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.
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
be a scalene triangle, 
be the median through 
, and 
be the incircle. Let touch 
at point 
and segment 
meet for the second time at point 
. Let 
be the triangle formed by lines and and the tangent to at . Prove that the incircle of triangle is tangent to the circumcircle of triangle .
be real numbers such that 
Prove that
Where 
Where 
are are the lengths of the sides of a 
gon such that
then![$$(n-2)\left[\sum_{i=1}^{n}{\frac{a_i^2}{(1-a_i)^2}}-\frac n{(n-1)^2}\right]\geq(2n-1)\left(\sum_{i=1}^{n}{\frac{a_i}{1-a_i}}-\frac n{n-1}\right)^2.$$](http://latex.artofproblemsolving.com/9/3/e/93e5529518807b92cd9e0382f5fb07a293128b04.png)
be an acute triangle. Tangents on the circumscribed circle of triangle at points 
and 
intersect at point . Let 
and 
be a foot of the altitudes from onto 
and 
and let 
be the midpoint of . Prove: is the orthocenter of the triangle 
.
cuts 
in half.
be a positive integer and let 
be an irrational number. Show that for every integer 
, there exists a positive integer 
such that 
.
and be positive integers with 
. In a group of people, each one or always, what is the parity of people who speak to is a subset of size 
of the set of people. The answer can only
” or “
”. and is it possible to determine which people speak the truth and
be pairwise different positive real numbers such that![\[a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\]](http://latex.artofproblemsolving.com/8/a/e/8ae0bd1f55eb2fb0b7209b62739a9b98f462ee56.png)
is an integer for every 
Prove that 
. Then we get that: 
,so 
and hence 
, which is a contradiction by 
. Then we get that: 
With the same way as case 
we get that 
and hence 
, which is a contradiction by 
.
. Then we get that: 
, and hence 
, so equivalently we have that 
, which has only 
as a positive integer solution. So, in this case 
,which is the only solution of the given equation.
and 
.
.
.
but, 
which is contradiction.
. then, 
.
.
. obviously contradicition.
.
yields 
is even, so 
. Now if 
we have ![\[2x\geq \nu_{2}(2020^x+2^y)=\min\{2x, 2y_1\}=\nu_{2}(2024^z)=3z.\]](http://latex.artofproblemsolving.com/e/6/3/e631c0799f33d41fc92bc875e0e69e3f038c1ba8.png)
However, 
as 
. Therefore 
, and the equation becomes 
. Modulo 
implies 
is odd, at which point 
. Comparing the 
's of both sides gives 
, so 
, 
for some nonnegative integer 
. Clearly, 
leads to the solution 
. When 
we derive a contradiction from:![\[1>\frac{2020^{3t+1}}{2024^{2t+1}}=\frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)^t\geq \frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)>1.\]](http://latex.artofproblemsolving.com/c/b/2/cb24c7a9ef36eabb403e6fa4982babe2ac46b8b8.png)
, which obviously works. We now prove it is the only one.
.
to get 
. Clearly, we must have 
, so is even and 
is odd. Let 
, for a positive integer 
. Then, the given equation becomes 
. Note that 
is always odd. Since 
and 
are both even, it follows if 
, then either 
or 
, neither of which are odd. Thus, 
. Because 
, we must have 
, and our claim follows. is .
. Note that when , 
is divisible by all the prime factors of 
, which are 
, , and 
. If 
, by Zsigmondy, there exists some 
that divides 
, and hence it is impossible that 
. Thus, , and we are done. 
, which clearly works.
gives 
. Since 
, if divided either one of or , then we have that one of 
is 
, which is absurd. Hence 
.
are both not multiples of , so they cannot be equal to 
. If 
, then we would have 
, which is absurd, so 
, so 
. Now we have ![\[ 2020^x + 4^x = 2024^z \]](http://latex.artofproblemsolving.com/4/b/e/4be984ca8fd9e9a9078af51ca885348c8cc030f5.png)
If 
, then by Zsigmondy there exists a prime 
not dividing 
that divides 
, absurd. Hence 
must hold.
Working in modulo 
, we obtain:
is 
for all 
.
is 
is 
and 
.
, where 
have the same parity 
.
Working in modulo , we obtain:
is odd number 
is odd number.
.
by 
results:
.
is an even non-negative number and results: the last digit of 
is 
.
must be , hence 
and we obtain
.
in the last relation, results:
is solution of the equation 
.
. by 
results:
.
is odd number; 
and 
are even numbers, hence the last equation has no solutions.
..
, odd number (see 
).
, even number, hence the equation has no solutions.
, even number, hence the equation has no solutions.
.
, where 
(see ).
.
.
.
, contradiction since
(proved many times in the previous posts).
has the unique solution .
. Because of this, we know y must be even and z must be odd. and 
. Prime factorizing the original equation, 
is the smallest factor, then 
, impossible. Similarly, if 2x is the largest factor of 2, 
is impossible. This means 
and the factor of 2 on the LHS is 
since the v2 of 
is at 2. (since its 0 mod 2 and not 0 mod 4). So we have 
. since 
is an integer, set 
Plugging in and dividing by largest factor of 2 we get 
k=0 works, but any larger k dosent work since 
so the only solution is k=0, which represents 
even and 
odd
we get:
if even then: 
which is a contradiction so 
even we get that 
so our equation transforms into the following:
Claim: 
which is a contradiction because we found that 
Claim: 
on both sides and using Lifting the Exponent Lemma (LTE) we get:
Since 
we get that 
Let 
also 
Now by taking we get:
but this isn't true for 
so 
clearly 
so 
is the only solution 
.
is even
gives is odd
is same as 
nonsense
nonsense
again we get is odd
(for a non-negetive integer k)
for 
for 
, so is even and is odd. Now consider of both sides. The RHS is 
, so it's odd. The LHS is 
, and it's odd if and only if . Now write the LHS as 
, and taking mod 4 on the inside sum gives 
. Now we have the relationship 
. We can easily check that 
is a solution, and we see that for , the LHS is strictly greater than the RHS, so there are no solutions 
.
and 
.
that satisfy the equation
we get:
![$2020^x=2024^z-8^z=2016[...]$](http://latex.artofproblemsolving.com/3/8/e/38eaa7c52b664c59165d40a2823dc5b2d0416c86.png)
contradiction since 
but not 
or 
we get that 
and then from inelyalites esily get 
,
and we claim there are no more solutions. Our strategy relies on that the powers of 
is the minimal among the three, then we have a contradiction via mod 
.
and 
.
.
,
.
(as 
for 
and 
for some positive integer 
.
,
,
.
.
is even, but not divisible by 
.
,
.
,
,
and we claim there are no more solutions. Have in mind that 
and 
.
:
and hence, by computations up to 
or similar, 
for some positive integer 
.
(we used 
by Fermat's Little Theorem) and hence, by computations up to 
or similar, 
:
,
.
.
now imply that 
and 
and later by 
)