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Difference between revisions of "2023 IMO Problems"

(Created page with "Problems of the 2023 IMO. ==Day I== ===Problem 1=== Determine all composite integers <math>n>1</math> that satisfy the following property: if <math>d_1,d_2,\dots,d_k</mat...")
 
 
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Let <math>n</math> be a positive integer. A Japanese triangle consists of <math>1 + 2 + \dots + n</math> circles arranged in an equilateral triangular shape such that for each <math>i = 1</math>, <math>2</math>, <math>\dots</math>, <math>n</math>, the <math>i^{th}</math> row contains exactly <math>i</math> circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of <math>n</math> circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with <math>n = 6</math>, along with a ninja path in that triangle containing two red circles.
 
Let <math>n</math> be a positive integer. A Japanese triangle consists of <math>1 + 2 + \dots + n</math> circles arranged in an equilateral triangular shape such that for each <math>i = 1</math>, <math>2</math>, <math>\dots</math>, <math>n</math>, the <math>i^{th}</math> row contains exactly <math>i</math> circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of <math>n</math> circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with <math>n = 6</math>, along with a ninja path in that triangle containing two red circles.
  
−
(image to be inserted)
+
<asy>
 +
// credit to vEnhance for the diagram (which was better than my original asy):
 +
size(4cm);
 +
  pair X = dir(240); pair Y = dir(0);
 +
  path c = scale(0.5)*unitcircle;
 +
  int[] t = {0,0,2,2,3,0};
 +
  for (int i=0; i<=5; ++i) {
 +
    for (int j=0; j<=i; ++j) {
 +
      filldraw(shift(i*X+j*Y)*c, (t[i]==j) ? lightred : white);
 +
      draw(shift(i*X+j*Y)*c);
 +
    }
 +
  }
 +
  draw((0,0)--(X+Y)--(2*X+Y)--(3*X+2*Y)--(4*X+2*Y)--(5*X+2*Y),linewidth(1.5));
 +
  path q = (3,-3sqrt(3))--(-3,-3sqrt(3));
 +
  draw(q,Arrows(TeXHead, 1));
 +
  label("$n = 6$", q, S);
 +
label("$n = 6$", q, S);
 +
</asy>
  
 
In terms of <math>n</math>, find the greatest <math>k</math> such that in each Japanese triangle there is a ninja path containing at least <math>k</math> red circles.
 
In terms of <math>n</math>, find the greatest <math>k</math> such that in each Japanese triangle there is a ninja path containing at least <math>k</math> red circles.

Latest revision as of 08:45, 5 July 2024

Problems of the 2023 IMO.

Day I

Problem 1

Determine all composite integers $n>1$ that satisfy the following property: if $d_1,d_2,\dots,d_k$ are all the positive divisors of $n$ with $1=d_1<d_2<\dots<d_k=n$, then $d_i$ divides $d_{i+1}+d_{i+2}$ for every $1\le i \le k-2$.

Solution

Problem 2

Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.

Solution

Problem 3

For each integer $k \geqslant 2$, determine all infinite sequences of positive integers $a_1, a_2, \ldots$ for which there exists a polynomial $P$ of the form $P(x)=x^k+c_{k-1} x^{k-1}+\cdots+c_1 x+c_0$, where $c_0, c_1, \ldots, c_{k-1}$ are non-negative integers, such that \[P\left(a_n\right)=a_{n+1} a_{n+2} \cdots a_{n+k}\] for every integer $n \geqslant 1$.

Solution

Day II

Problem 4

Let $x_1, x_2, \cdots , x_{2023}$ be pairwise different positive real numbers such that \[a_n = \sqrt{(x_1+x_2+ \text{···} +x_n)(\frac1{x_1} + \frac1{x_2} + \text{···} +\frac1{x_n})}\] is an integer for every $n = 1,2,\cdots,2023$. Prove that $a_{2023} \ge 3034$.

Solution

Problem 5

Let $n$ be a positive integer. A Japanese triangle consists of $1 + 2 + \dots + n$ circles arranged in an equilateral triangular shape such that for each $i = 1$, $2$, $\dots$, $n$, the $i^{th}$ row contains exactly $i$ circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row. Here is an example of a Japanese triangle with $n = 6$, along with a ninja path in that triangle containing two red circles.

[asy] // credit to vEnhance for the diagram (which was better than my original asy): size(4cm); pair X = dir(240); pair Y = dir(0); path c = scale(0.5)*unitcircle; int[] t = {0,0,2,2,3,0}; for (int i=0; i<=5; ++i) { for (int j=0; j<=i; ++j) { filldraw(shift(i*X+j*Y)*c, (t[i]==j) ? lightred : white); draw(shift(i*X+j*Y)*c); } } draw((0,0)--(X+Y)--(2*X+Y)--(3*X+2*Y)--(4*X+2*Y)--(5*X+2*Y),linewidth(1.5)); path q = (3,-3sqrt(3))--(-3,-3sqrt(3)); draw(q,Arrows(TeXHead, 1)); label("$n = 6$", q, S); label("$n = 6$", q, S); [/asy]

In terms of $n$, find the greatest $k$ such that in each Japanese triangle there is a ninja path containing at least $k$ red circles.

Solution

Problem 6

Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and \[\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ\]Let $BC_1$ and $CB_1$ meet at $A_2,$ let $CA_1$ and $AC_1$ meet at $B_2,$ and let $AB_1$ and $BA_1$ meet at $C_2.$

Prove that if triangle $A_1B_1C_1$ is scalene, then the three circumcircles of triangles $AA_1A_2, BB_1B_2$ and $CC_1C_2$ all pass through two common points.

Solution

See Also

2023 IMO (Problems) • Resources
Preceded by
2022 IMO 		1 • 2 • 3 • 4 • 5 • 6 Followed by
2024 IMO
All IMO Problems and Solutions
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