Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 IMO Problems/Problem 1"
(→‎Solution 2)
 
Line 45: Line 45:
 
For the statement to be true, there must be at least three pairs whose sum is each a perfect square. There must be p,q,r such that
 
For the statement to be true, there must be at least three pairs whose sum is each a perfect square. There must be p,q,r such that
  
−
p+q = x^2 and q+r = y^2, p+r = z^2.
+
<math>p+q = x^2</math> and <math>q+r = y^2</math>, <math>p+r = z^2</math>.
  
−
  WLOG n≤ p ≤ q ≤ r ≤ 2n ... Equation 1
+
  WLOG <math>n \le p \le q \le r \le 2n </math> ... Equation <math>1</math>
−
  p = (x^2 + z^2 – y^2) / 2
+
 
−
  q = (x^2 + y^2 – z^2) / 2
+
  <math>p = \frac{(x^2 + z^2 – y^2)}{2}</math>
−
  r = (y^2 + z^2 – x^2) / 2
+
  <math>q = \frac{(x^2 + y^2 – z^2)}{2}</math>
 +
  <math>r = \frac{(y^2 + z^2 – x^2)}{2}</math>
  
 
by equation 1
 
by equation 1
  
−
  2n ≤ x^2 + z^2 – y^2 ≤ 4n  ...(1)
+
  <math>2n \le x^2 + z^2 – y^2 \le 4n</math>   ...(1)
−
  2n ≤ x^2 + y^2 – z^2 ≤ 4n  ...(2)
+
  <math>2n \le x^2 + y^2 – z^2 \le 4n</math>   ...(2)
−
  2n ≤ y^2 + z^2 – z^2 ≤ 4n  ...(3)
+
  <math>2n \le y^2 + z^2 – z^2 \le 4n</math>   ...(3)
  
 
if we add (2) and (3) to (1),  
 
if we add (2) and (3) to (1),  
  
−
(1) + (2) + (3) =>
+
(1) + (2) + (3) <math>\Rightarrow</math>
 +
 
 +
<math>6n \le x^2 + y^2 + z^2 \le 12n</math>
  
−
6n ≤ x^2 + y^2 + z^2 ≤ 12n
 
 
  if we assume that x, y, and z is close enough,
 
  if we assume that x, y, and z is close enough,
−
6n ≤ 3x^2 ≤ 12n
 
−
2n ≤ x^2 ≤ 4n
 
−
√(2n) ≤ x ≤ 2√n
 
  
−
At this time 100 ≤ n, so let's put n = 100 to this
+
<math>6n \le 3x^2 \le 12n</math>
 +
<math>2n \le x^2 \le 4n</math>
 +
<math>\sqrt{(2n)} \le x \le 2\sqrt{n}</math>
 +
 
 +
At this time <math>100 \le n</math>, so let's put <math>n = 100</math> to this
  
−
  10√2 ≤ x,y,z ≤ 20
+
  <math>10\sqrt{2} \le (x,y,z) \le 20</math>
−
  15 ≤ x,y,z ≤ 20
+
  <math>15 \le (x,y,z) \le 20</math>
  
 
where
 
where
  
−
  2|x^2 + y^2 – z^2  
+
  <math>2|x^2 + y^2 - z^2</math>
−
  2|x^2 + z^2 – y^2  
+
  <math>2|x^2 + z^2 - y^2</math>
−
  2|y^2 + z^2 – z^2
+
  <math>2|y^2 + z^2 - z^2</math>
  
−
x = 16, y = 18, z = 20 fits perfectly
+
<math>x = 16, y = 18, z = 20</math> fits perfectly
  
−
therefore the minimum of n fits the proposition so the proposition is true
+
therefore the minimum of <math>n</math> fits the proposition so the proposition is true
  
 
~Mathhyhyhy
 
~Mathhyhyhy
 +
 +
~Kingfireboy
  
 
== See also ==
 
== See also ==

Latest revision as of 07:32, 13 April 2024

Problem

Let $n \geq 100$ be an integer. Ivan writes the numbers $n, n+1, \ldots, 2 n$ each on different cards. He then shuffles these $n+1$ cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.

