a real parameter. | Assuming $x \geq 0$, square the equation, obtaining
.
If we have $p + 4 \geq 4x^2$, we can square again, obtaining
We must have $4 - 2p > 0 \iff p < 2$, so we have
However, this is only a solution when
so we have $p\geq 0$ and $p \leq \frac {4}{3}$
and $x = \frac {4 - p}{2\sqrt {4 - 2p}}$ | null | null | null | null |
null | null | null | We claim $2^n$ is equivalent to $2, 4,$ and $1$ $\pmod{7}$ for $n$ congruent to $1$, $2$, and $0$ $\pmod{3}$, respectively.
(a) From the statement above, only $n$ divisible by $3$ will work.
(b) Again from the statement above, $2^n$ can never be congruent to $-1$ $\pmod{7}$, so there are no solutions for $n$.
| null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | (a) Find all positive integers $n$ for which $2^n-1$ is divisible by $7$.
(b) Prove that there is no positive integer $n$ for which $2^n+1$ is divisible by $7$. | null | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Determine all values $x$ in the interval $0\leq x\leq 2\pi$ which satisfy the inequality | We shall deal with the left side of the inequality first ($2\cos x \leq \left| \sqrt{1+\sin 2x} - \sqrt{1-\sin 2x } \right|$) and the right side after that.
It is clear that the left inequality is true when $\cos x$ is non-positive, and that is when $x$ is in the interval $[\pi/2, 3\pi/2]$. We shall now consider when $... | null | null | null | null |
null | null | null | In triangle $ABC$, let $BC=a$, $AC=b$, $AB=c$, $\angle ABC=\alpha$, and $\angle BAC=2\alpha$. Using the Law of Sines gives that
Therefore $\cos{\alpha}=\frac{a}{2b}$. Using the Law of Cosines gives that
This can be simplified to $a^2c=b(a^2+c^2-b^2)$. Since $a$, $b$, and $c$ are positive integers, $b|a^2c$. Note that... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | NO TRIGONOMETRY!!!
Let $a, b, c$ be the side lengths of a triangle in which $\angle C = 2\angle B.$
Extend $AC$ to $D$ such that $CD = BC = a.$ Then $\angle CDB = \frac{\angle ACB}{2} = \angle ABC$, so $ABC$ and $ADB$ are similar by AA Similarity. Hence, $c^2 = b(a+b)$. Then proceed as in Solution 2, as only algebraic ... | null | null | null | null | null | null | null | null | Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another. | null | [Incomplete, please edit]
In a given triangle $ABC$, let $A=2B$, $\implies C=180-3B$, and $\sin C=\sin 3B$.
Then
Hence, $(*)$ $a^2 = b(b+c)$ (with assumptions. This needs clearing up)
If $b$ is the shortest side, $(b+2)^2 = b^2 +b(b+1)$
$\implies (b-4)(b+1)=0$,
$\implies b=4, c=5, a=6$,
No other permutation of $a$, $b... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Prove that there are infinitely many natural numbers $a$ with the following property: the number $z = n^4 + a$ is not prime for any natural number $n$. | Suppose that $a = 4k^4$ for some $a$. We will prove that $a$ satisfies the property outlined above.
The polynomial $n^4 + 4k^4$ can be factored as follows:
Both factors are positive, because if the left one is negative, then the right one would also negative, which is clearly false.
It is also simple to prove that $n^... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $M$ be a point on the side $AB$ of $\triangle ABC$. Let $r_1, r_2$, and $r$ be the inscribed circles of triangles $AMC, BMC$, and $ABC$. Let $q_1, q_2$, and $q$ be the radii of the exscribed circles of the same triangles that lie in the angle $ACB$. Prove that | We use the conventional triangle notations.
Let $I$ be the incenter of $ABC$, and let $I_{c}$ be its excenter to side $c$. We observe that
and likewise,
Simplifying the quotient of these expressions, we obtain the result
Thus we wish to prove that
But this follows from the fact that the angles $AMC$ and $CBM$ are supp... | By similar triangles and the fact that both centers lie on the angle bisector of $\angle{C}$, we have $\frac{r}{q} = \frac{s-c}{s} = \frac{a + b - c}{a + b + c}$, where $s$ is the semi-perimeter of $ABC$. Let $ABC$ have sides $a, b, c$, and let $AM = c_1, MB = c_2, MC = d$. After simple computations, we see that the co... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Prove that the following assertion is true for $n=3$ and $n=5$, and that it is false for every other natural number $n>2:$
If $a_1, a_2,\cdots, a_n$ are arbitrary real numbers, then $(a_1-a_2)(a_1-a_3)\cdots (a_1-a_n)+(a_2-a_1)(a_2-a_3)\cdots (a_2-a_n)+\cdots+(a_n-a_1)(a_n-a_2)\cdots (a_n-a_{n-1})\ge 0.$ | Take $a_1 < 0$, and the remaining $a_i = 0$. Then $E_n = a_1(n-1) < 0$ for $n$ even,
so the proposition is false for even $n$.
Suppose $n \ge 7$ and odd. Take any $c > a > b$, and let $a_1 = a$, $a_2 = a_3 = a_4= b$,
and $a_5 = a_6 = ... = a_n = c$.
Then $E_n = (a - b)^3 (a - c)^{n-4} < 0$.
