Let a, b and c be the length of sides of a triangle ABC such that (a+b)/7 = (b+c)/8 = (c+a)/9. If r

Q: Why do we introduce the constant $\lambda$? Can't we just solve it directly?

Great question! While you could potentially manipulate the ratios directly, introducing $\lambda$ is a standard, robust technique in JEE problems. It converts a set of relative proportions into absolute algebraic expressions for $a$, $b$, and $c$. This makes the subsequent steps—like finding the perimeter and checking for a right-angled triangle—much cleaner and less prone to arithmetic errors.

Q: How did you know it was a right-angled triangle so quickly?

That is the beauty of recognizing patterns! When I saw the sides were $3\lambda$, $4\lambda$, and $5\lambda$, I immediately recognized the classic $3-4-5$ Pythagorean triplet. In competitive exams, always keep an eye out for these special ratios; they often signal that the triangle has a $90^\circ$ angle, which simplifies the geometry significantly.

Q: Why is the circumradius $R$ equal to $c/2$?

This is a fundamental property of right-angled triangles. The circumcenter of any right-angled triangle is always the midpoint of its hypotenuse. Since the circumradius $R$ is the distance from the circumcenter to any vertex, it must be half the length of the hypotenuse. It is a very handy shortcut!

Question

Let $a, b$ and $c$ be the length of sides of a triangle $ABC$ such that $\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}$. If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $ABC$, respectively, then the value of $\frac{R}{r}$ is equal to

Options:

(A) $\frac{5}{2}$

(B) $2$

(C) $\frac{3}{2}$

(D) $1$

Sigma Percentile · Accepted Answer

### The Geometry of Proportionality

Welcome, future engineers! Today, we are going to explore a problem that beautifully bridges the gap between algebraic manipulation and geometric intuition.

We are given a triangle $ABC$ with side lengths $a$, $b$, and $c$, governed by the elegant relationship:

$$
\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9}
$$

Our mission is to find the ratio of the circumradius $R$ to the inradius $r$. Let's embark on this journey.

### The Power of the Constant $\lambda$

Whenever you encounter a continuous ratio like this, the most powerful tool in your arsenal is to equate the entire expression to a single proportionality constant, $\lambda$. By setting:

$$
\frac{a+b}{7} = \frac{b+c}{8} = \frac{c+a}{9} = \lambda
$$

We transform a complex set of ratios into a system of three simple linear equations:

1. $a+b = 7\lambda$
2. $b+c = 8\lambda$
3. $c+a = 9\lambda$

This is the key to the lock. We have moved from abstract proportions to concrete algebraic expressions.

### Unveiling the Sides

Now, let's see what happens when we sum these equations. Adding them together, we get:

$$
2(a+b+c) = 7\lambda + 8\lambda + 9\lambda = 24\lambda
$$

Dividing by two, we find the sum of the sides: $a+b+c = 12\lambda$. This is a massive breakthrough!

To find the individual sides, we simply subtract the pairs from this total sum. For instance:

$$
a = (a+b+c) - (b+c) = 12\lambda - 8\lambda = 4\lambda
$$

Similarly, $b = 3\lambda$ and $c = 5\lambda$. We have successfully determined the side lengths in terms of $\lambda$.

### The Pythagorean Reveal

Look closely at these side lengths: $3\lambda$, $4\lambda$, and $5\lambda$. If we calculate the squares, we see that:

$$
a^2 + b^2 = (4\lambda)^2 + (3\lambda)^2 = 16\lambda^2 + 9\lambda^2 = 25\lambda^2 = c^2
$$

Our triangle is a **right-angled triangle** with the right angle at vertex $C$. This realization is a gift, as it makes calculating the radii much more straightforward.

### Calculating the Radii

For our right-angled triangle, the semi-perimeter $s$ is:

$$
s = \frac{12\lambda}{2} = 6\lambda
$$

The area $\Delta$ is:

$$
\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4\lambda 	imes 3\lambda = 6\lambda^2
$$

The inradius $r$ is given by:

$$
r = \frac{\Delta}{s} = \frac{6\lambda^2}{6\lambda} = \lambda
$$

Meanwhile, for a right-angled triangle, the circumradius $R$ is simply half the hypotenuse:

$$
R = \frac{c}{2} = \frac{5\lambda}{2}
$$

### The Final Connection

Finally, we calculate the ratio:

$$
\frac{R}{r} = \frac{\frac{5\lambda}{2}}{\lambda} = \frac{5}{2}
$$

The constant $\lambda$ cancels out, leaving us with a clean, elegant result. The final ratio is **$2.5$** or **$5:2$**.

Select Answer:

Visualized Solution

Visualizing the Given Triangle

Introducing the Proportionality Constant

Forming Linear Equations

Finding the Perimeter Sum

Calculating Individual Sides

Checking for a Right-Angled Triangle

Identifying the Right Angle

Setting up the Inradius Calculation

Calculating the Inradius $(r)$

Setting up the Circumradius Calculation

Calculating the Circumradius $(R)$

Final Ratio Calculation

The Sigma Insight: Properties of Triangles

The Geometry of Proportionality

The Power of the Constant $λ$

Unveiling the Sides

The Pythagorean Reveal

Calculating the Radii

The Final Connection

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