In a triangle ABC, let ∠C = π/2. If r is the inradius and R is the circumradius of the triangle, the

Q: Why is the circumcenter always at the midpoint of the hypotenuse?

In any triangle, the circumcenter is the intersection of perpendicular bisectors. For a right triangle, the angle at $C$ is $90^\circ$. By Thales' Theorem, any angle subtended by a diameter at the circumference is $90^\circ$. Thus, the hypotenuse must be the diameter of the circumcircle, and its midpoint is the center.

Q: Can I use the general formula $r = \frac{\Delta}{s}$ instead of $r = \frac{a+b-c}{2}$?

Yes, absolutely! For a right triangle, $\Delta = \frac{1}{2}ab$ and $s = \frac{a+b+c}{2}$. Substituting these, you get $r = \frac{ab}{a+b+c}$. This is equivalent to $\frac{a+b-c}{2}$ because $(a+b-c)(a+b+c) = (a+b)^2 - c^2 = a^2 + b^2 + 2ab - c^2$. Since $a^2 + b^2 = c^2$, this simplifies to $2ab$. So $r = \frac{2ab}{2(a+b+c)} = \frac{ab}{a+b+c}$.

Q: Is this result $2(r+R) = a+b$ true for all triangles?

No, this is a specific property of right-angled triangles. The relationship $2R = c$ is unique to right triangles. For general triangles, $2R = \frac{c}{\sin C}$.

Question

In a triangle $ABC$, let $\angle C = \pi/2$. If $r$ is the inradius and $R$ is the circumradius of the triangle, then $2(r + R)$ is equal to

Options:

(A) $a + b$

(B) $b + c$

(C) $c + a$

(D) $a + b + c$

Sigma Percentile · Accepted Answer

### The Geometry of Right Triangles: A Journey of Elegance

Welcome, fellow traveler on the path to JEE mastery! Today, we are not just solving a problem; we are uncovering a beautiful, hidden symmetry within the right-angled triangle.

Geometry is the poetry of logic, and in this problem, we will see how the circumcircle and the incircle dance together in perfect harmony.

### Phase 1: Visualizing the Right Triangle

Imagine you are standing before a right-angled triangle, $	riangle ABC$, with $\angle C = \frac{\pi}{2}$. The sides opposite to angles $A, B,$ and $C$ are $a, b,$ and $c$, respectively.

Here, $c$ is the hypotenuse—the longest, most powerful side of our triangle. This is our foundation. Keep this image clear in your mind as we proceed.

### Phase 2: The Circumcircle and the Magic of the Hypotenuse

The circumcircle is the circle that perfectly embraces all three vertices of our triangle. Its radius is $R$.

Now, here is a beautiful property of right-angled triangles: the circumcenter always lies exactly at the midpoint of the hypotenuse. Because the angle at $C$ is $90^{\circ}$, Thales' Theorem dictates that the hypotenuse $c$ must be the diameter of the circumcircle.

Mathematically, this gives us the elegant relation:
$$2R = c$$

### Phase 3: The Incircle and the Inradius

Next, let's bring in the incircle—the largest circle that can fit inside the triangle, touching all three sides. Its radius is $r$.

For a right-angled triangle, there is a direct and powerful formula for this inradius:
$$r = \frac{a + b - c}{2}$$

This formula is a gift, derived from the fact that the incircle touches the sides at specific points related to the semi-perimeter. If we rearrange this, we find the diameter of the incircle:
$$2r = a + b - c$$

### Phase 4: The Synthesis

Now, we arrive at the heart of the problem. We need to find the value of $2(r + R)$.

Let's expand this expression:
$$2(r + R) = 2r + 2R$$

We already have the expressions for $2r$ and $2R$ from our previous steps. Substituting them into our expression, we get:
$$2(r + R) = (a + b - c) + c$$

Look closely at this. The $-c$ and $+c$ terms cancel each other out with such satisfying precision!

We are left with the final result:
$$2(r + R) = a + b$$

### Conclusion

And there it is! The sum of the diameters of the incircle and circumcircle is simply the sum of the two perpendicular legs of the right triangle.

It is a result of profound simplicity. Remember, geometry is not about memorizing formulas; it is about visualizing the relationships between shapes. Keep practicing, stay curious, and you will find that the most complex problems often have the most elegant solutions.

Animated Solution for Mathematics - Trigonometry: In a triangle $A B C$ , let $∠ C = π /2$ . If $r$ is the inradius and $R$ is the circumradius of the triangle, then $2 (r + R)$ is equal to

Select Answer:

Visualized Solution

Visualizing the Right Triangle

The Circumcircle and Circumradius $R$

Relating $R$ to the Hypotenuse

The Incircle and Inradius $r$

Formula for Inradius $r$

Setting up $2 (r + R)$

Final Simplification

The Sigma Insight: Properties of Triangles

The Geometry of Right Triangles

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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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 ABC, let ∠C=π/2. If r is the inradius and R is the circumradius of the triangle, then 2(r+R) is equal to