Difference between revisions of "2019 AMC 10B Problems/Problem 4"
Kevinmathz (talk | contribs) (Undo revision 103400 by Sevenoptimus (talk)) (Tag: Undo) |
m (changed headers) |
||
Line 9: | Line 9: | ||
\qquad\textbf{(E) } (1,2)</math> | \qquad\textbf{(E) } (1,2)</math> | ||
− | ==Solution 1== | + | ==Solution== |
+ | ===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> | ||
Line 16: | Line 17: | ||
~IronicNinja | ~IronicNinja | ||
− | ==Solution 2== | + | ===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>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>. | 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>. | ||
Revision as of 19:15, 1 February 2020
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 .
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.