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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]
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0 replies
jlacosta
Mar 2, 2025
0 replies
MathPath 2025 form.
BraveCobra22aops   0
10 minutes ago
I created a form for people going to MathPath 2025: https://artofproblemsolving.com/community/c136h3528968_mathpath_2025.
0 replies
+2 w
BraveCobra22aops
10 minutes ago
0 replies
AMC- IMO preparation
asyaela.   9
N 42 minutes ago by Schintalpati
I'm a ninth grader, and I recently attempted the AMC 12, getting 18 questions correct and leaving 7 empty. I started working on Olympiad math in November and currently dedicate about two hours per day to preparation. I'm feeling a bit demotivated, but if it's possible for me to reach IMO level, I'd be willing to put in more time. How realistic is it for me to get there, and how much study would it typically take?
9 replies
asyaela.
4 hours ago
Schintalpati
42 minutes ago
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   29
N an hour ago by NashvilleSC
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AMC 12 level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
29 replies
TennesseeMathTournament
Mar 9, 2025
NashvilleSC
an hour ago
AIME score for college apps
Happyllamaalways   75
N an hour ago by hashbrown2009
What good colleges do I have a chance of getting into with an 11 on AIME? (Any chances for Princeton)

Also idk if this has weight but I had the highest AIME score in my school.
75 replies
+1 w
Happyllamaalways
Mar 13, 2025
hashbrown2009
an hour ago
Putnam 2017 B3
goveganddomath   34
N Today at 4:16 AM by OronSH
Source: Putnam
Suppose that $$f(x) = \sum_{i=0}^\infty c_ix^i$$is a power series for which each coefficient $c_i$ is $0$ or $1$. Show that if $f(2/3) = 3/2$, then $f(1/2)$ must be irrational.
34 replies
goveganddomath
Dec 3, 2017
OronSH
Today at 4:16 AM
3-dimensional matrix system
loup blanc   1
N Today at 3:33 AM by alexheinis
Let $A=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$.
i) Find the matrices $B\in M_3(\mathbb{R})$ s.t. $A^TA=B^TB,AA^T=BB^T$.
EDIT. ii) Show that each solution of i) is in $M_3(K)$, where $K=\mathbb{Q}[\cos(\dfrac{2}{3}\arctan(3\sqrt{3}))]$.
iii) Solve i) when $B\in M_3(\mathbb{C})$.
EDIT. In iii) we again consider the transpose of $B$ and not its conjugate transpose.
1 reply
loup blanc
Yesterday at 1:04 PM
alexheinis
Today at 3:33 AM
Square of a rational matrix of dimension 2
loup blanc   9
N Today at 1:28 AM by ysharifi
The following exercise was posted -two months ago- on the Website StackExchange; cf.
https://math.stackexchange.com/questions/5006488/image-of-the-squaring-function-on-mathcalm-2-mathbbq
There was no solution on Stack.

-Statement of the exercise-
We consider the matrix function $f:X\in M_2(\mathbb{Q})\mapsto X^2\in M_2(\mathbb{Q})$.
Find the image of $f$.
In other words, give a method to decide whether a given matrix has or does not have at least a square root
in $M_2(\mathbb{Q})$; if the answer is yes, then give a method to calculate at least one of its roots.
9 replies
loup blanc
Feb 17, 2025
ysharifi
Today at 1:28 AM
Differentiation Marathon!
LawofCosine   181
N Today at 12:24 AM by Levieee
Hello, everybody!

This is a differentiation marathon. It is just like an ordinary marathon, where you can post problems and provide solutions to the problem posted by the previous user. You can only post differentiation problems (not including integration and differential equations) and please don't make it too hard!

Have fun!

(Sorry about the bad english)
181 replies
LawofCosine
Feb 1, 2025
Levieee
Today at 12:24 AM
Integrals problems and inequality
tkd23112006   2
N Yesterday at 6:02 PM by PolyaPal
Let f be a continuous function on [0,1] such that f(x) ≥ 0 for all x ∈[0,1] and
$\int_x^1 f(t) dt \geq \frac{1-x^2}{2}$ , ∀x∈[0,1].
Prove that:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$
2 replies
tkd23112006
Feb 16, 2025
PolyaPal
Yesterday at 6:02 PM
real analysis
ay19bme   1
N Yesterday at 5:30 PM by alexheinis
.........
1 reply
ay19bme
Yesterday at 3:05 PM
alexheinis
Yesterday at 5:30 PM
Proving an inequality involving cosine functions
pii-oner   3
N Yesterday at 3:33 PM by pii-oner
Hello AoPS Community,

I am curious about how to demonstrate the following inequality:

[code] \sqrt{1 - |\cos(x \pm y)|^a} \leq \sqrt{1 - |\cos(x)|^a} + \sqrt{1 - |\cos(y)|^a}, \quad \text{for } a \geq 1. [/code]

I’ve plotted the functions, and the inequality seems to hold. However, I am looking for a rigorous way to prove it.

I’d greatly appreciate insights into how to break this down analytically and references to similar problems or techniques that might be helpful.

Thank you so much for your guidance and support!
3 replies
pii-oner
Jan 22, 2025
pii-oner
Yesterday at 3:33 PM
Linear algebra
Dynic   2
N Yesterday at 3:32 PM by loup blanc
Let A and B be two square matrices with the same size. Prove that if AB is an invertible matrix, then A and B are also invertible matrices
2 replies
Dynic
Yesterday at 2:48 PM
loup blanc
Yesterday at 3:32 PM
Very hard group theory problem
mathscrazy   3
N Yesterday at 9:50 AM by quasar_lord
Source: STEMS 2025 Category C6
Let $G$ be a finite abelian group. There is a magic box $T$. At any point, an element of $G$ may be added to the box and all elements belonging to the subgroup (of $G$) generated by the elements currently inside $T$ are moved from outside $T$ to inside (unless they are already inside). Initially $
T$ contains only the group identity, $1_G$. Alice and Bob take turns moving an element from outside $T$ to inside it. Alice moves first. Whoever cannot make a move loses. Find all $G$ for which Bob has a winning strategy.
3 replies
mathscrazy
Dec 29, 2024
quasar_lord
Yesterday at 9:50 AM
limit of u(pi/45)
EthanWYX2009   0
Yesterday at 7:08 AM
Source: 2025 Pi Day Challenge T5
Let \(\omega\) be a positive real number. Divide the positive real axis into intervals \([0, \omega)\), \([\omega, 2\omega)\), \([2\omega, 3\omega)\), \([3\omega, 4\omega)\), \(\ldots\), and color them alternately black and white. Consider the function \(u(x)\) satisfying the following differential equations:
\[
u''(x) + 9^2u(x) = 0, \quad \text{for } x \text{ in black intervals},
\]\[
u''(x) + 63^2u(x) = 0, \quad \text{for } x \text{ in white intervals},
\]with the initial conditions:
\[
u(0) = 1, \quad u'(0) = 1,
\]and the continuity conditions:
\[
u(x) \text{ and } u'(x) \text{ are continuous functions}.
\]Show that
\[
\lim_{\omega \to 0} u\left(\frac{\pi}{45}\right) = 0.
\]
0 replies
EthanWYX2009
Yesterday at 7:08 AM
0 replies
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
3001 posts
#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
3001 posts
#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
3001 posts
#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
3001 posts
#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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