In a triangle XYZ, let x, y, z be the lengths of sides opposite to the angles X, Y, Z, respectively,

Q: Why can we just set the ratio to a constant k?

This is a powerful property of equal ratios. If \frac{a}{b} = \frac{c}{d} = \frac{e}{f}, then each is equal to \frac{a+c+e}{b+d+f}. By summing the numerators, we get (s-x) + (s-y) + (s-z) = 3s - (x+y+z). Since x+y+z = 2s, this simplifies to s. It allows us to reduce three variables into one, which is the key to solving this problem.

Q: Why use r = \frac{\Delta}{s} instead of other formulas?

Because we have information about the incircle area (which gives r) and we can easily calculate s and \Delta in terms of k. It is the most direct bridge between the side lengths and the circle properties. Always look for the path of least resistance in geometry!

Q: How do we handle the half-angle identities?

In JEE problems, always look for the relationship between the half-angles and the sides. Using identities like \cos^2 \frac{Z}{2} = \frac{s(s-z)}{xy} is much faster than calculating the angles themselves. It transforms a trigonometric problem into a simple algebraic one.

Question

In a triangle $XYZ$, let $x, y, z$ be the lengths of sides opposite to the angles $X, Y, Z$, respectively, and $2s = x + y + z$. If $\frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2}$ and area of incircle of the triangle $XYZ$ is $\frac{8\pi}{3}$, then

Options:

(A) area of the triangle $XYZ$ is $6\sqrt{6}$

(B) the radius of circumcircle of the triangle $XYZ$ is $\frac{35}{6}\sqrt{6}$

(C) $\sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2} = \frac{4}{35}$

(D) $\sin^2 \left(\frac{X+Y}{2}\right) = \frac{3}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a triangle $XYZ$ with sides $x, y, z$ and semi-perimeter $s$. The problem provides the following relationship:

$$\frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2}$$

Let this ratio be equal to a constant $k$. By the property of ratios, we can sum the numerators and denominators:

$$\frac{(s-x) + (s-y) + (s-z)}{4+3+2} = \frac{3s - (x+y+z)}{9}$$

Since $x+y+z = 2s$, the numerator simplifies to $3s - 2s = s$. Thus, we have $k = \frac{s}{9}$, which implies $s = 9k$.

### Expressing Sides in Terms of $k$

Using the constant $k$, we can express each side of the triangle:

$s - x = 4k \Rightarrow x = 9k - 4k = 5k$

$s - y = 3k \Rightarrow y = 9k - 3k = 6k$

$s - z = 2k \Rightarrow z = 9k - 2k = 7k$

We have successfully reduced the geometry of the triangle to a single variable, $k$.

### Connecting the Incircle and Area

The area of the incircle is given as $\frac{8\pi}{3}$. Since the area of a circle is $\pi r^2$, we find:

$$r^2 = \frac{8}{3}$$

Next, we apply Heron's formula to find the area of the triangle $\Delta$:

$$\Delta = \sqrt{s(s-x)(s-y)(s-z)} = \sqrt{(9k)(4k)(3k)(2k)} = \sqrt{216k^4} = 6\sqrt{6}k^2$$

Using the relation $r = \frac{\Delta}{s}$, we square both sides to get $r^2 = \frac{\Delta^2}{s^2}$:

$$\frac{8}{3} = \frac{(6\sqrt{6}k^2)^2}{(9k)^2} = \frac{216k^4}{81k^2} = \frac{8k^2}{3}$$

This simplifies to $k^2 = 1$, hence **$k = 1$**. The sides of the triangle are therefore **$5, 6,$ and $7$**.

### Evaluating the Properties

**1. Area of the Triangle:**
Since $\Delta = 6\sqrt{6}k^2$ and $k=1$, the area is **$6\sqrt{6}$**.

**2. Circumradius $R$:**
Using the formula $R = \frac{xyz}{4\Delta}$:

$$R = \frac{5 \cdot 6 \cdot 7}{4 \cdot 6\sqrt{6}} = \frac{35}{4\sqrt{6}} = \frac{35\sqrt{6}}{24}$$

**3. Product of Sines of Half-Angles:**
Using the identity $r = 4R \sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2}$, the product is $\frac{r}{4R}$:

$$\frac{r}{4R} = \frac{\sqrt{8/3}}{4 \cdot \frac{35}{4\sqrt{6}}} = \frac{2\sqrt{2}/\sqrt{3}}{35/\sqrt{6}} = \frac{2\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{2}\sqrt{3}}{35} = \frac{4}{35}$$

**4. Value of $\sin^2 \left(\frac{X+Y}{2}\right)$:**
Since $X+Y+Z = 180^\circ$, this expression is equivalent to $\cos^2 \frac{Z}{2}$. Using the half-angle formula:

$$\cos^2 \frac{Z}{2} = \frac{s(s-z)}{xy} = \frac{9(9-7)}{5 \cdot 6} = \frac{18}{30} = \frac{3}{5}$$

