A,B', C'; A',B, C'; A',B', C are collinear (IMO SL 1987-P12)

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

A,B', C'; A',B, C'; A',B', C are collinear (IMO SL 1987-P12)

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

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

5452 posts

#1 h Aug 19, 2010, 10:27 AM • 2 Y

Y by wiseman, Adventure10

Given a nonequilateral triangle $ABC$ , the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $A'B'C'$ (the vertices listed counterclockwise) for which the triples of points $A,B', C'; A',B, C';$ and $A',B', C$ are collinear.

Proposed by Poland.

Z K Y

The post below has been deleted. Click to close.

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

4149 posts

#2 h Aug 20, 2010, 10:22 PM • 4 Y

Y by Arab, Adventure10, Mango247, MS_asdfgzxcvb

Fixed circumcircles $(X),(Y),(Z)$ of $\triangle A'BC , \triangle C'AB , \triangle B'CA$ meet at the Miquel point of $\triangle A'B'C' \cup ABC,$ i.e. the 1st Fermat point $F$ of $\triangle ABC.$ Let $T$ be the midpoint of $B'C',$ running on midcircle $(K)$ of $(Y),(Z),$ and let $L$ be the 2nd intersection of $A'T$ with $(X);$ midpoint of its arc $BFC.$ Since $AF$ bisects $\angle BFC,$ it follows that $FL$ is the external bisector of $\angle BFC$ $\Longrightarrow$ $\angle AFL=90^{\circ}$ $\Longrightarrow$ $L \in (K).$

Let $G$ and $U$ be the centroids of $\triangle ABC$ and $\triangle A'B'C'.$ Since $G$ is also the centroid of $\triangle XYZ,$ then $\frac{_{\overline{GX}}}{^{\overline{GK}}}=-2 \ (*).$ But $\frac{_{\overline{UA'} \cdot \overline{UL}}}{^{\overline{UL} \cdot \overline{UT}}}=-2$ $\Longrightarrow$ powers of $U$ WRT circles $(X),(K)$ are in the same ratio $\Longrightarrow$ $U$ is on circle $\Omega$ coaxal with $(X),(K),$ but from $(*)$ we deduce that $G$ is the center of $\Omega.$ Therefore, locus of $U$ is the circle $\Omega$ centered at the centroid $G$ of $\triangle ABC$ and passing through its 1st Fermat point $F.$

Z K Y

The post below has been deleted. Click to close.

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

418 posts

skytin

#3 h Nov 6, 2011, 7:12 AM • 1 Y

Y by Adventure10

Hint :
take midpoints of arc's AB , BC , CA of (ABC') , (BCA') , (CAB')

Z K Y

The post below has been deleted. Click to close.

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

4149 posts

#4 h Dec 25, 2014, 11:10 PM • 3 Y

Y by Adventure10, Mango247, MS_asdfgzxcvb

Another proof, following skytin's idea:

Circumcircles $(X),(Y),(Z)$ of $\triangle A'BC,$ $\triangle B'CA$ and $\triangle C'AB$ concur at the Fermat point $F$ of $\triangle ABC.$ Since $U$ is the incenter of $\triangle A'B'C',$ then $UA',UB',UC'$ cut $(X),(Y),(Z)$ again at the midpoints $A_0,B_0,C_0$ of its arcs $FBC,FCA,FAB,$ these are none other than the reflections of $X,Y,Z$ on $BC,CA,AB.$ Thus $\triangle A_0B_0C_0$ is the inner equilateral Napoleon triangle of $\triangle ABC,$ whose center is the centroid $G$ of $\triangle ABC.$ Its circumcircle passes through $F$ and $U,$ since $\angle (FC_0,FB_0)=60^{\circ}$ and $\angle (UC_0,UB_0)=60^{\circ}$ $\pmod\pi.$ 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.

