Prove that a triangle ABC is equilateral if and only if tan A + tan B + tan C = 3√3.

Q: Wait, isn't AM-GM only for positive numbers? What if the triangle is obtuse?

Excellent observation! In an obtuse triangle, one tangent is negative, which would make the sum negative. However, the identity $\sum \tan A = \prod \tan A$ still holds. For the specific case where the sum is $3\sqrt{3}$ (a positive value), the angles must be acute. If the triangle were obtuse, the sum of tangents would be negative or less than $3\sqrt{3}$, so the condition $3\sqrt{3}$ forces the triangle to be acute, making the AM-GM application valid.

Q: Why is the identity $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ true?

It comes from the fact that $A+B+C = 180^\circ$. Thus $A+B = 180^\circ - C$. Taking the tangent of both sides: $\tan(A+B) = \tan(180^\circ - C) = -\tan C$. Expanding $\tan(A+B)$ using the formula $\frac{\tan A + \tan B}{1 - \tan A \tan B} = -\tan C$ and rearranging gives you this beautiful identity.

Q: Why does equality in AM-GM imply $A=B=C$?

The AM-GM inequality states that the arithmetic mean is greater than or equal to the geometric mean, with equality holding if and only if all the terms are equal. Since we found the sum is exactly $3\sqrt{3}$, we are at the equality condition, forcing $\tan A = \tan B = \tan C$, which implies $A=B=C$.

Question

Prove that a triangle $ABC$ is equilateral if and only if $	an A + 	an B + 	an C = 3\sqrt{3}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a trigonometry problem; we are uncovering a fundamental truth about the geometry of triangles. We are tasked with proving that a triangle $ABC$ is equilateral if and only if $	an A + 	an B + 	an C = 3\sqrt{3}$.

This "if and only if" statement is a powerful mathematical bridge. It tells us that the property of being equilateral is perfectly mirrored by this specific trigonometric sum. Let us embark on this journey in two distinct phases: the Forward Path and the Backward Path.

### Phase 1: The Forward Path

Imagine you are standing in an equilateral triangle. Every angle is $60^\circ$. The symmetry is perfect.

If $A = B = C = 60^\circ$, then the sum of the tangents is simply $	an 60^\circ + 	an 60^\circ + 	an 60^\circ$. Since $	an 60^\circ = \sqrt{3}$, the sum becomes $\sqrt{3} + \sqrt{3} + \sqrt{3}$, which is $3\sqrt{3}$.

This direction is the "easy" part, but it gives us the target. It tells us that $3\sqrt{3}$ is the "magic number" for symmetry.

### Phase 2: The Backward Path

This is where the real work begins. We start with the assumption that $	an A + 	an B + 	an C = 3\sqrt{3}$. How do we prove the triangle is equilateral?

We need a bridge. That bridge is the fundamental identity for any triangle $ABC$:

$$	an A + 	an B + 	an C = 	an A 	an B 	an C$$

This identity is a masterpiece of trigonometry. It arises from the fact that $A + B + C = 180^\circ$. When you expand $	an(A+B) = 	an(180^\circ - C)$, you derive this beautiful relationship.

Whenever you see a sum and a product of variables, your mind should immediately jump to the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Phase 3: The Engine of Inequality

The AM-GM inequality states that for positive real numbers, the arithmetic mean is always greater than or equal to the geometric mean:

$$\frac{	an A + 	an B + 	an C}{3} \ge (	an A 	an B 	an C)^{1/3}$$

Let us substitute our identity into this inequality. Replace the product $	an A 	an B 	an C$ with the sum $	an A + 	an B + 	an C$. Let $S = 	an A + 	an B + 	an C$.

The inequality becomes:

$$\frac{S}{3} \ge S^{1/3}$$

If we cube both sides, we get:

$$\frac{S^3}{27} \ge S$$

Assuming $S > 0$ (which is true for acute triangles), we divide by $S$ to get $S^2 \ge 27$, which simplifies to $S \ge 3\sqrt{3}$.

### Conclusion: The Moment of Revelation

This is the moment of revelation. We have proven that for any triangle, the sum of the tangents is at least $3\sqrt{3}$. But our problem statement gives us the condition that the sum is exactly $3\sqrt{3}$.

This means we are at the absolute minimum value of the function. In the AM-GM inequality, the equality holds if and only if all the terms are equal.

Therefore, $	an A = 	an B = 	an C$. This forces $A = B = C = 60^\circ$. The triangle is **equilateral**. We have traversed the path, used the identity, applied the inequality, and arrived at the truth. Keep this logic in your toolkit; it is a weapon for many JEE problems.

Animated Solution for Mathematics - Trigonometry: Prove that a triangle $A B C$ is equilateral if and only if $tan A + tan B + tan C = 33$ .

Visualized Solution

Visualizing Triangle $A B C$

Forward Case: Equilateral Assumption

Substituting $6 0^{\circ}$

Calculating the Sum

Backward Case: Starting with the Sum

Introducing AM-GM Inequality

Substituting the Identity

Solving the Inequality

The Minimum Value Condition

Final Conclusion

The Sigma Insight: Conditional Identities

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

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If $α + β + γ = 2 π$ , then

Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that a triangle ABC is equilateral if and only if tanA+tanB+tanC=33​.