blog bump
by MathStudent2002, Aug 7, 2018, 3:18 AM
blog bump 
Problem Set #3
by Tommy2000, Dec 23, 2016, 2:38 AM
This is overdue, but I have been rather busy recently... Let's start with some jokes to alleviate the mood.
-1) [minimario] Prove that
.
-2) [Classic] Prove that for
, ![$\sqrt[n]{2} \not\in \mathbb Q$](//latex.artofproblemsolving.com/e/e/6/ee6f13c79d9b7b87a55c6afcb090f7c5cb950369.png)
.
Ok now some real stuff:
1) USACO Dec 16 #1-2
2) [EGMO] Let
be a triangle, and 
be such that 
is a 
triangle with obtuse angle at and 
. Furthermore, let 
be the midpoint of 
and 
such that 
. Prove that 
.
I'll add more as I remember them (I kind of forgot to record stuff)
-1) [minimario] Prove that
.
.
to obtain the result.-2) [Classic] Prove that for
, ![$\sqrt[n]{2} \not\in \mathbb Q$](http://latex.artofproblemsolving.com/e/e/6/ee6f13c79d9b7b87a55c6afcb090f7c5cb950369.png)
.
,
,Ok now some real stuff:
1) USACO Dec 16 #1-2
2) [EGMO] Let
be a triangle, and 
be such that 
is a 
triangle with obtuse angle at and 
. Furthermore, let 
be the midpoint of 
and 
such that 
. Prove that 
.
be the other trisection point on 
.
is cyclic, so it follows that 
.
,
.
,
.
is a midline of 
,I'll add more as I remember them (I kind of forgot to record stuff)
This post has been edited 1 time. Last edited by Tommy2000, Dec 23, 2016, 2:42 AM
Reason: Fun :D
Reason: Fun :D
Problem Set #2
by Tommy2000, Dec 7, 2016, 2:51 AM
1) USACO 2016 Jan Plat #1
2) CF 587C
3) Find all
such that 
.
4) Let
be cyclic with 
. Let 
. Let 
be the midpoints of 
, respectively. The bisector of 
intersects 
at 
and the bisector of 
intersects 
at 
. Prove 
.
does in fact work.
solution is as follows: Consider the starting and ending row for the rectangle. They key observation is given a valid rectangle, you would never move the right edge leftwards. Now, keep moving the right most boundary until you can't, then skip a block until you can do a 
rectangle, and repeat (Line Sweep).2) CF 587C
,
people up to its the 
-
.3) Find all
such that 
.
to get 
.
.
and 
,![\[ 2b + f(f(y) - b) = f(f(b) + y) = f(f(a) + y) = 2a + f(f(y) - a) \]](http://latex.artofproblemsolving.com/2/f/b/2fb306913feaaa21ea00fcd992f1d3e83e38db9a.png)
To show 
,
for some 
.
,
,
,
.
,
.4) Let
be cyclic with 
. Let 
. Let 
be the midpoints of 
, respectively. The bisector of 
intersects 
at 
and the bisector of 
intersects 
at 
. Prove 
.
.![\[ \frac{BP}{DP} = \frac{BC}{DC} \cdot \frac{\sin BCA}{\sin DCA} = \frac{BC}{DC} \cdot \frac{\sin BDA}{\sin DBA} = \frac{BC}{DC} \cdot \frac{BA}{DA} = 1\]](http://latex.artofproblemsolving.com/8/e/d/8ed205c78efaeaef664a750725d496c8f218b124.png)
so 
is the midpoint as desired.
in complex numbers. Since 
and 
lie on 
concur.
.This post has been edited 1 time. Last edited by Tommy2000, Dec 7, 2016, 5:01 AM
Reason: Added #3
Reason: Added #3
Problem Set #1
by Tommy2000, Nov 29, 2016, 5:29 AM
Hmm, I guess I'll record some problems I recently did/had explained to me word for word '-' I'll post sols later if I have time...
1) Show that if
is prime for 
, then 
are primitive roots.
2) Prove that for
,
![\[ \sum_{m = 1} ^ p \left( \sum_{n = 1} ^ h \left( \frac{m + n}{p} \right) \right) ^ 2 = h (p - h) \]](//latex.artofproblemsolving.com/4/f/e/4fe29ff7873efd8744184b1018d3c742090f6c81.png)
3) USAMO 2005/3
This problem isn't that fun...so I'm too lazy to post a solution.
4) WOOT 11/25 PoTD
5) USACO Dec 15 Plat #1-3
6) 2016 PUMaC Finals #1-3
7) Find all
satisfying 
and 
.
8) (2016 Turkey TST #5) Find all functions
such that 
and 
.
It's also worth noting this nice result in fractals' solution. It was something I tried to prove, but ultimately could not.
1) Show that if
is prime for 
, then 
are primitive roots.
is not a primitive root. Then 
.
.
,
is not a primitive root. Then 
.
.
