2021 AMC 10 A Problem 21

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To help you master 2021 AMC 10 A Problem 21, this page includes:

• The original AMC problem to try, plus the correct answer

• Clear, step-by-step solutions using different methods

• Key ideas, useful formulas, and takeaways you can apply to similar problems

• Related problems from past AMC exams, plus new practice problems that build the same skills

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2021 AMC10 A Problem 21

Solution

Solution 1

Coordinate-Vector Perimeter Method

Use the fixed side directions of an equiangular hexagon to locate the intersections forming the two equilateral triangles. This produces a direct identity between the hexagon’s perimeter and the two triangle side lengths.

An equiangular hexagon has exterior angles of $60^{\circ }$, so consecutive sides have directions separated by $60^{\circ }$. Also, an equilateral triangle of side length $s$ has area $\frac{\sqrt{3}}{4}{s}^2$.

Step 1: Label the side lengths and choose coordinates

Let the six side lengths follow the vertex order. Place $C$ at the origin and make $CD$ horizontal. Since each interior angle is $120^{\circ }$, each turn through an exterior angle is $60^{\circ }$, giving the listed unit direction vectors.

$$AB=a,BC=b,CD=c,DE=d,EF=e,FA=f$$

$$C=(0,0),D=(c,0)$$

$$\overrightarrow{AB}=a\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$$

$$\overrightarrow{BC}=b\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$$

$$\overrightarrow{CD}=c(1,0)$$

$$\overrightarrow{DE}=d\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$$

$$\overrightarrow{EF}=e\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$$

$$\overrightarrow{FA}=f(-1,0)$$

Step 2: Use the closure of the hexagon

Walking once around a closed polygon gives zero total horizontal displacement and zero total vertical displacement. Applying this to the six side vectors gives two relations. Together they show that the last two side lengths have the same sum as two earlier side lengths.

$$-a+b+2c+d-e-2f=0$$

$$-a-b+d+e=0$$

$$e=a+b-d$$

$$f=c+d-a$$

$$e+f=b+c$$

Step 3: Find the side length of the first outer triangle

The lines containing $AB$, $CD$, and $EF$ meet at angles of $60^{\circ }$, so they form an equilateral triangle. On the horizontal line containing $CD$, the intersection with line $AB$ lies a distance $b$ to the left of $C$, while the intersection with line $EF$ lies a distance $d$ to the right of $D$. Therefore, the horizontal side of this triangle consists of three consecutive lengths.

$$AB\cap CD=(-b,0)$$

$$CD\cap EF=(c+d,0)$$

$$x=(c+d)-(-b)=b+c+d$$

Step 4: Find the side length of the second outer triangle

The lines containing $BC$, $DE$, and $FA$ also have pairwise angles of $60^{\circ }$, so they form another equilateral triangle. Intersecting lines $BC$ and $DE$ with the horizontal line containing $FA$ gives the endpoints of one side of this triangle.

$$h=\frac{\sqrt{3}}{2}(a+b)$$

$$BC\cap FA=\left(-\frac{a+b}{2},h\right)$$

$$DE\cap FA=\left(c+\frac{a+b}{2},h\right)$$

$$y=\left(c+\frac{a+b}{2}\right)-\left(-\frac{a+b}{2}\right)=a+b+c$$

Step 5: Relate the two triangle sides to the hexagon perimeter

Add the two outer triangle side lengths. The closure relation replaces the repeated pair of side lengths with the two missing hexagon side lengths, so the result is exactly the sum of all six hexagon sides.

$$x+y=(b+c+d)+(a+b+c)$$

$$x+y=a+b+c+d+(b+c)$$

$$x+y=a+b+c+d+e+f$$

$$P=x+y$$

Step 6: Use the two given triangle areas

Apply the equilateral-triangle area formula to each outer triangle. Side lengths are positive, so take the positive square roots.

$$\frac{\sqrt{3}}{4}{x}^2=192\sqrt{3}$$

$$x^2=768=256\cdot 3$$

$$x=16\sqrt{3}$$

$$\frac{\sqrt{3}}{4}{y}^2=324\sqrt{3}$$

$$y^2=1296$$

$$y=36$$

Step 7: Compute the requested expression

The perimeter is the sum of the two calculated side lengths. It is already in the required radical form, and the radicand is squarefree.

$$P=x+y=36+16\sqrt{3}$$

$$m=36,n=16,p=3$$

$$m+n+p=36+16+3=55$$

Correct answer: C. $55$
The answer is $\boxed{55}$, corresponding to choice $\boxed{\text{C}}$. For verification, a side of $16\sqrt{3}$ gives area $\frac{\sqrt{3}}{4}(16\sqrt{3}{)}^2=192\sqrt{3}$, and a side of $36$ gives area $\frac{\sqrt{3}}{4}(36{)}^2=324\sqrt{3}$, exactly matching both given areas.

Similar Problems in Real Exams

The following five AMC 10 A problems are closely related to AMC 10 A 2021 Question 21. These real AMC 10 A problems test the same key concept—Geometry. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.

2017 AMC 10 A Problem 10

2025 AMC 10 A Problem 3

2025 AMC 10 A Problem 6

2025 AMC 10 A Problem 10

2025 AMC 10 A Problem 15

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

• Exterior angle of an equiangular hexagon:

This applies because every interior angle of the hexagon is $120^{\circ }$. It shows that successive side directions differ by $60^{\circ }$, so each set of three alternating side-lines forms an equilateral triangle.

• Equilateral-triangle area:

$A$ is the area and $s>0$ is the side length of an equilateral triangle. Both triangles determined by alternating side-lines are equilateral, so their given areas determine their side lengths.

• Vector closure of a polygon:

The side vectors must be taken in order around the polygon, with horizontal and vertical components tracked consistently. After traveling from $A$ around all six sides, one returns to $A$. The component equations yield the needed relation $e+f=b+c$.

• Perimeter identity for this equiangular hexagon:

Where $x$ and $y$ are the side lengths of the two equilateral triangles formed by alternating side-lines. The hexagon is equiangular with interior angles $120^{\circ }$, and the two triangles are formed by the lines containing alternating sides. Coordinate intersections give $x=b+c+d$ and $y=a+b+c$, while polygon closure gives $e+f=b+c$. Thus their sum contains every hexagon side exactly once.

Key Takeaway: Remember This Strategy for Next Time

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