The ex-radii r1, r2, r3 of Δ ABC are in H.P. Show that its sides a, b, c are in A.P.

Q: Wait, why do we use the reciprocal property? Can't we just solve for $r_1, r_2, r_3$ directly?

That is a fantastic question! While you technically could work with the H.P. definition directly, it involves complex fractions that are prone to algebraic errors. The beauty of the reciprocal property is that it transforms a non-linear relationship (H.P.) into a linear one (A.P.). In JEE, we don't just want to solve the problem; we want to solve it elegantly. Transforming to A.P. simplifies the algebra significantly, allowing the $\Delta$ terms to cancel out naturally.

Q: Is it always true that if $x, y, z$ are in A.P., then $x+k, y+k, z+k$ are in A.P.?

Yes, absolutely! If you have an arithmetic progression with a common difference $d$, adding a constant $k$ to every term simply shifts the entire sequence. The difference between consecutive terms remains $(y+k) - (x+k) = y - x = d$. This property is a cornerstone of sequence manipulation and is exactly what we use to remove the semi-perimeter $s$ from our expressions.

Q: What if the triangle is equilateral? Does the proof still hold?

An equilateral triangle is a special case where $a = b = c$. In this case, $r_1 = r_2 = r_3$. A sequence where all terms are equal is technically both an A.P. (with common difference $0$) and an H.P. (with common difference $0$). So, yes, the proof holds perfectly! The logic remains robust even for the most symmetric of triangles.

Question

The ex-radii $r_1, r_2, r_3$ of $\Delta ABC$ are in H.P. Show that its sides $a, b, c$ are in A.P.

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Triangles: A Journey into Ex-radii

Welcome, future engineer. Today, we are not just solving a problem; we are uncovering a beautiful, hidden symmetry within the geometry of triangles.

Often, when we look at a triangle, we see only the sides $a$, $b$, and $c$ and the angles $A$, $B$, and $C$. But there is a deeper layer—the world of ex-circles.

Imagine a triangle $ABC$. If you extend its sides, you can draw three circles, each tangent to one side and the extensions of the other two. These are the ex-circles, and their radii, $r_1$, $r_2$, and $r_3$, are the keys to a profound algebraic truth.

### Phase 1: The Geometry of Ex-circles

Before we touch the algebra, let us ground ourselves in the geometry. The ex-radii are not random; they are intimately connected to the triangle's area $\Delta$ and its semi-perimeter $s = \frac{a+b+c}{2}$.

The standard formulas are our bedrock:

$$r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c}$$

These formulas are elegant. They tell us that as a side length increases, the corresponding ex-radius changes in a very specific, inverse manner.

The problem asks us to consider a scenario where these three radii are in Harmonic Progression (H.P.). A sequence is in H.P. if the reciprocals of its terms form an Arithmetic Progression (A.P.). That is the secret door we must unlock.

### Phase 2: The Algebraic Transformation

We are given that $r_1, r_2, r_3$ are in H.P. By the definition we just discussed, this implies that the sequence of their reciprocals, $\frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3}$, must be in A.P.

Let us look at the reciprocals using our formulas:

$$\frac{1}{r_1} = \frac{s-a}{\Delta}, \quad \frac{1}{r_2} = \frac{s-b}{\Delta}, \quad \frac{1}{r_3} = \frac{s-c}{\Delta}$$

By taking the reciprocal, we have moved the side-dependent terms from the denominator to the numerator. We now have a sequence: $\frac{s-a}{\Delta}, \frac{s-b}{\Delta}, \frac{s-c}{\Delta}$, which is in A.P.

### Phase 3: The Elegance of Cancellation

In mathematics, we love it when things cancel out. In an Arithmetic Progression, if you multiply every term by a non-zero constant, the sequence remains an A.P.

So, let us multiply our entire sequence by $\Delta$. The result is breathtakingly simple:

$$(s-a), (s-b), (s-c) 	ext{ are in A.P.}$$

We have stripped away the area $\Delta$. We are left with the semi-perimeter $s$ and the sides $a, b, c$.

We use the property that subtracting a constant from every term of an A.P. preserves the progression. Let us subtract $s$ from each term:

$$(s-a) - s, (s-b) - s, (s-c) - s \Rightarrow -a, -b, -c 	ext{ are in A.P.}$$

### Phase 4: The Final Reveal

We are almost at the finish line. We have $-a, -b, -c$ in A.P.

To get to $a, b, c$, we simply multiply the entire sequence by $-1$. Multiplying an A.P. by a constant (even a negative one) keeps it an A.P.

Thus, **$a, b, c$ are in A.P.**

Think about what we have just achieved. We started with a condition on the ex-radii and, through a series of logical, elegant steps, we proved a fundamental property about the sides of the triangle itself.

Whenever you face a complex problem, remember this journey: visualize the geometry, identify the core algebraic property, and trust the process of simplification. You have the tools; now go forth and conquer.

Animated Solution for Mathematics - Trigonometry: The ex-radii $r_{1}, r_{2}, r_{3}$ of $Δ A B C$ are in H.P. Show that its sides $a, b, c$ are in A.P.

Visualized Solution

Visualizing Ex-radii

The Ex-radii Formulas

H.P. to A.P. Transformation

Substituting the Formulas

Simplifying the Progression

Isolating the Sides

Final Conclusion

The Sigma Insight: Properties of Triangles

The Hidden Symmetry of Triangles

Phase 1

Phase 2

Phase 3

Phase 4

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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; } The ex-radii r1​,r2​,r3​ of ΔABC are in H.P. Show that its sides a,b,c are in A.P.