AMC 10 · 2012 · #3

Easy mode Grade 5

Problem

On a grid, there is a point at $(1000, 2012)$ . There is also a flat, side-to-side line at height $y = 2000$ , like a mirror lying across the grid. Reflect the point across this line to its mirror spot on the other side. What are the coordinates of the mirrored point?

Pick an answer.

(A)

(998,2012)

(B)

(1000,1988)

(C)

(1000,2024)

(D)

(1000,4012)

(E)

(1012,2012)

AMC 10 2012 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.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A point sits at $(1000, 2012)$. Fold the plane along the horizontal line $y = 2000$. Find where the point lands after that fold.

Givens: The point is at $(1000, 2012)$.; The mirror line is $y = 2000$, a horizontal line.; Reflection means the mirror line is exactly halfway between a point and its image.

Unknowns: The coordinates of the reflected point.

Understand

Restated: A point sits at $(1000, 2012)$. Fold the plane along the horizontal line $y = 2000$. Find where the point lands after that fold.

Givens: The point is at $(1000, 2012)$.; The mirror line is $y = 2000$, a horizontal line.; Reflection means the mirror line is exactly halfway between a point and its image.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

Reflection is about position, so a quick sketch of the point and the horizontal mirror line makes the mirror image easy to see. Then split the work into two small questions: what happens to $x$, and what happens to $y$. A picture also lets you throw out choices that move in the wrong direction.

Execute — Answer: B

#1 Draw a Diagram 4.G.A.3 Step 1

Sketch the point and the mirror line

Draw the horizontal line $y = 2000$ and mark the point $(1000, 2012)$ above it.
A horizontal mirror flips things up and down, not left and right, so the left-right position ($x = 1000$) does not change.
Only the height ($y$) moves.

$$x = 1000 \text{ stays the same}$$

💡 Folding along a horizontal crease only swaps up for down, so the sideways position is untouched.

#7 Identify Subproblems 4.NBT.B.4 Step 2

Measure the gap to the line

Find how far the point is above the mirror line by subtracting the two heights.
The point is at height $2012$ and the line is at height $2000$, so the gap is $2012 - 2000 = 12$.

$$2012 - 2000 = 12$$

💡 The distance from a point to a horizontal line is just the difference of their heights.

#7 Identify Subproblems 4.NBT.B.4 Step 3

Drop the same distance below

A reflection keeps the same distance on the other side of the line.
The point was $12$ above the line, so its image is $12$ below the line.
Start at the line's height $2000$ and go down $12$: $2000 - 12 = 1988$.

$$2000 - 12 = 1988$$

💡 The mirror line sits exactly halfway, so the image is the same gap on the far side.

#3 Eliminate Possibilities 5.G.A.1 Step 4

Write the reflected coordinates

Put the unchanged $x$ together with the new $y$.
The reflected point is $(1000, 1988)$.
Choices that change $x$ or move the point up can be crossed out, since a horizontal mirror only moves the point straight down here.
This matches answer (B).

$$(1000,\ 1988)$$

💡 An ordered pair pins the point down: keep the steady coordinate, swap in the mirrored one.

[1] #1 4.G.A.3 Draw the horizontal line $y = 2000$ and mark the point $(1000, 2012)$ above it.

[2] #7 4.NBT.B.4 Find how far the point is above the mirror line by subtracting the two heights.

[3] #7 4.NBT.B.4 A reflection keeps the same distance on the other side of the line. The point wa

[4] #3 5.G.A.1 Put the unchanged $x$ together with the new $y$. The reflected point is $(1000,

Review

Reasonableness: The line $y = 2000$ should be the midpoint of the point and its image in height. The two heights are $2012$ and $1988$, and their average is $(2012 + 1988) / 2 = 2000$, exactly the mirror line. The $x$-coordinate stayed $1000$, as a horizontal mirror requires. So $(1000, 1988)$, answer (B), is consistent.

Alternative: Use the midpoint idea directly: the mirror height $2000$ must be halfway between the old height $2012$ and the new height $h$, so $2000 = (2012 + h)/2$, giving $h = 4000 - 2012 = 1988$. Same result without measuring the gap first.

CCSS standards used (min grade 5)

4.G.A.3 Recognize a line of symmetry for a two-dimensional figure (Seeing that a horizontal mirror line keeps the left-right position and flips only the height.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Computing the distance $2012 - 2000 = 12$ and the new height $2000 - 12 = 1988$.)
5.G.A.1 Use a pair of perpendicular number lines forming a coordinate system (Reading and writing the point as an ordered pair $(x, y)$ in the coordinate plane.)

⭐ To mirror a point across a flat line, keep the sideways number and put the height the same distance on the other side.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Sketch the point and the mirror line

Measure the gap to the line

Drop the same distance below

Write the reflected coordinates

Review

CCSS standards used (min grade 5)

More like this