In a triangle with sides a, b, c, r1 r2 r3 (which are the ex-radii) then (a) a b c (b) a

Q: Wait, why are we allowed to just divide by $\Delta$ and invert the fractions? Is there a risk of division by zero?

That is a brilliant question! In any non-degenerate triangle, the area $\Delta$ is strictly positive. Furthermore, $s-a$, $s-b$, and $s-c$ are all positive because the triangle inequality states that the sum of any two sides is greater than the third side (e.g., $b+c > a \implies b+c-a > 0 \implies 2s-2a > 0 \implies s-a > 0$). Since all these terms are strictly positive, we are perfectly safe to divide and invert without any risk of division by zero.

Q: I always mess up the inequality sign flips. How do I keep track?

Think of it as a balance scale. If you have a heavy weight on the left, and you take the reciprocal of both sides, the lighter side becomes heavier. Every time you perform an operation that involves a reciprocal of positive numbers, or multiplication by a negative number, pause and flip the sign. It is a habit you must build. Treat the inequality sign as a living, breathing entity that reacts to your algebraic operations!

Q: Can I just memorize that $r_1$ corresponds to $a$?

While memorization might get you through a quick test, in JEE Advanced, you want to understand the symmetry. $r_1$ is opposite to $A$, $r_2$ to $B$, and $r_3$ to $C$. If you understand the formula $r_1 = \frac{\Delta}{s-a}$, you never have to memorize the order; you can derive it in seconds. Always aim for the derivation, not the rote memory.

Question

In a triangle with sides $a, b, c, r_1 > r_2 > r_3$ (which are the ex-radii) then Options: (A) $a > b > c$ (B) $a < b < c$ (C) $a > b$ and $b < c$ (D) $a < b$ and $b > c$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Ex-circles Welcome, future engineer. Today, we are not just solving an inequality; we are uncovering the hidden symmetry of a triangle. When we look at a triangle with sides $a$, $b$, and $c$, we often focus on the interior. But there is a whole world outside—the world of ex-circles. These circles, which touch one side of the triangle and the extensions of the other two, hold a beautiful, inverse relationship with the sides of the triangle. Let us embark on this journey to understand why $r_1 > r_2 > r_3$ implies $a > b > c$. ### The Algebraic Bridge To solve this, we must first translate our geometric intuition into the language of algebra. We recall the standard formulas for the ex-radii of a triangle: $$r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c}$$ Here, $\Delta$ is the area of the triangle, and $s$ is the semi-perimeter. Notice the elegance here: the ex-radius is inversely proportional to the difference between the semi-perimeter and the side length. The problem gives us the condition $r_1 > r_2 > r_3$. By substituting our formulas, we get the inequality: $$\frac{\Delta}{s-a} > \frac{\Delta}{s-b} > \frac{\Delta}{s-c}$$ ### The Inequality Dance Now, we must be careful. We have a chain of inequalities. Since the area $\Delta$ is a positive constant, we can divide the entire expression by $\Delta$ without any fear. This leaves us with: $$\frac{1}{s-a} > \frac{1}{s-b} > \frac{1}{s-c}$$ This is where many students stumble. We need to isolate the side lengths $a$, $b$, and $c$. To do this, we must take the reciprocal of each term. Remember the golden rule: when you take the reciprocal of positive numbers, the inequality signs must flip. So, the expression transforms into: $$s-a < s-b < s-c$$ ### The Final Revelation We are almost there. We have $s-a < s-b < s-c$. To isolate the sides, we subtract the semi-perimeter $s$ from every part of the inequality. This gives us: $$-a < -b < -c$$ Finally, we multiply the entire inequality by $-1$. As you know, multiplying an inequality by a negative number reverses the inequality signs once again. Thus, $-a < -b < -c$ becomes **$a > b > c$**. Look at that! The order of the ex-radii perfectly mirrors the order of the sides. It is a beautiful, symmetric result. You have successfully navigated the geometry, the algebra, and the traps of inequality manipulation. Keep this clarity of thought, and you will conquer any problem JEE throws at you.

Animated Solution for Mathematics - Trigonometry: In a triangle with sides $a, b, c, r_{1} > r_{2} > r_{3}$ (which are the ex-radii) then

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Visualized Solution

Visualizing the Triangle and Ex-radii

The Ex-radii Formulae

Setting up the Inequality

Canceling the Area $Δ$

Inverting the Fractions

Isolating the Sides

Final Conclusion

The Sigma Insight: Properties of Triangles

Analyzing the Geometry of Ex-circles

The Algebraic Bridge

The Inequality Dance

The Final Revelation

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle with sides a,b,c,r1​>r2​>r3​ (which are the ex-radii) then