Select Answer:

Visualized Solution

Initial Setup of Triangle $X Y Z$

Relating $s$ to a Constant $k$

Expressing Sides in Terms of $k$

Analyzing the Incircle

Area of Triangle via Heron's Formula

Solving for the Constant $k$

Checking Option (A): Area of Triangle

Checking Option (B): Circumradius $R$

Checking Option (C): Sine Product

Checking Option (D): Half-Angle Sine Squared

The Sigma Insight: Properties of Triangles

Analyzing the Setup

Expressing Sides in Terms of $k$

Connecting the Incircle and Area

Evaluating the Properties

Similar Questions

Let $a, b$ and $c$ be the length of sides of a triangle $A B C$ 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 $A B C$ , respectively, then the value of $\frac{R}{r}$ is equal to

If in a triangle $A B C$ , $A B = 5$ units, $∠ B = cos^{- 1} (\frac{3}{5})$ and radius of circumcircle of $Δ A B C$ is 5 units, then the area (in sq. units) of $Δ A B C$ is:

In a triangle PQR, let $∠ P QR = 3 0^{\circ}$ and the sides PQ and QR have lengths $103$ and 10, respectively. Then, which of the following statement(s) is (are) TRUE ?

In a triangle $A B C$ , $a : b : c = 4 : 5 : 6$ . The ratio of the radius of the circumcircle to that of the incircle is ..................

In a triangle, the sum of lengths of two sides is $x$ and the product of the lengths of the same two sides is $y$ . If $x^{2} - c^{2} = y$ , where $c$ is the length of the third side of the triangle, then the circumradius of the triangle is :

In a triangle the sum of two sides is $x$ and the product of the same sides is $y$ . If $x^{2} - c^{2} = y$ , where $c$ is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is

One angle of an isosceles $Δ$ is $12 0^{\circ}$ and radius of its incircle $= 3$ . Then the area of the triangle in sq. units is

Comprehension Passage

Let $a$ be the area of the triangle $A B C$ . Then the value of $(64 a)^{2}$ is

Then the inradius of the triangle $A B C$ is

In $△ A B C$ , the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of $△ A B C$ is 30 cm $^{2}$ and R and r are respectively the radii of circumcircle and incircle of $△ A B C$ then the value of $2 R + r$ (in cm ) is equal to

Select Answer:

Visualized Solution

Initial Setup of Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } XYZ

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s to a Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Expressing Sides in Terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Analyzing the Incircle

Area of Triangle via Heron's Formula

Solving for the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Checking Option (A): Area of Triangle

Checking Option (B): Circumradius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Checking Option (C): Sine Product

Checking Option (D): Half-Angle Sine Squared

The Sigma Insight: Properties of Triangles

Analyzing the Setup

Expressing Sides in Terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Connecting the Incircle and Area

Evaluating the Properties

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If in a triangle ABC, AB=5 units, ∠B=cos−1(53​) and radius of circumcircle of ΔABC is 5 units, then the area (in sq. units) of ΔABC is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle PQR, let ∠PQR=30∘ and the sides PQ and QR have lengths 103​ and 10, respectively. Then, which of the following statement(s) is (are) TRUE ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle ABC, a:b:c=4:5:6. The ratio of the radius of the circumcircle to that of the incircle is ..................

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } One angle of an isosceles Δ is 120∘ and radius of its incircle =3​. Then the area of the triangle in sq. units is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a be the area of the triangle ABC. Then the value of (64a)2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Then the inradius of the triangle ABC is

Initial Setup of Triangle $X Y Z$

Relating $s$ to a Constant $k$

Expressing Sides in Terms of $k$

Solving for the Constant $k$

Checking Option (B): Circumradius $R$

Expressing Sides in Terms of $k$

If in a triangle $A B C$ , $A B = 5$ units, $∠ B = cos^{- 1} (\frac{3}{5})$ and radius of circumcircle of $Δ A B C$ is 5 units, then the area (in sq. units) of $Δ A B C$ is:

In a triangle PQR, let $∠ P QR = 3 0^{\circ}$ and the sides PQ and QR have lengths $103$ and 10, respectively. Then, which of the following statement(s) is (are) TRUE ?

In a triangle $A B C$ , $a : b : c = 4 : 5 : 6$ . The ratio of the radius of the circumcircle to that of the incircle is ..................

One angle of an isosceles $Δ$ is $12 0^{\circ}$ and radius of its incircle $= 3$ . Then the area of the triangle in sq. units is

Let $a$ be the area of the triangle $A B C$ . Then the value of $(64 a)^{2}$ is

Then the inradius of the triangle $A B C$ is