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Difference between revisions of "2024 AMC 8 Problems/Problem 23"

m (→‎Solution 1)
(→‎Solution 1)
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~mathkiddus
 
~mathkiddus
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==Proof of This Claim==
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\textbf{Lemma 1 for Problem 23:}\\
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Let <math>p</math> and <math>q</math> be relatively prime positive integers. When a <math>p\times q</math> rectangle is split up into <math>pq</math> unit squares, exactly <math>p + q - 1</math> unit squares are crossed by the diagonal of this rectangle.
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\textbf{Proof:}\\
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First, we claim that the diagonal does not cross the corner of a unit square. \\\\
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To prove this claim we proceed by way of contradiction. Plot the rectangle on the Cartesian plane at the vertices <math>(0,0),(p,0),(q,p),(0,q).</math> The diagonal has endpoints at <math>(0,0),(q,p)</math>, so its slope is <math>\frac{p}{q}.</math>  Now, suppose the diagonal goes through the corner point <math>(a,b)</math>, where <math>a<q</math> and <math>b<p</math>. The slope of this line is <math>\frac{b}{a}</math>, which must be equal to <math>\frac{p}{q},</math> implying that <math>\frac{p}{q}</math> can be reduced, contradicting the fact that <math>p</math> and <math>q</math> are relatively prime. We conclude that no corner points of a grid entry (unit square) are crossed. \\\\
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Since no corner points are crossed, each time the diagonal crosses either a horizontal or vertical grid line, exactly one more unit square is touched by the diagonal. There are <math>p-1</math> horizontal lines and <math>q-1</math> vertical lines, so there are <math>p + q - 2</math> total lines crossed by the diagonal. This doesn't include the square in the bottom left corner, crossed initially. Therefore, there are <math>p + q-2 + 1 = p + q - 1</math> unit squares crossed by the diagonal and our claim is proven.
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\textbf{Lemma 2 for Problem 23:}\\
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Let <math>p</math> and <math>q</math> be positive integers. When a <math>p\times q</math> rectangle is split up into <math>pq</math> units squares, exactly <math>p + q - \gcd(p, q)</math> unit squares are crossed by the diagonal of this rectangle.
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If <math>\gcd (p,q) = 1</math>, then we are done by Lemma 1. \\\\
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Suppose <math>\gcd(p,q) = k>1</math>, i.e <math>p = ak</math> and <math>q = bk</math>, for positive integers <math>a</math> and <math>b</math>. We can then split the <math>p\times q</math> rectangle up into <math>k</math> <math>\frac{p}{k} \times \frac{q}{k}</math> rectangles, strung together at the diagonal. An example for <math>(p,q)=(4,6)</math> is shown below, where two <math>2\times 3</math> rectangles are strung together:
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<asy>
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unitsize(1cm);
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draw((0,0)--(6,4),linewidth(1));
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currentpen = linewidth(.5);
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for (real i = 0; i <= 6; ++i) {
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    draw((i, 0)--(i, 4));
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}
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for (real i = 0; i < 5; ++i) {
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    draw((0, i)--(6, i));
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}
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currentpen = linewidth(1.5);
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for (real i = 0; i <= 3; ++i) {
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    draw((i, 0)--(i, 2));
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}
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for (real i = 0; i <= 2; ++i) {
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    draw((0, i)--(3, i));
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}
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for (real i = 3; i <= 6; ++i) {
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    draw((i, 2)--(i, 4));
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}
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for (real i = 3; i <= 5; ++i) {
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    draw((3, i-1)--(6, i-1));
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}
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</asy>
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\\\\
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After the diagonal crosses the corner point of a square, the pattern repeats itself with the next one. By Lemma 1, there are <math>\frac{p}{k}+ \frac{q}{k} - 1</math> diagonals crossed in each rectangle. There are <math>k = \gcd (p, q)</math> rectangles, so the number of crossed diagonals in total is
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<cmath>k\left( \frac{p}{k}+ \frac{q}{k} - 1 \right) = p + q - \gcd(p,q).</cmath>
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-Benedict T (countmath1)
  
 
==Video Solution by Power Solve (crystal clear!)==
 
==Video Solution by Power Solve (crystal clear!)==

Revision as of 22:45, 26 January 2024

Problem

Rodrigo has a very large sheet of graph paper. First he draws a line segment connecting point $(0,4)$ to point $(2,0)$ and colors the $4$ cells whose interiors intersect the segment, as shown below. Next Rodrigo draws a line segment connecting point $(2000,3000)$ to point $(5000,8000)$. How many cells will he color this time?

[asy] draw((-1,5)--(-1,-1),gray(.8)); draw((0,5)--(0,-1),gray(.8)); draw((1,5)--(1,-1),gray(.8)); draw((2,5)--(2,-1),gray(.8)); draw((3,5)--(3,-1),gray(.8)); draw((4,5)--(4,-1),gray(.8)); draw((5,5)--(5,-1),gray(.8)); draw((-1,5)--(5, 5),gray(.8)); draw((-1,4)--(5,4),gray(.8)); draw((-1,3)--(5,3),gray(.8)); draw((-1,2)--(5,2),gray(.8)); draw((-1,1)--(5,1),gray(.8)); draw((-1,0)--(5,0),gray(.8)); draw((-1,-1)--(5,-1),gray(.8)); dot((0,4)); label("$(0,4)$",(0,4),NW); dot((2,0)); label("$(2,0)$",(2,0),SE); draw((0,4)--(2,0)); draw((-1,0) -- (5,0), arrow=Arrow); draw((0,-1) -- (0,5), arrow=Arrow); filldraw((0,4)--(1,4)--(1,3)--(0,3)--cycle, gray(.8), gray(.8)+linewidth(1)); [/asy]

