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.
denote the set of all real numbers. Find all functions 
such that ![\[ f\left( x^{2}+f(y)\right) =y+\left( f(x)\right) ^{2}\hspace{0.2in}\text{for all}\,x,y\in \mathbb{R}. \]](http://latex.artofproblemsolving.com/8/5/c/85c37ee472f918eace12c8cd7bef6766e617d2aa.png)
(where 
are the digits and 
) we define the \textit{permutation sum} as the value 
For example the permutation sum of 20253 is 
Let 
and 
be two fivedigit integers with the same permutation sum.
.
pairwise distinct positive integer numbers 
, find the maximum possible number of equal numbers among the fractions of the form![\[
\frac{a_i^2 + a_j^2}{a_i + a_j}
\]](http://latex.artofproblemsolving.com/3/5/3/3537ddec6d793d3831ba82a1e13b98d43b1c29bc.png)
be the incircle of triangle 
. Line 
is parallel to side 
and tangent to . Line 
is parallel to side 
and tangent to . It turned out that the intersection point of and lies on circumcircle of 
, we have:
satisfy all conditions below:
for all 
for all 
for all
that are solutions of the equation 
.
![\[
2^a + 3^b + 1 = 6^c
\]](http://latex.artofproblemsolving.com/1/9/e/19eb5efa2ccdf54cd277339f8d3b99f75168779d.png)
are such that 
, find the minimum value of expression 
of prime positive integers 
such that number 
equals to a square of an integer.
be real number greater than 
. Let![\[S=\log_a {bc}+\log_b {ca}+\log_c {ab}\]](http://latex.artofproblemsolving.com/5/9/e/59e1a33017116963165f744b49b8515f126c64ef.png)
.
for which:
for all 
.
intersect at 
. Since 
, 
is cyclic, so 
(well-known). Thus, 
is cyclic, so 
, so since the circumcenter 
lies on the perpendicular bisector of , we must have 
.
meet 
at 
. Then 
are concyclic, therefore , the Simson-line of 
w.r.t. the triangle 
, passes through the midpoint of .
, and 
. Then 
, 
.
, , 
collinear, 
for 
and since 
, ![\[ \frac{1}{k-a}+ \frac{1}{k-\frac{1}{a}}=0 \implies p=\frac{1}{2} (a+\frac{1}{a}) \]](http://latex.artofproblemsolving.com/4/9/c/49ccf30703a265f7ea09ce00056f7ef4592fc89f.png)
Consider the transformation 
. Then it suffices to show 
, 
, 
are collinear. ![\[ \frac{\frac{b}{a}-1}{b-\frac{1}{a}}=\frac{\frac{a}{b}-1}{a-\frac{1}{b}} \implies \frac{b-a}{ab-1}=\frac{a-b}{1-ab} \]](http://latex.artofproblemsolving.com/2/3/1/2317f9cbc20ffda5fba86a202ddb5a1ebf6252f1.png)
which is true.
be unit circle then 
, 
, 
and 
, so we need 
colinear but by the composed shifting 
and colinearity lemma all we need is 
.
thus bisects as desired, and done 
.
and 
,
is right-angled at 
.
all lie on the circle with diameter ![\[ \angle BHP = \angle CAP = \angle B = \angle HBA \quad \quad \text{(1)} \]](http://latex.artofproblemsolving.com/8/3/0/830a1ba5da967f7e79eb2354bdebd3d7b0a4ced6.png)
Now let 
,
shows that 
.
also implies that ![\[ \angle HAB = \frac{\pi}{2} - \angle HBA = \frac{\pi}{2} - \angle BHP = \angle PHA = \angle MHA \]](http://latex.artofproblemsolving.com/c/d/4/cd45bb734fcf01545cf54cfaeca16ce43d4e1368.png)
This implies that 
.
as required.
is a cyclic quad. Angle chasing yields 
.
Now, we proceed with complex numbers. Let the circumcircle of 
be the unit circle and let the complex numbers 
and 
be complex numbers representing 
and 
Then, by the Complex Foot formula, we get 
Also, note that the midpoint of 
and 
We wish to show that 
and 
The latter can be verified by expanding algebraically, so we are done. 
.
which is equivalent to
It is obviously true.