AMC 10 · 2021 · #20

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

equilateral-trianglearea-trianglesthirty-sixty-ninety-trianglearea-difference identify-subproblemsarea-difference ↑ Prerequisites: area-trianglesequilateral-triangle

📏 Long solution 💡 4 insights 📊 Diagram

Problem

The figure is constructed from $11$ line segments, each of which has length $2$ . The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$ , where $m$ and $n$ are positive integers. What is $m + n ?$

Pick an answer.

(A)

~20

(B)

~21

(C)

~22

(D)

~23

(E)

~24

AMC 10 2021 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 pentagon $ABCDE$ is drawn together with two interior points $F$ and $G$, using $11$ line segments — every segment has length $2$. The area of the pentagon can be written as $\sqrt{m} + \sqrt{n}$ for positive integers $m, n$. Find $m + n$.

Givens: $11$ segments, each of length $2$: pentagon edges $AB, BC, CD, DE, EA$ plus $AF, AG, BF, CF, DG, EG$; Vertex layout (from the asy code): $A$ on top, $B$ upper-left, $C$ lower-left, $D$ lower-right, $E$ upper-right; $F, G$ are interior points; Equal-length segments at every vertex $\Rightarrow$ four embedded equilateral triangles $ABF, BCF, AEG, EDG$; Area is in the form $\sqrt{m} + \sqrt{n}$; Choices: (A) $20$, (B) $21$, (C) $22$, (D) $23$, (E) $24$

Unknowns: $m + n$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships

Tool #1 (Draw a Diagram) — sketch the pentagon and mark all $11$ length-$2$ segments. The drawing makes the four equilateral sub-triangles ($\triangle ABF, \triangle BCF, \triangle AGE, \triangle GDE$) jump out, and from there the $120^\circ$ angles at $B$ and $E$ are visible (two adjacent $60^\circ$ angles). Tool #7 (Identify Subproblems) — split the pentagon into the three triangles $\triangle ABC$, $\triangle AED$, and $\triangle ACD$ along the diagonals $AC$ and $AD$. Each sub-area is computable on its own with elementary triangle facts, and the three add to the answer. Tool #17 (Visualize) supports the symmetry observation that lets us only compute $\triangle ABC$ once (since $\triangle AED$ is its mirror image).

Execute — Answer: D

#1 Draw a Diagram 4.G.A.2 Step 1

Identify the equilateral triangles.
Each of $\triangle ABF, \triangle BCF, \triangle AEG, \triangle EDG$ has all three sides equal to $2$ (every segment in the figure is length $2$), so each is equilateral with all interior angles equal to $60^\circ$.

$AB = BF = AF = 2 \Rightarrow \triangle ABF$ equilateral, $\angle ABF = 60^\circ$; similarly for the other three.

💡 Same side length on all three edges of a triangle $\Rightarrow$ equilateral $\Rightarrow$ $60^\circ$ angles.

#17 Visualize Spatial Relationships 4.MD.C.7 Step 2

At vertex $B$ the pentagon angle $\angle ABC$ is split by the segment $BF$ into $\angle ABF$ and $\angle FBC$ — both $60^\circ$ from the equilateral triangles.
So $\angle ABC = 60^\circ + 60^\circ = 120^\circ$.
By the symmetric construction at $E$, $\angle AED = 120^\circ$ as well.

$$\angle ABC = 60^\circ + 60^\circ = 120^\circ;\quad \angle AED = 120^\circ$$

💡 Two adjacent $60^\circ$ angles share a side, so they sum to $120^\circ$.

#7 Identify Subproblems 7.G.B.6 Step 3

Split the pentagon by the diagonals $AC$ and $AD$ into three triangles: $\triangle ABC$ (left ear), $\triangle AED$ (right ear), $\triangle ACD$ (middle).
The pentagon's area equals the sum.

$$[ABCDE] = [ABC] + [ACD] + [AED]$$

💡 Diagonals from the apex carve the pentagon into three simpler triangles.

