Circumcentre is Incentre

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

Circumcentre is Incentre

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: China Mathematical Olympiad 2014

The post below has been deleted. Click to close.

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

252 posts

61plus

#1 h Dec 21, 2013, 12:21 PM • 4 Y

Y by Davi-8191, Adventure10, Mango247, and 1 other user

Let $ABC$ be a triangle with $AB>AC$ . Let $D$ be the foot of the internal angle bisector of $A$ . Points $F$ and $E$ are on $AC,AB$ respectively such that $B,C,F,E$ are concyclic. Prove that the circumcentre of $DEF$ is the incentre of $ABC$ if and only if $BE+CF=BC$ .

This post has been edited 2 times. Last edited by WakeUp, Jul 5, 2015, 1:47 AM

Z K Y

The post below has been deleted. Click to close.

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

36 posts

#2 h Dec 21, 2013, 1:58 PM • 1 Y

Y by Adventure10

$EF$ is always parallel to a particular line, so there is only one pair of E and F such that $BE+CF=BC$ .
Let I be the incenter of ABC, now we prove that $(I, ID)$ intersect AB,AC at 4 points, and out of the 4 points there are 2 points that $BCFE$ is concyclic and $BE+CF=BC$ .
Let $X,Y$ be the symmetric image of D through the internal angle bisector of B and C.
Then X,Y is on $(I, ID)$ . and using the ratio $\frac{AX}{AB}=\frac{AY}{AC}$ , we get that $XY//BC$ .
Now let X', Y' be the symmetric image of X,Y through the internal angle bisector of A.
Then $X'Y'//EF$ (a familiar result), and X',Y' also on $(I, ID)$ .
$AX'+AY'=AX+AY=AB+AC-BC \Rightarrow CX'+BY'=BC$ .
So $X',Y'$ are the $E,F$ we find.
Out of any other 4 pairs of points of intersection, it is easily seen that no other pairs than $X',Y'$ satisfy $BCY'X'$ concyclic.
Thus the conclusion follows.

Z K Y

The post below has been deleted. Click to close.

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

341 posts

#3 h Dec 27, 2013, 5:03 PM • 2 Y

Y by Adventure10, Mango247

Pick a point $J$ on $BC$ such that $BJ = BE$ and $CJ = CF$ , or $BE + CF = BC$ . Let $I$ be the incenter of triangle $ABC$ . It is easily seen that the perpendicular bisectors of $EJ$ and $FJ$ meet at $I$ . Now let the circle with center $I$ and radius $IE = IF = IJ$ meet $AB$ and $AC$ again at $L$ and $K$ , respectively with $AL > AE$ and $CK > CF$ and meet $BC$ at $J$ and $D'$ . We will prove that $D'$ is $D$ in our problem.

It is seen that $\angle BD'I = \angle BLI$ (symmetry across $BI$ ) and $\angle BLI = \angle AEI$ , or $\angle BD'I = \angle AEI$ , and $BD'IE$ is cyclic. Furthermore, we are also given that $BCFE$ is cyclic. Therefore, $\angle D'IE = \angle CFE$ .

Note that the center angle $\angle EIK$ subtending arc $EK$ is twice the angle $\angle EFK = \angle AFE$ , or $\angle AIE$ = $\frac{1}{2} \angle EIK = \angle AFE$ . However, $\angle CFE + \angle AFE = 180^\circ$ ; therefore, $\angle D'IE + \angle AIE = 180^\circ$ , and point $D'$ is point $D$ in our problem, or the incenter of triangle $ABC$ is also the circumcenter of triangle $DEF$ .

Z K Y

The post below has been deleted. Click to close.

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

252 posts

MMEEvN

#4 h Dec 10, 2014, 3:07 PM • 2 Y

Y by Adventure10, Mango247

For the only if part draw a line through $D$ parallel to $EF$ which will intersect $AC$ and $AB$ at $K$ and $L$ respectively It is easily observed that $KL =CD$ ( $\Delta ADK \equiv \Delta ADB$ ) $\Rightarrow EL+FK=KL$ .But $\frac{EL}{FK}=\frac{AL}{AK}=\frac{DL}{DK} \Rightarrow EL=DL$ and $FK=DK \Rightarrow \angle FED= 90-C/2$ .Now the circle $DEF$ intersects $AB ,BC , CA$ at $P,Q ,R$ again .We easily get $PE=QD=FR$ .The rest is easy

This post has been edited 1 time. Last edited by MMEEvN, Dec 10, 2014, 6:14 PM

Z K Y

The post below has been deleted. Click to close.

