AoPS Academy
be a cyclic quadrilateral. A circle centered at 
passes through 
and 
and meets lines 
and 
again at points 
and 
(distinct from 
). Let 
denote the orthocenter of triangle 
Prove that if lines 
are concurrent, then triangle 
and 
are similar.
such that 
, where 
are non-negative reals such that 
.
.
such that: 
be an odd prime number. Prove that there exist sequence such that: 
)=(
)
=1(mod p) then 
=2(mod p) with all odd prime p
such that: 
?
such that: ?
such that: ?
. The other part is your
such that: be an odd prime number. Prove that there exist sequence such that: 
. Choose 
where 
. We get that 
which is impossible if both 
are even and so one of is odd, let us say without loss of generality that 
is odd. Then choose 
and we see that 
which means that necessarily 
is always odd, now choose 
and see that 
which is a contradiction and hence our assumption is wrong as desired. is an odd prime, take 
such that 
. We see that 
which is a contradiction, implying that 
for all 
which is the desired contradiction.
such that: be an odd prime number. Prove that there exist sequence such that: 
.
is odd then as we need 
So 
must be even.
then 
must be odd. Now we need 
such that 
but observe if is even then 
and similarly if is odd then 
. Hence contradiction. is even then 
so 
is even and 
so is odd. But then 
Hence contradiction. be an odd prime number. Prove that there exist sequence such that: we can always have a sequence 
which indeed satisfy 
. 
's I hope my sequence works 
where 
is fixed and 
varies We have![\[ \text{gcd}(a_{2m}+(2t+1), a_{2t+1}+(2m) ) = 1\]](http://latex.artofproblemsolving.com/4/d/e/4defa4fe4fba22b7997b0cdf25ab82f1c026ec9c.png)
where we take 
![\[ (a_1, a_2, a_3, a_4, ... ) = (0,0,0,0, ... ) \]](http://latex.artofproblemsolving.com/7/5/0/75054a4dc8a22b74906743ae36bc6b59a98b5021.png)
![\[ (a_1,a_2,a_3,a_4, ... ) = (1,1,1,1, ... ) \]](http://latex.artofproblemsolving.com/5/c/2/5c2537793c4063e43cf125732eb5e04efa1f8d47.png)
and we have 
, contradiction.
and we have 
, contradiction.
exists.
with an even index , as otherwise, if there is another integer with an even index, say 
, then 
. Similarly, there can be at most one odd number 
with an odd index 
. Hence, for all the other numbers, the paritity of and 
are different. Among those, choose an even and any odd 
. This also immediatily implies that 
.
works for any odd prime . First note if does not divide 
, then can't divide both of 
and 
, and hence cannot divide their gcd. Now,
so we are done.
is odd.![\[\gcd(a_1 + 2k + 1, a_{2k + 1} + 1) = 1\]](http://latex.artofproblemsolving.com/c/b/c/cbc33daa897dc568fa6aaae39636a754929432f3.png)
So, 
is odd. Consider is odd. Then,![\[\gcd(a_2 + 3, a_3 + 2) = 1\]](http://latex.artofproblemsolving.com/2/8/d/28d6eff02cd0d08c8b0cf5a5e3ce35db3b6272a2.png)
is assumed odd, 
. Hence, 
and 
. Contradiction. So, is even.![\[\gcd(a_2 + 4, a_4 + 2) = 1\]](http://latex.artofproblemsolving.com/c/4/a/c4a43e374290b67db615a9974fe399562103f18b.png)
, so 
is odd 
is odd.![\[\gcd(a_4 + 5, a_5 + 4) = 1\]](http://latex.artofproblemsolving.com/8/2/1/8217bbb6e1b86c5a554bc3cf961303936d17abd1.png)
is even and so does 
. Check that, 
, contradiction! Now, we assumed to be odd. Now, we are left with the part to be even.![\[\gcd(a_1 + 2k, a_{2k} + 1) = 1\]](http://latex.artofproblemsolving.com/8/2/d/82dfe8fea3743c325ee45e85ba2ae3b01246eaa9.png)
must be odd. So, 
is even. Hence, is even. Let's assume is even too.![\[\gcd(a_3 + 1, a_1 + 3) = 1\]](http://latex.artofproblemsolving.com/8/b/f/8bfe1db16995502989be046bc5c7f7efc871b408.png)
Contradiction. So, is odd.![\[\gcd(a_3 + 2k + 1, a_{2k + 1} + 3) = 1\]](http://latex.artofproblemsolving.com/7/d/3/7d3847edcda41862b5d4a8abec64e37568218e82.png)
Hence, 
is even. Now, note that,![\[d \mid a_1 + 5 \text{ and } d \mid a_5 + 1\]](http://latex.artofproblemsolving.com/9/0/8/90852f05c0536ffb8f522751ad3f1bcee957688f.png)
![\[d \mid (a_5 - a_1) - 4\]](http://latex.artofproblemsolving.com/2/5/8/258fdcc469e711feec0ceefd2f82a4ea04ca02b6.png)
and both are even, hence 
, contradiction! 
