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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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Burak0609
Burak0609   0
11 minutes ago
$a^7(a-1)=19b(19b+2) \implies a^7(a-1)+1=(19b+1)^2$.
So we can see $(19b+1)^2=a^8-a^7+1=(a^2-a+1)(a^6-a^4-a^3+a+1$ and $gcd(a^2-a+1,a^6-a^4-a^3+a+1)=1,19$ but $gcd(a^2-a+1,a^6-a^4-a^3+a+1)=1$ because $(19b+1)^2 \equiv 0(mod 19)$. I mean $a^2-a+1$ and $a^6-a^4-a^3+a+1$ are perfect squares. $a^2 \le a^2-a+1 \le (a+1)^2$. a should be 0 or 1 because of $a^2 \le a^2-a+1 \le (a+1)^2$. We have two solution. These are $(a,b)=(0,0),(1,0)
0 replies
Burak0609
11 minutes ago
0 replies
Can Euclid solve this geo ?
S.Ragnork1729   31
N an hour ago by PeterZeus
Source: INMO 2025 P3
Euclid has a tool called splitter which can only do the following two types of operations :
• Given three non-collinear marked points $X,Y,Z$ it can draw the line which forms the interior angle bisector of $\angle{XYZ}$.
• It can mark the intersection point of two previously drawn non-parallel lines .
Suppose Euclid is only given three non-collinear marked points $A,B,C$ in the plane . Prove that Euclid can use the splitter several times to draw the centre of circle passing through $A,B$ and $C$.

Proposed by Shankhadeep Ghosh
31 replies
S.Ragnork1729
Jan 19, 2025
PeterZeus
an hour ago
Answer is Year
solasky   2
N an hour ago by AshAuktober
Source: Japan MO Preliminary 2021/1
For all relatively prime positive integers $m$, $n$ satisfying $m + n = 90$, what is the maximum possible value of $mn$?
2 replies
solasky
Jun 15, 2024
AshAuktober
an hour ago
series and factorials?
jenishmalla   8
N an hour ago by Maximilian113
Source: 2025 Nepal ptst p4 of 4
Find all pairs of positive integers \( n \) and \( x \) such that
\[
1^n + 2^n + 3^n + \cdots + n^n = x!
\]
(Petko Lazarov, Bulgaria)
8 replies
jenishmalla
Mar 15, 2025
Maximilian113
an hour ago
Collinear Centers and Midarcs
Miku3D   34
N an hour ago by lelouchvigeo
Source: 2021 APMO P3
Let $ABCD$ be a cyclic convex quadrilateral and $\Gamma$ be its circumcircle. Let $E$ be the intersection of the diagonals of $AC$ and $BD$. Let $L$ be the center of the circle tangent to sides $AB$, $BC$, and $CD$, and let $M$ be the midpoint of the arc $BC$ of $\Gamma$ not containing $A$ and $D$. Prove that the excenter of triangle $BCE$ opposite $E$ lies on the line $LM$.
34 replies
Miku3D
Jun 9, 2021
lelouchvigeo
an hour ago
Bashing??
John_Mgr   0
an hour ago
I have learned little about what bashing mean as i am planning to start geo, feels like its less effort required and doesnt need much knowledge about the synthetic solutions?
what do you guys recommend ? also state the major difference of them... especially of bashing pros and cons..
0 replies
John_Mgr
an hour ago
0 replies
1 area = 2025 points
giangtruong13   1
N an hour ago by kiyoras_2001
In a plane give a set $H$ that has 8097 distinct points with area of a triangle that has 3 points belong to $H$ all $ \leq 1$. Prove that there exists a triangle $G$ that has the area $\leq 1 $ contains at least 2025 points that belong to $H$( each of that 2025 points can be inside the triangle or lie on the edge of triangle $G$)X
1 reply
giangtruong13
Today at 8:31 AM
kiyoras_2001
an hour ago
A board with crosses that we color
nAalniaOMliO   2
N an hour ago by CHESSR1DER
Source: Belarusian National Olympiad 2025
In some cells of the table $2025 \times 2025$ crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least $k$ crosses.
Find the minimal $k$ for which it is always possible to color crosses in two colors such that any balanced set has crosses of both colors.
2 replies
nAalniaOMliO
Mar 28, 2025
CHESSR1DER
an hour ago
Geometry Finale: Incircles and concurrency
lminsl   173
N 2 hours ago by Parsia--
Source: IMO 2019 Problem 6
Let $I$ be the incentre of acute triangle $ABC$ with $AB\neq AC$. The incircle $\omega$ of $ABC$ is tangent to sides $BC, CA$, and $AB$ at $D, E,$ and $F$, respectively. The line through $D$ perpendicular to $EF$ meets $\omega$ at $R$. Line $AR$ meets $\omega$ again at $P$. The circumcircles of triangle $PCE$ and $PBF$ meet again at $Q$.

Prove that lines $DI$ and $PQ$ meet on the line through $A$ perpendicular to $AI$.

Proposed by Anant Mudgal, India
173 replies
lminsl
Jul 17, 2019
Parsia--
2 hours ago
Problem 1
blug   2
N 2 hours ago by kjhgyuio
Source: Polish Math Olympiad 2025 Finals P1
Find all $(a, b, c, d)\in \mathbb{R}$ satisfying
\[\begin{aligned}
\begin{cases}
    a+b+c+d=0,\\
    a^2+b^2+c^2+d^2=12,\\
    abcd=-3.\\
\end{cases}
\end{aligned}\]
2 replies
blug
3 hours ago
kjhgyuio
2 hours ago
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!
G H J
G H BBookmark kLocked kLocked NReply
Source: 2021 Fall AMC10B P10
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Mathdreams
1448 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
133 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
2070 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
11158 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
5553 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
1075 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
3302 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
2962 posts
#21
Y by
i dare someone to do 2021 imo/3 using 2nd-grade level steps
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asdf334
7586 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
125 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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