AMC 10 · 2014 · #23
Grade 8 geometry-2dA rectangular piece of paper whose length is 3 times the width has area A. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area B. What is the ratio AB?
Pick an answer.
AMC 10 2014 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: A rectangle is $\sqrt3$ times as long as it is wide and has area $A$. Its two long edges are each split into three equal parts, and a straight crease is drawn from the first split point on one long edge to the second split point on the opposite long edge. The sheet is folded flat along this crease, and the folded silhouette has area $B$. Find the ratio $\frac{B}{A}$.
Givens: The rectangle's length is $\sqrt3$ times its width, and its area is $A$.; Each long edge is divided into three equal segments.; The crease runs from the first divider on one long edge to the second divider on the opposite long edge.; Folding flat along the crease produces a new shape of area $B$.; Answer choices: (A) $\frac{1}{2}$, (B) $\frac{3}{5}$, (C) $\frac{2}{3}$, (D) $\frac{3}{4}$, (E) $\frac{4}{5}$.
Unknowns: The ratio $\frac{B}{A}$ of the folded shape's area to the original area.
Understand
Restated: A rectangle is $\sqrt3$ times as long as it is wide and has area $A$. Its two long edges are each split into three equal parts, and a straight crease is drawn from the first split point on one long edge to the second split point on the opposite long edge. The sheet is folded flat along this crease, and the folded silhouette has area $B$. Find the ratio $\frac{B}{A}$.
Givens: The rectangle's length is $\sqrt3$ times its width, and its area is $A$.; Each long edge is divided into three equal segments.; The crease runs from the first divider on one long edge to the second divider on the opposite long edge.; Folding flat along the crease produces a new shape of area $B$.; Answer choices: (A) $\frac{1}{2}$, (B) $\frac{3}{5}$, (C) $\frac{2}{3}$, (D) $\frac{3}{4}$, (E) $\frac{4}{5}$.
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #1 Draw a Diagram, #7 Identify Subproblems, #4 Introduce a Variable
The crease and the fold are a physical, spatial action, so Tool #17 (Visualize Spatial Relationships) is the anchor: you must picture the flap flipping over the crease before any number helps. Tool #1 (Draw a Diagram) turns that mental picture into coordinates, so 'where does the corner land' becomes something you can compute. Tool #7 (Identify Subproblems) splits the task into two clean pieces: first see that $B=A-\text{overlap}$, then measure the overlap. Tool #4 (Introduce a Variable) fixes a convenient width of $1$ so every area is a concrete number.
Execute — Answer: C
6.NS.C.8 Step 1 Fix a width and set coordinates
- The ratio $B/A$ does not depend on the sheet's real size, so choose the width to be $1$.
- Then the length is $\sqrt3$ and the area is $A=\sqrt3\cdot 1=\sqrt3$.
- Place the sheet with corners $(0,0)$, $(\sqrt3,0)$, $(\sqrt3,1)$, $(0,1)$.
- The long bottom and top edges have length $\sqrt3$, so their dividers sit at $x=\tfrac{\sqrt3}{3}$ and $x=\tfrac{2\sqrt3}{3}$.
- The crease runs from the first bottom divider $P=\left(\tfrac{\sqrt3}{3},0\right)$ up to the second top divider $Q=\left(\tfrac{2\sqrt3}{3},1\right)$.
💡 Choosing width $1$ and dropping the figure onto axes turns every corner and crease into an exact number you can measure.
8.G.A.1 Step 2 Turn folding into area minus overlap
- Folding along the crease reflects the flap on one side onto the other side.
- A reflection never stretches paper, so the total paper area is still $\sqrt3$; but the flap now lies on top of the sheet, and wherever two layers stack the outline gains nothing new.
- So the folded area is the full area minus the doubly-covered region: $B=A-\text{overlap}$.
- To find that region, see where the folded corner lands.
- The crease is the perpendicular bisector of the diagonal joining the top-left corner $D=(0,1)$ and the bottom-right corner $R=(\sqrt3,0)$: the crease $PQ$ and that diagonal share the midpoint $\left(\tfrac{\sqrt3}{2},\tfrac12\right)$, and their slopes $\sqrt3$ and $-\tfrac{1}{\sqrt3}$ multiply to $-1$.
- Reflecting over the perpendicular bisector of $DR$ therefore sends $D$ exactly onto $R$.
💡 A crease is a mirror line, and a mirror line that perpendicularly bisects two corners swaps one for the other.