,2) Prove that for
,![\[ \sum_{m = 1} ^ p \left( \sum_{n = 1} ^ h \left( \frac{m + n}{p} \right) \right) ^ 2 = h (p - h) \]](http://latex.artofproblemsolving.com/4/f/e/4fe29ff7873efd8744184b1018d3c742090f6c81.png)
.![\[ h \sum_{i = 1} ^ p \left( \frac{i}{p} \right) ^ 2 + 2 \sum_{k = 1} ^ h (h - k) S_k \]](http://latex.artofproblemsolving.com/c/4/4/c44f103ebd911d969aa45349cead8d1d1a482846.png)
In the first part, all are equal to 
,
.
to the sum.
for all 
.![\[ S_k = \sum_{i = 1} ^ p \left( \frac{i ^ 2 + ik}{p} \right) = \sum_{i = 1} ^ p \left( \frac{(i + k/2) ^ 2 - (k / 2) ^ 2}{p} \right) \]](http://latex.artofproblemsolving.com/9/4/9/9492612a13b46fe382d0fe4ed1e11a83eb7f01d3.png)
It suffices to find when 
for some 
.
(it's nonzero) to obtain an equation of the form 
where 
in this are bijective with the 
in the previous equation. We can write 
and 
for some 
.
or else 
.
yield the same pair, we see there are 
values of 
which yield 
values of 
values of 
.
as desired.
as desired.3) USAMO 2005/3
This problem isn't that fun...so I'm too lazy to post a solution.
4) WOOT 11/25 PoTD
.
and 
so 
is cyclic, so 
lies on the perpendicular bisector of 
.
,
,
,
.
.5) USACO Dec 15 Plat #1-3
6) 2016 PUMaC Finals #1-3
is inside the quadrilateral, we have 
iff
Hence, the two angles are equal.7) Find all
satisfying 
and 
.
is odd. Suppose 
,
and 
.
,
is either 
,
and this contradicts 
.
.
such that 
.
.
is negative. Since 
,
,
,![$f(c) \in [0, 1]$](http://latex.artofproblemsolving.com/d/6/9/d69c715e5e86ee1cd3f5364e1fd7ea53d1237f83.png)
.![$f(k) \in [\lfloor k \rfloor, \lceil k \rceil]$](http://latex.artofproblemsolving.com/0/3/2/032622cb8582feb4cfa8533d59ac1b49bf94b55e.png)
.
is increasing. Suppose 
and 
.
.
.
for integers 
are dense in reals, so we're done.8) (2016 Turkey TST #5) Find all functions
such that 
and 
.
by plugging 
in (2). Furthermore, it's clear 
for all 
and 
by (1).
for 
for prime 
such that 
.
.![\[ 1 + pm = qn \mid f(1) + f(pm) = 1 + f(pm) \]](http://latex.artofproblemsolving.com/5/8/9/589fe59975b6800a310e14f0e31623f7282fb017.png)
However, we know 
,
for 
with 
and 
.
.
.
,![\[ p + q ^ x \mid p ^ a + \left( q ^ x \right) ^ b \]](http://latex.artofproblemsolving.com/f/3/f/f3feb186f8518dcdcb19ea05c264ac05aef11d31.png)
Applying the euclidean algorithm, we see![\[ p + q ^ x \mid p ^ {a - b} - 1 \]](http://latex.artofproblemsolving.com/1/6/5/165b8622fabc0a1eb15f5f096ae407c306365631.png)
However, 
is arbitrary, so this is a contradiction for sufficiently large It's also worth noting this nice result in fractals' solution. It was something I tried to prove, but ultimately could not.
This post has been edited 15 times. Last edited by Tommy2000, Dec 7, 2016, 2:51 AM
Reason: Subject name
Reason: Subject name
CSS Attempt
by Tommy2000, Aug 28, 2015, 4:04 PM
Oops, I kinda wanted to change the color of stuff, but I'm not too good at CSS yet 
Oh well, let me know if you like it or if it's too bright (which I expect it is, but idk)
BTW: In case you couldn't tell, I wasn't joking about pink being my favorite color
EDIT: I changed the title to "Blazer". However, this is NOT something related to 420, my school mascot is actually the "Blazer"
Oh well, let me know if you like it or if it's too bright (which I expect it is, but idk)
BTW: In case you couldn't tell, I wasn't joking about pink being my favorite color
EDIT: I changed the title to "Blazer". However, this is NOT something related to 420, my school mascot is actually the "Blazer"
This post has been edited 2 times. Last edited by Tommy2000, Aug 28, 2015, 4:07 PM
27. 2014 ISL C2
by Tommy2000, Aug 25, 2015, 5:28 PM
We have 
sheets of paper, with the number 
written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are 
and 
, then we erase these numbers and write the number 
on both sheets. Prove that after 
steps, the sum of the numbers on all the sheets is at least 
.
sheets of paper, with the number 
written on each of them. We perform the following operation. In every step we choose two distinct sheets; if the numbers on the two sheets are 
and 
, then we erase these numbers and write the number 
on both sheets. Prove that after 
steps, the sum of the numbers on all the sheets is at least 
.
times since 
.
steps, the product is at least 
.![$\sum \ge 2^m \sqrt[2^m]{\prod}=2^m*\sqrt[2^m]{2^{m*2^m}}=4^m$](http://latex.artofproblemsolving.com/d/2/8/d28401e173136828207612f89eb1bcf3a55defea.png)
.This post has been edited 1 time. Last edited by Tommy2000, Aug 25, 2015, 5:28 PM
Reason: typo
Reason: typo
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