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Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
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Problem 2
delegat   147
N an hour ago by math-olympiad-clown
Source: 0
Let $n\ge 3$ be an integer, and let $a_2,a_3,\ldots ,a_n$ be positive real numbers such that $a_{2}a_{3}\cdots a_{n}=1$. Prove that
\[(1 + a_2)^2 (1 + a_3)^3 \dotsm (1 + a_n)^n > n^n.\]

Proposed by Angelo Di Pasquale, Australia
147 replies
delegat
Jul 10, 2012
math-olympiad-clown
an hour ago
Coloring points of a square, finding a monochromatic hexagon
goodar2006   6
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P1
Prove that for each coloring of the points inside or on the boundary of a square with $1391$ colors, there exists a monochromatic regular hexagon.
6 replies
goodar2006
Sep 15, 2012
quantam13
2 hours ago
Van der Warden Theorem!
goodar2006   7
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part 2-P2
Suppose $W(k,2)$ is the smallest number such that if $n\ge W(k,2)$, for each coloring of the set $\{1,2,...,n\}$ with two colors there exists a monochromatic arithmetic progression of length $k$. Prove that


$W(k,2)=\Omega (2^{\frac{k}{2}})$.
7 replies
goodar2006
Sep 15, 2012
quantam13
2 hours ago
Maxi-inequality
giangtruong13   0
2 hours ago
Let $a,b,c >0$ and $a+b+c=2abc$. Find max: $$P= \sum_{cyc} \frac{a+2}{\sqrt{6(a^2+2)}}$$
0 replies
giangtruong13
2 hours ago
0 replies
Isosceles triangles among a group of points
goodar2006   2
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part1-P2
Consider a set of $n$ points in plane. Prove that the number of isosceles triangles having their vertices among these $n$ points is $\mathcal O (n^{\frac{7}{3}})$. Find a configuration of $n$ points in plane such that the number of equilateral triangles with vertices among these $n$ points is $\Omega (n^2)$.
2 replies
1 viewing
goodar2006
Jul 27, 2012
quantam13
2 hours ago
APMO Number Theory
somebodyyouusedtoknow   12
N 2 hours ago by math-olympiad-clown
Source: APMO 2023 Problem 2
Find all integers $n$ satisfying $n \geq 2$ and $\dfrac{\sigma(n)}{p(n)-1} = n$, in which $\sigma(n)$ denotes the sum of all positive divisors of $n$, and $p(n)$ denotes the largest prime divisor of $n$.
12 replies
somebodyyouusedtoknow
Jul 5, 2023
math-olympiad-clown
2 hours ago
My Unsolved Problem
ZeltaQN2008   0
2 hours ago
Source: IDK
Let \( P(x) = x^{2024} + a_{2023}x^{2023} + \cdots + a_1x + a_0 \) be a polynomial with real coefficients.

(a) Suppose that \( 2023a_{2023}^2 - 4048a_{2022} < 0 \). Prove that the polynomial \( P(x) \) cannot have 2024 real roots.

(b) Suppose that \( a_0 = 1 \) and \( 2023(a_1^2 + a_2^2 + \cdots + a_{2023}^2) \leq 4 \). Prove that \( P(x) \geq 0 \) for all real numbers \( x \).
0 replies
ZeltaQN2008
2 hours ago
0 replies
Points of a grid
goodar2006   2
N 2 hours ago by quantam13
Source: Iran 3rd round 2012-Special Lesson exam-Part1-P4
Prove that from an $n\times n$ grid, one can find $\Omega (n^{\frac{5}{3}})$ points such that no four of them are vertices of a square with sides parallel to lines of the grid. Imagine yourself as Erdos (!) and guess what is the best exponent instead of $\frac{5}{3}$!
2 replies
goodar2006
Jul 27, 2012
quantam13
2 hours ago
Classical NT FE
Kimchiks926   6
N 3 hours ago by math-olympiad-clown
Source: Baltic Way 2022, Problem 16
Let $\mathbb{Z^+}$ denote the set of positive integers. Find all functions $f:\mathbb{Z^+} \to \mathbb{Z^+}$ satisfying the condition
$$ f(a) + f(b) \mid (a + b)^2$$for all $a,b \in \mathbb{Z^+}$
6 replies
Kimchiks926
Nov 12, 2022
math-olympiad-clown
3 hours ago
Hagge circle, Thomson cubic, coaxal
kosmonauten3114   0
3 hours ago
Source: My own (maybe well-known)
Let $\triangle{ABC}$ be a scalene triangle, $\triangle{M_AM_BM_C}$ its medial triangle, and $P$ a point on the Thomson cubic (= $\text{K002}$) of $\triangle{ABC}$. (Suppose that $P \notin \odot(ABC)$ ).
Let $\triangle{A'B'C'}$ be the circumcevian triangle of $P$ wrt $\triangle{ABC}$.
Let $\triangle{P_AP_BP_C}$ be the pedal triangle of $P$ wrt $\triangle{ABC}$.
Let $A_1$ be the reflection in $BC$ of $A'$. Define $B_1$, $C_1$ cyclically.
Let $A_2$ be the reflection in $M_A$ of $A'$. Define $B_2$, $C_2$ cyclically.
Let $A_3$ be the reflection in $P_A$ of $A'$. Define $B_3$, $C_3$ cyclically.

