Why does $\theta_1 + \theta_2 + \theta_3 = \pi$? Shouldn't it be $2\pi$?

You are absolutely right to question that! The sum of the *full* angles around the incenter $I$ is indeed $2\pi$. However, we defined the angles as $\angle FIE = 2\theta_1$, $\angle FID = 2\theta_2$, and $\angle DIE = 2\theta_3$. Therefore, $2\theta_1 + 2\theta_2 + 2\theta_3 = 2\pi$. When you divide the entire equation by $2$, you get $\theta_1 + \theta_2 + \theta_3 = \pi$. It is a classic trap, but you caught it!

JEE Advanced 2000

LEVELJEE Advanced

Animated Solution for Mathematics - Trigonometry: Let $A B C$ be a triangle with incentre $I$ and inradius $r$ . Let $D, E, F$ be the feet of the perpendiculars from $I$ to the sides $B C, C A$ and $A B$ respectively. If $r_{1}, r_{2}$ and $r_{3}$ are the radii of circles inscribed in the quadrilaterals $A F I E, B D I F$ and $C E I D$ respectively, prove that $\frac{r _{1}}{r - r _{1}} + \frac{r _{2}}{r - r _{2}} + \frac{r _{3}}{r - r _{3}} = \frac{r _{1} r _{2} r _{3}}{( r - r _{1} ) ( r - r _{2} ) ( r - r _{3} )}$ .

Progressive Hints

Hint 1

Hint 1: Visualize the quadrilateral

C E I D

. The perpendiculars

I D

and

I E

are radii of the incircle. Notice that the small circle inscribed in this quadrilateral is tangent to

I D

and

I E

. What does this imply about the distance from the center of the small circle to the sides

I D

and

I E

Hint 2

Hint 2: Construct a right-angled triangle involving the center of the small circle, the incenter

I

, and the tangency point on

I D

. If the small circle has radius

r_{3}

, can you express the distance from the incenter

I

to the center of the small circle in terms of

r

and

r_{3}

Hint 3

Hint 3: You should arrive at the relation

tan θ_{3} = \frac{r _{3}}{r - r _{3}}

, where

2 θ_{3} = ∠ D I E

. Now, consider the sum of the angles around the incenter

I

. How do

∠ F I E

∠ F I D

, and

∠ D I E

relate to the identity

tan A + tan B + tan C = tan A tan B tan C

Visualized Solution

The Geometric Setup

Triangle $A B C$ with incenter $I$ and inradius $r$ .
Feet of perpendiculars from $I$ to sides are $D, E, F$ .
$I D = I E = I F = r$ .

The Three Quadrilaterals

The perpendiculars divide the triangle into three quadrilaterals: $A F I E$ , $B D I F$ , and $C E I D$ .
We are given three smaller circles inscribed in these quadrilaterals.
Let their radii be $r_{1}, r_{2}$ , and $r_{3}$ respectively.

Analyzing Quadrilateral $C E I D$

Focus on quadrilateral $C E I D$ .
The circle with radius $r_{3}$ is tangent to $I D, I E, C D$ , and $C E$ .
Let $∠ D I E = 2 θ_{3}$ .
By symmetry, the center $O_{3}$ lies on the angle bisector of $∠ D I E$ .

Geometry of the Inscribed Circle

Let the circle touch $I D$ at $T_{3}$ .
Since $∠ I D C = 9 0^{\circ}$ , the figure formed by $D, T_{3}, O_{3}$ , and the tangency point on $C D$ is a square of side $r_{3}$ .
Therefore, the distance $D T_{3} = r_{3}$ .

The Right Triangle $△ I T_{3} O_{3}$

In $△ I T_{3} O_{3}$ , $∠ I T_{3} O_{3} = 9 0^{\circ}$ .
The distance $I T_{3} = I D - D T_{3} = r - r_{3}$ .
The opposite side $O_{3} T_{3} = r_{3}$ .

Trigonometric Relation for $r_{3}$

Using trigonometry in $△ I T_{3} O_{3}$ :
$tan θ_{3} = \frac{Opposite}{Adjacent}$
$tan θ_{3} = \frac{r _{3}}{r - r _{3}}$

Generalizing for All Quadrilaterals

By applying the exact same logic to quadrilaterals $A F I E$ and $B D I F$ :
$tan θ_{1} = \frac{r _{1}}{r - r _{1}}$
$tan θ_{2} = \frac{r _{2}}{r - r _{2}}$

Sum of Angles Around the Incenter

The sum of all angles around the incenter $I$ is $36 0^{\circ}$ or $2 π$ .
$∠ F I E + ∠ F I D + ∠ D I E = 2 π$
$2 θ_{1} + 2 θ_{2} + 2 θ_{3} = 2 π$
Dividing by $2$ : $θ_{1} + θ_{2} + θ_{3} = π$

The Tangent Identity

For any three angles such that $A + B + C = π$ :
$tan A + tan B + tan C = tan A tan B tan C$
Applying this to our angles:
$tan θ_{1} + tan θ_{2} + tan θ_{3} = tan θ_{1} tan θ_{2} tan θ_{3}$