6.G.A.1 Step 3 Measure the doubly-covered triangle
- The overlap is the region the flap covers when it lands.
- Its slanted side is the crease $PQ$; the folded top edge $QD$ comes down onto $QR$, and the bottom edge closes the shape from $R$ back to $P$.
- So the double layer is triangle $PQR$ with $P=\left(\tfrac{\sqrt3}{3},0\right)$, $Q=\left(\tfrac{2\sqrt3}{3},1\right)$, and $R=(\sqrt3,0)$.
- Its base $PR$ lies flat on the bottom edge with length $\sqrt3-\tfrac{\sqrt3}{3}=\tfrac{2\sqrt3}{3}$, and the apex $Q$ stands at height $1$.
- Area of a triangle is half the base times the height: $\tfrac12\cdot\tfrac{2\sqrt3}{3}\cdot 1=\tfrac{\sqrt3}{3}$.
💡 Once the overlap is a plain triangle sitting on the bottom edge, base times height over two finishes it.
6.RP.A.1 Step 4 Take the ratio B over A
- Subtract the overlap from the whole area: $B=\sqrt3-\tfrac{\sqrt3}{3}=\tfrac{2\sqrt3}{3}$.
- Then divide by $A=\sqrt3$, and the shared $\sqrt3$ cancels: $\dfrac{B}{A}=\dfrac{2\sqrt3/3}{\sqrt3}=\dfrac{2}{3}$.
- The ratio is $\tfrac{2}{3}$, which is choice (C).
💡 Because both areas carry the same $\sqrt3$, that messy radical cancels and a clean ratio is left.
6.NS.C.8 The ratio $B/A$ does not depend on the sheet's real size, so choose the width to 8.G.A.1 Folding along the crease reflects the flap on one side onto the other side. A re 6.G.A.1 The overlap is the region the flap covers when it lands. Its slanted side is the 6.RP.A.1 Subtract the overlap from the whole area: $B=\sqrt3-\tfrac{\sqrt3}{3}=\tfrac{2\s Review
Reasonableness: The fold only tucks one corner inward, so it should erase a modest slice of paper, not most of it; a ratio comfortably above $\tfrac12$ and below $1$ is expected, and $\tfrac23$ fits. Cross-check the overlap triangle by its sides: the two slanted sides $PQ$ and $QR$ each measure $\sqrt{\left(\tfrac{\sqrt3}{3}\right)^2+1^2}=\tfrac{2}{\sqrt3}=\tfrac{2\sqrt3}{3}$, and the base $PR$ is also $\tfrac{2\sqrt3}{3}$, so triangle $PQR$ is equilateral. That is exactly the symmetric shape a $60^\circ$ crease should fold up, which supports the overlap area $\tfrac{\sqrt3}{3}$ and the final $\tfrac23$.
Alternative: Skip coordinates and use the equilateral triangle directly. The crease length is $\sqrt{\left(\tfrac{\sqrt3}{3}\right)^2+1^2}=\tfrac{2}{\sqrt3}$, so the doubly-layered piece is equilateral with that side and area $\tfrac{\sqrt3}{4}\left(\tfrac{2}{\sqrt3}\right)^2=\tfrac{\sqrt3}{3}$. The rectangle's area $\sqrt3$ is exactly three copies of this triangle; folding buries one copy under a second layer, leaving two copies visible, so $\tfrac{B}{A}=\tfrac{2}{3}$.
CCSS standards used (min grade 8)
6.NS.C.8Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane (Placing the sheet and its crease on axes so the corners $P$, $Q$, $D$, and $R$ become exact points to measure between.)8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Treating the fold as a reflection that preserves area and sends the top-left corner $D$ onto the bottom-right corner $R$.)6.G.A.1Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles (Computing the doubly-covered triangle $PQR$ as half base times height, giving $\tfrac{\sqrt3}{3}$.)6.RP.A.1Understand the concept of a ratio and use ratio language to describe a relationship between two quantities (Forming the final ratio $B/A$ and simplifying it to $\tfrac{2}{3}$.)
⭐ Folding a corner over just hides one overlapping triangle, and here that lost triangle is one-third of the sheet, so two-thirds of the paper is left: $B/A=\tfrac{2}{3}$.
⭐ Folding a corner over just hides one overlapping triangle, and here that lost triangle is one-third of the sheet, so two-thirds of the paper is left: $B/A=\tfrac{2}{3}$.
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