Let A₀A₁A₂A₃A₄A₅ be a regular hexagon inscribed in a circle of unit radius. Then the product of the

Q: Wait, why is the interior angle of a regular hexagon $120^\circ$?

Great question! The sum of interior angles of any polygon with $n$ sides is $(n-2) \times 180^\circ$. For a hexagon, $n=6$, so the sum is $(6-2) \times 180^\circ = 4 \times 180^\circ = 720^\circ$. Since it is a regular hexagon, all six interior angles are equal, so each angle is $720^\circ / 6 = 120^\circ$.

Q: Why can't we just use the distance formula with coordinates?

You absolutely can! If you place the center at $(0,0)$, the vertices are at $(\cos(k \cdot 60^\circ), \sin(k \cdot 60^\circ))$ for $k=0, 1, ..., 5$. However, using geometry and the Law of Cosines is often faster and less prone to calculation errors in an exam setting. Both methods will lead you to the same elegant result.

Q: How do I know $A_0 A_4 = A_0 A_2$ without calculating it?

Look at the symmetry! A regular hexagon is highly symmetric. If you reflect the hexagon across the diameter passing through $A_0$ and $A_3$, the vertex $A_2$ maps to $A_4$. Because this reflection is an isometry (a distance-preserving transformation), the length of the segment $A_0 A_2$ must be equal to the length of the segment $A_0 A_4$.

Question

Let $A_0 A_1 A_2 A_3 A_4 A_5$ be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments $A_0 A_1, A_0 A_2$ and $A_0 A_4$ is

Options:

(A) $3/4$

(B) $3\sqrt{3}$

(C) 3

(D) $3\sqrt{3}/2$

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: Unlocking the Hexagon

Imagine standing at the center of a perfect, unit-radius circle. Around you, six points are spaced with absolute precision, forming a regular hexagon $A_0 A_1 A_2 A_3 A_4 A_5$.

This is not just a shape; it is a masterclass in geometric harmony. Our mission is to find the product of three specific segments: $A_0 A_1$, $A_0 A_2$, and $A_0 A_4$.

At first glance, this might seem like a tedious exercise in coordinate geometry, but let us pause and look for the elegance hidden within the structure.

### Phase 1: The Foundation ($A_0 A_1$)

Let us start with the simplest segment: $A_0 A_1$. If we connect the center of the circle, $O$, to the vertices $A_0$ and $A_1$, we form a triangle $\Delta A_0 O A_1$.

Since $O$ is the center and $A_0, A_1$ are on the circle, $OA_0 = OA_1 = R = 1$. The central angle $\angle A_0 O A_1$ is simply $360^{\circ} / 6 = 60^{\circ}$.

An isosceles triangle with a $60^{\circ}$ angle is, by definition, an equilateral triangle. Thus, all sides are equal, and we immediately find that $A_0 A_1 = 1$.

This is a fundamental property: the side of a regular hexagon inscribed in a circle is equal to the radius of that circle.

### Phase 2: The Diagonal ($A_0 A_2$)

Now, let us tackle the segment $A_0 A_2$. This is a diagonal of the hexagon. Consider the triangle $\Delta A_0 A_1 A_2$.

We know $A_0 A_1 = 1$ and $A_1 A_2 = 1$. The interior angle of a regular hexagon is $120^{\circ}$, which is the angle $\angle A_0 A_1 A_2$.

We have two sides and the included angle—the perfect setup for the Law of Cosines:

$$A_0 A_2^2 = A_0 A_1^2 + A_1 A_2^2 - 2(A_0 A_1)(A_1 A_2) \cos(120^{\circ})$$

Substituting our values, we get:

$$A_0 A_2^2 = 1^2 + 1^2 - 2(1)(1) \cos(120^{\circ})$$

Recall that $\cos(120^{\circ}) = -\frac{1}{2}$. The calculation becomes:

$$A_0 A_2^2 = 1 + 1 - 2\left(-\frac{1}{2}\right) = 2 + 1 = 3$$

Taking the square root, we find $A_0 A_2 = \sqrt{3}$.

### Phase 3: The Power of Symmetry ($A_0 A_4$)

Finally, we look at $A_0 A_4$. Do we need to repeat the Law of Cosines? Absolutely not!

A regular hexagon is a marvel of symmetry. If you draw a line of symmetry through $A_0$ and $A_3$, you will see that $A_2$ and $A_4$ are mirror images of each other.

Therefore, the distance $A_0 A_4$ must be identical to $A_0 A_2$. Thus, $A_0 A_4 = \sqrt{3}$.

### The Final Synthesis

We have all our pieces: $A_0 A_1 = 1$, $A_0 A_2 = \sqrt{3}$, and $A_0 A_4 = \sqrt{3}$. The product is:

$$A_0 A_1 \cdot A_0 A_2 \cdot A_0 A_4 = 1 \cdot \sqrt{3} \cdot \sqrt{3} = 3$$

And there it is—a result as clean and satisfying as the geometry itself. We didn't need complex coordinates or heavy algebra; we simply listened to the geometry, respected the symmetry, and let the math unfold.

Keep this perspective, and you will find that even the most daunting problems become beautiful journeys. The final answer is **3**.

Animated Solution for Mathematics - Trigonometry: Let $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}$ be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments $A_{0} A_{1}, A_{0} A_{2}$ and $A_{0} A_{4}$ is

Select Answer:

Visualized Solution

Visualizing the Hexagon

Analyzing Side $A_{0} A_{1}$

Central Angle and Equilateral Triangle

Length of $A_{0} A_{1}$

Setting up for $A_{0} A_{2}$

Applying the Law of Cosines

Substituting Values

Evaluating the Cosine

Calculating $A_{0} A_{2}$

Finding $A_{0} A_{4}$ via Symmetry

Setting up the Final Product

Final Calculation

The Sigma Insight: Properties of Triangles

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

The Final Synthesis

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