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.
the point 
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
at 
, and to the lines 
and 
at 
and 
, respectively. The lines 
and 
meet at 
, and the lines 
and 
meet at 
Let 
be the point of intersection of the lines 
and , and let 
be the point of intersection of the lines 
and 
Prove that is the midpoint of 
opposite the vertex 
is the circle that is tangent to the line segment , to the ray beyond 
, and to the ray beyond 
.)
such that if 
and 
are distinct rational numbers satisfying 
or 
, then 
? Justify your answer.
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
, solve the given equation in 
., the equation has at most 
real solutions.
and 
intersect at two points 
and 
. Let 
be the line tangent to these circles at 
and 
, respectively, so that lies closer to than . Let 
be the line parallel to and passing through the point , with 
on and 
on . Lines 
and 
meet at 
; lines 
and meet at 
; lines 
and meet at 
. Show that 
.
is such that 
for any positive integers 
. Assume there exists a positive integer 
such that 
for all positive integers 
. Determine all possible values of 
.
satisfy that 
. Prove that: 
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of are either all odd or all even.
be real numbers . Prove that : There exist positive integer 
such that 
Where 
be an isosceles triangle (
) with incenter 
. Circle 
passes through and and is tangent to 
. intersects and circumcircle of at 
and 
, respectively. Let be the midpoint of and 
be the midpoint of 
. Prove that 
, 
and are concurrent.
. 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 
)