So the proposition is false ... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum. | There are $2^{10}-2=1022$ distinct subsets of our set of 10 two-digit numbers. The sum of the elements of any subset of our set of 10 two-digit numbers must be between $10$ and $90+91+92+93+94+95+96+97+98+99 < 10 \cdot 100 = 1000$. (There are fewer attainable sums.) As $1000 < 1022$, the Pigeonhole Principle implies th... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Point $O$ lies on line $g;$ $\overrightarrow{OP_1}, \overrightarrow{OP_2},\cdots, \overrightarrow{OP_n}$ are unit vectors such that points $P_1, P_2, \cdots, P_n$ all lie in a plane containing $g$ and on one side of $g.$ Prove that if $n$ is odd, Here $\left|\overrightarrow{OM}\right|$ denotes the length of vector $\o... | We prove it by induction on the number $2n+1$ of vectors. The base step (when we have one vector) is clear, and for the induction step we use the hypothesis for the $2n-1$ vectors obtained by disregarding the outermost two vectors. We thus get a vector with norm $\ge 1$ betwen two with norm $1$. The sum of the two vect... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Three players $A, B$ and $C$ play the following game: On each of three cards an integer is written. These three numbers $p, q, r$ satisfy $0 < p < q < r$. The three cards are shuffled and one is dealt to each player. Each then receives the number of counters indicated by the card he holds. Then the cards are shuffled a... | Answer: player $C$.
Let $n$ be the number of rounds played, then obviously $n (p + q + r) = 20 + 10 + 9 = 39$. So $n$ must be a divisor of 39, i. e. $n \in \{ 1, 3, 13, 39 \}$. But $p \geq 1,$ $q \geq p + 1 \geq 2,$ $r \geq q + 1 \geq 3,$ so $p + q + r \geq 1 + 2 + 3 = 6$ and $n = \frac{39}{p + q + r} \leq \frac{39}{6}... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $x_i, y_i$ $(i=1,2,\cdots,n)$ be real numbers such that Prove that, if $z_1, z_2,\cdots, z_n$ is any permutation of $y_1, y_2, \cdots, y_n,$ then | We can expand and simplify the inequality a bit, and using the fact that $z$ is a permutation of $y$, we can cancel some terms.
Consider the pairing $x_1 \rightarrow y_1$, $x_2 \rightarrow y_2$, ... $x_n \rightarrow y_n$. By switching around some of the $y$ values, we have obtained the pairing $x_1 \rightarrow z_1$, ... | We can rewrite the summation as
Since $\sum^n_{i=1} y_i^2 = \sum^n_{i=1} z_i^2$, the above inequality is equivalent to
We will now prove that the left-hand side of the inequality is the greatest sum reached out of all possible values of $\sum^n_{i=1}x_iz_i$. Obviously, if $x_1 = x_2 = \ldots = x_n$ or $y_1 = y_2 = ... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | In a convex quadrilateral (in the plane) with the area of $32 \text{ cm}^{2}$ the sum of two opposite sides and a diagonal is $16 \text{ cm}$. Determine all the possible values that the other diagonal can have. | Label the vertices $A$, $B$, $C$, and $D$ in such a way that $AB + BD + DC = 16$, and $\overline{BD}$ is a diagonal.
The area of the quadrilateral can be expressed as $BD \cdot ( d_1 + d_2 ) / 2$, where $d_1$ and $d_2$ are altitudes from points $A$ and $C$ onto $\overline{BD}$. Clearly, $d_1 \leq AB$ and $d_2 \leq DC$.... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | In the interior of a square $ABCD$ we construct the equilateral triangles $ABK, BCL, CDM, DAN.$ Prove that the midpoints of the four segments $KL, LM, MN, NK$ and the midpoints of the eight segments $AK, BK, BL, CL, CM, DM, DN, AN$ are the 12 vertices of a regular dodecagon. | Just use complex numbers, with $a = 1$, $b = i$, $c = - 1$ and $d = - i$.
With some calculations, we have $k = \frac {\sqrt {3} - 1}{2}( - 1 - i)$, $l = \frac {\sqrt {3} - 1}{2}(1 - i)$, $m = \frac {\sqrt {3} - 1}{2}(1 + i)$ and $n = \frac {\sqrt {3} - 1}{2}( - 1 + i)$.
Now it's an easy job to calculate the twelve midp... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $m$ and $n$ be positive integers such that $1 \le m < n$. In their decimal representations, the last three digits of $1978^m$ are equal, respectively, to the last three digits of $1978^n$. Find $m$ and $n$ such that $m + n$ has its least value. | We have $1978^m\equiv 1978^n\pmod {1000}$, or $978^m-978^n=1000k$ for some positive integer $k$ (if it is not positive just do $978^n-978^m=-1000k$). Hence $978^n\mid 1000k$. So dividing through by $978^n$ we get $978^{m-n}-1=\frac{1000k}{978^n}$. Observe that $2\nmid LHS$, so $2\nmid RHS$. So since $2|| 978^n$, clearl... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | If $p$ and $q$ are natural numbers so thatprove that $p$ is divisible with $1979$. | We first write
Now, observe that
and similarly $\frac{1}{661}+\frac{1}{1318}=\frac{1979}{661\cdot 1318}$ and $\frac{1}{662}+\frac{1}{1317}=\frac{1979}{662\cdot 1317}$, and so on. We see that the original equation becomes
where $s=660\cdot 661\cdots 1319$ and $r=\frac{s}{660\cdot 1319}+\frac{s}{661\cdot 1318}+\cdots+\fr... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | $P$ is a point inside a given triangle $ABC$. $D, E, F$ are the feet of the perpendiculars from $P$ to the lines $BC, CA, AB$, respectively. Find all $P$ for which
is least. | We note that $BC \cdot PD + CA \cdot PE + AB \cdot PF$ is twice the triangle's area, i.e., constant. By the Cauchy-Schwarz Inequality,
with equality exactly when $PD = PE = PF$, which occurs when $P$ is the triangle's incenter, Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to... | null | null | null | null |
null | null | null | Clearly $f(1) \ge 1 \Rightarrow f(m+1) \ge f(m)+f(1) \ge f(m)+1$ so $f(9999) \ge 9999$.Contradiction!So $f(1)=0$.This forces $f(3)=1$.Hence $f(3k+3) \ge f(3k)+f(3)>f(3k)$ so the inequality $f(3)<f(6)<\cdots<f(9999)=3333$ forces $f(3k)=k \forall k \le 3333$.Now $f(3k+2) \ge k+1 \Rightarrow f(6k+4) \ge 2k+2 \Rightarrow f... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | We show that $f(n) = [n/3]$ for $n <= 9999$, where [ ] denotes the integral part. We show first that $f(3) = 1$. $f(1)$ must be $0$, otherwise $f(2) - f(1) - f(1)$ would be negative. Hence $f(3) = f(2) + f(1) + 0$ or $1$ = $0$ or $1$. But we are told $f(3) > 0$, so $f(3) = 1$. It follows by induction that $f(3n) >= n$.... | null | null | null | null | null | null | Similar to solution 3.
