AoPS Academy
with circumcenter 
and orthocenter 
, let 
be an arbitrary point on the circumcircle of triangle such that does not lie on line 
and that line 
is not parallel to line 
. Let 
be the point on the circumcircle of triangle such that 
is perpendicular to , and let 
be the point on line 
such that 
. Let 
and 
be the points on the circumcircle of triangle such that 
is a diameter, and 
and 
are parallel. Let 
be the midpoint of 
.
and 
are perpendicular.
and 
are perpendicular.
with 
, a tangent to the circumcircle of at point 
intersects the extension of 
at point 
. 
is the midpoint of 
, and 
intersects the circumcircle of at 
. Prove that 
.
![\[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}\]](http://latex.artofproblemsolving.com/4/d/2/4d21982ba412c0bb4a7985798e81bc5b01ac0e43.png)
given that 
, 
, 
, 
are nonnegative real numbers such that 
.
be the orthocenter of acute triangle , let be the foot of the altitude from 
to 
, and let be the reflection of across . Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
satisfy 
?
such that 
is a multiple of 
?
is written as the sum of not necessarily distinct two-digit numbers. What is the least number of two-digit numbers needed to write this sum?
where and are constants, 
is the time in minutes, 
is the length of the trail in miles, and 
is the altitude gain in feet. The model estimates that it will take 
minutes to hike to the top if a trail is 
miles long and ascends 
feet, as well as if a trail is 
miles long and ascends 
feet. How many minutes does the model estimate it will take to hike to the top if the trail is 
miles long and ascends 
feet?
distinct primes?
![$7^i \bmod 234: [1, 7, 49, 109, 61, 193, 181, 97, 211, 73, 43, 67], 1, ...$](http://latex.artofproblemsolving.com/1/8/b/18b20bd29ead625064263f360d5ce1eb216173db.png)
![$3^i \bmod 234: 1, 3, [9, 27, 81], 9, ...$](http://latex.artofproblemsolving.com/b/f/a/bfa5a68ab3df4e8c5a064774de62b405196ed2c6.png)
has the only solution 
.
for some small 
for some small b
or 
(so the solution won't be repeated in the periods)
or 
, e.g.,
if somebody would be so nice 
and explane this a little bit 
will not appear in the periods
such that ![\[2^x+3^y=5^z??\]](http://latex.artofproblemsolving.com/5/b/d/5bdbb5423d0805cb641f457c2bcdd21d9b5d4872.png)
and we want to show that there are no others.
. And surely 
doesn't work - look carefully at periods 
. and :![$2^i \bmod 8160: 1, 2, 4, 8, 16, [32, 64, 128, 256, 512, 1024, 2048, 4096], 32, \dots$](http://latex.artofproblemsolving.com/9/8/d/98d68b76e461496c87dcb7d8b2eaa1e87c162177.png)
![$3^i \bmod 8160: 1, [3, 9, 27, 81, 243, 729, 2187, 6561, 3363, 1929, 5787, 1041, 3123, 1209, 3627, 2721], 3, \dots$](http://latex.artofproblemsolving.com/c/1/4/c149c1c1edd54e116c5e6b76ed61762ee10f595c.png)
![$5^i \bmod 8160: 1, [5, 25, 125, 625, 3125, 7465, 4685, 7105, 2885, 6265, 6845, 1585, 7925, 6985, 2285, 3265], 5, \dots$](http://latex.artofproblemsolving.com/d/f/4/df48bd8526e68b029c9345a5c63bf645693305ba.png)
, how did you get this 
and in general case
the modulus should be divisible by either 
or 
or 
., thus, there is no residue 
in the period of the powers of 2 and no solution with 
can be obtained from there.
and :
for every 
.
![\[2^a\cdot3^b-5^c\cdot7^d=1\]](http://latex.artofproblemsolving.com/9/5/7/95729be47c89c59bdf0850e25b383899d168554b.png)
![$2^i \bmod 48: 1, 2, 4, 8, [16, 32], 16, \dots$](http://latex.artofproblemsolving.com/6/e/a/6ea9a35f6ca96ff9d893428829fa4aeb75075459.png)
![$3^i \bmod 48: 1, [3, 9, 27, 33], 3, \dots$](http://latex.artofproblemsolving.com/3/1/3/313b76c40446b8e822d10729dd6726735a6a5dc5.png)
![$5^i \bmod 48: [1, 5, 25, 29], 1, \dots$](http://latex.artofproblemsolving.com/8/3/8/838f077054e81261d3d2180de813113258631999.png)
![$7^i \bmod 48: [1, 7], 1, \dots$](http://latex.artofproblemsolving.com/7/f/f/7ff0effd270a2213628800356d6ac965271abf3e.png)
with each power in the corresponding period modulo 48. Therefore, to form a solution at least one of the powers 
must be out of the period modulo 48.
has a finite number of solutions it's enough to prove the same for the following equations:
(this one obviously has no solution)
with each power in the corresponding period modulo 48. Therefore, to form a solution at least one of the powers must be out of the period modulo 48. has a finite number of solutions it's enough to prove the same for the following equations: (this one obviously has no solution)
.
we see that 
and 
must be of same parity.
.) and are odd, thus 
and 
.
.
.
so 
!
thus 
so 
thus 
, but 
, contradiction with , thus 
, which yields 
or .
prove that![$\frac{a}{b} + \sqrt {\frac{b}{c}} + \sqrt[3]{{\frac{c}{a}}} > \frac{5}{2}$](http://latex.artofproblemsolving.com/6/a/6/6a657cb647e279b0745e3189c3f9d54aa25bf976.png)
just because the solution given in a book is totally wrong and since it does not seem to follow without a little work. Am I wrong?
as you can see you have one mistake so... 
has only a finite number of integer solutions.
just because the solution given in a book is totally wrong and since it does not seem to follow without a little work. Am I wrong?