Note that for the latter condition to hold, we must have at most 
value of for which 
is even. Therefore, 
From here, we can proceed with casework.
However, since is odd, 
is even, and hence 
must be odd, so 
is odd.
However, since 
is odd, 
is even, and hence 
must be odd, so 
is odd.
we cannot have both and be odd, since then there would be more than one value of for which is even, and hence 
a contradiction.
holds since is an odd prime.
and 
. Then, 
, so this means that at least one of 
and 
is odd. 
and this means that at least one of 
and 
is odd. Now, by comparing 
that is odd and even, we arrive at a contradiction as gcd has to be 2.
works where is a variable that makes positive.
such that: be an odd prime number. Prove that there exist sequence such that: as an odd number we will get a contradiction because we could pair all fo and we will see that all of 
will be odd so we could not pair two odd because we will get an 
of a multiple of 2. Therefore, is even, and also because of the previous temp, we get all even terms odd and every odd ters even and we will get a contradiction if we pair 1 even term and 1 odd term.
and 
. Then, 
and 
can't both be even, so one of and 
is clearly odd. Say it is . Then, for all odd integers 
, 
is even. Thus, 
cannot be even, which implies that 
is odd for all odd integers . But then, considering two odd integers 
and 
we have that 
and 
are both even, and thus there 
is greater than 1. Contradiction! Thus, there cannot exist such a sequence.
for all positive integers . In particular, here 
for all indices . Then, say there exists two indices and which violate the desired condition. Then, 
and 
. But, if![\[p \mid (p+1)-i+j \implies p \mid j-i +1 \text{ and } p \mid (p+1)-j+i \implies p\mid i-j+1\]](http://latex.artofproblemsolving.com/a/8/3/a83fceba5835669531d647dea141ea3c8c9c61b6.png)
summing implies that 
which is a clear contradiction. Thus, this sequence indeed satisfies the given condition and we are done. (This actually solves the extension in #5 as well).
, it follows that or is odd. WLOG is odd. Then 
so is odd if 
is odd. But 
and 
are both even, a contradiction. 
such that 
. Then 
is not divisible by since 
. 
: 
implies that either or is odd. WLOG let to be odd. Then: 
implies 
is odd for all integers 
. Notice that: 
implies 
to be even, contradiction. There doesnt exists such a sequence.
evidently works and we are done.)
thus: 
. Taking any sequence which satisfies these conditions would work and we are done.
: implies that either or is odd. WLOG let to be odd. Then: implies is odd for all integers . Notice that: implies to be even, contradiction. There doesnt exists such a sequence. evidently works and we are done.) thus: . Taking any sequence which satisfies these conditions would work and we are done.
for some odd . If is even, take even (any even), which gives that must be even. Note 
a contradiction since 
. If is odd, consider any other odd, . This gives 
odd which means even, and we finish like earlier.
, which finishes.
or 
is odd. Let it is 
is odd for every 
- contradiction
and 
conclude 
odd. Take 
to get 
odd. Take 
odd to get 
this works, since 
.
s.t. 
)
s.t. 
.
.
must be even, therefore for the 
of the two terms to be 1, 
must be odd 
is odd.
we see that 
is also odd.
.
and 
works.
gives that 
and finish.
.
such that 
will be enough.
and easily steer clear out of this problem 
Similarly, we get 
Since 
S
For 
Clearly this works.
for any 
then 
or 
for all 
Then, 
a contradiction. Hence, we assume 
for all 
a contradiction.
and assume FTSOC that 
Then, 
or 