Proof:
Lemma 1: $f(mn)\geq nf(m)$
Let, $P(m,m)$ be assertion.
Similarly,we can induct to get $f(nm)\geq nf(m)$.
Lemma proved.
Then we see that,
Then,
Then we can easily get,by assertion $P(1980,2)$
Hence, $\boxed{f(1980)=660,661}$.And, we are done. $\blacksquare$
This solution was posted and copy... | null | The function $f(n)$ is defined on the positive integers and takes non-negative integer values. $f(2)=0,f(3)>0,f(9999)=3333$ and for all $m,n:$Determine $f(1982)$. | null | First observe that
Since $f(3)$ is a positive integer, we need $f(3)=1$. Next, observe that
\begin{align*}
3333=f(9999)\geq 5f(1980)+33f(3)=5f(1980)+33\quad\Longrightarrow\quad f(1980)\leq 660
\end{align*}On the other hand, $f(1980)\geq 660f(3)=660$, so combine the two inequalities we obtain $f(1980)=660$. Finally, wri... | null | null | null |
null | null | null | Let $x=y=1$ and we have $f(f(1))=f(1)$. Now, let $x=1,y=f(1)$ and we have $f(f(f(1)))=f(1)f(1)\Rightarrow f(1)=[f(1)]^2$ since $f(1)>0$ we have $f(1)=1$.
Plug in $y=x$ and we have $f(xf(x))=xf(x)$. If $a=1$ is the only solution to $f(a)=a$ then we have $xf(x)=1\Rightarrow f(x)=\frac{1}{x}$. We prove that this is the... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Find all functions $f$ defined on the set of positive reals which take positive real values and satisfy the conditions:
(i) $f(xf(y))=yf(x)$ for all $x,y$;
(ii) $f(x)\to0$ as $x\to \infty$. | null | Let $x=1$ so If $f(a)=f(b),$ then $af(1)=f(f(a))=f(f(b))=bf(1)\implies a=b$ because $f(1)$ goes to the real positive integers, not $0.$ Hence, $f$ is injective. Let $x=y$ so
so $xf(x)$ is a fixed point of $f.$ Then, let $y=1$ so $f(xf(1))=f(x)\implies f(1)=1$ as $x$ can't be $0$ so $1$ is a fixed point of $f.$ ... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $x$, $y$, $z$ be nonnegative real numbers with $x + y + z = 1$. Show that
$0 \leq xy+yz+zx-2xyz \leq \frac{7}{27}$ | Note that this inequality is symmetric with x,y and z.
To prove note that $x+y+z=1$ implies that at most one of $x$, $y$, or $z$ is greater than $\frac{1}{2}$. Suppose $x \leq \frac{1}{2}$, WLOG. Then, $xy+yz+zx-2xyz=yz(1-2x)+xy+zx\geq 0$ since $(1-2x)\geq 0$, implying all terms are positive.
To prove $xy+yz+zx-2xyz \... | null | null | null | null |
null | Lemma. Let $I$ be the in-center of $ABC$ and points $P$ and $Q$ be on the lines $AB$ and $BC$ respectively. Then $BP + CQ = BC$ if and only if $APIQ$ is a cyclic quadrilateral.
Solution. Assume that rays $AD$ and $BC$ intersect at point $P$. Let $S$ be the center od circle touching $AD$, $DC$ and $CB$. Obviosuly $S$ is... | null | Let $O$ be the center of the circle mentioned in the problem. Let $T$ be the second intersection of the circumcircle of $CDO$ with $AB$. By measures of arcs, $\angle DTA = \angle DCO = \frac{\angle DCB}{2} = \frac{\pi}{2} - \frac{\angle DAB}{2}$. It follows that $AT = AD$. Likewise, $TB = BC$, so $AD + BC = AB$, as... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let the circle have center $O$ and radius $r$, and let its points of tangency with $BC, CD, DA$ be $E, F, G$, respectively. Since $OEFC$ is clearly a cyclic quadrilateral, the angle $COE$ is equal to half the angle $GAO$. Then
Likewise, $DG = OB - EB$. It follows that
Q.E.D. | null | null | null | This solution is incorrect. The fact that $BC$ is tangent to the circle does not necessitate that $B$ is its point of tangency. -Nitinjan06
From the fact that AD and BC are tangents to the circle mentioned in the problem, we have
$\angle{CBA}=90\deg$
and
$\angle{DAB}=90\deg$.