#7 Identify Subproblems 8.G.B.7 Step 4

$\triangle ABC$ has $AB = BC = 2$ and $\angle ABC = 120^\circ$.
Drop the altitude from $B$ to $AC$; it bisects $\angle ABC$ into two $60^\circ$ angles and creates a $30$-$60$-$90$ right triangle with hypotenuse $2$.
So the altitude has length $2 \sin 60^\circ = \sqrt{3}$?
No — easier: drop altitude from $B$ doesn't help; instead use the SAS area formula.
Area $= \tfrac{1}{2} \cdot AB \cdot BC \cdot \sin(\angle ABC) = \tfrac{1}{2} \cdot 2 \cdot 2 \cdot \sin 120^\circ = 2 \cdot \tfrac{\sqrt{3}}{2} = \sqrt{3}$.

$$[ABC] = \tfrac{1}{2}(2)(2)\sin 120^\circ = 2 \cdot \tfrac{\sqrt{3}}{2} = \sqrt{3}$$

💡 Two sides and the included angle $\Rightarrow$ SAS area $= \tfrac{1}{2}ab\sin\theta$, with $\sin 120^\circ = \sqrt{3}/2$.

#17 Visualize Spatial Relationships 8.G.A.2 Step 5

By the figure's left-right symmetry, $\triangle AED$ is congruent to $\triangle ABC$ (same two sides of length $2$, same $120^\circ$ apex angle).
So $[AED] = \sqrt{3}$ also.

$$[AED] = \sqrt{3}$$

💡 Mirror image $\Rightarrow$ same area.

#7 Identify Subproblems 8.G.B.7 Step 6

For $\triangle ACD$ we need $AC, AD, CD$.
We know $CD = 2$ (it's one of the $11$ segments).
For $AC$, use the law of cosines on $\triangle ABC$: $AC^2 = 2^2 + 2^2 - 2(2)(2)\cos 120^\circ = 4 + 4 - 8(-\tfrac{1}{2}) = 12$.
So $AC = \sqrt{12} = 2\sqrt{3}$.
By symmetry $AD = 2\sqrt{3}$ too.

$$AC^2 = 4 + 4 + 4 = 12,\quad AC = AD = 2\sqrt{3}$$

💡 Law of cosines with $\cos 120^\circ = -\tfrac{1}{2}$ gives $AC^2 = 12$ — i.e., $AC = 2\sqrt{3}$.

#7 Identify Subproblems 8.G.B.7 Step 7

$\triangle ACD$ is isoceles with $AC = AD = 2\sqrt{3}$ and base $CD = 2$.
Drop the altitude from $A$ to the midpoint $M$ of $CD$; then $CM = MD = 1$.
By the Pythagorean theorem in $\triangle AMD$, $AM^2 = AD^2 - MD^2 = 12 - 1 = 11$, so $AM = \sqrt{11}$.
Area $= \tfrac{1}{2} \cdot CD \cdot AM = \tfrac{1}{2}(2)(\sqrt{11}) = \sqrt{11}$.

$$AM = \sqrt{12 - 1} = \sqrt{11};\quad [ACD] = \tfrac{1}{2}(2)(\sqrt{11}) = \sqrt{11}$$

💡 Isoceles + Pythagorean $\Rightarrow$ altitude $\sqrt{11}$, area $\sqrt{11}$.

#7 Identify Subproblems 8.EE.A.2 Step 8

Sum: $[ABCDE] = \sqrt{3} + \sqrt{3} + \sqrt{11} = 2\sqrt{3} + \sqrt{11} = \sqrt{12} + \sqrt{11}$ (since $2\sqrt{3} = \sqrt{4 \cdot 3} = \sqrt{12}$).
So $\{m, n\} = \{11, 12\}$ and $m + n = 23$.
Choice (D).

$$[ABCDE] = \sqrt{12} + \sqrt{11} \Rightarrow m + n = 12 + 11 = 23\ \Rightarrow\ (D)$$

💡 Combine the three triangle areas, fold $2\sqrt{3}$ into $\sqrt{12}$, add $m + n$.