Prove that $\odot(A_1B_1C_1)$, $\odot(A_2B_2C_2)$, $\odot(A_3B_3C_3)$ and the orthocentroidal circle of $\triangle{ABC}$ are coaxal.
0 replies
kosmonauten3114
3 hours ago
0 replies
Logical guessing game!
Mathdreams   22
N Nov 15, 2024 by JH_K2IMO
Source: 2021 Fall AMC10B P10
Fourty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$
22 replies
Mathdreams
Nov 17, 2021
JH_K2IMO
Nov 15, 2024
Logical guessing game!
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G H BBookmark kLocked kLocked NReply
Source: 2021 Fall AMC10B P10
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Mathdreams
1472 posts
#1 • 2 Y
Y by HWenslawski, megarnie
Fourty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$
This post has been edited 1 time. Last edited by Mathdreams, Nov 17, 2021, 4:15 PM
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Toinfinity
603 posts
#2 • 1 Y
Y by wamofan
Was it (A) 27, 2 and 25
This post has been edited 1 time. Last edited by Toinfinity, Nov 17, 2021, 4:10 PM
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Mathlete12345654
163 posts
#3
Y by
Can confirm @above
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mathboy100
675 posts
#4
Y by
Mathdreams wrote:
Fourty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$

If Alice does not know who has the largest number, then she doesn't have $1$ or $40$. Thus, Bob must have $2$ or $39$. $2$ is prime, so Bob has $2$ and Alice has $25$. The answer is $27$.
This post has been edited 1 time. Last edited by mathboy100, Nov 17, 2021, 4:18 PM
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mewto
677 posts
#5
Y by
Yes, I got A. I'm so relieved
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amopuri
139 posts
#6
Y by
oops i kept thinking that bob had to have a bigger number so i kept getting stuck with $39$

skipped it, came back to it, and got it tho
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StopSine
475 posts
#7
Y by
Why did I think that 225 was not a perfect square for the first 3 minutes of the problem.
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buddy2007
2071 posts
#8
Y by
Mathdreams wrote:
Fourty slips of paper numbered $1$ to $40$ are placed in a hat. Alice and Bob each draw one number from the hat without replacement, keeping their numbers hidden from each other. Alice says, "I can't tell who has the larger number." Then Bob says, "I know who has the larger number." Alice says, "You do? Is your number prime?" Bob replies, "Yes." Alice says, "In that case, if I multiply your number by $100$ and add my number, the result is a perfect square. " What is the sum of the two numbers drawn from the hat?

$\textbf{(A) }27\qquad\textbf{(B) }37\qquad\textbf{(C) }47\qquad\textbf{(D) }57\qquad\textbf{(E) }67$
I got 2 answers, 27 (225) and 67 (3136) but i had to use a bit of logic to get (A)
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IceMatrix
399 posts
#9 • 1 Y
Y by FIREDRAGONMATH16
I liked this one. Thought it was fairly standard logic application for experienced solvers, and the solution process felt satisfying. Those uninitiated(first or second time test takers) probably had more difficulty though.
If you are curious: https://youtu.be/RyN-fKNtd3A?t=1474
This post has been edited 1 time. Last edited by IceMatrix, Nov 17, 2021, 5:12 PM
Reason: Added link
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jasperE3
11386 posts
#10 • 1 Y
Y by megarnie
I agree with #9.
For Alice to not know who has the larger number, she must not have either $1$ or $40$. Thus, Bob needs to have either $1,2,39$, or $40$. The only prime out of these is $2$. Alice's number is $x$, then $200+x$ is a square. The only such number from $1$ to $40$ is $25$, so the sum is $2+25=\boxed{\textbf{(A)}~27}$.
This post has been edited 1 time. Last edited by jasperE3, Nov 17, 2021, 5:23 PM
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HighWater
377 posts
#11
Y by
mewto wrote:
Yes, I got A. I'm so relieved

*PHEW!* I talked to a lot of people and some got E, so I was worried I got it wrong, but then I explained my logic and they immediately realized their mistakes.
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asimov
1443 posts
#12
Y by
Can confirm A. It was a very nice and fun problem.
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megarnie
5611 posts
#13
Y by
We claim that Bob gets a $1$, $2$, $39$, or $40$.
Proof: If not, then Alice could've gotten a number from either side of Bob.

