two triangles have equal circumradii

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

two triangles have equal circumradii

G H J

Reveal topic tags

geometric transformation

power of a point

L

G H BBookmark kLocked kLocked NReply

Source: Italy TST 2009 p5

The post below has been deleted. Click to close.

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

761 posts

#1 h Mar 10, 2012, 5:34 AM • 2 Y

Y by Adventure10, Mango247

Two circles $O_1$ and $O_2$ intersect at $M,N$ . The common tangent line nearer to $M$ of the two circles touches $O_1,O_2$ at $A,B$ respectively. Let $C,D$ be the symmetric points of $A,B$ with respect to $M$ respectively. The circumcircle of triangle $DCM$ intersects circles $O_1$ and $O_2$ at points $E,F$ respectively which are distinct from $M$ . Prove that the circumradii of the triangles $MEF$ and $NEF$ are equal.

Z K Y

The post below has been deleted. Click to close.

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

1347 posts

WakeUp

#2 h Mar 10, 2012, 12:21 PM • 2 Y

Y by Adventure10, Mango247

Clearly the result is equivalent to proving $\angle EMF+\angle ENF=180^{\circ}$ . Let $MN$ meet $AB$ at $L$ . Then $MN$ is the radical axis of $O_1$ and $O_2$ so $L$ has equal power with respect to $O_1$ and $O_2$ i.e. $LA^2=LB^2$ (since $AB$ is a common tangent) hence $L$ is the midpoint of $AB$ . Now clearly $ABCD$ is a parallelogram. Since $M$ is the midpoint of $BD$ , and $L$ is the midpoint of $AB$ it follows that $ML||AD||BC$ .

Next we show $B,C,F$ are collinear. Note that $\angle CFM=\angle CDM=\angle ABD=\angle ABM=\angle BFM$ . It follows $B,C,F$ are collinear and hence so are $A,D,E$ by a symmetric argument. But now $AMNE$ and $BMNF$ are cyclic trapeziums, and hence they must be isosceles cyclic trapeziums. Then $\angle MNE=\angle NMA$ and $\angle MNP=\angle NMB$ . Therefore $\angle AMB=\angle ENF$ . So $\angle EMF+\angle ENF=180^{\circ}\iff \angle EMF=\angle BMC$ . We will in fact prove $\triangle MBC$ is similar to $\triangle MEF$ which will prove the result. Indeed, $\angle MBC=\angle ADB=\angle MFE$ and of course $\angle MCB=\angle MEF$ , so $\triangle MBC$ is similar to $\triangle MEF$ .

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

1886 posts

Potla

#3 h Jun 8, 2012, 10:54 AM • 2 Y

Y by Adventure10, Mango247

A very similar solution

Z K Y

The post below has been deleted. Click to close.

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

Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#4 h Jan 8, 2022, 7:45 AM

Y by

Note that if radius of $MEF$ and $NEF$ are equal then Reflection of $N$ must lie an $MEF$ so we need to prove $\angle EMF + \angle ENG = \angle 180$ .
we will prove $EF$ is common tangent of our two circles. Let $E,F$ be tangency points of common tangent we will prove $E,F$ lie on $MCD$ .

Step1 : $A,D,E$ and $B,C,F$ are collinear.
$MN$ bisects both $AB$ and $CD$ so $MN || AD || BC$ and we have $MN || AE || BF$ so $A,D,E$ and $B,C,F$ are collinear.

Step2 : $E,F$ lie on $MCD$ .
$\angle MED = \angle MAB = \angle MCD \implies MDEC$ is cyclic.
$\angle MFC = \angle MBA = \angle MDC \implies MCFD$ is cyclic.
so $MCFED$ is cyclic as wanted.

Now that we have $EF$ is tangent to circles let's prove our main goal : $\angle MEF + \angle NEF = \angle 180$ .
$\angle EMF + \angle ENF = \angle FEN + \angle EFN + \angle EFN = \angle 180$ .
we're Done.

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Mar 10, 2022, 5:02 AM

Z K Y

The post below has been deleted. Click to close.

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

Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#5 h Jan 8, 2022, 8:07 AM

Y by

Another way to show $EF$ is common tangent. Let $O_1$ and $O_2$ be circles of two circles.
$\angle MED = \angle DCM = \angle MAB = \angle MEA \implies E,D,A$ are collinear.
$\angle MFC = \angle MDC = \angle MBA = \angle MFB \implies F,C,B$ are collinear.
Let $E'$ and $F'$ be tangency points of other common tangent we have $AE' \perp O_1O_2$ and $BF' \perp O_1O_2$ and $MN \perp O_1O_2$ so $AE' || MN || BF'$ . But $MN$ bisects $AB$ and $CD$ so $AE || MN || BF$ so $E$ is $E'$ and $F$ is $F'$ .

This post has been edited 1 time. Last edited by Mahdi_Mashayekhi, Mar 10, 2022, 5:05 AM

Z K Y

The post below has been deleted. Click to close.

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

3 posts

#6 h Mar 15, 2025, 5:28 AM

Y by

Since $AM=CM$ and $BM=DM$ , we get that $ABCD$ is a parallelogram. using that we get that $\angle ACD = \angle BAC = \angle AEM$ since $AB$ is tangent to $\triangle AEM$ and $\angle MCD= \angle MED$ since $CMDE$ is cyclic. so $\angle MED = \angle AEM$ then $A,D,E$ are collinear.
similarly we get that $B,C,F$ are collinear.
using sinus law we know that $\frac {EF}{sin \angle ENF} = 2R$ where $R$ is the circumradii of $(ENF)$ . so it suffices to prove that $\frac {EF}{sin \angle ENF} = \frac {EF}{sin \angle EMF}$ or $sin \angle ENF = sin \angle EMF$
$AMNE$ is cyclic so $\angle MAE = 180 - \angle MNE$ and since $BMNF$ is cyclic we get that $\angle MBF = 180 - \angle MNF$ so $\angle ENF = \angle MAE + \angle MBF$
$MDEFC$ is cyclic so $\angle MCE = \angle MDA$ and $\angle EMF =\angle ECF = 180 - \angle BCE = 180 - \angle MCE - \angle BCA = 180 - \angle ADB - \angle ACB = 180 - \angle DAC - \angle DBC = 180 - \angle MAE -\angle MBF$
so $\angle ENF = 180 - \angle EMF$ so $sin \angle ENF = sin \angle EMF$ and we are done

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a