Find the ratio of radii

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

No tags match your search

M

Find the ratio of radii

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: INMO 1995 Problem 4

The post below has been deleted. Click to close.

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

1592 posts

Rushil

#1 h Oct 6, 2005, 12:19 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle and a circle $\Gamma'$ be drawn lying outside the triangle, touching its incircle $\Gamma$ externally, and also the two sides $AB$ and $AC$ . Show that the ratio of the radii of the circles $\Gamma'$ and $\Gamma$ is equal to $\tan^ 2 { \left( \dfrac{ \pi - A }{4} \right) }.$

Z K Y

The post below has been deleted. Click to close.

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

2643 posts

yetti

#2 h Oct 10, 2005, 12:36 PM • 2 Y

Y by Adventure10, Mango247

Let $I$ be the triangle incenter and $\Gamma \equiv (I)$ the incircle We follow the usual construction of the circle $\Gamma' \equiv (J)$ tangent to the lines $b = CA, c = AB$ and externally tangent to the incircle $(I)$ with the help of an expansion transformation. Consider the circles $(I), (J)$ and the tangents $b, c$ directed. For example, let the circle $(I)$ be directed clockwise, i.e., its radius $-r < 0$ is negative. Then the circle $(J)$ then has to be directed counter-clockwise with a positive radius $r_0 > 0$ . The directed lines $b, c$ are tangents of the directed circle $(J)$ , but not of the directed circle $(I)$ . Perform an expansion transformation by the distance $r > 0$ . The directed circle $(I)$ is carried into a circle of zero radius, i.e., into the point $I$ . The directed lines $b, c$ are carried into the directed lines $b', c'$ parallel to $b, c$ , respectively, but shifted by the distace $r$ to the left when looking down their directions. The directed circle $(J)$ is carried into the directed circle $(J')$ with the same center $J$ and radius $r_0 + r$ , tangent to the directed lines $b', c'$ , and passing through the point $I$ . To construct the center $J$ of the circle $(J')$ , we draw an arbitrary circle tangent to the lines $b', c'$ centered on the bisector of the angle formed by the lines $b', c'$ , which is identical with the bisector of the angle formed by the lines $b, c$ , i.e., with the internal bisector $AI$ of the $\angle A$ . To simplify things, we choose the circle $(A)$ centered at the point $A$ , which has radius $r$ , i.e., it is congruent to the circle $(I)$ . This circle is centrally similar to the desired circle $(J')$ with the external homothety center at the intersection $A' \equiv b' \cap c'$ of the lines $b', c'$ . Using this homothety, the tangency points $U, K$ of the circles $(A), (J')$ with the line $b'$ and the intersections $V, I$ of these 2 circles with the angle bisector $AI$ are centrally cimilar with the same homothety coefficient. Hence, the lines $UV \parallel KI$ are parallel. This allows to find the tangency point $K$ of the circle $(J')$ with the line $b'$ . The center $J$ of this circle is simply the intersection of a normal to the line $b'$ at the tangency point $K$ with the angle bisector $AI$ . Knowing the common center $J$ of the circles $(J'), (J)$ , we can easily construct the desired circle $(J)$ .

Let $S$ be the tangency point of the incircle $(I)$ with the triangle side $CA$ and let $P, Q$ be the intersections of the angle bisector $AI$ with the incircle, so that the points $A, P, I, Q$ follow on the angle bisector $AI$ in this order. Obviously, the point $P$ is the tangency point of the circles $(I), (J)$ . The isosceles triangles $\triangle AUV \sim \triangle JKI$ are centrally similar with the homothety center $A'$ , hence, their corresponding sides are parallel. The isosceles triangles $\triangle AUV \cong \triangle ISQ$ and their sides are parallel. The angles $\angle QIS = \angle IJK$ , $\angle ISQ = \angle JKI$ are easily calculated:

$\angle IJK = \angle QIS = \pi - \angle AIS = \pi - \left(\frac \pi 2 - \frac{\angle A}{2}\right) = \frac{\pi + \angle A}{2}$

$\angle JKI = \angle ISQ = \frac{\pi - \angle QIS}{2} = \frac{\pi - \angle A}{4}$

Let $S'$ be the intersection of the normal $IS \perp b, b'$ with the line $b'$ . From the right angle triangle $\triangle IS'K$ wit the side $IS' = 2r$ and the angle $\angle KIS' = \angle ISQ = \frac{\pi - \angle A}{4}$ , we get

$IK = \frac{IS'}{\cos \widehat{KIS'}} = \frac{2r}{\cos \frac{\pi - \widehat A}{4}}$

and from the isosceles triangle $\triangle JKI$ with the sides $JI = JK = r + r_0$ and the base angles $\angle JKI = \angle JIK = \frac{\pi - \angle A}{4}$ , we get