Now, from the fact that ABCD is cyclic, we ... | null | null | We use the notation of the previous solution. Let $X$ be the point on the ray $AD$ such that $AX = AO$. We note that $OF = OG = r$; $\angle OFC = \angle OGX = \frac{\pi}{2}$; and $\angle FCO = \angle GXO = \frac{\pi - \angle BAD}{2}$; hence the triangles $OFC, OGX$ are congruent; hence $GX = FC = CE$ and $AO = AG + G... | null | A circle has center on the side $AB$ of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD + BC = AB$. | null | Let $O$ be the center of the circle mentioned in the problem, and let $T$ be the point on $AB$ such that $AT = AD$. Then $\angle DTA = \frac{ \pi - \angle DAB}{2} = \angle DCO$, so $DCOT$ is a cyclic quadrilateral and $T$ is in fact the $T$ of the previous solution. The conclusion follows. | null | null | null |
null | null | null | We do casework with modular arithmetic.
$d\equiv 0,3 \pmod{4}: 13d-1$ is not a perfect square.
$d\equiv 2\pmod{4}: 2d-1$ is not a perfect square.
Therefore, $d\equiv 1, \pmod{4}.$ Now consider $d\pmod{16}.$
$d\equiv 1,13 \pmod{16}: 13d-1$ is not a perfect square.
$d\equiv 5,9\pmod{16}: 5d-1$ is not a perfect square.
As... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $d$ be any positive integer not equal to $2, 5$ or $13$. Show that one can find distinct $a,b$ in the set $\{2,5,13,d\}$ such that $ab-1$ is not a perfect square. | null | Proof by contradiction:
Suppose $p^2=2d-1$, $q^2=5d-1$ and $r^2=13d-1$. From the first equation, $p$ is an odd integer. Let $p=2k-1$. We have $d=2k^2-2k+1$, which is an odd integer. Then $q^2$ and $r^2$ must be even integers, denoted by $4n^2$ and $4m^2$ respectively, and thus $r^2-q^2=4m^2-4n^2=8d$, from which
can be... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $p_n (k)$ be the number of permutations of the set $\{ 1, \ldots , n \} , \; n \ge 1$, which have exactly $k$ fixed points. Prove that
(Remark: A permutation $f$ of a set $S$ is a one-to-one mapping of $S$ onto itself. An element $i$ in $S$ is called a fixed point of the permutation $f$ if $f(i) = i$.) | The sum in question simply counts the total number of fixed points in all permutations of the set. But for any element $i$ of the set, there are $(n-1)!$ permutations which have $i$ as a fixed point. Therefore
as desired.
Slightly Clearer Solution
For any $k$, if there are $p_n(k)$ permutations that have $k$ fixed po... | The probability of any number $i$ where $1\le i\le n$ being a fixed point is $\frac{1}{n}$. Thus, the expected value of the number of fixed points is $n\times \frac{1}{n}=1$.
The expected value is also $\sum_{k=0}^{n} \frac{k \cdot p_n (k)}{n!}$.
Thus, or | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Prove that in the set $\{1,2, \ldots, 1989\}$ can be expressed as the disjoint union of subsets $A_i, \{i = 1,2, \ldots, 117\}$ such that
i.) each $A_i$ contains 17 elements
ii.) the sum of all the elements in each $A_i$ is the same.
| Let us start pairing numbers in the following fashion, where each pair sums to $1990$:
$(1, 1990-1), (2, 1990-2), \ldots (936, 1054)$
There are a total of $117*8$ pairs above. Let us start putting these pairs into each of the $117$ subsets starting with the first pair $(1, 1990-1)$ going into $A_1$, $(2, 1990-2)$ into ... | null | null | null | null |
null | null | null | With simple angle chasing, we find that triangles $CEG$ and $MDB$ are similar.
so, $GE/EC = MD/MB$. (*)
again with simple angle chasing, we find that triangles $CEF$ and $AMD$ are similar.
so, $EF/DM = CE/AM$. (**)
so, by (*) and (**), we have $GE/EF = MA/MB = t/(t-1)$.
This solution was posted and copyrighted by e.lop... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Chords $AB$ and $CD$ of a circle intersect at a point $E$ inside the circle. Let $M$ be an interior point of the segment $\overline{EB}$. The tangent line at $E$ to the circle through $D, E$, and $M$ intersects the lines $\overline{BC}$ and ${AC}$ at $F$ and $G$, respectively.
If $\frac{AM}{AB} = t$, find $\frac{EG}{E... | null | This problem can be bashed with PoP and Ratio Lemma. Rewriting the given ratio gets $\frac{MA}{MB}=\frac{t}{1-t}$.