Video Solutions

https://youtu.be/cI9p-Z4-Sc8 [Video contains solutions to all day 1 problems]

https://youtu.be/0Vd4ZBEr3o4

https://www.youtube.com/watch?v=a3L9O7b1WYg [disclaimer: only a sketch of the solution]

Solution 1

If we can guarantee that there exist $3$ cards such that every pair of them sum to a perfect square, then we can guarantee that one of the piles contains $2$ cards that sum to a perfect square. Assume the perfect squares $p^2$, $q^2$, and $r^2$ satisfy the following system of equations: \usepackage{amsmath} \begin{align*} a+b &= p^2 \\ b+c &= q^2 \\ a+c &= r^2 \end{align*} where $a$, $b$, and $c$ are numbers on three of the cards. Solving for $a$, $b$, and $c$ in terms of $p$, $q$, and $r$ tells us that $a = \frac{p^2 + r^2 - q^2}{2}$, $b=\frac{p^2 + q^2 - r^2}{2}$, and $c=\frac{q^2 + r^2 - p^2}{2}$. We can then substitute $p^2 = (2e-1)^2$, $q^2 = (2e)^2$, and $r^2 = (2e+1)^2$ to cancel out the $2$s in the denominatior, and simplifying gives $a = 2e^2 + 1$, $b = 2e(e-2)$, and $c = 2e(e+2)$. Now, we have to prove that there exists three numbers in these forms between $n$ and $2n$ when $n \ge 100$. Notice that $b$ will always be the least of the three and $c$ will always be the greatest of the three. So it is sufficient to prove that there exists numbers in the form $2e(e-2)$ and $2e(e+2)$ between $n$ and $2n$.


For two numbers in the form of $2e(e-2)$ and $2e(e+2)$ to be between $n$ and $2n$, the inequalities \usepackage{amsmath} \begin{align*} 2e(e-2) &\ge n \\ 2e(e+2) &\le 2n \\ \end{align*} must be satisfied. We can then expand and simplify to get that \usepackage{amsmath} \begin{align*} e^2 - 2e - \frac{n}{2} &\ge 0 \\ e^2 + 2e - n &\le 0. \\ \end{align*} Then, we can complete the square on the left sides of both inequalities and isolate $e$ to get that \usepackage{amsmath} \begin{align*} e &\ge \sqrt{1 + \frac{n}{2}} + 1 \\ e &\le \sqrt{1 + n} - 1 \\ \end{align*} Notice that $e$ must be an integer, so there must be an integer between $\sqrt{1 + n} - 1$ and $\sqrt{1 + \frac{n}{2}} + 1$. If $\sqrt{1 + n} - 1$ and $\sqrt{1 + \frac{n}{2}} + 1$ differ by at least $1$, then we can guarantee that there is an integer between them (and those integers are the possible values of $e$). Setting up the inequality $\sqrt{1 + n} - \sqrt{1 + \frac{n}{2}} - 2 \ge 1$ and solving for $n$ tells us that $n \in [107, \infty)$ always works. Testing the remaining $7$ numbers ($100$ to $106$) manually tells us that there is an integer between $\sqrt{1 + n} - 1$ and $\sqrt{1 + \frac{n}{2}} + 1$ when $n \ge 100$. Therefore, there exists a triplet of integers $(a,b,c)$ with $a, b, c \in \{n, n+1, ..., 2n\}$ when $n \ge 100$ such that every pair of the numbers sum to a perfect square. By the pigeonhole principle, we know that $2$ of the numbers must be on cards in the same pile, and hence, when $n \ge 100$, there will always be a pile with $2$ numbers that sum to a perfect square. $\square$

~Mathdreams

Solution 2

For the statement to be true, there must be at least three pairs whose sum is each a perfect square. There must be p,q,r such that

$p+q = x^2$ and $q+r = y^2$, $p+r = z^2$. 

WLOG $n \le p \le q \le r \le 2n$ ...  Equation $1$

$p = \frac{(x^2 + z^2 – y^2)}{2}$
$q = \frac{(x^2 + y^2 – z^2)}{2}$
$r = \frac{(y^2 + z^2 – x^2)}{2}$

by equation 1 

$2n \le x^2 + z^2 – y^2 \le 4n$   ...(1)
$2n \le x^2 + y^2 – z^2 \le 4n$   ...(2)
$2n \le y^2 + z^2 – z^2 \le 4n$   ...(3)

if we add (2) and (3) to (1),

(1) + (2) + (3) $\Rightarrow$ 

$6n \le x^2 + y^2 + z^2 \le 12n$

if we assume that x, y, and z is close enough,

$6n \le 3x^2 \le 12n$
$2n \le x^2 \le 4n$
$\sqrt{(2n)} \le x \le 2\sqrt{n}$

At this time $100 \le n$, so let's put $n = 100$ to this 

$10\sqrt{2} \le (x,y,z) \le 20$
$15 \le (x,y,z) \le 20$

where 

$2|x^2 + y^2 - z^2$ 
$2|x^2 + z^2 - y^2$ 
$2|y^2 + z^2 - z^2$

$x = 16, y = 18, z = 20$ fits perfectly

therefore the minimum of $n$ fits the proposition so the proposition is true

~Mathhyhyhy

~Kingfireboy

See also

2021 IMO (Problems) • Resources
Preceded by
First Problem 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 2
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_IMO_Problems/Problem_1&oldid=218145"