Consider the following statements concerning a triangle ABC: (i) The sides a, b, c and area Δ are ra

Q: Wait, why do we use $\tan \frac{B}{2}$? Why not just use $\sin B$ directly?

Great question! While $\sin B$ is useful, $\tan \frac{B}{2}$ is the 'golden key' because it directly incorporates the area $\Delta$ and the semi-perimeter $s$. It bridges the gap between the geometry of the area and the trigonometry of the angles, making the algebraic manipulation much cleaner.

Q: Is the circumradius $R$ always rational in this problem?

Yes! If $a$ and $\sin A$ are rational, then $2R = \frac{a}{\sin A}$ must be rational. This is a powerful realization because it allows us to express any side length as a product of rational numbers, effectively proving the rationality of the other sides.

Q: Do I need to worry about the triangle being degenerate?

In the context of JEE Advanced, we generally assume non-degenerate triangles where $A+B+C = 180^\circ$ and the area $\Delta > 0$. Focus on the algebraic relationships; the geometry will hold up perfectly!

Question

Consider the following statements concerning a triangle $ABC$: (i) The sides $a, b, c$ and area $\Delta$ are rational. (ii) $a, 	an \frac{B}{2}, 	an \frac{C}{2}$ are rational. (iii) $a, \sin A, \sin B, \sin C$ are rational. Prove that (i) $\implies$ (ii) $\implies$ (iii) $\implies$ (i).

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a logical journey. We are going to explore the deep, interconnected nature of a triangle's properties.

We have three statements, and we are going to prove they are all equivalent. Think of this as a relay race where each runner hands the baton to the next, eventually returning to the start.

### Phase 1: The Area-Side Connection $(i) \implies (ii)$

Let us begin by assuming statement (i) is true: the sides $a, b, c$ and the area $\Delta$ are rational. Our goal is to prove statement (ii), which involves the rationality of $a$, $	an \frac{B}{2}$, and $	an \frac{C}{2}$.

First, we define the semi-perimeter $s = \frac{a+b+c}{2}$. Since $a, b, c$ are rational, their sum is rational, and thus $s$ is rational.

Now, we invoke the classic half-angle formula:

$$ 	an \frac{B}{2} = \frac{\Delta}{s(s-b)} $$

Look at this expression. $\Delta$ is rational, $s$ is rational, and $b$ is rational. Since the set of rational numbers is closed under subtraction and division, the entire right-hand side must be rational.

The same logic applies to:

$$ 	an \frac{C}{2} = \frac{\Delta}{s(s-c)} $$

We have successfully bridged the gap!

### Phase 2: The Trigonometric Bridge $(ii) \implies (iii)$

Now, we assume statement (ii) is true. We know $a$, $	an \frac{B}{2}$, and $	an \frac{C}{2}$ are rational. We need to reach statement (iii): the rationality of $a$, $\sin A$, $\sin B$, and $\sin C$.

We use the powerful double-angle identity:

$$ \sin 	heta = \frac{2	an(	heta/2)}{1+	an^2(	heta/2)} $$

By substituting our rational tangents into this formula, we immediately see that $\sin B$ and $\sin C$ must be rational.

But what about $\sin A$? We don't have $	an \frac{A}{2}$ yet. However, we know that $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^\circ$.

This implies:

$$ 	an \frac{A}{2} = \cot\left(\frac{B}{2} + \frac{C}{2}ight) = \frac{1 - 	an(B/2)	an(C/2)}{	an(B/2) + 	an(C/2)} $$

Since $	an \frac{B}{2}$ and $	an \frac{C}{2}$ are rational, $	an \frac{A}{2}$ is rational. Plugging this back into our double-angle identity for $\sin A$ completes the second leg of our journey.

### Phase 3: Closing the Loop $(iii) \implies (i)$

Finally, we assume statement (iii) is true: $a$, $\sin A$, $\sin B$, and $\sin C$ are rational. We must return to the start and prove $b, c$, and $\Delta$ are rational.

We use the Sine Rule:

$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$

Since $a$ and $\sin A$ are rational, the circumdiameter $2R$ is rational. Consequently, $b = (2R)\sin B$ and $c = (2R)\sin C$ are products of rational numbers, making them rational.

Finally, the area is calculated as:

$$ \Delta = \frac{1}{2}bc \sin A $$

Since $b, c, \sin A$ are all rational, $\Delta$ is also a product of rational numbers. The loop is closed! We have proven that these three descriptions of a triangle are mathematically identical.

Keep this elegance in mind—often, the most complex problems are just simple truths viewed from different angles.

Visualized Solution

Understanding the Cyclic Proof Strategy

Step 1: Assume Statement (i) is True

Step 2: The Half-Angle Formula

Step 3: Completing $(i) ⟹ (ii)$

Step 4: Assume Statement (ii) is True

Step 5: Rationality of $sin B$ and $sin C$

Step 6: Finding $tan \frac{A}{2}$

Step 7: Completing $(ii) ⟹ (iii)$

Step 8: Assume Statement (iii) is True

Step 9: Rationality of Sides $b$ and $c$

Step 10: Completing the Loop

The Sigma Insight: Properties of Triangles

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Visualized Solution

Understanding the Cyclic Proof Strategy

Step 1: Assume Statement (i) is True

Step 2: The Half-Angle Formula

Step 3: Completing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (i)⟹(ii)

Step 4: Assume Statement (ii) is True

Step 5: Rationality of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinB and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinC

Step 6: Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan2A​

Step 7: Completing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (ii)⟹(iii)