such that if any 
elements are removed from the set 
,
be a triangle with incenter 
,
and circumcircle 
.
be the midpoints of sides 
,
,
and let 
be the tangency points of 
be the intersections of line 
with line 
and line 
,
be the midpoint of arc 
of 
.
bisects 
.
be the set of integers, and let 
be a function. Prove that there are infinitely many integers 
such that the function 
defined by 
is not bijective.
,
such that 
.
![\[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}?\]](http://latex.artofproblemsolving.com/2/d/2/2d248ce9756f0f09265dd538de3b16b22a944c46.png)
on the board with 
,
.
and 
Distinction: 100.5 
*Ties are (unofficially) broken by the soonest submission time.
[rule]
*Ties are (unofficially) broken by the soonest submission time.
Y by Robomania_534, sargamsujit, rayfish, Jndd, tennisrules, mainstain, ATGY, brainfertilzer, r00tsOfUnity, bjump, skyguy88, BraindeadAIMErs, EpicBird08, rohan.sp, Bolt_386, selenium_e, justJen, Bradygho, megarnie, huashiliao2020, Rounak_iitr
Howdy,
Winter is not a Season it is a Celebration! You know what that means... it’s that time of year to celebrate how Or(z) you are. We,
Arrowhead575 , Asdf334, ezpotd, hurdler, rayfish, sdash314, Serena_Xu, squareman, superagh, Taco12, and vsamc ,
would like to announce
Origin of Z Mathematics Series
Origin of Z will feature the Or(z)MC 10/12 (already concluded), Or(z)IME, and a team round with a level similar to that of the
USAJMO.
Feel free to sign-up with to stay up-to-date and to bump the thread.
This isn’t required to take the test nor does it force you to take the test. But it’s a great way to show support for us and bump the thread to the top of the forums, so we do appreciate it!
Prizes:
The top 5 individual contestants for the AMC 10, the top 3 individual contestants for the AMC 12, and the top 6 individual contestants from the AIME (who did not place top 5 in the AMC10/12) will receive a one-year subscription to Mathematica (valued at $88). From basic arithmetic to integral calculus, Mathematica covers a broad range of mathematics for high school and beyond. With over 6,000 built-in functions covering all areas of technical computing, Mathematica builds unprecedentedly powerful algorithms across all areas- including neural networks, machine learning, geometry, data science, and much more.
The 4 individuals on the
place team will each receive AoPS coupons valued at $25 each. (I am sure we all know what AoPS is.)Many more prizes to come!
Submitting: Feel free to take whenever you want, though please time yourself for 3 hours. There will be a google form including signups and submitting. If you have any issues, please PM Arrowhead575 and Rayfish (though you must PM us when you are starting, and when you are ending).
Additionally, a huge shout-out to P_Groudon, greenturtle, Natmath, bluelinfish, mahalermath, mathnerd101, Bobby R, and Jacob D. for assisting in the problem-writing and test-solving process.
Or(z)10/12:
Release Date: 10/1/2022
Deadline: 10/31/2022
Test Booklets:
Or(z)10
Or(z)12
Solutions Booklets:
Or(z)10 Solutions
Or(z)12 Solutions
Leaderboards
Or(z)IME:
Release Date: 1/11/2023
Test Deadline: 2/6/2023
Long Test Booklet:
Or(z)IME
Pamphlet Test Booklet
Or(z)IME Pamphlet
Submission Form:
Submission Form
Link to Private Discussion Forum
Discussion Forum
Or(z)(J)MO:
Release Date: TBD
If you have any additional questions, I'm sure they can be answered on our website,
https://originofz.wixsite.com/theorz , and feel free to ask in this thread as well.
Enjoy the Test!
Origin of Z team
This post has been edited 14 times. Last edited by Arrowhead575, Feb 3, 2023, 4:16 PM
-
.
.
,
representing the two-digit value for the minute hand, then find 
.
positive integers on a sheet of paper. He then takes every pair of positive integers on the paper and writes their greatest common divisor on a blackboard. Let 
be the number of positive integers on the blackboard that are divisible by 
.
be the sum of all possible values of 
.
be on ray 
past 
,
past 
be on ray 
past 
,
be on ray 
past 
,
,
,
,
.
![[asy]
size(10cm);
real x, y;
real[] hist;
x = 0.7;
y = 0.1;
hist[1] = 71;
hist[2] = 66;
hist[3] = 70;
hist[4] = 44;
hist[5] = 74;
hist[6] = 46;
hist[7] = 55;
hist[8] = 33;
hist[9] = 43;
hist[10] = 29;
hist[11] = 16;
hist[12] = 23;
hist[13] = 28;
hist[14] = 2;
hist[15] = 5;
draw((x,0)--(16*x,0));
for(int i = 1; i <= 15; ++i) {
draw((i*x,0)--(i*x,hist[i]*y)--((i + 1)*x,hist[i]*y)--((i + 1)*x,0));
label(string(i),((i + 0.5)*x,0),S,fontsize(10));
label(string(hist[i]),((i + 0.5)*x,hist[i]*y),N,fontsize(8));
}
[/asy]](http://latex.artofproblemsolving.com/7/0/c/70cf5d8b6579c313d77f99b9c6ab43aa4de21b2e.png)