Final Substitution

Substitute the expressions for $tan θ_{i}$ into the identity:
$\frac{r _{1}}{r - r _{1}} + \frac{r _{2}}{r - r _{2}} + \frac{r _{3}}{r - r _{3}} = (\frac{r _{1}}{r - r _{1}}) (\frac{r _{2}}{r - r _{2}}) (\frac{r _{3}}{r - r _{3}})$
$\frac{r _{1}}{r - r _{1}} + \frac{r _{2}}{r - r _{2}} + \frac{r _{3}}{r - r _{3}} = \frac{r _{1} r _{2} r _{3}}{( r - r _{1} ) ( r - r _{2} ) ( r - r _{3} )}$
Hence Proved.

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The Sigma Insight: Properties of Triangles

Solution Diagram

Analyzing the Geometry of the Incenter

The incenter

I

serves as the point equidistant from all sides of a triangle. When we drop perpendiculars

I D

I E

, and

I F

to the sides, we effectively partition the triangle into three distinct quadrilaterals.

Consider the quadrilateral

C E I D

. Because the incenter is equidistant from the sides, the circle inscribed within this region with radius

r_{3}

is constrained by the right angles at

D

and

E

This configuration creates a symmetry that simplifies the geometry significantly. By defining the angle

∠ D I E = 2 θ_{3}

, we can utilize trigonometric relationships to bridge the gap between local and global properties.

The Master Equation

The relationship between the radius of the small circle

r_{3}

, the inradius

r

, and the angle

θ_{3}

is given by:

tan θ_{3} = \frac{r _{3}}{r - r _{3}}

This expression acts as the key to connecting the geometry of the small inscribed circle to the global geometry of the triangle. By repeating this process for all three quadrilaterals, we derive three distinct expressions involving

θ_{1}

θ_{2}

, and

θ_{3}

The Final Synthesis

The critical realization is that the sum of these half-angles satisfies the condition:

θ_{1} + θ_{2} + θ_{3} = π

This identity leads us to the fundamental trigonometric relationship:

tan θ_{1} + tan θ_{2} + tan θ_{3} = tan θ_{1} tan θ_{2} tan θ_{3}

When you substitute the ratios derived from the radii into this identity, the proof resolves elegantly. By connecting these simple truths, you uncover the fundamental harmony of the triangle.

Common Traps & FAQs

Wait, why is the quadrilateral

C E I D

related to a square? Isn't it just a general quadrilateral?

That is a fantastic observation! It is not a square in the general sense, but look at the local geometry around the tangency points. The circle is tangent to

I D

and

I E

. Since

I D

and

I E

are perpendicular to the sides

B C

and

A C

respectively, and the circle is tangent to these segments, the distance from the center of the small circle to

I D

and

I E

r_{3}

. Because the angle at

D

9 0^{\circ}

and the angle at

E

9 0^{\circ}

, the region near the vertex

D

and

E

behaves like a square of side

r_{3}

. This local square property is exactly what allows us to define the distance

I T_{3} = r - r_{3}

Why does

θ_{1} + θ_{2} + θ_{3} = π

? Shouldn't it be

2 π

You are absolutely right to question that! The sum of the *full* angles around the incenter

I

is indeed

2 π

. However, we defined the angles as

∠ F I E = 2 θ_{1}

∠ F I D = 2 θ_{2}

, and

∠ D I E = 2 θ_{3}

. Therefore,

2 θ_{1} + 2 θ_{2} + 2 θ_{3} = 2 π

. When you divide the entire equation by

2

, you get

θ_{1} + θ_{2} + θ_{3} = π

. It is a classic trap, but you caught it!

Is there a way to solve this without trigonometry?

While you could potentially use pure geometric similarity or coordinate geometry, it would be significantly more tedious. The beauty of this problem lies in the trigonometric identity for angles summing to

π

. In JEE Advanced, recognizing when to switch from pure geometry to trigonometry is a superpower. It turns a complex algebraic nightmare into a sleek, elegant proof.

Visualized Solution

The Geometric Setup

The Three Quadrilaterals

Analyzing Quadrilateral $C E I D$

Geometry of the Inscribed Circle

The Right Triangle $△ I T_{3} O_{3}$

Trigonometric Relation for $r_{3}$

Generalizing for All Quadrilaterals

Sum of Angles Around the Incenter

The Tangent Identity

Final Substitution

The Sigma Insight: Properties of Triangles

Analyzing the Geometry of the Incenter

The Master Equation

The Final Synthesis

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Visualized Solution

The Geometric Setup

The Three Quadrilaterals

Analyzing Quadrilateral .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CEID

Geometry of the Inscribed Circle

The Right Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △IT3​O3​

Trigonometric Relation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r3​

Generalizing for All Quadrilaterals

Sum of Angles Around the Incenter

The Tangent Identity

Final Substitution

The Sigma Insight: Properties of Triangles

Analyzing the Geometry of the Incenter

The Master Equation

The Final Synthesis

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