Since Bob got a prime, Bob got a $2$.

So Alice got a $25,56,89$, but $25$ is the only one less than $40$.

Thus, the answer is $2+25=\boxed{\textbf{(A)}\ 57}$.
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Overlord123
799 posts
#14
Y by
Solution
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exp-ipi-1
1074 posts
#15
Y by
got it wrong because i forgot that 2 was a prime :wacko:
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pog
4917 posts
#16
Y by
My bestie DeToasty3 had a lot of fun looking at the solution for this problem

Solution
This post has been edited 2 times. Last edited by pog, Jan 8, 2022, 1:20 AM
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pog
4917 posts
#17 • 2 Y
Y by HrishiP, DeToasty3
Should say "Forty" and not "Fourty" lol

$\underbrace{2 + 2 + \cdots + 2 + 2}_{100\text{ times}}$
This post has been edited 1 time. Last edited by pog, Jan 8, 2022, 1:30 AM
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hansenhe
3390 posts
#18
Y by
Post #13 by megarnie

@megarnie 2+25=27...
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DeToasty3
596 posts
#19 • 19 Y
Y by fuzimiao2013, pog, pandabearcat, centslordm, bissue, nikenissan, v4913, mahaler, RedFlame2112, megarnie, RP3.1415, asdf334, HrishiP, math31415926535, rayfish, DankBasher619, john0512, Bryan013, Jack_w
We start with a miraculous claim which is the pith of this problem:

Claim 1. Bob's number must be $2$.

Proof. Before we can get our hands dirty, we shall prove a few subclaims:

Subclaim 1.1. Alice cannot have the number $1$.

Proof. If she did, then Alice would know that Bob's number is greater than hers.

Subclaim 1.2. Alice cannot have the number $40$.

Proof. If she did, then Alice would know that Bob's number is less than hers.

Subclaim 1.3. $1$ is a possibility for Bob's number.

Proof. If $1$ is Bob's number, then he knows that Alice's number is larger than his.

Subclaim 1.4. $2$ is a possibility for Bob's number.

Proof. If $2$ is Bob's number, then he knows that, because Alice's number cannot be $1$, Alice's number is larger than his.

Subclaim 1.5. $39$ is a possibility for Bob's number.

Proof. If $39$ is Bob's number, then he knows that, because Alice's number cannot be $40$, Alice's number is less than his.

Subclaim 1.6. $40$ is a possibility for Bob's number.

Proof. If $40$ is Bob's number, then he knows that Alice's number is less than his.

We have deduced that Bob's number can be either $1$, $2$, $39$, or $40$. Of these four numbers, only $2$ is prime, hence we have proved the claim. $\square$

Now that we know that Bob's number is $2$, we may use Alice's final sentence to finish the problem. Multiplying $2$ by $100$ is merely equivalent to computing the expression $\underbrace{2 + 2 + 2 + \cdots + 2 + 2}_{100\text{ times}}$, from which we produce the number $200$. Now, we must search for perfect squares. Before we do this, we have to bound the range of possible values of this perfect square. To do this, we introduce another claim:

Claim 2. The perfect square must be contained in the interval $(202,240)$.

To prove this claim, we introduce two subclaims.

Subclaim 2.1. The perfect square must be strictly greater than $202$.

Proof. As Alice cannot have the number $1$, she must have at least the number $2$. However, since Bob has the number $2$, Alice must have at least the number $3$. Thus, the perfect square must be at least $200+3=203$, hence proven.

Subclaim 2.2. The perfect square must be strictly less than $240$.

Proof. As Alice cannot have the number $40$, she must have at most the number $39$. Thus, the perfect square must be at most $200+39=239$, hence proven.

Combining these two subclaims, we have that the perfect square must be contained in the interval $(202,240)$, as desired. $\square$

Now, we must search for perfect squares in the interval $(202,240)$. Upon squaring $15$, we find that the number obtained is $225$, which, by sheer luck, is in the interval $(202,240)$. However, upon squaring $14$ and $16$, we obtain the numbers $196$ and $256$, neither of which are in the interval. It is well-known that $f(x)=x^2$ is strictly increasing as $x$ increases from $0$. Thus, if $14^2$ and $16^2$ are out of the interval $(202,240)$ from opposite sides of the interval, we may conclude that $15^2=225$ is the only possible perfect square. This means that Alice's number is $225-200=25$.

