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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
1 viewing
jlacosta
Mar 2, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
questions from a first-time applicant to math camps
akliu   23
N 4 minutes ago by John_Mgr
hey!! im a first time applicant for a lot of math camps (namely: usa-canada mathcamp, PROMYS, Ross, MathILY, HCSSiM), and I was just wondering:

1. how much of an effect would being a first-time applicant have on making these math camps individually?
2. I spent a huge amount of effort (like 50 or something hours) on the USA-Canada Mathcamp application quiz in particular, but I'm pretty worried because supposedly almost no first-time applicants get into the camp. Are there any first-time applicants that you know of, and what did their applications (as in, qualifying quiz solutions) look like?
3. Additionally, a lot of people give off the impression that not doing the full problem set will screw your application over, except in rare cases. How much do you think a fakesolve would impact my PROMYS application chances?

thanks in advance!
23 replies
akliu
Mar 12, 2025
John_Mgr
4 minutes ago
combo j3 :blobheart:
rhydon516   14
N 5 minutes ago by aliz
Source: USAJMO 2025/3
Let $m$ and $n$ be positive integers, and let $\mathcal R$ be a $2m\times 2n$ grid of unit squares.

A domino is a $1\times2$ or $2\times1$ rectangle. A subset $S$ of grid squares in $\mathcal R$ is domino-tileable if dominoes can be placed to cover every square of $S$ exactly once with no domino extending outside of $S$. Note: The empty set is domino tileable.

An up-right path is a path from the lower-left corner of $\mathcal R$ to the upper-right corner of $\mathcal R$ formed by exactly $2m+2n$ edges of the grid squares.

Determine, with proof, in terms of $m$ and $n$, the number of up-right paths that divide $\mathcal R$ into two domino-tileable subsets.
14 replies
rhydon516
Yesterday at 12:08 PM
aliz
5 minutes ago
high tech FE as J1?!
imagien_bad   48
N 13 minutes ago by Super_AA
Source: USAJMO 2025/1
Let $\mathbb Z$ be the set of integers, and let $f\colon \mathbb Z \to \mathbb Z$ be a function. Prove that there are infinitely many integers $c$ such that the function $g\colon \mathbb Z \to \mathbb Z$ defined by $g(x) = f(x) + cx$ is not bijective.
Note: A function $g\colon \mathbb Z \to \mathbb Z$ is bijective if for every integer $b$, there exists exactly one integer $a$ such that $g(a) = b$.
48 replies
imagien_bad
Yesterday at 12:00 PM
Super_AA
13 minutes ago
Anyone LFT for SMT?
Mathdreams   0
27 minutes ago
Hi everyone,

Is there anyone willing to join an SMT (Stanford Math Tournament) team?

I have a team looking for one more person.