By Ratio Lemma, $\frac{FB}{FC}=\frac{BE}{CE} \cdot \frac{\sin{\angle{FEB}}}{\sin{\angle{FEC}}}=\frac{DE}{AE} \cdot \frac{\sin{\angle{EDM}}}{\sin{\angle{DME}}}=\frac{DE}{AE} \cdot \frac{EM}{DE}=\frac{ME}{EA... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Find all integers $a$, $b$, $c$ satisfying $1 < a < b < c$ such that $(a - 1)(b -1)(c - 1)$ is a divisor of $abc - 1$. | $1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\frac{abc}{(a-1)(b-1)(c-1)}$
$1<\frac{abc-1}{(a-1)(b-1)(c-1)}<\left(\frac{a}{a-1}\right) \left(\frac{b}{b-1}\right) \left(\frac{c}{c-1}\right)$
With $1<a<b<c$ it implies that $a \ge 2$, $b \ge 3$, $c \ge 4$
Therefore, $\frac{a}{a-1}=1+\frac{1}{a-1}$
which for $a$ gives: $\frac{a}{a-1} \... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $f\left(x\right)=x^n+5x^{n-1}+3$, where $n>1$ is an integer. Prove that $f\left(x\right)$ cannot be expressed as the product of two non-constant polynomials with integer coefficients. | For the sake of contradiction, assume that $f\left(x\right)=g\left(x\right)h\left(x\right)$ for polynomials $g\left(x\right)$ and $h\left(x\right)$ in $\mathbb{R}$. Furthermore, let $g\left(x\right)=b_mx^m+b_{m-1}x^{m-1}+\ldots+b_1x+b_0$ with $b_i=0$ if $i>m$ and $h\left(x\right)=c_{n-m}x^{n-m}+c_{n-m-1}x^{n-m-1}+\ldot... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $AM$ and $XY$ intersect at $Z$. Because $\angle AMC = \angle BND = \angle APT = 90^\circ$, we have quadrilaterals $AMPT$ and $DNPT$ cyclic. Therefore, $Z$ lies on the radical axis of the two circumcircles of these quadrilaterals. But $Z$ also lies on radical axis $XY$ of the original two circles, so the power of $Z... | null | null | null | null | null | null | null | null | Let $A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $AC$ and $BD$ intersect at $X$ and $Y$. The line $XY$ meets $BC$ at $Z$. Let $P$ be a point on the line $XY$ other than $Z$. The line $CP$ intersects the circle with diameter $AC$ at $C$ and $M$, and the line $BP$ intersects the ... | Since $M$ is on the circle with diameter $AC$, we have $\angle AMC=90$ and so $\angle MCA=90-A$. We similarly find that $\angle BND=90$. Also, notice that the line $XY$ is the radical axis of the two circles with diameters $AC$ and $BD$. Thus, since $P$ is on $XY$, we have $PN\cdot PB=PM\cdot PC$ and so by the conve... | Let $AM$ and $PT$ (a subsegment of $XY$) intersect at $Z$. Now, assume that $Z, N, P$ are not collinear. In that case, let $ZD$ intersect the circle with diameter $BD$ at $N'$ and the circle through $D, P, T$ at $N''$.
We know that $\angle AMC = \angle BND = \angle ATP = 90^\circ$ via standard formulae, so quadrilatera... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | We are given a positive integer $r$ and a rectangular board $ABCD$ with dimensions $|AB|=20$, $|BC|=12$. The rectangle is divided into a grid of $20 \times 12$ unit squares. The following moves are permitted on the board: one can move from one square to another only if the distance between the centers of the two squa... | First we define the rectangular board in the cartesian plane with centers of the unit squares as integer coordinates and the following coordinates for the squares at the corners of $A$, $B$, $C$, $D$, as follows: $A=(1,1)$, $B=(20,1)$, $C=(20,12)$, $D=(1,12)$
Let $(x_i,y_i)$ be the coordinates of the piece after move $... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | In the plane the points with integer coordinates are the vertices of unit squares. The squares are colored alternatively black and white (as on a chessboard).
For any pair of positive integers $m$ and $n$, consider a right-angled triangle whose vertices have integer coordinates and whose legs, of lengths $m$ and $n$, ... | For any pair of positive integers $m$ and $n$, consider a rectangle $ABCD$ whose vertices have integer coordinates and whose legs, of lengths $m$ and $n$, lie along edges of the squares.
Let $A$, $B$, $C$, and $D$, be the lower left vertex, lower right vertex, upper right vertex, and upper left vertex of rectangle $ABC... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | In the convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ are perpendicular and the opposite sides $AB$ and $DC$ are not parallel. Suppose that the point $P$, where the perpendicular bisectors of $AB$ and $DC$ meet, is inside $ABCD$. Prove that $ABCD$ is a cyclic quadrilateral if and only if the triangles $ABP$ a... | This problem needs a solution. If you have a solution for it, please help us out by adding it. | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Determine all finite sets $S$ of at least three points in the plane which satisfy the following condition:
For any two distinct points $A$ and $B$ in $S$, the perpendicular bisector of the line segment $AB$ is an axis of symmetry of $S$. | Upon reading this problem and drawing some points, one quickly realizes that the set $S$ consists of all the vertices of any regular polygon.
Now to prove it with some numbers:
Let $S=\left\{ P_{0},P_{1},P_{2},...,P_{n-1} \right\}$, with $n\ge 3$, where $P_{i}$ is a vertex of a polygon which we can define their $xy$ co... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Two circles $G_1$ and $G_2$ intersect at two points $M$ and $N$. Let $AB$ be the line tangent to these circles at $A$ and $B$, respectively, so that $M$ lies closer to $AB$ than $N$. Let $CD$ be the line parallel to $AB$ and passing through the point $M$, with $C$ on $G_1$ and $D$ on $G_2$. Lines $AC$ and $BD$ meet at ... | $\textbf{Proof of problem:}$ Let ray $NM$ intersect $AB$ at $X$. By our lemma, $\textit{(the two circles are tangent to AB)}$, $X$ bisects $AB$. Since $\triangle{NAX}$ and $\triangle{NPM}$ are similar, and $\triangle{NBX}$ and $\triangle{NQM}$ are similar implies $M$ bisects $PQ$.