Finally, we may take the sum of Alice's and Bob's numbers, which are $25$ and $2$, respectively, for a final answer of $25+2=\boxed{\textbf{(A) }27}$, and we are done. $\blacksquare$

Remark. This problem is clearly unsuitable for the AMC 10. Proving even one of these claims by itself proves to be a significant challenge, perhaps too much for tenth graders.
This post has been edited 2 times. Last edited by DeToasty3, Jan 8, 2022, 2:19 AM
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fuzimiao2013
3312 posts
#20
Y by
I'm laughing because it's an overkill, but I honestly would LOVE to see more solutions like this - all the intermediate steps, the thought processes - you can get a lot more if you're a beginner.
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Math4Life2020
2966 posts
#21
Y by
i dare someone to do 2021 imo/3 using 2nd-grade level steps
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asdf334
7585 posts
#22
Y by
DeToasty3 wrote:
We start with a miraculous claim which is the pith of this problem:

Claim 1. Bob's number must be $2$.

Proof. Before we can get our hands dirty, we shall prove a few subclaims:

Subclaim 1.1. Alice cannot have the number $1$.

Proof. If she did, then Alice would know that Bob's number is greater than hers.

Subclaim 1.2. Alice cannot have the number $40$.

Proof. If she did, then Alice would know that Bob's number is less than hers.

Subclaim 1.3. $1$ is a possibility for Bob's number.

Proof. If $1$ is Bob's number, then he knows that Alice's number is larger than his.

Subclaim 1.4. $2$ is a possibility for Bob's number.

Proof. If $2$ is Bob's number, then he knows that, because Alice's number cannot be $1$, Alice's number is larger than his.

Subclaim 1.5. $39$ is a possibility for Bob's number.

Proof. If $39$ is Bob's number, then he knows that, because Alice's number cannot be $40$, Alice's number is less than his.

Subclaim 1.6. $40$ is a possibility for Bob's number.

Proof. If $40$ is Bob's number, then he knows that Alice's number is less than his.

We have deduced that Bob's number can be either $1$, $2$, $39$, or $40$. Of these four numbers, only $2$ is prime, hence we have proved the claim. $\square$

Now that we know that Bob's number is $2$, we may use Alice's final sentence to finish the problem. Multiplying $2$ by $100$ is merely equivalent to computing the expression $\underbrace{2 + 2 + 2 + \cdots + 2 + 2}_{100\text{ times}}$, from which we produce the number $200$. Now, we must search for perfect squares. Before we do this, we have to bound the range of possible values of this perfect square. To do this, we introduce another claim:

Claim 2. The perfect square must be contained in the interval $(202,240)$.

To prove this claim, we introduce two subclaims.

Subclaim 2.1. The perfect square must be strictly greater than $202$.

Proof. As Alice cannot have the number $1$, she must have at least the number $2$. However, since Bob has the number $2$, Alice must have at least the number $3$. Thus, the perfect square must be at least $200+3=203$, hence proven.

Subclaim 2.2. The perfect square must be strictly less than $240$.

Proof. As Alice cannot have the number $40$, she must have at most the number $39$. Thus, the perfect square must be at most $200+39=239$, hence proven.

Combining these two subclaims, we have that the perfect square must be contained in the interval $(202,240)$, as desired. $\square$

Now, we must search for perfect squares in the interval $(202,240)$. Upon squaring $15$, we find that the number obtained is $225$, which, by sheer luck, is in the interval $(202,240)$. However, upon squaring $14$ and $16$, we obtain the numbers $196$ and $256$, neither of which are in the interval. It is well-known that $f(x)=x^2$ is strictly increasing as $x$ increases from $0$. Thus, if $14^2$ and $16^2$ are out of the interval $(202,240)$ from opposite sides of the interval, we may conclude that $15^2=225$ is the only possible perfect square. This means that Alice's number is $225-200=25$.

Finally, we may take the sum of Alice's and Bob's numbers, which are $25$ and $2$, respectively, for a final answer of $25+2=\boxed{\textbf{(A) }27}$, and we are done. $\blacksquare$

Remark. This problem is clearly unsuitable for the AMC 10. Proving even one of these claims by itself proves to be a significant challenge, perhaps too much for tenth graders.

Truly a marvelous solution, the likes of which have not been seen in a century.
This post has been edited 2 times. Last edited by asdf334, Jan 8, 2022, 2:22 AM
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JH_K2IMO
131 posts
#24
Y by
If Alice had drawn either 1 or 40, she would have immediately known who drew the larger number.
Therefore, Alice’s number is not 1 or 40.
Bob, on the other hand, says that he knows who has the larger number, so Bob must have drawn one of the numbers 1, 2, 39, or 40.
Since Bob's number is a prime number, his number must be 2.
Given that 2 x 100 + (Alice's number) must be a perfect square, Alice’s number must be 25.
Therefore, the sum of their numbers is 2 + 25 = 27 .
The answer is A.
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