[1] #1 4.G.A.2 Identify the equilateral triangles. Each of $\triangle ABF, \triangle BCF, \tria

[2] #17 4.MD.C.7 At vertex $B$ the pentagon angle $\angle ABC$ is split by the segment $BF$ into

[3] #7 7.G.B.6 Split the pentagon by the diagonals $AC$ and $AD$ into three triangles: $\triang

[4] #7 8.G.B.7 $\triangle ABC$ has $AB = BC = 2$ and $\angle ABC = 120^\circ$. Drop the altitud

[5] #17 8.G.A.2 By the figure's left-right symmetry, $\triangle AED$ is congruent to $\triangle

[6] #7 8.G.B.7 For $\triangle ACD$ we need $AC, AD, CD$. We know $CD = 2$ (it's one of the $11$

[7] #7 8.G.B.7 $\triangle ACD$ is isoceles with $AC = AD = 2\sqrt{3}$ and base $CD = 2$. Drop t

[8] #7 8.EE.A.2 Sum: $[ABCDE] = \sqrt{3} + \sqrt{3} + \sqrt{11} = 2\sqrt{3} + \sqrt{11} = \sqrt{

Review

Reasonableness: Sanity-check magnitudes. $\sqrt{12} \approx 3.46$ and $\sqrt{11} \approx 3.32$, so the pentagon area is about $6.78$. The pentagon has "width" roughly $|BE| \approx 3$ (centers of two equilateral triangles separated horizontally) and "height" roughly $2$ — so an area near $7$ is in the right ballpark. Also each ear triangle (area $\sqrt{3} \approx 1.73$) is plausible for a triangle with two sides of $2$ and a $120^\circ$ apex. The form $\sqrt{m} + \sqrt{n}$ with $m, n$ both small integers near $12$ matches the answer-choice spread ($20$ to $24$), and $23$ is precisely (D).

Alternative: Tool #1 (Coordinate Geometry) — drop coordinates onto the figure. Place $M$ (the midpoint of $CD$) at the origin and $CD$ along the $x$-axis, so $C = (-1, 0), D = (1, 0)$. The equilateral triangle conditions place $F = (-\tfrac{1}{2}, \sqrt{3})$ (above the midpoint of $BC$ — wait, more carefully, $F$ is the apex of $\triangle BCF$, etc.). Compute $A$'s coordinates, then use the shoelace formula on the five pentagon vertices. Same answer $\sqrt{12} + \sqrt{11}$ but more arithmetic — the SAS / isoceles split is cleaner.

CCSS standards used (min grade 8)

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines (Identifying equilateral triangles from three equal sides.)
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems (Adding two $60^\circ$ angles at vertex $B$ to get $\angle ABC = 120^\circ$, and similarly at $E$.)
7.G.B.6 Solve real-world problems involving area, surface area, and volume (Decomposing the pentagon into three triangles and summing their areas.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing $AC = \sqrt{12}$ via the law of cosines (cosine of $120^\circ$ + Pythagorean reasoning), then the altitude $AM = \sqrt{11}$ in $\triangle AMD$.)
8.G.A.2 Understand that a two-dimensional figure is congruent to another using transformations (Using the figure's left-right symmetry to copy $[ABC] = [AED]$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Rewriting $2\sqrt{3}$ as $\sqrt{12}$ to match the requested form $\sqrt{m} + \sqrt{n}$.)

⭐ This AMC 10 problem only needs Grade 8 Pythagorean reasoning you already know! Cut the pentagon with diagonals $AC$ and $AD$ — the two outer triangles have $120^\circ$ apexes and area $\sqrt{3}$ each, and the middle isoceles triangle has altitude $\sqrt{11}$ and area $\sqrt{11}$. Total $= 2\sqrt{3} + \sqrt{11} = \sqrt{12} + \sqrt{11}$, so $m + n = 23$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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