Solution 1

Let $f(x, y)$ be the number of cells the line segment from $(0, 0)$ to $(x, y)$ passes through. The problem is then equivalent to finding \[f(5000-2000, 8000-3000)=f(3000, 5000).\] Sometimes the segment passes through lattice points in between the endpoints, which happens $\text{gcd}(3000, 5000)-1=999$ times. This partitions the segment into $1000$ congruent pieces that each pass through $f(3, 5)$ cells, which means the answer is \[1000f(3, 5).\] Note that a new square is entered when the lines pass through one of the lines in the coordinate grid, which for $f(3, 5)$ happens $3-1+5-1=6$ times. Because $3$ and $5$ are relatively prime, no lattice point except for the endpoints intersects the line segment from $(0, 0)$ to $(3, 5).$ This means that including the first cell closest to $(0, 0),$ The segment passes through $f(3, 5)=6+1=7$ cells. Thus, the answer is $\boxed{7000}.$ Alternatively, $f(3, 5)$ can be found by drawing an accurate diagram, leaving you with the same answer.

~BS2012

Note: A general form for finding $f(x, y)$ is $x+y-\text{gcd}(x, y).$ We subtract $\text{gcd}(x, y)$ to account for overlapping, when the line segment goes through a lattice point.

~mathkiddus

Proof of This Claim

\textbf{Lemma 1 for Problem 23:}\\ Let $p$ and $q$ be relatively prime positive integers. When a $p\times q$ rectangle is split up into $pq$ unit squares, exactly $p + q - 1$ unit squares are crossed by the diagonal of this rectangle.


\textbf{Proof:}\\ First, we claim that the diagonal does not cross the corner of a unit square. \\\\ To prove this claim we proceed by way of contradiction. Plot the rectangle on the Cartesian plane at the vertices $(0,0),(p,0),(q,p),(0,q).$ The diagonal has endpoints at $(0,0),(q,p)$, so its slope is $\frac{p}{q}.$ Now, suppose the diagonal goes through the corner point $(a,b)$, where $a<q$ and $b<p$. The slope of this line is $\frac{b}{a}$, which must be equal to $\frac{p}{q},$ implying that $\frac{p}{q}$ can be reduced, contradicting the fact that $p$ and $q$ are relatively prime. We conclude that no corner points of a grid entry (unit square) are crossed. \\\\ Since no corner points are crossed, each time the diagonal crosses either a horizontal or vertical grid line, exactly one more unit square is touched by the diagonal. There are $p-1$ horizontal lines and $q-1$ vertical lines, so there are $p + q - 2$ total lines crossed by the diagonal. This doesn't include the square in the bottom left corner, crossed initially. Therefore, there are $p + q-2 + 1 = p + q - 1$ unit squares crossed by the diagonal and our claim is proven.

\textbf{Lemma 2 for Problem 23:}\\ Let $p$ and $q$ be positive integers. When a $p\times q$ rectangle is split up into $pq$ units squares, exactly $p + q - \gcd(p, q)$ unit squares are crossed by the diagonal of this rectangle.


If $\gcd (p,q) = 1$, then we are done by Lemma 1. \\\\ Suppose $\gcd(p,q) = k>1$, i.e $p = ak$ and $q = bk$, for positive integers $a$ and $b$. We can then split the $p\times q$ rectangle up into $k$ $\frac{p}{k} \times \frac{q}{k}$ rectangles, strung together at the diagonal. An example for $(p,q)=(4,6)$ is shown below, where two $2\times 3$ rectangles are strung together:

[asy] unitsize(1cm); draw((0,0)--(6,4),linewidth(1)); currentpen = linewidth(.5); for (real i = 0; i <= 6; ++i) { draw((i, 0)--(i, 4)); } for (real i = 0; i < 5; ++i) { draw((0, i)--(6, i)); } currentpen = linewidth(1.5); for (real i = 0; i <= 3; ++i) { draw((i, 0)--(i, 2)); } for (real i = 0; i <= 2; ++i) { draw((0, i)--(3, i)); } for (real i = 3; i <= 6; ++i) { draw((i, 2)--(i, 4)); } for (real i = 3; i <= 5; ++i) { draw((3, i-1)--(6, i-1)); } [/asy] \\\\ After the diagonal crosses the corner point of a square, the pattern repeats itself with the next one. By Lemma 1, there are $\frac{p}{k}+ \frac{q}{k} - 1$ diagonals crossed in each rectangle. There are $k = \gcd (p, q)$ rectangles, so the number of crossed diagonals in total is \[k\left( \frac{p}{k}+ \frac{q}{k} - 1 \right) = p + q - \gcd(p,q).\]

-Benedict T (countmath1)

Video Solution by Power Solve (crystal clear!)

https://www.youtube.com/watch?v=fzgWcEz4K_A

Video Solution 1 by Math-X (First fully understand the problem!!!)

https://www.youtube.com/watch?v=dqqAk-Cd_5M

~Math-X

Video Solution 2 by OmegaLearn.org

https://youtu.be/wNymnFQfN_k

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=x8Zo7QOB-us

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=w8zha2ijVQQ

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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