Edit: If you are interested, please PM me, and I'll answer any questions there :)
0 replies
Mathdreams
27 minutes ago
0 replies
USA 97 [1/(b^3+c^3+abc) + ... >= 1/(abc)]
Maverick   45
N 2 hours ago by Marcus_Zhang
Source: USAMO 1997/5; also: ineq E2.37 in Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu
Prove that, for all positive real numbers $ a$, $ b$, $ c$, the inequality
\[ \frac {1}{a^3 + b^3 + abc} + \frac {1}{b^3 + c^3 + abc} + \frac {1}{c^3 + a^3 + abc} \leq \frac {1}{abc}
\]
holds.
45 replies
1 viewing
Maverick
Sep 12, 2003
Marcus_Zhang
2 hours ago
The prime inequality learning problem
orl   137
N 2 hours ago by Marcus_Zhang
Source: IMO 1995, Problem 2, Day 1, IMO Shortlist 1995, A1
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Prove that
\[ \frac {1}{a^{3}\left(b + c\right)} + \frac {1}{b^{3}\left(c + a\right)} + \frac {1}{c^{3}\left(a + b\right)}\geq \frac {3}{2}.
\]
137 replies
orl
Nov 9, 2005
Marcus_Zhang
2 hours ago
hard ............ (2)
Noname23   2
N 3 hours ago by mathprodigy2011
problem
2 replies
Noname23
Yesterday at 5:10 PM
mathprodigy2011
3 hours ago
Abelkonkurransen 2025 3a
Lil_flip38   5
N 3 hours ago by ariopro1387
Source: abelkonkurransen
Let \(ABC\) be a triangle. Let \(E,F\) be the feet of the altitudes from \(B,C\) respectively. Let \(P,Q\) be the projections of \(B,C\) onto line \(EF\). Show that \(PE=QF\).
5 replies
Lil_flip38
Yesterday at 11:14 AM
ariopro1387
3 hours ago
Inequality by Po-Ru Loh
v_Enhance   54
N 3 hours ago by Marcus_Zhang
Source: ELMO 2003 Problem 4
Let $x,y,z \ge 1$ be real numbers such that \[ \frac{1}{x^2-1} + \frac{1}{y^2-1} + \frac{1}{z^2-1} = 1. \] Prove that \[ \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \le 1. \]
54 replies
v_Enhance
Dec 29, 2012
Marcus_Zhang
3 hours ago
Problem 5
Functional_equation   14
N 3 hours ago by ali123456
Source: Azerbaijan third round 2020(JBMO Shortlist 2019 N6)
$a,b,c$ are non-negative integers.
Solve: $a!+5^b=7^c$

Proposed by Serbia
14 replies
Functional_equation
Jun 6, 2020
ali123456
3 hours ago
a^12+3^b=1788^c
falantrng   6
N 4 hours ago by ali123456
Source: Azerbaijan NMO 2024. Junior P3
Find all the natural numbers $a, b, c$ satisfying the following equation:
$$a^{12} + 3^b = 1788^c$$.
6 replies
falantrng
Jul 8, 2024
ali123456
4 hours ago
stuck on a system of recurrence sequence
Nonecludiangeofan   0
4 hours ago
Please guys help me solve this nasty problem that i've been stuck for the past month:
Let \( (a_n) \) and \( (b_n) \) be two sequences defined by:
\[
a_{n+1} = \frac{1 + a_n + a_n b_n}{b_n} \quad \text{and} \quad b_{n+1} = \frac{1 + b_n + a_n b_n}{a_n}
\]for all \( n \ge 0 \), with initial values \( a_0 = 1 \) and \( b_0 = 2 \).

Prove that:
\[
a_{2024} < 5.
\]
(btw am still not comfortable with system of recurrence sequences)
0 replies
Nonecludiangeofan
4 hours ago
0 replies
A huge group of children compare their heights
Tintarn   5
N 4 hours ago by InCtrl
Source: All-Russian MO 2024 9.8
$1000$ children, no two of the same height, lined up. Let us call a pair of different children $(a,b)$ good if between them there is no child whose height is greater than the height of one of $a$ and $b$, but less than the height of the other. What is the greatest number of good pairs that could be formed? (Here, $(a,b)$ and $(b,a)$ are considered the same pair.)
Proposed by I. Bogdanov
5 replies
Tintarn
Apr 22, 2024
InCtrl
4 hours ago
Iran Inequality
mathmatecS   15
N 5 hours ago by Marcus_Zhang
Source: Iran 1998
When $x(\ge1),$ $y(\ge1),$ $z(\ge1)$ satisfy $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2,$ prove in equality.
$$\sqrt{x+y+z}\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$
15 replies
mathmatecS
Jun 11, 2015
Marcus_Zhang
5 hours ago
Official Mock AMC C Solutions
JSRosen3   33
N Aug 5, 2004 by Pork_Chop8
Let's not post solutions until beta says it's OK. Maybe after he's posted the results.
33 replies
JSRosen3
Aug 3, 2004
Pork_Chop8
Aug 5, 2004
Official Mock AMC C Solutions
G H J
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JSRosen3
335 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let's not post solutions until beta says it's OK. Maybe after he's posted the results.
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joml88
6343 posts
#2 • 2 Y
Y by Adventure10, Mango247
1. Since the three values that are squared must be $\geq 0$ (since they are squared) the only way the whole thing can equal 0 is if each of those things squared is 0. So we have

