Congratulations Wolstenholme!
by jellymoop, Aug 11, 2016, 10:15 PM
Congratulations Wolstenholme! Good job for staying alive!! This is actually impressive considering your eating / sleeping habits. This summer I have been privileged to watch you reach new heights such as teaching Geo 3, publishing a book, roasting me to unbelievable levels, and turning into a massive weeb. You have also become a personal mentor of sorts, and I'd like to describe the positive impact you've had on my life in hopes that you understand how much I appreciate your presence in my life.
First off, I'd like to thank you for opening my eyes and heart to people who transcend the gender spectrum. I am so disappointed in myself for ever asking why you use "aura" as a pronoun. I cannot believe I ever doubted you when you claimed not to be a singular entity. But with your kind guidance and reminders that I am a 'privileged piece of cisgender scum,' you taught me to be empathetic. Now I am ready to encounter people with new age spiritual identities in college, because they can't be much worse than you.
Second, I'd like to thank you for teaching me to be patient. An ambitious Trans-Earth like you has a busy schedule, and I understand if you prioritize every living being underneath league of legends. Those minutes, hours, days spent outside your door, waiting for your league game to slow down so you could get up and let me in--- I really treasured that time to myself. I think those moments were when I realized I wanted to become Exactly Like You. And don't get me started on how fuzzy I felt when you would open the door and go running back to your desk because you let me in mid-game.
Third, I'd like to thank you for being vulnerable and human around me. Sometimes I worry that our generation is becoming too emotionless, but seeing the sadness in your eyes when you requested that I stop eating your Milanos really erased all worries for me. I have never felt so strongly moved before, and I realize I may not see such raw emotion again.
Finally, I'd like to thank you for showing me how to think critically. You have transformed my perspective on the world, as I will never high five another person again. I was a fool to assume it'd be safe to high five you. Also, your constant stream of roasts has helped me develop a better understanding of who I am today. Thank you, Wolstenholme. I look forward to living in the same city as you in a month!!!
I am sorry that there are only four points; you have taught me all I know. However, my brain capacity is limited at age 14. Everyone else should share what they have learned from the great and mighty Wolstenholme!!
First off, I'd like to thank you for opening my eyes and heart to people who transcend the gender spectrum. I am so disappointed in myself for ever asking why you use "aura" as a pronoun. I cannot believe I ever doubted you when you claimed not to be a singular entity. But with your kind guidance and reminders that I am a 'privileged piece of cisgender scum,' you taught me to be empathetic. Now I am ready to encounter people with new age spiritual identities in college, because they can't be much worse than you.
Second, I'd like to thank you for teaching me to be patient. An ambitious Trans-Earth like you has a busy schedule, and I understand if you prioritize every living being underneath league of legends. Those minutes, hours, days spent outside your door, waiting for your league game to slow down so you could get up and let me in--- I really treasured that time to myself. I think those moments were when I realized I wanted to become Exactly Like You. And don't get me started on how fuzzy I felt when you would open the door and go running back to your desk because you let me in mid-game.
Third, I'd like to thank you for being vulnerable and human around me. Sometimes I worry that our generation is becoming too emotionless, but seeing the sadness in your eyes when you requested that I stop eating your Milanos really erased all worries for me. I have never felt so strongly moved before, and I realize I may not see such raw emotion again.
Finally, I'd like to thank you for showing me how to think critically. You have transformed my perspective on the world, as I will never high five another person again. I was a fool to assume it'd be safe to high five you. Also, your constant stream of roasts has helped me develop a better understanding of who I am today. Thank you, Wolstenholme. I look forward to living in the same city as you in a month!!!
I am sorry that there are only four points; you have taught me all I know. However, my brain capacity is limited at age 14. Everyone else should share what they have learned from the great and mighty Wolstenholme!!
centered at the point 
.
and 
,
require the Law of Sines/Cosines which is just dumb. So, BOGTRO and I (with Chris Shao also aiding in the conception of this beautiful idea) present the following "clearly-intended" solution - prepare your bodies!
(we could have easily used 
instead of 
is sent to a half-space whose boundary lies 
above the 
-
is sent to the point 
where![\[ u = \frac{r^2x}{x^2 + y^2 + z^2} \: \: v = \frac{r^2y}{x^2 + y^2 + z^2} \: \: w = \frac{r^2z}{x^2 + y^2 + z^2} \]](http://latex.artofproblemsolving.com/7/1/d/71d826c5c024b80f3986498a7c8587f703f3a50b.png)
![\[ \frac{d(u, v, w)}{d(x, y, z)} = \frac{r^6}{(x^2 + y^2 + z^2)^6}\begin{vmatrix}y^2 + z^2 - x^2 && -2xy && -2xz\\-2yx && z^2 + x^2 - y^2 && -2yz\\-2xz && -2yz && x^2 + y^2 - z^2\end{vmatrix} = -\frac{r^6}{(x^2 + y^2 + z^2)^3} \]](http://latex.artofproblemsolving.com/9/1/2/912accce53ee6d3d6a3ffac6905973f2fda584f0.png)
so 
and also 
.
Let 
,
,
be integers for 
.
,
such that 
is odd for at least 
values of 
except for 
and count the number of times 
odd results in total and so by the pigeonhole principle one of your seven choices of 
works. Probabilistic method yay!
