AMC 10 · 2011 · #20
Grade 8 geometry-2dRhombus ABCD has side length 2 and ∠B=120°. Region R consists of all points inside the rhombus that are closer to vertex B than any of the other three vertices. What is the area of R?
Pick an answer.
AMC 10 2011 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 rhombus $ABCD$ has every side equal to $2$, and the angle at vertex $B$ is $120^\circ$. Inside the rhombus, region $R$ is the set of points that lie closer to corner $B$ than to any of the other three corners $A$, $C$, or $D$. Find the area of $R$.
Givens: Rhombus $ABCD$ with all four sides equal to $2$; $\angle B = 120^\circ$, so the neighboring angles $\angle A$ and $\angle C$ are each $60^\circ$; $R$ is every interior point closer to $B$ than to $A$, $C$, or $D$; Answer choices: $\dfrac{\sqrt{3}}{3},\ \dfrac{\sqrt{3}}{2},\ \dfrac{2\sqrt{3}}{3},\ 1+\dfrac{\sqrt{3}}{3},\ 2$
Unknowns: The area of region $R$
Understand
Restated: A rhombus $ABCD$ has every side equal to $2$, and the angle at vertex $B$ is $120^\circ$. Inside the rhombus, region $R$ is the set of points that lie closer to corner $B$ than to any of the other three corners $A$, $C$, or $D$. Find the area of $R$.
Givens: Rhombus $ABCD$ with all four sides equal to $2$; $\angle B = 120^\circ$, so the neighboring angles $\angle A$ and $\angle C$ are each $60^\circ$; $R$ is every interior point closer to $B$ than to $A$, $C$, or $D$; Answer choices: $\dfrac{\sqrt{3}}{3},\ \dfrac{\sqrt{3}}{2},\ \dfrac{2\sqrt{3}}{3},\ 1+\dfrac{\sqrt{3}}{3},\ 2$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
The phrase "closer to $B$" is about position, so Tool #1 (Draw a Diagram) is the anchor: draw the rhombus, notice the $120^\circ$ angle splits it into two equilateral triangles, and picture where the border of $R$ must run. Tool #4 (Introduce a Variable) pins the picture down with coordinates so "closer to $B$" becomes a distance you can compare. The border between "closer to $B$" and "closer to a neighbor" is the perpendicular bisector of the segment joining them, so $R$ turns out to be a pentagon. Tool #7 (Identify Subproblems) then finds that pentagon's corners one bisector at a time and adds up its area.
Execute — Answer: C
8.G.A.5 Step 1 Split the rhombus into two triangles
- Because $\angle B = 120^\circ$, the neighboring angle $\angle A = 180^\circ - 120^\circ = 60^\circ$.
- In triangle $ABD$ the two sides $AB$ and $AD$ are both $2$ with a $60^\circ$ angle between them, which forces the third side $BD = 2$ and makes triangle $ABD$ equilateral.
- The same holds for triangle $CBD$.
- So the rhombus is just two equilateral triangles of side $2$ glued along the diagonal $BD$.
💡 Two equal sides meeting at $60^\circ$ leave no room for the triangle to be anything but equilateral.
6.NS.C.8 Step 2 Put the figure on a grid
- Place $B$ at the origin and open its two sides symmetrically about the $x$-axis.
- Using the $30$-$60$-$90$ shape of each equilateral half (height $\sqrt{3}$), the corners land at $B=(0,0)$, $A=(1,\sqrt{3})$, $C=(1,-\sqrt{3})$, and $D=(2,0)$.
- Quick check: $BA=\sqrt{1^2+(\sqrt{3})^2}=\sqrt{4}=2$, exactly the side length, so the placement is right.
💡 Coordinates turn the fuzzy words "closer to" into a distance you can actually measure.
8.G.B.8 Step 3 Turn "closer to B" into borders
- A point is closer to $B$ than to a neighbor exactly when it sits on $B$'s side of the perpendicular bisector of the segment joining $B$ to that neighbor.
