2019 AMC 10B Problem 4

P1-10~2 minPrint

Coordinate Geometry Sequences & Series Set up the equation/formula and compute; no special trick needed Test small/specific values or special cases to find or verify the answer

All lines with equation $ax+by=c$ such that $a,b,c$ form an arithmetic progression pass through a common point. What are the coordinates of that point?

Show solution without answering

Key insight

If every such line passes through one point, any two of them will find it. Use the easy progressions 1, 1, 1 and 1, 2, 3.

Solution

The problem promises that all these lines meet at one point. So two of them are enough to find it. Pick the simplest arithmetic progressions.

$a, b, c = 1, 1, 1$ gives the line $x + y = 1$ .
$a, b, c = 1, 2, 3$ gives the line $x + 2y = 3$ .
Subtract the first equation from the second: $y = 2$ .
Put $y = 2$ into $x + y = 1$ : $x = -1$ .

So the point should be $(-1, 2)$ .

Check it with a third line, just to be safe. Take $a, b, c = 2, 5, 8$ (common difference $3$ ): $2(-1) + 5(2) = -2 + 10 = 8$ . Yes, $(-1, 2)$ is on $2x + 5y = 8$ too.

Why it always works: in an arithmetic progression the middle term is the average of the other two, so $c = 2b - a$ . Plugging in $x = -1$ , $y = 2$ gives $a(-1) + b(2) = 2b - a$ , which is exactly $c$ .

The answer is $\boxed{\textbf{(A)}\ (-1,2)}$ .

Why this works

When a problem says "all of these pass through a common point," you do not need the general case to find it. Two simple members of the family pin the point down, and a third member is a quick check. The general reason is that the progression rule $c = 2b - a$ looks exactly like the line equation with $x = -1$ and $y = 2$ .

Alternative approach

Test the answer choices in a single line of the family, such as $x + 2y = 3$ (from $1, 2, 3$ ). The correct point must satisfy it.

$(-1, 2)$ : $-1 + 4 = 3$ . Fits.
$(0, 1)$ : $0 + 2 = 2$ . No.
$(1, -2)$ : $1 - 4 = -3$ . No.
$(1, 0)$ : $1 + 0 = 1$ . No.
$(1, 2)$ : $1 + 4 = 5$ . No.

Only (A) survives.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

If every such line passes through one point, then any two of them locate it. Take two easy progressions.

$(a, b, c) = (1, 1, 1)$ : the line $x + y = 1$ .
$(a, b, c) = (1, 2, 3)$ : the line $x + 2y = 3$ .

Subtracting the first equation from the second gives $y = 2$ , and then $x = -1$ . So the point is $(-1, 2)$ .

Confirm it works in general: an arithmetic progression satisfies $b = \frac{a + c}{2}$ , i.e. $c = 2b - a$ . At $(-1, 2)$ the left side is $a(-1) + b(2) = 2b - a = c$ , so the point lies on every line in the family.

The answer is $\boxed{\textbf{(A)}\ (-1,2)}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 8: Graphing Lines
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)