AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
natural numbers such that there exist natural numbers 
, 
such that 
is a prime number.
be a positive integer and let 
be positive integers. Define for all 
, 
to be the least occurring integer in 
(if the smallest number of occurrences is shared between two distinct integers 
, let be the smaller of the two).
is eventually periodic.
, for
We get: 
and 
be tangents from point 
to a circle with center 
, and let 
be a chord through the point of tangency. If 
, find the measure of 
., points , 
, and 
lie on the circumference. If the measure of the minor arc is 
, what is the measure of the inscribed angle 
?, , , and 
lie on a circle in that order. It is given that 
, and 
. Find 
.
given that
be a cyclic quadrilateral. Let 
be the midpoint of the arc of its circumcircle which does not contain or . Let the lines 
and meet at 
and the lines 
and 
meet at 
. Prove that the lines 
and are parallel.
be the set of all positive integers. Determine all subsets 
of 
for which there exists a function 
such that![\[\mathcal{S}=\{f(a+b)-f(a)-f(b)\mid a,b\in\mathbb{Z}_{>0}\}.\]](http://latex.artofproblemsolving.com/6/1/2/612e4c41bfb19f78f1373529495473e6ad26c30b.png)
Pitchayut Saengrungkongka, Thailand
such that the following equality holds for any real numbers and .
such that for all 
,![\[xf(x+f(y))=(y-x)f(f(x)).\]](http://latex.artofproblemsolving.com/e/4/a/e4a3bbb8b91d2aa62d699c24df342fa59be71915.png)
be positive real numbers. Prove that
, suppose
What is the measure of angle ?
of real numbers such that the following conditions hold for all 
:
. Let be a subset of 
such that no sum of three distinct elements of is divisible by 
. What is the maximum value of 
.
, we can include elements that are congruent to 
and 
, that is, numbers of the form 
and 
. is 40.
can be 
cyclically with period 
.
,
Thus the terms from 
up to 
split into 
sums of 
.
.
,
Hence the entire sum equals 
.
and 
be tangents to the circle at points 
and 
,
is isosceles with 
,
.
,![\[
\angle ABC = \angle ACB = \frac{180^\circ - 50^\circ}{2} = 65^\circ.
\]](http://latex.artofproblemsolving.com/3/e/3/3e3996be325fffc7a44fd02c43d188515320d3df.png)
be the center of the circle. Since 
and 
are radii, triangle 
is isosceles.
![\[
\angle BOC = 180^\circ - 25^\circ - 25^\circ = \boxed{130^\circ}.
\]](http://latex.artofproblemsolving.com/2/2/3/223e2b34799b0c77aa2e6c7a48c9328053cdb354.png)
![\[
\angle BAC
= \tfrac12 \times (\text{measure of arc }BC)
= \tfrac12 \times 100^\circ
= \boxed{50^\circ}.
\]](http://latex.artofproblemsolving.com/3/1/0/31064ba09c9691d54f4ac0d49ffd2691b1365ec5.png)
,
,
,
lie on a circle, and we're given 
,![\[
\angle ABC \text{ and } \angle ADC \text{ intercept the same arc } AC.
\]](http://latex.artofproblemsolving.com/8/8/7/88741f33c2a91f6055e2319d29304eb21da7ac30.png)
Therefore,![\[
\angle ABC = \angle ADC = 40^\circ.
\]](http://latex.artofproblemsolving.com/b/6/3/b630ae628445ecb7c5d7e1d55829077653938c19.png)
,
is isosceles with ![\[
\angle ABC = \angle BAC.
\]](http://latex.artofproblemsolving.com/1/c/e/1ce9d3657c30d242d0b7284a56e073da97d27f6d.png)
Let 
.![\[
\angle ACB = 180^\circ - 2x.
\]](http://latex.artofproblemsolving.com/a/5/4/a546a645c9a99d23f35c6f991414e938538c6012.png)
,
,![\[
\angle ADC = \frac{1}{2} \cdot \text{arc } AC = 40^\circ \Rightarrow \text{arc } AC = 80^\circ.
\]](http://latex.artofproblemsolving.com/d/c/a/dca6a82d5b783d038d72eb9b24c7c0faa76a9f0c.png)
,![\[
\angle ABC = \frac{1}{2} \cdot 80^\circ = \boxed{40^\circ}.
\]](http://latex.artofproblemsolving.com/0/8/5/085866d1587eabfb35d0181eb06a7958d8489327.png)
![\[
\frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta}
\]](http://latex.artofproblemsolving.com/e/6/1/e618ab30bf3fcdb952f92d2c0bdb7d6ff8593391.png)
![\[
\sin(2\theta) = 2\sin\theta\cos\theta \Rightarrow \sin^2(2\theta) = 4\sin^2\theta\cos^2\theta
\Rightarrow \sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2(2\theta)
\]](http://latex.artofproblemsolving.com/b/2/d/b2d779116d132d79446c983e8f2473583f518901.png)
![\[
\frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} = \frac{4}{\sin^2(2\theta)}
\]](http://latex.artofproblemsolving.com/1/c/4/1c4fd8393d35c1125b8eebdefec338168c012eed.png)
,![\[
\sin(2\theta) = \frac{\tan(2\theta)}{1 + \tan^2(2\theta)} = \frac{4}{\sqrt{17}}
\]](http://latex.artofproblemsolving.com/a/c/6/ac63c4b53e60424e0df8c3026df1ca0c1febbcda.png)
![\[
\frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = \frac{1}{[\frac{1}{4}][\sin^2(2\theta)]}=\frac{4}{(\frac{4}{\sqrt{17}})^2} = \boxed{\frac{17}{4}}
\]](http://latex.artofproblemsolving.com/5/d/b/5db53b8e0befb44ed07e17933d3fa854a44c7ab9.png)
Expanding gives
Regroup terms: 
Since 
,
,
this becomes
so 
Thus 
or 
To prove 
then 
so 
and 
which implies 
contradicting the first equation. Hence, 
and 
.
elements such that the sum would be divisible by 
(mod 5) and 
(mod 5) are compatible with each other in terms of the constraints. In fact, this is the only residue duo that satisfies the constraints given for A. Thus, it maximizes the amount of elements.
elements in 