\begin{eqnarray*}
a^2-1 &=& 0 \\
b^2-4 &=& 0\\
c^2-9 &=& 0
\end{eqnarray*}

So $a=\pm 1, b=\pm 2, and \ c=\pm 3$ So there are 2*2*2=8 ordered pairs. D

2. For this one note that \[1+3+5+\ldots+(2n-1)=n^2\] So we then have it equaling 21^2=441 C
This post has been edited 1 time. Last edited by joml88, Aug 3, 2004, 10:17 PM
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ThAzN1
867 posts
#3 • 2 Y
Y by Adventure10, Mango247
3. How many ordered pairs (x,y) satisfy x+4y=2004?

Note that y>0, and x>0 implies y<501. Thus y=1,2,...500 yield all of the solutions and there are 500 pairs


4. A charity sells 140 benefit tickets for a total of 2001 dollars. Some tickets sell for full price (a whole dollar amount),
and the rest for half price. How much money (in dollars) is raised by the full-price tickets?

Set up an equation... let x = number of full price tickets, d = full price of a ticket.

We have,

xd + (140-x)d/2 = 2001
2xd + 140d - xd = 4002
xd + 140d = 4002
d(x+140) = 4002

Since x+140|4002 and 140 < x+140 < 280, we factor 4002=2*3*23*29, try some numbers, and conclude that x=2*3*29=174, so x=34, d=23, xd=782.
This post has been edited 3 times. Last edited by ThAzN1, Aug 3, 2004, 11:09 PM
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dts
469 posts
#4 • 2 Y
Y by Adventure10, Mango247
Rep123max wrote:
23. Basically, we have to get $\cos5\theta$ in terms of $\cos\theta$ and $\sin\theta$.

By DeMoivere's, we have:
$\cos5\theta+i\sin\theta=(\cos\theta+i\sin\theta)^5$.

So basically, $\cos5\theta$ will be the real parts of $(\cos\theta+i\sin\theta)^5$.

When we expand it, the real parts are $\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$.

We know that $\cos\theta=\frac{1}{4}$ and by the Pythagorean Identity, $\sin\theta=\sqrt{\frac{15}{16}}$.

Subbing this into what we got above, it turns out to be $\frac{61}{64}$.

Because calculator use is permitted on the AMC, there is an easier way: evaluate Arccos(1/4), multiply by 5, and take the cosine of that :D. Not very elegant, but probably quicker.
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joml88
6343 posts
#5 • 2 Y
Y by Adventure10, Mango247
11. $\cot{10^\circ}+\tan{5^\circ}=$


Express cot and tan in terms of sin and cos:
\[\begin{eqnarray*}
&=& \frac{\cos{10}}{\sin{10}}+\frac{\sin{5}}{\cos{5}}\\
&=& \frac{\cos{10} \cos{5}+\sin{10}\sin{5}}{\sin{10}\cos{5}}\\
&=& \frac{\cos{5}}{\sin{10}\cos{5}}\\
&=& \frac{1}{\sin{10}}\\
&=& \csc{10}\\
\end{eqnarray}\]

E is the answer.

14. Using change of base we get:

\[\log_{100!}{2}+\log_{100!}{3}+\ldots+\log_{100!}{100}\]

which by the property of adding logs equals:

\[\log_{100!}{100!}=1\]

C

16. I liked this one.

Square both the equations we are given to get:

\[\sin^2{x}-\sin^2{y}-2\sin{x}\sin{y}=\frac{1}{9}\]

and

\[\cos^2{x}-\cos^2{y}-2\cos{x}\cos{y}=\frac{1}{4}\]

Adding them and using the fact that $\sin^2{x}+\cos^2{x}=1$ and the cosine of the sum of angles formula we get(and moving a few terms):

\[\cos{x-y}=\frac{59}{72}\]
A
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interesting_move
249 posts
#6 • 2 Y
Y by Adventure10, Mango247
Rep123max wrote:
23. Basically, we have to get $\cos5\theta$ in terms of $\cos\theta$ and $\sin\theta$.

