2019 AMC 10B Problems/Problem 4
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
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.