Difference between revisions of "2021 AMC 12B Problems/Problem 25"

Revision as of 10:26, 11 July 2021

The following problem is from both the 2021 AMC 10B #25 and 2021 AMC 12B #25, so both problems redirect to this page.

Problem

Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$

$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$

Solution 4

As the procedure shown in the Solution 1, the lower bound of $m$ is $\frac{2}{3}.$ Here I give a more logic way to show how to find the upper bound of $m.$ Denote N= $\sum_{x=1}^{30}(\lfloor mx \rfloor)$ as the number of lattice points in $S$ .

$N = \lfloor m \rfloor+\lfloor 2m \rfloor+\lfloor 3m \rfloor+\cdots+\lfloor 30m \rfloor = 300 .$

Let $m = \frac{2}{3}+k$ . for $\forall x_{i}\le 30, x\in N^{*}, \lfloor mx_{i} \rfloor = \lfloor \frac{2}{3}x+xk \rfloor.$

Our target is finding the minimum value of $k$ which can increase one unit of $\lfloor mx_{i} \rfloor .$

Notice that:

When $x_{i} = 3n, \frac{2}{3}x_{i}=2n$

When $x_{i} = 3n+1, \lfloor mx_{i} \rfloor = \lfloor 2n+\frac{2}{3}+(3n+1)k \rfloor$

When $x_{i} = 3n+2, \lfloor mx_{i} \rfloor = \lfloor 2n+1+\frac{1}{3}+(3n+2)k \rfloor$ . Here $n\in N^{*}, n \le 9.$

Denote $k_{min}=min\left \{k_{min1},k_{min2} \right \}.$

Obviously $k_{min1}$ and $k_{min2}$ are decreasing. Let's considering the situation when $n=9.$

$k_{min}=min\left\{\frac{1}{84},\frac{2}{87}\right\}=\frac{1}{84}.$

So $m_{max}=\frac{2}{3}+\frac{1}{84}=\frac{19}{28}.$

Above all, $\frac{b}{a}=\frac{19}{28}-\frac{2}{3}=\frac{1}{84}, a+b = 85.$ Choose $\boxed{C}.$

~PythZhou.

Solution 2

I know that I want about $\frac{2}{3}$ of the box of integer coordinates above my line. There are a total of 30 integer coordinates in the desired range for each axis which gives a total of 900 lattice points. I estimate that the slope, m, is $\frac{2}{3}$ . Now, although there is probably an easier solution, I would try to count the number of points above the line to see if there are 600 points above the line. The line $y=\frac{2}{3}x$ separates the area inside the box so that $\frac{2}{3}$ of the are is above the line.

I find that the number of coordinates with $x=1$ above the line is 30, and the number of coordinates with $x=2$ above the line is 29. Every time the line $y=\frac{2}{3}x$ hits a y-value with an integer coordinate, the number of points above the line decreases by one. I wrote out the sum of 30 terms in hopes of finding a pattern. I graphed the first couple positive integer x-coordinates, and found that the sum of the integers above the line is $30+29+28+28+27+26+26 \ldots+ 10$ . The even integer repeats itself every third term in the sum. I found that the average of each of the terms is 20, and there are 30 of them which means that exactly 600 above the line as desired. This give a lower bound because if the slope decreases a little bit, then the points that the line goes through will be above the line.

To find the upper bound, notice that each point with an integer-valued x-coordinate is either $\frac{1}{3}$ or $\frac{2}{3}$ above the line. Since the slope through a point is the y-coordinate divided by the x-coordinate, a shift in the slope will increase the y-value of the higher x-coordinates. We turn our attention to $x=28, 29, 30$ which the line $y=\frac{2}{3}x$ intersects at $y= \frac{56}{3}, \frac{58}{3}, 20$ . The point (30,20) is already counted below the line, and we can clearly see that if we slowly increase the slope of the line, we will hit the point (28,19) since (28, $\frac{56}{3}$ ) is closer to the lattice point. The slope of the line which goes through both the origin and (28,19) is $y=\frac{19}{28}x$ . This gives an upper bound of $m=\frac{19}{28}$ .

Taking the upper bound of m and subtracting the lower bound yields $\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$ . This is answer $1+84=$ $\boxed{\textbf{(E)} ~85}$ .

~theAJL

Diagram

$[asy] /* Created by Brendanb4321 */ import graph; size(16cm); defaultpen(fontsize(9pt)); xaxis(0,30,Ticks(1.0)); yaxis(0,25,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30/28*19), dotted); for (int i = 1; i<=30; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); label("$m=2/3$", (32,20)); label("$m=19/28$", (32.3,20.8)); [/asy]$

Solution 3

An alternative approach with the same methodology as Solution 1 can be done using Pick's Theorem. Wikipedia page: https://en.wikipedia.org/wiki/Pick%27s_theorem It's a formula to find the amount of lattice points strictly inside a polygon. Approximation of the lower bound is still necessary.

Video Solution , Very Easy

https://youtu.be/PC8fIZzICFg ~hippopotamus1

(Video solution is in Chinese) ~jhu08

Video Solution by Interstigation (In-Depth, Straight-forward)

https://youtu.be/aAogEAOcL2Y

~Interstigation

@@ Line 29: / Line 29: @@
 Obviously <math>k_{min1} </math> and <math>k_{min2}</math> are decreasing. Let's considering the situation when <math>n=9.</math>
-<math>k_{min}=min{\frac{1}{84},\frac{2}{87}}=\frac{1}{84}.</math>
+<math>k_{min}=min\left\{\frac{1}{84},\frac{2}{87}\right\}=\frac{1}{84}.</math>
 So <math>m_{max}=\frac{2}{3}+\frac{1}{84}=\frac{19}{28}. </math>

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last problem
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 AMC 12 Problems and Solutions

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last problem
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 AMC 10 Problems and Solutions

Page

Toolbox

Search