By DeMoivere's, we have:
$\cos5\theta+i\sin\theta=(\cos\theta+i\sin\theta)^5$.

So basically, $\cos5\theta$ will be the real parts of $(\cos\theta+i\sin\theta)^5$.

When we expand it, the real parts are $\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$.

We know that $\cos\theta=\frac{1}{4}$ and by the Pythagorean Identity, $\sin\theta=\sqrt{\frac{15}{16}}$.

Subbing this into what we got above, it turns out to be $\frac{61}{64}$.

Why is $\cos5\theta+i\sin\theta=(\cos\theta+i\sin\theta)^5$.?

-interesting_move
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dts
469 posts
#7 • 2 Y
Y by Adventure10, Mango247
#10: Because $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}}{a-b}$, the expression becomes $\sqrt9-\sqrt8+\sqrt8-\sqrt7+\sqrt7-\sqrt6+\sqrt6-\sqrt5+\sqrt5-\sqrt4+\sqrt4-\sqrt3=3-\sqrt3$, and we are done.
This post has been edited 1 time. Last edited by dts, Aug 3, 2004, 11:13 PM
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ThAzN1
867 posts
#8 • 2 Y
Y by Adventure10, Mango247
Shouldn't it be
\cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5
?
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joml88
6343 posts
#9 • 2 Y
Y by Adventure10, Mango247
I fixed it. I was having a lot of trouble getting the eqnarray to work. I still have to get used to using that :)
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white_horse_king88
1779 posts
#10 • 2 Y
Y by Adventure10, Mango247
Number 7 - The Tricky One.

Realize that you form a 30 degree angle when you turn 150 degrees left, so you have a triangle with sides 3 and rt3, but that is it. The other side can be rt3, or it can be 2rt3 depending on whether or not the triangle is a 30-60-90 or a 30-120-30. So (E). Undetermined.
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white_horse_king88
1779 posts
#11 • 2 Y
Y by Adventure10, Mango247
Number 9) Find the unit digit of the sum
1 + 2^5 + 3^5 + ... 2005^5
.

Notice, that the continuation pattern for the units digit is at the most in groups of 4, so the 5th will always be the original digit. So, the last digit can be determined by 1+2+3+...2005 = 2005(2006)/2 = ... 5. The last digit is 5, or (C).
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beta
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#12 • 2 Y
Y by Adventure10, Mango247
15)
Let the tetrahedron be ABCD, with BC and AD opposite, and let their midpoints be M and N, respectively. Note that $\Delta ABC  \Delta BCD$ are equilateral triangles. Therefore length of $AM=\frac{\sqrt3}{2} =BD$. Point M must lie on the perpendicular bisector of AD, so $MN \perp AD$ , ANM is a right triangle, $AN=\frac{1}{2} $ . By Pythagorean Theorem, $MN=\frac{\sqrt2}{2}$
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white_horse_king88
1779 posts
#13 • 2 Y
Y by Adventure10, Mango247
Aren't you the Mock AMC C creator? Why are you posting the solutions - just wondering.

For Number 13) To find the number of digits in 5^89, you take log_5^89. You know that log_2 = 0.3010, and you know that log_2 + log_5 = log_10 = 1, so log_5 = 0.699. log_5^89 = 89(log_5) = 89(0.699). This is if you don't have a scientific calculator on you. You get 62.something, and since it is 10^62.some, you have 63 digits. If you have a scientific calc, you can just plug in the numbers.
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beta
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#14 • 2 Y
Y by Adventure10, Mango247
white_horse_king88 wrote:
Aren't you the Mock AMC C creator? Why are you posting the solutions - just wondering.