By simple parallel line rules, $\angle... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Consider an acute triangle $\triangle ABC$. Let $P$ be the foot of the altitude of triangle $\triangle ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $\triangle ABC$. Assume that $\angle C \geq \angle B+30^{\circ}$. Prove that $\angle A+\angle COP < 90^{\circ}$. | Take $D$ on the circumcircle with $AD \parallel BC$. Notice that $\angle CBD = \angle BCA$, so $\angle ABD \ge 30^\circ$. Hence $\angle AOD \ge 60^\circ$. Let $Z$ be the midpoint of $AD$ and $Y$ the midpoint of $BC$. Then $AZ \ge R/2$, where $R$ is the radius of the circumcircle. But $AZ = YP$ (since $AZYP$ is a recta... | Notice that because $\angle{PCO} = 90^\circ - \angle{A}$, it suffices to prove that $\angle{POC} < \angle{PCO}$, or equivalently $PC < PO.$
Suppose on the contrary that $PC > PO$. By the triangle inequality, $2 PC = PC + PC > PC + PO > CO = R$, where $R$ is the circumradius of $ABC$. But the Law of Sines and basic trig... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | $S$ is the set of all $(h,k)$ with $h,k$ non-negative integers such that $h + k < n$. Each element of $S$ is colored red or blue, so that if $(h,k)$ is red and $h' \le h,k' \le k$, then $(h',k')$ is also red. A type $1$ subset of $S$ has $n$ blue elements with different first member and a type $2$ subset of $S$ has $n$... | This problem needs a solution. If you have a solution for it, please help us out by adding it. | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BM... |
Let $\angle{ACB} = a$, $\angle{CBA} = b$, and $\angle{ONR} = k$. Call $\omega$ the circle with diameter $BC$ and $\odot{AMN}$ the circumcircle of $\triangle{AMN}$.
Our ultimate goal is to show that $\angle{CNR} + \angle{RMB} = 180^\circ$. To show why this solves the problem, assume this statement holds true. Call $K$ ... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1, A_2$ on $BC$, $B_1$, $B_2$ on $CA$ and $C_1$, $C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths. Prove that the lines $A_1B_2, B_1C_2$ and $C_1A_2$ are concurrent. | This problem needs a solution. If you have a solution for it, please help us out by adding it. | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies $\angle PBA+\angle PCA = \angle PBC+\angle PCB$. Show that $AP \geq AI$, and that equality holds if and only if $P=I.$ | We have
and similarly
Since $\angle PBA + \angle PCA = \angle PBC + \angle PCB$, we have $\angle PCB - \angle PCA = \angle PBA - \angle PBC.$
It follows that Hence, $B,P,I,$ and $C$ are concyclic.
Let ray $AI$ meet the circumcircle of $\triangle ABC\,$ at point $J$. Then, by the Incenter-Excenter Lemma, $JB=JC=JI=JP... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Real numbers $a_1, a_2, \dots , a_n$ are given. For each $i$ ($1\le i\le n$) define
and let
.
(a) Prove that, for any real numbers $x_1\le x_2\le \cdots\le x_n$,
(b) Show that there are real numbers $x_1\le x_2\le x_n$ such that equality holds in (*)
| Since $d_i=\max\{a_j:1\le j\le i\}-\min\{a_j:i\le j\le n\}$, all $d_i$ can be expressed as $a_p-a_q$, where $1\le p\le i\le q \le n$.
Thus, $d$ can be expressed as $a_p-a_q$ for some $p$ and $q$, $1\le p\le q\le n$
Lemma) $d\ge 0$
Assume for contradiction that $d<0$, then for all $i$, $a_i \le \max\{a_j:1\le j\le i\}... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | An acute-angled triangle $ABC$ has orthocentre $H$. The circle passing through $H$ with centre the midpoint of $BC$ intersects the line $BC$ at $A_1$ and $A_2$. Similarly, the circle passing through $H$ with centre the midpoint of $CA$ intersects the line $CA$ at $B_1$ and $B_2$, and the circle passing through $H$ with... | Let $M_A$, $M_B$, and $M_C$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. It's not hard to see that $M_BM_C\parallel BC$. We also have that $AH\perp BC$, so $AH \perp M_BM_C$. Now note that the radical axis of two circles is perpendicular to the line connecting their centers. We know that $H$ is on the ... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $n$ be a positive integer and let $a_1,\ldots,a_k (k\ge2)$ be distinct integers in the set $\{1,\ldots,n\}$ such that $n$ divides $a_i(a_{i+1}-1)$ for $i=1,\ldots,k-1$. Prove that $n$ doesn't divide $a_k(a_1-1)$.
Author: Ross Atkins, Australia | Let $n=pq$ such that $p\mid a_1$ and $q\mid a_2-1$. Suppose $n$ divides $a_k(a_1-1)$.
Note $q\mid a_2-1$ implies $(q,a_2)=1$ and hence $q\mid a_3-1$. Similarly one has $q\mid a_i-1$ for all $i$'s, in particular, $p\mid a_1$ and $q\mid a_1-1$ force $(p,q)=1$. Now $(p,a_1-1)=1$ gives $p\mid a_k$, similarly one has $p\mi... | Let $n = p_1^{b_1}p_2^{b_2} \cdots p_s^{b_s}$. Then after toying around with the $p_i^{b_i}$ and what they divide, we have that $p_i^{b_i} \nmid a_k$, and so in particular, $n \nmid a_k$.
