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2023 IMO Problems/Problem 5: Difference between revisions

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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; also available in solution video]
<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.


==Solution==
==Solution==
<math>k=[log2(n)]+1</math>
https://www.youtube.com/watch?v=jZNIpapyGJQ [Video contains solutions to all day 2 problems]
https://www.youtube.com/watch?v=jZNIpapyGJQ [Video contains solutions to all day 2 problems]


Latest revision as of 08:46, 5 July 2024

Problem

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

https://www.youtube.com/watch?v=jZNIpapyGJQ [Video contains solutions to all day 2 problems]

See Also

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