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== | |
− | |||
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> | 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> | ||
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~IronicNinja | ~IronicNinja | ||
− | + | ==Solution 2== | |
− | 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> | + | 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>a</math> and <math>d</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 | 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 | ||
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~Starshooter11 | ~Starshooter11 | ||
+ | ==Solution 5 (Solution 1 but faster and easier)== | ||
+ | Since the problem doesn't specify any further conditions other than an arithmetic sequence (i.e., that the numbers have to be increasing, or positive or something like that) we choose the sequences <math>(0, 1, 2)</math> and <math>(1, 0, -1)</math> which correspond to the equations <math>0x+y=2</math> and <math>1x+0y=-1</math>. These just simplify to <math>x=-1</math> and <math>y=2</math>, so the coordinate is <math>(-1, 2)</math>. | ||
+ | |||
+ | ~JH. L | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/9QZlOagfFMs | ||
− | + | ~Education, the Study of Everything | |
− | |||
==Video Solution== | ==Video Solution== |
Latest revision as of 09:24, 24 June 2023
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 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
Solution 5 (Solution 1 but faster and easier)
Since the problem doesn't specify any further conditions other than an arithmetic sequence (i.e., that the numbers have to be increasing, or positive or something like that) we choose the sequences and which correspond to the equations and . These just simplify to and , so the coordinate is .
~JH. L
Video Solution
~Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.