Assume by way of contradiction that $n \mid a_k(a_1 - 1)$. Then $n \mid a_1 - 1$. Now we shift our view towards the $a_i(a_{i + 1} -... | null | null | null |
null | null | null | Put $x=y=0$. Then $f(0)=0$ or $\lfloor f(0) \rfloor=1$.
$\bullet$ If $\lfloor f(0) \rfloor=1$, putting $y=0$ we get $f(x)=f(0)$, that is f is constant.
Substituing in the original equation we find $f(x)=0, \ \forall x \in \mathbb{R}$ or $f(x)=a, \ \forall x \in \mathbb{R}$, where $a \in [1,2)$.
$\bullet$ If $f(0)=0$, ... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $y=0$, then $f(0)=f(x)\left\lfloor f(0)\right\rfloor$.
Case 1: $\left\lfloor f(0)\right\rfloor\neq 0$
Then $f(x)=\frac{f(0)}{\left\lfloor f(0)\right\rfloor}$ is a constant. Let $f(x)=k$, then $k=k\left\lfloor k \right\rfloor \Leftrightarrow k=0 \vee 1\leq k<2$. It is easy to check that this are solutions.
Case 2: $... | null | null | null | null | null | null | Clearly $f(\left\lfloor x\right\rfloor y) = f(\left\lfloor \lfloor x \rfloor \right\rfloor y) = f(\lfloor x \rfloor)\left\lfloor f(y)\right\rfloor$, so $(f(x) - f(\lfloor x \rfloor))\left\lfloor f(y)\right\rfloor = 0$ for all $x,y\in\mathbb{R}$.
If $\left\lfloor f(y)\right\rfloor = 0$ for all $y \in \mathbb{R}$, then b... | null | Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds
$f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor$
where $\left\lfloor a\right\rfloor$ is greatest integer not greater than $a.$
Author: Pierre Bornsztein, France | null | Substituting $y=0$ we have $f(0) = f(x) [f(0)]$.
If $[f0)] \ne 0$ then $f(x) = \frac{f(0)}{[f(0)]}$. Then $f(x)$ is constant. Let $f(x)=c$. Then substituting that in (1) we have $c=c[c] \Rightarrow c(1-[c])=0 \Rightarrow c=0$, or $[c]=1$. Therefore $f(x)=c$ where $c=0$ or $c \in [1,2)$
If $[f(0)] = 0$ then $f(0)... | null | null | null |
null | null | null | Firstly, if we order $a_1 \ge a_2 \ge a_3 \ge a_4$, we see $2(a_3 + a_4) \ge (a_1+a_2)+(a_3+a_4) = s_A \geq 0$, so $(a_3, a_4)$ isn't a couple that satisfies the conditions of the problem. Also, $2(a_4 + a_2) = (a_4 + a_4) + (a_2 + a_2) \ge (a_4+a_3)+(a_2+a_1) = s_A \ge 0$, so again $(a_2, a_4)$ isn't a good couple. W... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Given any set $A = \{a_1, a_2, a_3, a_4\}$ of four distinct positive integers, we denote the sum $a_1 +a_2 +a_3 +a_4$ by $s_A$. Let $n_A$ denote the number of pairs $(i, j)$ with $1 \leq i < j \leq 4$ for which $a_i +a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers which achieve the largest pos... | null | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Same as Solution 2, except noticing that (letting $s = \dfrac{a + b + c}{2}$ be the semi-perimeter):
--Suli 18:21, 8 February 2015 (EST) | null | null | null | null | null | null | As before in Solution 2, we find that $\angle{JFL} = \dfrac{\angle{A}}{2}.$ But it is clear that $AJ$ bisects $\angle{KAL}$, so $\angle{JAL} = \dfrac{\angle{A}}{2} = \angle{JFL}$ and hence $AFJL$ is cyclic. In particular, $\angle{AFJ} = \angle{ALJ} = 90^\circ$, and continue as in Solution 2. | null | Given triangle $ABC$ the point $J$ is the centre of the excircle opposite the vertex $A.$ This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G.$ Let $S$ be the point of intersection of t... | First, $BK = BM$ because $BK$ and $BM$ are both tangents from $B$ to the excircle $J$. Then $BJ \bot KM$. Call the $X$ the intersection between $BJ$ and $KM$. Similarly, let the intersection between the perpendicular line segments $CJ$ and $LM$ be $Y$. We have $\angle XBM = \angle XBK = \angle FBA$ and $\angle XMB ... | For simplicity, let $A, B, C$ written alone denote the angles of triangle $ABC$, and $a$, $b$, $c$ denote its sides.
Let $R$ be the radius of the A-excircle. Because $CM = CL$, we have $CML$ isosceles and so $\angle{CML} = \dfrac{\angle{C}}{2}$ by the Exterior Angle Theorem. Then because $\angle{FBS} = 90^\circ - \dfra... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1,m_2,...,m_k$ (not necessarily different) such that
$1+\frac{2^k-1}{n}=(1+\frac{1}{m_1})(1+\frac{1}{m_2})...(1+\frac{1}{m_k})$. | We prove the claim by induction on $k$.
Base case: If $k = 1$ then $1 +\frac{2^1-1}{n} = 1 + \frac{1}{n}$, so the claim is true for all positive integers $n$.