Why can't I post the solutions :maybe:
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Ultimatemathgeek
111 posts
#15 • 2 Y
Y by Adventure10, Mango247
Rep123max wrote:
23. Basically, we have to get $\cos5\theta$ in terms of $\cos\theta$ and $\sin\theta$.

By DeMoivere's, we have:
$\cos5\theta+i\sin5\theta=(\cos\theta+i\sin\theta)^5$.

So basically, $\cos5\theta$ will be the real parts of $(\cos\theta+i\sin\theta)^5$.

When we expand it, the real parts are $\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$.

We know that $\cos\theta=\frac{1}{4}$ and by the Pythagorean Identity, $\sin\theta=\sqrt{\frac{15}{16}}$.

Subbing this into what we got above, it turns out to be $\frac{61}{64}$.

Any short cut to expand and finding the real part?
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MysticTerminator
3697 posts
#16 • 2 Y
Y by Adventure10, Mango247
that IS the shortcut :D
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white_horse_king88
1779 posts
#17 • 2 Y
Y by Adventure10, Mango247
I don't think you really need a shortcut to expanding it unless you intend to expand it the long way. You derive the coefficients for isin :theta: to the 0, 2, 4th powers, when the outcome is real. It's not that difficult - a little binomial to assist the coefficients part, and out comes the final polinomial by De Moivre's theorem.[/tex]
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Ultimatemathgeek
111 posts
#18 • 2 Y
Y by Adventure10, Mango247
beta wrote:
15)
Let the tetrahedron be ABCD, with BC and AD opposite, and let their midpoints be M and N, respectively. Note that $\Delta ABC  \Delta BCD$ are equilateral triangles. Therefore length of $AM=\frac{\sqrt3}{2} =BD$. Point M must lie on the perpendicular bisector of AD, so $MN \perp AD$ , ANM is a right triangle, $AN=\frac{1}{2} $ . By Pythagorean Theorem, $MN=\frac{\sqrt2}{2}$

How do you know M must lie on the perpendicular bisector of AD?
It took 3 minutes for me to prove that. Is it some sort of theorem, or did you also have to figure it out?
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white_horse_king88
1779 posts
#19 • 2 Y
Y by Adventure10, Mango247
It's an old theorem that you learn in most Euclidean Geometry classes. That if N is the midpoint of AD, and AM = MD, then M must lie on the perpendicular bisector of AD. Also, all points on the perpendicular bisector of AD are equidistant from A and D. :)
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Ultimatemathgeek
111 posts
#20 • 2 Y
Y by Adventure10, Mango247
white_horse_king88 wrote:
It's an old theorem that you learn in most Euclidean Geometry classes. That if N is the midpoint of AD, and AM = MD, then M must lie on the perpendicular bisector of AD. Also, all points on the perpendicular bisector of AD are equidistant from A and D. :)

Oh, I get it. It is the isoceles triangle thing. The perpendicular bisector of the base contains the non-base corner. Right?
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h_s_potter2002
1306 posts
#21 • 2 Y
Y by Adventure10, Mango247
white_horse_king88 wrote:
For Number 13)you know that log_2 + log_5 = log_10 = 1

How do you figure this out?
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beta
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#22 • 2 Y
Y by Adventure10, Mango247
Anyone wants to do my favorite ones (20, 21, 24, 25)?
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white_horse_king88
1779 posts
#23 • 2 Y
Y by Adventure10, Mango247
You figure it out by loga_b + loga_c = loga_bc. You know by fact that log10_10 = 1.
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h_s_potter2002
1306 posts
#24 • 3 Y
Y by Adventure10, Adventure10, Mango247
Rep123max wrote:
#24. I wish I had really attempted it during the actual contest. I just started to look at simpler cases or \[\sum_{k=0}^n(-1)^k{n\choose k}k\]

For n=1, it is -1. And then for everyone after that, it is 0, so the answer is C. I'll try to prove it later.

What exactly does Sigma mean, in this case? I've seen it being used, but never really understood it.
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Pork_Chop8
179 posts
#25 • 2 Y
Y by Adventure10, Mango247
#24. I also looked at simpler cases on the test, which is a very effective way to cheat, but here is the way to do it.

