Difference between revisions of "2019 AMC 10B Problems/Problem 4"
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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 | ||
− | + | =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>. | 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>. | ||
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<math>\boxed{\textbf{(A) } (-1, 2)}</math>. | <math>\boxed{\textbf{(A) } (-1, 2)}</math>. | ||
~Starshooter11 | ~Starshooter11 | ||
− | |||
==Solution 5 (Solution 1 but faster and easier)== | ==Solution 5 (Solution 1 but faster and easier)== |
Revision as of 02:28, 9 June 2022
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
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
~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.