Inductive hypothesis: Suppose that for some $m \in \mathbb{Z}^{+}$ the claim is true for $k = m$, for all $n \in \mathbb{Z}^{+}$.
Inductive step: Let $n$ be arbi... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $a_0<a_1<a_2<\cdots \quad$ be an infinite sequence of positive integers, Prove that there exists a unique integer $n\ge1$ such that | Define $f(n) = a_0 + a_1 + \dots + a_n - n a_{n+1}$. (In particular, $f(0) = a_0.$) Notice that because $a_{n+2} \ge a_{n+1}$, we have
Thus, $f(n) > f(n+1)$; i.e., $f$ is monotonic decreasing. Therefore, because $f(0) > 0$, there exists a unique $N$ such that $f(N-1) > 0 \ge f(N)$. In other words,
This rearranges to... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | We say that a finite set $\mathcal{S}$ in the plane is balanced
if, for any two different points $A$, $B$ in $\mathcal{S}$, there is
a point $C$ in $\mathcal{S}$ such that $AC=BC$. We say that
$\mathcal{S}$ is centre-free if for any three points $A$, $B$, $C$ in
$\mathcal{S}$, there is no point $P$ in $\mathcal{S}$ s... | Part (a): We explicitly construct the sets $\mathcal{S}$. For
odd $n$, $\mathcal{S}$ can be taken to be the vertices of
regular polygons $P_n$ with $n$ sides: given any two vertices $A$ and
$B$, one of the two open half-spaces into
which $AB$ divides $P_n$ contains an odd number of $k$ of vertices of
$P_n$. The $((k+1... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Le... | This problem needs a solution. If you have a solution for it, please help us out by adding it. | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2, \ldots$ for $n \geq 0$ as
Determine all values of $a_0$ such that there exists a number $A$ such that $a_n = A$ for infinitely many values of $n$. | First we observe the following:
When we start with $a_0=3$, we get $a_1=6$, $a_2=9$, $a_3=3$ and the pattern $3,6,9$ repeats.
When we start with $a_0=6$, we get $a_1=9$, $a_2=3$, $a_3=6$ and the pattern $3,6,9$ repeats.
When we start with $a_0=9$, we get $a_1=3$, $a_2=6$, $a_3=9$ and the pattern $3,6,9$ repeats.
When w... | null | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel ... | http://wiki-images.artofproblemsolving.com/5/5d/FB_IMG_1531446409131.jpg
The diagram is certainly not to scale, but the argument is sound (I believe) and involves re-ordering the construction as specified in the original problem so that an identical state of affairs results, yet in so doing differently it is made clear... | The essence of the proof is using a rhombus formed by the perpendicular bisectors of the segments $BD, CE, AB$ and $AC$ and the parallelism of its diagonal and the base of the triangle formed by the perpendiculars from one vertex of the rhombus.
The perpendicular bisectors of the segments $BD, CE, AB$ and $AC$ interse... | null | null | null |
null | null | null | Let us substitute $0$ in for $a$ to get
Now, since the domain and range of $f$ are the same, we can let $x = f(b)$ and $f(0)$ equal some constant $c$ to get
Therefore, we have found that all solutions must be of the form $f(x) = 2x + c.$
Plugging back into the original equation, we have: $4a + c + 4b + 2c = 4a + 4b + ... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $\mathbb{Z}$ be the set of integers. Determine all functions $f : \mathbb{Z} \to \mathbb{Z}$ such that, for all
integers $a$ and $b$, | null | We plug in $a=-b=x$ and $a=-b=x+k$ to get
respectively.
Setting them equal to each other, we have the equation and moving "like terms" to one side of the equation yields Seeing that this is a difference of outputs of $f,$ we can relate this to slope by dividing by $2k$ on both sides. This gives us which means tha... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Consider the convex quadrilateral $ABCD$. The point $P$ is in the interior of $ABCD$. The following ratio equalities hold:
Prove that the following three lines meet in a point: the internal bisectors of angles $\angle ADP$ and $\angle PCB$ and the perpendicular bisector of segment $\overline{AB}$. | null | null | null | null | null |
null | null | null | If we can guarantee that there exist $3$ cards such that every pair of them sum to a perfect square, then we can guarantee that one of the piles contains $2$ cards that sum to a perfect square. Assume the perfect squares $p^2$, $q^2$, and $r^2$ satisfy the following system of equations:
where $a$, $b$, and $c$ are nu... | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | Let $n \geq 100$ be an integer. Ivan writes the numbers $n, n+1, \ldots, 2 n$ each on different cards. He then shuffles these $n+1$ cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square. | null | For the statement to be true, there must be at least three pairs whose sum is each a perfect square. There must be p,q,r such that
p+q = x^2 and q+r = y^2, p+r = z^2.
by equation 1
if we add (2) and (3) to (1),
(1) + (2) + (3) =>
At this time 100 ≤ n, so let's put n = 100 to this
where
x = 16, y = 18, z = 20 fits perfe... | null | null | null |
null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | null | The Bank of Oslo issues two types of coin: aluminium (denoted A) and bronze (denoted B). Marianne has $n$ aluminium coins and $n$ bronze coins, arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer $k\le 2n$, Marianne repeated... | https://www.youtube.com/watch?v=nYD-qIOdi_c [Video contains solutions to all day 1 problems]
https://youtu.be/KHn3xD6wS3A
[Video contains problem 1 discussion]
We call a chain basic when it is the largest possible for the coins it consists of. Let \(A=[i,j]\) be the basic chain with the \(i\)-th and \(j\)-th coins bein... | null | null | null | null |
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