$\displaystyle\sum_{k=0}^n(-1)^k{n\choose k}k=\displaystyle\sum_{k=1}^n(-1)^k{n\choose k}k+\displaystyle(-1)^{0}{n\choose 0}\cdot 0=\displaystyle\sum_{k=1}^n(-1)^k\displaystyle\frac{n!}{(n-k)!(k)!}k$
$=\displaystyle\sum_{k=1}^n(-1)^k\displaystyle\frac{n!}{(n-k)!(k-1)!}=\displaystyle\sum_{k=1}^n(-1)^k\displaystyle\frac{n\cdot (n-1)!}{(n-k)!(k-1)!}$
$=\displaystyle\sum_{k=1}^n(-1)^k {n-1\choose k-1}\cdot n=\left(\displaystyle\sum_{k=0}^{n-1}(-1)^{k+1}{n-1\choose k}\right)\cdot n=-(1-1)^{n-1}\cdot n = 0$(C)

Edit: By binomial theorem, $(1-1)^{n-1}$ becomes $(1)^{n-1}(-1)^{0}\displaystyle{n-1\choose 0}+(1)^{n-2}(-1)^{1}\displaystyle{n-1\choose 1}+(1)^{n-3}(-1)^{2}\displaystyle{n-1\choose 2}+...+(1)^{1}(-1)^{n-2}\displaystyle{n-1\choose n-2}+(1)^{0}(-1)^{n-1}\displaystyle{n-1\choose n-1}=\displaystyle\sum_{k=0}^{n-1}(-1)^{k}{n-1\choose k}=(-1)(-1)\displaystyle\sum_{k=0}^{n-1}(-1)^{k}{n-1\choose k}=(-1)\left(\displaystyle\sum_{k=0}^{n-1}(-1)^{k+1}{n-1\choose k}\right)$


#25, consult my non-scale drawing. I solved using coordinates :lol:. Always an ugly way to bash problems you are having trouble with. Anyways, $B = (0,0), C=(\sqrt{40},0), A=(\sqrt{10},\sqrt{40})$. To find D, note that BDC is a right triangle, so the altitude from D times BC times one half = BD times DC times one half, calculating areas in different ways, so we can conclude the y-coordinate of D $=\displaystyle\frac{12}{\sqrt{40}}=\displaystyle\frac{6}{\sqrt{10}}$. By pythagorean theorem, $(x-coord of D)^{2} + (\displaystyle\frac{6}{\sqrt{10}})^{2} = 2^{2} \Rightarrow x = \displaystyle\frac{2}{\sqrt{10}}$. So $D=(\displaystyle\frac{2}{\sqrt{10}},\displaystyle\frac{6}{\sqrt{10}})$ and the distance between A and D is $\sqrt{(\sqrt{10}-\displaystyle\frac{2}{\sqrt{10}})^{2}+(\sqrt{40}-\displaystyle\frac{6}{\sqrt{10}})^{2}}=\sqrt{26}$(B)
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beta
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#26 • 2 Y
Y by Adventure10, Mango247
A good trig solution to #25:
Draw OM, where M is the midpoint of AC. Since $MA=\sqrt{10}$, Pythagorean Theorem we have $OM=\sqrt{40}, \tan OAC=2, \tan BAC=\frac{1}{3}$.
$\tan(OAC-BAC)=\frac{2-\frac{1}{3}}{1+2 \cdot \frac{1}{3}}=1=\tan OAB$.
So $\angle OAB=45^{\circ}, \cos OAB=\frac{\sqrt2}{2}$
By Law of Cosines,
$(\sqrt{50})^2+(6^2)-2(\sqrt{50})(6)\frac{\sqrt2}{2}=OB^2=26$.
$OB=\sqrt{26}$
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Ultimatemathgeek
111 posts
#27 • 2 Y
Y by Adventure10, Mango247
Rep123max wrote:
For the question about if there was a shortcut for expanding it, you won't have to write down any of the terms where $i\sin\theta$ is taken to an odd power, because that would have an $i$ in it. And if you were trying to say find $\sin5\theta$, you wouln't write down any of the terms where $i\sin\theta$ is taken to an even power.

