AMC 10 · 2019 · #4
Grade 6 algebraProblem
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?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Consider every line of the form $a x + b y = c$ where $a$, $b$, $c$ are three numbers in arithmetic progression (each next term differs from the previous by the same amount). All such lines pass through one common point. Find the coordinates of that point.
Givens: Line equation form: $a x + b y = c$; $a$, $b$, $c$ form an arithmetic progression: $b - a = c - b$ (i.e., $c = 2b - a$); Every choice of arithmetic-progression $(a, b, c)$ gives a line passing through the same point; Answer choices: (A) $(-1, 2)$, (B) $(0, 1)$, (C) $(1, -2)$, (D) $(1, 0)$, (E) $(1, 2)$
Unknowns: The common point $(x, y)$
Understand
Restated: Consider every line of the form $a x + b y = c$ where $a$, $b$, $c$ are three numbers in arithmetic progression (each next term differs from the previous by the same amount). All such lines pass through one common point. Find the coordinates of that point.
Givens: Line equation form: $a x + b y = c$; $a$, $b$, $c$ form an arithmetic progression: $b - a = c - b$ (i.e., $c = 2b - a$); Every choice of arithmetic-progression $(a, b, c)$ gives a line passing through the same point; Answer choices: (A) $(-1, 2)$, (B) $(0, 1)$, (C) $(1, -2)$, (D) $(1, 0)$, (E) $(1, 2)$
Plan
Primary tool: #3 Eliminate Possibilities
Secondary: #9 Solve an Easier Related Problem, #6 Guess and Check
Tool #9 (Easier Problem): we don't need to handle every AP — pick the two simplest ones, say $(a, b, c) = (1, 2, 3)$ and $(1, 3, 5)$. These give two specific lines $x + 2y = 3$ and $x + 3y = 5$. Their common point must include the answer. Tool #3 (Eliminate) + Tool #6 (Guess and Check): plug each of the five answer choices into both lines; the only point that satisfies BOTH equations is the answer. No algebra needed.
Execute — Answer: A
4.OA.C.5 Step 1 - Pick two simple arithmetic progressions and write down the resulting lines.
- Use $(a, b, c) = (1, 2, 3)$: equation $1 \cdot x + 2 \cdot y = 3$.
- Use $(a, b, c) = (1, 3, 5)$: equation $1 \cdot x + 3 \cdot y = 5$.
- The common point must satisfy BOTH.
💡 Two concrete APs give two concrete lines — easier to test.
6.EE.A.2 Step 2 - Test each answer choice in Line 1: $x + 2y = 3$.
- (A) $(-1, 2)$: $-1 + 4 = 3$ ✓.
- (B) $(0, 1)$: $0 + 2 = 2 \ne 3$ ✗.
- (C) $(1, -2)$: $1 - 4 = -3 \ne 3$ ✗.
- (D) $(1, 0)$: $1 + 0 = 1 \ne 3$ ✗.
- (E) $(1, 2)$: $1 + 4 = 5 \ne 3$ ✗.
- Only (A) survives.
💡 Plug each candidate into the first line; eliminate the ones that fail.
6.EE.A.2 Step 3 - Confirm (A) $(-1, 2)$ also satisfies Line 2: $x + 3y = 5$.
- Substitute: $-1 + 3(2) = -1 + 6 = 5$ ✓.
- Both lines passed — so $(-1, 2)$ is on both.
💡 Double-check the surviving point on the second line.
5.G.A.1 Step 4 - Verify with a third AP to be sure $(-1, 2)$ works for every AP, not just the two we picked.
- Try $(a, b, c) = (2, 5, 8)$ (common difference $3$): $2(-1) + 5(2) = -2 + 10 = 8 = c$ ✓.
- Try $(a, b, c) = (3, 2, 1)$ (common difference $-1$): $3(-1) + 2(2) = -3 + 4 = 1 = c$ ✓.
- The pattern holds, so the common point is $(-1, 2)$.
💡 If a point survives on every AP we test, it's the universal common point.
4.OA.C.5 Pick two simple arithmetic progressions and write down the resulting lines. Use 6.EE.A.2 Test each answer choice in Line 1: $x + 2y = 3$. (A) $(-1, 2)$: $-1 + 4 = 3$ ✓. 6.EE.A.2 Confirm (A) $(-1, 2)$ also satisfies Line 2: $x + 3y = 5$. Substitute: $-1 + 3(2 5.G.A.1 Verify with a third AP to be sure $(-1, 2)$ works for every AP, not just the two Review
Reasonableness: Sanity: an AP gives $c = 2b - a$. Plug $(x, y) = (-1, 2)$ into $a x + b y$: $a(-1) + b(2) = -a + 2b = 2b - a = c$ ✓. So for ANY arithmetic-progression triple $(a, b, c)$, the line $ax + by = c$ passes through $(-1, 2)$. The answer holds universally — not just for our test cases.
Alternative: Tool #13 (Algebra): write $c = 2b - a$ from the AP condition. Substitute into $ax + by = c$: $ax + by = 2b - a$, regroup as $a(x + 1) + b(y - 2) = 0$. For this to hold for all choices of $a$ and $b$, both coefficients must vanish: $x + 1 = 0$ and $y - 2 = 0$, giving $(x, y) = (-1, 2)$.
CCSS standards used (min grade 6)
4.OA.C.5Generate a number or shape pattern following a given rule (Recognizing an arithmetic progression and picking simple AP triples like $(1, 2, 3)$ and $(1, 3, 5)$.)5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Reading each answer choice as a coordinate pair $(x, y)$ to test in the line equations.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating $ax + by$ for each candidate $(x, y)$ and concrete $(a, b)$ — e.g. $-1 + 2(2) = 3$.)
⭐ This AMC 10 problem only needs Grade 6 expression-evaluation you already know — pick two simple APs $(1, 2, 3)$ and $(1, 3, 5)$, plug each answer-choice point into both lines, and only $(-1, 2)$ survives. Answer: (A)!
⭐ This AMC 10 problem only needs Grade 6 expression-evaluation you already know — pick two simple APs $(1, 2, 3)$ and $(1, 3, 5)$, plug each answer-choice point into both lines, and only $(-1, 2)$ survives. Answer: (A)!
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