Difference between revisions of "2019 AMC 10B Problems/Problem 4"
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\qquad\textbf{(E) } (1,2)</math> | \qquad\textbf{(E) } (1,2)</math> | ||
− | ==Solution 1== | + | ==Solution== |
+ | ===Solution 1=== | ||
− | If all lines satisfy the | + | If all lines satisfy the condition, then we can just plug in values for <math>a</math>, <math>b</math>, and <math>c</math> that form an arithmetic progression. Let's use <math>a=1</math>, <math>b=2</math>, <math>c=3</math>, and <math>a=1</math>, <math>b=3</math>, <math>c=5</math>. Then the two lines we get are: <cmath>x+2y=3</cmath> <cmath>x+3y=5</cmath> |
− | Use elimination | + | Use elimination to deduce <cmath>y = 2</cmath> and plug this into one of the previous line equations. We get <cmath>x+4 = 3 \Rightarrow x=-1</cmath> Thus the common point is <math>\boxed{\textbf{(A) } (-1,2)}</math>. |
− | + | ~IronicNinja | |
− | ==Solution 2== | + | ===Solution 2=== |
− | We know that <math>a,b,c</math> | + | We know that <math>a</math>, <math>b</math>, and <math>c</math> form an arithmetic progression, so if the common difference is <math>d</math>, we can say <math>a,b,c = a, a+d, a+2d.</math> Now we have <math>ax+ (a+d)y = a+2d</math>, and expanding gives <math>ax + ay + dy = a + 2d.</math> Factoring gives <math>a(x+y-1)+d(y-2) = 0</math>. Since this must always be true (regardless of the values of <math>x</math> and <math>y</math>), we must have <math>x+y-1 = 0</math> and <math>y-2 = 0</math>, so <math>x,y = -1, 2,</math> and the common point is <math>\boxed{\textbf{(A) } (-1,2)}</math>. |
+ | |||
+ | |||
+ | ===Solution 3=== | ||
+ | We use process of elimination. <math>\textbf{B}</math> doesn't necessarily work because <math>b = c</math> isn't always true. <math>\textbf{C, D, E}</math> also doesn't necessarily work because the x-value is <math>1</math>, but the y-value is an integer. So by process of elimination, <math>\boxed{\textbf{(A) } (-1, 2)}</math> is our answer. ~Baolan | ||
+ | |||
+ | ==Solution 4== | ||
+ | We know that in <math>ax + by = c</math>, <math>a</math>, <math>b</math>, and <math>c</math> are in an arithmetic progression. We can simplify any arithmetic progression to be <math>0</math>, <math>1</math>, <math>2</math>, and <math>-1</math>, <math>0</math>, <math>1</math>. | ||
+ | |||
+ | For example, the progression <math>2</math>, <math>4</math>, <math>6</math> can be rewritten as <math>0</math>, <math>2</math>, <math>4</math> by going back by one value. We can then divide all 3 numbers by 2 which gives us <math>0</math>, <math>1</math>, <math>2</math>. | ||
+ | |||
+ | Now, we substitute <math>a</math>, <math>b</math>, and <math>c</math> with <math>0</math>, <math>1</math>, <math>2</math>, and <math>-1</math>, <math>0</math>, <math>1</math> respectively. This gives us | ||
+ | |||
+ | <math>y = 2</math> and <math>-x = 1</math> which can be written as <math>x = -1</math>. The only point of intersection is <math>(-1,2)</math>. So, our answer is | ||
+ | |||
+ | <math>\boxed{\textbf{(A) } (-1, 2)}</math>. | ||
+ | ~Starshooter11 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/kB_dR5H7Pzw | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 11:49, 17 June 2020
Contents
Problem
All lines with equation such that form an arithmetic progression pass through a common point. What are the coordinates of that point?
Solution
Solution 1
If all lines satisfy the condition, then we can just plug in values for , , and that form an arithmetic progression. Let's use , , , and , , . Then the two lines we get are: Use elimination to deduce and plug this into one of the previous line equations. We get Thus the common point is .
~IronicNinja
Solution 2
We know that , , and form an arithmetic progression, so if the common difference is , we can say Now we have , and expanding gives Factoring gives . Since this must always be true (regardless of the values of and ), we must have and , so and the common point is .
Solution 3
We use process of elimination. doesn't necessarily work because isn't always true. also doesn't necessarily work because the x-value is , but the y-value is an integer. So by process of elimination, is our answer. ~Baolan
Solution 4
We know that in , , , and are in an arithmetic progression. We can simplify any arithmetic progression to be , , , and , , .
For example, the progression , , can be rewritten as , , by going back by one value. We can then divide all 3 numbers by 2 which gives us , , .
Now, we substitute , , and with , , , and , , respectively. This gives us
and which can be written as . The only point of intersection is . So, our answer is
. ~Starshooter11
Video Solution
~savannahsolver
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
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