I got it, don't worry about it. I was slightly confused at that moment.
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Ultimatemathgeek
111 posts
#28 • 2 Y
Y by Adventure10, Mango247
Pork_Chop8,
I don't get the last step for the first problem you did; could you explain?


What number was the second problem?
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beta
3001 posts
#29 • 2 Y
Y by Adventure10, Mango247
Binomial Theorem,$\displaystyle \sum_{k=0}^{n} {n \choose k}x^ky^{n-k}=(x+y)^k$.

Sub in x=1, y=-1, we have $\displaystyle \sum_{k=0}^{n} (-1)^k(1)^{n-k}{n \choose k}=(1-1)^n=0$. Pork_Chop8's solution was the intended solution, I made #24 up.

The second problem is #25
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Ultimatemathgeek
111 posts
#30 • 2 Y
Y by Adventure10, Mango247
beta wrote:
Binomial Theorem,$\displaystyle \sum_{k=0}^{n} {n \choose k}x^ky^{n-k}=(x+y)^k$.

Sub in x=1, y=-1, we have $\displaystyle \sum_{k=0}^{n} (-1)^k(1)^{n-k}{n \choose k}=(1-1)^n=0$. Pork_Chop8's solution was the intended solution, I made #24 up.

The second problem is #25

X= -1, y=1?

By the way, beta, how did you know MA was sqrt(10) for #25
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Ultimatemathgeek
111 posts
#31 • 2 Y
Y by Adventure10, Mango247
Rep123max wrote:
From the Pythagorean Theorem, he got $2^2+6^2=AC^2$ so $AC=\sqrt{40}=2\sqrt{10}$.

Since its the midpoint, it is half of that, or $\sqrt{10}$.

Got it. I think my brain cells died somehow :(
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crayshot97
7 posts
#32 • 2 Y
Y by Adventure10, Mango247
20. n^2-3n-126=k^2, where k is an integer. First, note that if n=x is a solution, then n=3-x is also a solution. Therefore, when adding these we get 3, so the solution must be a positive multiple of 3. The only choice that satisfies this is 12(A). Now for an honest solution, we solve the quadratic n^2-3n-(126+k^2)=0. The discriminant must be a perfect square so we get 4k^2+513=q^2, where q is also an integer. Therefore (q-2k)(q+2k)=513=3^3*19. Therefore, we get 4 possible values for k. Each value of k gives 2 values for n which sum to 3. Therefore, the answer is 4*3=12.
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Ultimatemathgeek
111 posts
#33 • 2 Y
Y by Adventure10, Mango247
crayshot97 wrote:
20. n^2-3n-126=k^2, where k is an integer. First, note that if n=x is a solution, then n=3-x is also a solution. Therefore, when adding these we get 3, so the solution must be a positive multiple of 3. The only choice that satisfies this is 12(A). Now for an honest solution, we solve the quadratic n^2-3n-(126+k^2)=0. The discriminant must be a perfect square so we get 4k^2+513=q^2, where q is also an integer. Therefore (q-2k)(q+2k)=513=3^3*19. Therefore, we get 4 possible values for k. Each value of k gives 2 values for n which sum to 3. Therefore, the answer is 4*3=12.

Good solution. Especially the first one, I thought it was very interesting. Now I have to ask you, how do you know if n=x is a solution, then n=3-x is a solution? I know it works but how did you figure that out in the first place?
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Pork_Chop8
179 posts
#34 • 2 Y
Y by Adventure10, Mango247
$n^{2}-3n-126=k^{2} \Rightarrow n(n-3)=126 + k^{2}$
Sub in $n = 3 - n'$, you get $(3-n')(3-n'-3)=(n')^{2} - 3n' = 126 + k^{2}$
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