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

2312 posts

#5 h Dec 10, 2014, 3:36 PM • 4 Y

Y by Fermat_Theorem, Adventure10, and 2 other users

My solution:

Let $I$ be the incenter of $\triangle ABC$ .
Let $D', E', F'$ be the projection of $I$ on $BC, AB, CA$ , respectively .

Since $B, C, E, F$ are concyclic ,
so from $AB>AC$ we get $AF>AE$ ,
hence $I$ is the circumcenter of $\triangle DEF \Longleftrightarrow ID=IE=IF$

$\Longleftrightarrow Rt\triangle IDD' \cong Rt\triangle IEE' \cong Rt\triangle IFF' \Longleftrightarrow DD'=EE'=FF'$

$\Longleftrightarrow CF+BE=CF'+BE'=CD'+BD'=BC$ .

Q.E.D

Z K Y

The post below has been deleted. Click to close.

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

3212 posts

#6 h Dec 17, 2014, 11:34 AM • 1 Y

Y by Adventure10

Let $I$ be the incenter of triangle $ABC$ . If $I$ is the circumcircle of $DEF$ . This follows $ID=IE$ and $\angle IBE= \angle IBD$ . We have $\frac{IE}{\sin \angle IBE}= \frac{ID}{ \sin \angle IBD}$ so $\frac{IB}{\sin \angle BEI}= \frac{IB}{\sin \angle BDI}$ . This follows $\angle BEI= \angle BDI$ . Therefore $\triangle BEI= \triangle BDI \; ( \text{A.S.A})$ or $BE=BD$ . Similarly, we obtain $FC=CD$ . Thus, $BE+CF=BC$ .
If $BE+CF=BC$ . Let $D'$ be a point on $BC$ such that $BD'=BE,D'C=CF$ . It is easy to prove that $\triangle IBE= \triangle IBD' \; ( \text{S.A.S})$ so $IE=ID$ . Similarly, we get $ID=IF$ . Hence, $I$ is the circumcircle of triangle $DFE$ .

Z K Y

The post below has been deleted. Click to close.

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

143 posts

#7 h Mar 4, 2015, 1:22 PM • 3 Y

Y by RideTheLightning84, Adventure10, Mango247

shinichiman: $sin a= sin b <-> a=b$ is not equivanlent

Z K Y

The post below has been deleted. Click to close.

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

bobthesmartypants

4337 posts

bobthesmartypants

#8 h Aug 29, 2016, 9:04 PM • 2 Y

Y by Adventure10, Mango247

It is clear that as we move the circle passing through $B,C$ , the value $BE+CF$ is monotonic, so there exists exactly one circle passing through $B, C$ such that $BE+CF=BC$ .
Let $I$ be the incenter. Let $E', F'$ distinct from $B, C$ be defined as $E'=AB\cap (IDB)$ and $F'=AC\cap (IDC)$ . By Radical Lemma on $(CF'ID), (BE'ID)$ , and the concurrency $A=CF'\cap ID\cap BE'$ , $B, C, E, F$ is cyclic so $E'=E$ and $F'=F$ . If we let $D'$ be the reflection of $D$ about the perpendicular from $I$ to $BC$ , then $BE=BD'$ and $CF=CD'$ , so $BE+CF=BC$ . Finally, since $\angle IDB=\angle IEA$ and $I$ is the incenter, then $ID=IE$ . Similarly, $ID=IF$ , so $I$ is the circumcenter of $\triangle DEF$ . The conditions are all satisfied so we are almost done.

It remains to prove that for all other possible circles through $B, C$ , the circumcenter of $\triangle DEF$ is not $I$ . But this is easy; if $E, F$ are made closer to $A$ than in the above conditions, then $IE > ID$ contradiction, while if $E, F$ are made further to $A$ than in the above conditions, then $IF > ID$ again contradiction, so we are done.

Z K Y

The post below has been deleted. Click to close.

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

93 posts

#9 h Jul 7, 2019, 12:08 PM • 2 Y

Y by Adventure10, Mango247

Note that if $E$ and $F$ are points on $AB$ and $AC$ , respectively. $AEIF$ is cyclic if and only if $BE+CF=BC$ . This pretty much finishes the problem

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a