- The bisector of $BD$ (from $(0,0)$ to $(2,0)$) is the vertical line $x=1$.
- The bisector of $BA$ meets side $BA$ at its midpoint $\left(\tfrac12,\tfrac{\sqrt{3}}{2}\right)$, and by the mirror symmetry the bisector of $BC$ meets side $BC$ at $\left(\tfrac12,-\tfrac{\sqrt{3}}{2}\right)$.
- Region $R$ is the part of the rhombus on $B$'s side of all three bisectors.
💡 The exact halfway line between two corners is the fair boundary — cross it and the nearer corner switches.
8.G.B.8 Step 4 Find the pentagon's five corners
- Trace the border of $R$.
- Along side $BA$ it runs from $B$ out to the midpoint $\left(\tfrac12,\tfrac{\sqrt{3}}{2}\right)$, then follows the $BA$-bisector to the point equally far from $A$, $B$, and $D$ — the center of equilateral triangle $ABD$, which is $\left(1,\tfrac{\sqrt{3}}{3}\right)$.
- It drops down the line $x=1$ to the twin point $\left(1,-\tfrac{\sqrt{3}}{3}\right)$ (equally far from $B$, $C$, $D$), then back along the $BC$-bisector to $\left(\tfrac12,-\tfrac{\sqrt{3}}{2}\right)$ and home to $B$.
- So $R$ is a pentagon with these five corners.
💡 A corner of $R$ is a spot tied for nearest between three corners at once, so it sits where two bisectors cross.
6.G.A.1 Step 5 Add up the pentagon's area
- Run the shoelace formula on the five corners in order.
- The cross-products add to $\tfrac{4\sqrt{3}}{3}$, and half of that is the area.
- This gives $\dfrac{2\sqrt{3}}{3}$, which is choice $\textbf{(C)}$.
💡 The shoelace formula sweeps around the corners and totals the enclosed area in one pass.
8.G.A.5 Because $\angle B = 120^\circ$, the neighboring angle $\angle A = 180^\circ - 12 6.NS.C.8 Place $B$ at the origin and open its two sides symmetrically about the $x$-axis. 8.G.B.8 A point is closer to $B$ than to a neighbor exactly when it sits on $B$'s side o 8.G.B.8 Trace the border of $R$. Along side $BA$ it runs from $B$ out to the midpoint $\ 6.G.A.1 Run the shoelace formula on the five corners in order. The cross-products add to Review
Reasonableness: The whole rhombus has area $2\times\dfrac{\sqrt{3}}{4}\cdot 2^2 = 2\sqrt{3}\approx 3.46$, and $\dfrac{2\sqrt{3}}{3}\approx 1.15$ is exactly one-third of it. That fits the numbers: choice (C) $\approx 1.15$ sits sensibly between (A) $\approx 0.58$ and (D) $\approx 1.58$, so $R$ is a believable chunk of the rhombus, not the whole thing and not a sliver.
Alternative: Skip coordinates. Inside one equilateral triangle $ABD$, the three perpendicular bisectors meet at the center and cut the triangle into three congruent kites, one hugging each vertex; so the part nearest $B$ is exactly $\tfrac13$ of that triangle's area $\sqrt{3}$, namely $\dfrac{\sqrt{3}}{3}$. The triangle $CBD$ contributes another $\dfrac{\sqrt{3}}{3}$ near $B$. Adding the two gives $\dfrac{2\sqrt{3}}{3}$, confirming (C) without any grid.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Using $\angle B=120^\circ$ to get $\angle A=60^\circ$ and conclude each half of the rhombus is an equilateral triangle.)6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Placing the four corners at coordinates so that distances to $B$ can be compared.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Comparing distances to each corner via perpendicular bisectors and locating the pentagon's corner points.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of the pentagon $R$ with the shoelace formula.)
⭐ The fair border between two corners is the line halfway between them, so "closest to $B$" is a small pentagon whose area you can just add up.
⭐ The fair border between two corners is the line halfway between them, so "closest to $B$" is a small pentagon whose area you can just add up.
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