Prove that the expression ![\[ \dfrac {\text {gcd}(m, n)}{n} \dbinom {n}{m} \]](http://latex.artofproblemsolving.com/c/0/a/c0a6877cd49e904728b24e7743fe8021efd6d312.png)
is an integer for all pairs of integers 
.
such that 
so our expression is equal to![\[ \frac{am + bn}{n}\binom{n}{m} = b\binom{n}{m} + a\binom{n - 1}{m - 1} \]](http://latex.artofproblemsolving.com/0/6/d/06d84438ad75704bc64860571133424d4385b0f5.png)
Let 
be a continuous function such that 
for all 
.
for 
.
we have that 
and so since 
is continuous we must have 
.
.
be a number such that 
.
and iterating we have that if 
for any 
then 
.
for 
are clearly dense in 
(since we can make 
arbitrarily large) so since 
by continuity we have that ![$[-1, 1]$](http://latex.artofproblemsolving.com/5/c/3/5c3818b9565a33fd3aadba10026d32c5e3eea90f.png)
as desired.
Let 
be a finite set of positive integers. We define finite sets 
of positive integers as follows: the integer 
if and only if exactly one of 
or 
.
for which 
.
and let 
.
hence 
.
is a power of two by the Frobenius endomorphism we have 
which is clearly equivalent to the problem condition so since there are infinitely many powers of two we are done.
Let 
be a set of more than 
distinct points with coordinates of the form 
in 
.
consider the 
away from each other. Now we note that the expected value of 
so for some 
.
Let 
be a positive real number. What are the possible values of 
given that 
are positive numbers for which 
?![\[ 0 < \displaystyle\sum_{j=0}^{\infty} x_j^2 < \left(\displaystyle\sum_{j=0}^{\infty} x_j\right)^2 = A^2. \]](http://latex.artofproblemsolving.com/8/e/3/8e3729694c0748f6ae69708c19dd403aa7284bb6.png)
let 
for some 
.![\[ \displaystyle\sum_{j=0}^{\infty} x_j^2 = A^2\left(\frac{1 - r}{1 + r}\right)\]](http://latex.artofproblemsolving.com/6/5/1/651310b387cdccdff605ffcea21922b3d3761882.png)
takes on all values in the interval 
we have that the desired interval is precisely 
Prove that there exist infinitely many integers 
,
are each the sum of the squares of two integers. [Example: 
,
,
.
for some integer 
.
are clearly the sum of two squares, so it suffices to show that there exist infinitely many 
is the sum of two squares. I claim even more - there exist infinitely many 
for some positive integer 
and note that 
is a solution. Hence with standard Pell equation theory every pair 
where 
for positive integers 
is a solution and hence we have found infinitely many 
The octagon 
is inscribed in a circle with the vertices around the circumference in the given order. Given that the polygon 
is a square of area 
,
is a rectangle of area 
,
so if 
and 
.
)
and 
.
and 
and 
and 
.
with 
.
and 
and 
.![\[ [P_2P_3P_4] = [P_6P_7P_8] = \frac{\sqrt{2}}{2}\left(x - \sqrt{2}\right) \]](http://latex.artofproblemsolving.com/2/7/e/27ed8cfe15c413694755e20e00e592ded41c450f.png)
![\[ [P_4P_5P_6] = [P_8P_1P_2] = \sqrt{2}\left(x - \frac{\sqrt{2}}{2}\right) \]](http://latex.artofproblemsolving.com/2/e/4/2e4ae28f95f98415d5edc3771352dfdae52104c3.png)
![\[ 3\sqrt{2}x. \]](http://latex.artofproblemsolving.com/2/a/5/2a544355dbac8c7ba1d1faf6bc603bdf5c7d2e22.png)
(the radius of the circle) hence the maximum area of the octagon is 
.
Show that the improper integral ![\[ \lim_{B \rightarrow \infty} \displaystyle\int_{0}^{B} \sin (x) \sin (x^2) dx \]](http://latex.artofproblemsolving.com/8/3/5/83584525a5af2b8544d1c2fa084b81f41ebdf47d.png)
converges.
and 
and that 
converges.
Three distinct points with integer coordinates lie in the plane on a circle of radius 
.
.
and that the triangle doesn't contain the center of the circle (or else one side would subtend an arc greater or equal to one-third the circle's circumference and we would be trivially done). Let 
be the longest side of our triangle. It's clear then that the altitude to side 
so the area of the triangle if equal to or less than![\[ \frac{s}{2}\left(r - \sqrt{r^2 - \frac{s^2}{4}}\right) = \frac{s^3}{8\left(r + \sqrt{r^2 - \frac{s^2}{4}}\right)} \le \frac{s^3}{8r} \]](http://latex.artofproblemsolving.com/b/1/6/b16bd4ab9c9656f637b1a94703db6e9bd99cf5e7.png)
hence we have 
and thus ![$s \ge \sqrt[3]{4r} > \sqrt[3]{r}$](http://latex.artofproblemsolving.com/f/3/b/f3bff6c7d1601eb8788debdf92be2bfeeb160370.png)
as desired.
Let 
of integers such that 
and 
for all 
.
then either 
or 
.
so since 
and since the sequence of differences 
is also purely periodic each difference must be equal in absolute value to 
.
holds. Then it's clear by periodicity that 
must all be distinct and now it's easy to see that we must have 
as desired or else the previous statement will be contradicted.
August 2016
June 2016