Areas of inscribed circles

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

w

18 users

TOPICS

No tags match your search

M

No tags match your search

M

Areas of inscribed circles

G H J

Reveal topic tags

area of a triangle

Heron's formula

L

G H BBookmark kLocked kLocked NReply

Source: IMO 1964, Day 1, Problem 3

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1398 posts

DPopov

#1 h Oct 14, 2005, 1:40 AM • 2 Y

Y by Adventure10, Mango247

A circle is inscribed in a triangle $ABC$ with sides $a,b,c$ . Tangents to the circle parallel to the sides of the triangle are contructe. Each of these tangents cuts off a triagnle from $\triangle ABC$ . In each of these triangles, a circle is inscribed. Find the sum of the areas of all four inscribed circles (in terms of $a,b,c$ ).

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

777 posts

#2 h Oct 18, 2005, 11:52 AM • 2 Y

Y by Adventure10, Mango247

Let $(I,r)$ the inscribed circle of $\triangle ABC$

The tangent parallel to the side $BC$ intersects the sides $AB,AC$ at the points $B_1,C_1$ , forming in this way a triangle $T_1=\triangle AB_1C_1$ .

Since $B_1C_1 \parallel BC$ , we have $\triangle AB_1C_1 \sim \triangle ABC$ , with ratio of similarity $m_1=\frac{B_1C_1}{BC}$

Similarly, the other formed triangles $T_2=\triangle A_2BC_2, \ \ T_3 = \triangle A_3B_3C$ are also similar to $\triangle ABC$ , with ratio $m_2$ and $m_3$ respectively.

The distance of the parallel lines $BC,B_1C_1$ is $2r$

Let $AD$ the altitude of $\triangle ABC$ , and $AD_1$ the altitude of $\triangle AB_1C_1$
We bring a parallel line $L$ from $A$ to $BC$ .

From the parallel lines $L, B_1C_1,BC$ we get:

$\frac{AD_1}{AD} = \frac{AB_1}{AB} = m_1 \Rightarrow$

$AD_1 = m_1 AD$

But $DD_1$ is the diameter of $(I,r)$ , so $DD_1=2r\Rightarrow AD=AD_1+2r$

$AD = m_1 AD + 2r \Rightarrow$

$(1-m_1)AD = 2r \Rightarrow$

$\frac{AD}{r} = \frac{2}{1-m_1} \ \ \ (1)$

Let $s = \frac{a+b+c}{2}$ , the semi-perimeter of $\triangle ABC$

The area of $ABC$ can be expressed in two different ways:

$(ABC) = \frac{AD \cdot a}{2} = s \cdot r \Rightarrow$
$\frac{AD}{r} = \frac{2s}{a} \ \ \ (2)$

$(1),(2) \Rightarrow$

$\frac{2}{1-m_1} = \frac{2s}{a}$

So we find that $m_1 = \frac{s-a}{s}$ , and similarly $m_2 = \frac{s-b}{s}, m_3 = \frac{s-c}{s}$

The area of $(I,r)$ is $\pi \cdot r^2$ .

Each of the triangles $T_1,T_2,T_3$ is similar to the triangle $ABC$ , so the incircle of $T_k$ is similar to $(I,r)$ with the same ratio of similarity $m_k$ . So we have:

Inradius of $T_1 = m_1 \cdot r$
Inradius of $T_2 = m_2 \cdot r$
Inradius of $T_3 = m_3 \cdot r$

Hence, the areas of their inscribed circles are $\pi {m_1}^2 r^2,\ \ \pi {m_2}^2 r^2,\ \ \pi {m_3}^2 r^2$

So we want the sum $S = \pi \cdot r^2 \cdot (1+{m_1}^2+{m_2}^2+{m_3}^2)$

The problem is already solved, because $m_1,m_2,m_3$ are functions of $a,b,c$ .

But I'll try to find another expression, not containing $m_1,m_2,m_3$

${m_1}^2 = (\frac{s-a}{s})^2 = (1-\frac{a}{s})^2 = 1-2\frac{a}{s}+\frac{a^2}{s^2}$

${m_2}^2 = (\frac{s-b}{s})^2 = (1-\frac{b}{s})^2 = 1-2\frac{b}{s}+\frac{b^2}{s^2}$

${m_3}^2 = (\frac{s-c}{s})^2 = (1-\frac{c}{s})^2 = 1-2\frac{c}{s}+\frac{c^2}{s^2}$

$\Rightarrow$

${m_1}^2 +{m_2}^2 +{m_3}^2 = 3 -2\cdot \frac{a+b+c}{s} +\frac{a^2+b^2+c^2}{s^2}$

${m_1}^2 +{m_2}^2 +{m_3}^2 = 3 -2\cdot 2 +\frac{a^2+b^2+c^2}{s^2} = -1 +\frac{a^2+b^2+c^2}{s^2}$

Hence,

$1+{m_1}^2 +{m_2}^2 +{m_3}^2 = \frac{a^2+b^2+c^2}{s^2}$

Finally,

$S = \pi \cdot r^2 \cdot \frac{a^2+b^2+c^2}{s^2}$

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

9001 posts

#3 h Nov 5, 2010, 7:13 PM • 2 Y

Y by Adventure10, Mango247

Addition
$S =\pi\cdot r^{2}\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{2}}$
We know that the area of $ABC$ can be expressed as:
$(ABC) = s\cdot r\Rightarrow r=\frac{(ABC)}{s}$
$S =\pi\cdot (\frac{(ABC)}{s})^{2}\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{2}}$
$S =\pi\cdot (ABC)^{2}\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{4}}$
We know also Heron's formula:
$(ABC) = \sqrt{s(s-a)(s-b)(s-c)}$
$S =\pi\cdot s(s-a)(s-b)(s-c)\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{4}}$
$S =\pi\cdot (s-a)(s-b)(s-c)\cdot\frac{a^{2}+b^{2}+c^{2}}{s^{3}}$
Because $s=\frac{a+b+c}{2}$ , $s-a=\frac{-a+b+c}{2}$ , $s-b=\frac{a-b+c}{2}$ and $s-c=\frac{a+b-c}{2}$ :
$S =\pi\cdot \frac{(a+b-c)(b+c-a)(c+a-b)(a^{2}+b^{2}+c^{2})}{(a+b+c)^{3}}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1893 posts

#4 h Feb 1, 2023, 5:33 PM

Y by

Obviously the three smaller triangles are similar to triangle $ABC$ .

We employ barycentric coordinates with reference triangle $ABC$ . Since $I=\left(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)$ , the ratios of the side lengths to the side lengths of $ABC$ are
$\frac{a+b-c}{a+b+c},\frac{a-b+c}{a+b+c},\frac{-a+b+c}{a+b+c},\frac{a+b+c}{a+b+c}.$ Summing the squares, the ratio of the final area to the area of the inscribed circle of triangle $ABC$ is just
$\frac{4(a^2+b^2+c^2)}{(a+b+c)^2}.$ So our answer is just
$\frac{\pi(a^2+b^2+c^2)(a+b-c)(a-b+c)(-a+b+c)}{(a+b+c)^3}.$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a