$r + r_0 = JK = \frac{IK}{2 \cos \widehat{JKI}} = \frac{IK}{2 \cos \frac{\pi - \widehat A}{4}} = \frac{r}{\cos^2 \frac{\pi - \widehat A}{4}}$

$r_0 = \frac{r}{\cos^2 \frac{\pi - \widehat A}{4}} - r = r\ \frac{1 - \cos^2 \frac{\pi - \widehat A}{4}}{\cos^2 \frac{\pi - \widehat A}{4}} = r\ \frac{\sin^2 \frac{\pi - \widehat A}{4}}{\cos^2 \frac{\pi - \widehat A}{4}} = r\ \tan^2 \frac{\pi - \widehat A}{4}$

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

7054 posts

#3 h Aug 31, 2010, 3:02 AM • 2 Y

Y by Adventure10, Mango247

Quote:

Let $\Gamma (J,\rho )$ be a circle which is tangent externally to the incircle $w=C(I,r)$ of $\triangle ABC$ and to the rays $(AB$ , $(AC$ . Find the ratio $\frac {\rho}{r}$ .

Proof. In the interior of the angle $\widehat {BAC}$ exist two circles, $\Gamma_1(J_1,\rho_1 )$ and $\Gamma_2(J_2,\rho_2 )$ with mentioned properties, where $\rho_1<r<\rho_2$ and $\rho_1\rho_2=r^2$ . Prove easily that $\boxed{\frac {s-a}{r}=\frac {2\sqrt {r\rho}}{|\rho -r|}}\ (1)$ , where $a+b+c=2s$ . The relation $(1)$ is equivalently with $(s-a)^2\cdot\rho^2-2r[(s-a)^2+2r^2]\cdot\rho+r^2(s-a)^2=0$ $\iff$ $\frac {\rho}{r}\in\{\frac {(s-a)^2+2r^2\pm 2r\sqrt{(s-a)^2+r^2}}{(s-a)^2}\}$ . Since $IA^2=(s-a)^2+r^2$ obtain $\frac {\rho}{r}\in \{\frac {IA^2+r^2\pm 2r\cdot IA}{(s-a)^2}\}$ $\iff$ $\frac {\rho}{r}\in\{(\frac {IA\pm r}{s-a})^2\}$ . Since $\frac {IA\pm r}{s-a}=\frac {1\pm \frac {r}{IA}}{\frac {s-a}{IA}}=$ $\frac {1\pm \sin\frac A2}{\cos\frac A2}=$ $\frac {1\pm \cos\frac {\pi -A}{2}}{\sin \frac {\pi -A}{2}}$ . In conclusion, $\{\begin{array}{c} \rho_1=r\cdot \tan^2\frac {\pi -A}{4}\\\\ \rho_2=r\cdot\cot^2\frac {\pi -A}{4}\end{array}$ $\iff$ $\frac {\rho}{r}\in \{\tan^2\frac {\pi -A}{4}\ ,\ \cot^2\frac {\pi -A}{4}\}$ .

Quote:

Extension. Let $w=C(I,r)$ , $w_1=C(I_1,r_1)$ , $w_2=C(I_2,r_2)$ be three circles which are internally to the angle $\widehat{BAC}$ , are tangent to
rays $(AB$ , $(AC$ and the circle $w$ is externally tangent to the circles $w_1$ , $w_2$ . Then $\{r_1,r_2\}=\{r\cdot \tan^2\frac {\pi -A}{4}\ ,\ r\cdot\cot^2\frac {\pi -A}{4}\}$ .

Z K Y

The post below has been deleted. Click to close.

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

102 posts

#4 h Jan 3, 2014, 6:07 AM • 2 Y

Y by Adventure10, Mango247

Rushil wrote:

Let $ABC$ be a triangle and a circle $\Gamma'$ be drawn lying outside the triangle, touching its incircle $\Gamma$ externally, and also the two sides $AB$ and $AC$ . Show that the ratio of the radii of the circles $\Gamma'$ and $\Gamma$ is equal to $\tan^ 2 { \left( \dfrac{ \pi - A }{4} \right) }.$

Shouldn't the question say that $\Gamma'$ be drawn lying inside the triangle??

Z K Y

The post below has been deleted. Click to close.

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

43 posts

#5 h Aug 23, 2024, 10:09 AM • 1 Y

Y by MrdiuryPeter

Unknown-6174 wrote:

Rushil wrote:

Let $ABC$ be a triangle and a circle $\Gamma'$ be drawn lying outside the triangle, touching its incircle $\Gamma$ externally, and also the two sides $AB$ and $AC$ . Show that the ratio of the radii of the circles $\Gamma'$ and $\Gamma$ is equal to $\tan^ 2 { \left( \dfrac{ \pi - A }{4} \right) }.$

Shouldn't the question say that $\Gamma'$ be drawn lying inside the triangle??

Yeah

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a