216 posts

#5 h Dec 26, 2014, 11:59 AM • 3 Y

Y by GammaBetaAlpha, Adventure10, Mango247

A more generalized decision: Let $K$ be an arbitrary point in the plane.Draw circles $\odot(BKC),\odot(CKA),\odot(AKB)$ . We'll prove that the locus of the centroid $G$ of triangles like $\triangle{MNP}$ such that $P \in \odot(BKC),N \in \odot(CKA),M \in \odot(AKB)$ and $MN,MP,NP$ passes through $A,B,C$ respectively, is a circle centered at the centroid $L$ of triangle $\triangle{A'B'C'}$ passing through $K$ where $A',B',C'$ are the circumcircles of $\odot(BKC),\odot(CKA),\odot(AKB)$ respectively.

$\rightarrow \vec{LG} = \frac{1}{3}.(\vec{MC'}+\vec{NB'}+\vec{PA'})$ .
$\rightarrow$ Now let $M$ traverses a distance equal to $\alpha$ on the perimeter of $\odot(AKB)$ and makes the point $M'$ ( $\widehat{M'C'M}=2\alpha$ ). WLOG suppose the movement is clockwise; Then if we call the intersection points of $M'A$ and $M'B$ with $\odot(CKA)$ and $\odot(BKC)$ as $N',P'$ respectively, we obviously have $\widehat{M'C'M}=\widehat{N'B'N}=\widehat{P'A'P}=2\alpha \Rightarrow$ If $G'$ be the centroid of $\triangle{M'N'P'}$ then $\vec{GG'}$ is the rotatated vector of $\vec{LG}$ with angle $2\alpha$ (Note that $N'$ and $P'$ are also traversing clockwise). So the length of $\vec{LG}$ is constant $\Longrightarrow$ The locus of $G$ is a circle centered at the centroid $L$ of triangle $\triangle{A'B'C'}$ passing through $K \blacksquare$ .

Z K Y

The post below has been deleted. Click to close.

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

2312 posts

#6 h Dec 26, 2014, 12:44 PM • 6 Y

Y by wiseman, GammaBetaAlpha, Adventure10, Mango247, MS_asdfgzxcvb, and 1 other user

wiseman wrote:

A more generalized decision: Let $K$ be an arbitrary point in the plane.Draw circles $\odot(BKC),\odot(CKA),\odot(AKB)$ . We'll prove that the locus of the centroid $G$ of triangles like $\triangle{MNP}$ such that $P \in \odot(BKC),N \in \odot(CKA),M \in \odot(AKB)$ and $MN,MP,NP$ passes through $A,B,C$ respectively, is a circle centered at the centroid $L$ of triangle $\triangle{A'B'C'}$ passing through $K$ where $A',B',C'$ are the circumcircles of $\odot(BKC),\odot(CKA),\odot(AKB)$ respectively.

Another proof of wiseman's generalization:

Let $P', N', M'$ be the midpoint of $NM, MP, PN$ , respectively .
Let $\triangle A_1B_1C_1$ be the anti-pedal triangle of $K$ WRT $\triangle ABC$ .
Let $G_1$ be the centroid of $\triangle A_1B_1C_1$ .

Easy to see $A', B', C'$ is the midpoint of $KA_1, KB_1, KC_1$ , respectively .

Since all $\triangle PNM$ are similar,
so $PP'$ pass through a fixed point $P_1 \in \odot (KBC)$ .
Similarly, $NN', MM'$ pass through a fixed point $N_1 \in \odot (KCA), M_1 \in \odot (KAB)$ , respectively .

Since $PP', NN', MM'$ are concurrent at $G$ ,
so we get $G, K, P_1, N_1, M_1$ are concyclic .
ie. the locus of $G$ is a circle $\mathcal{C}$ passing through $K$

Consider the case $P \equiv A_1, N \equiv B_1, M \equiv C_1$ .

Since $KA_1$ is the diameter of $\odot (KBC)$ ,
so we get $P_1$ is the projection of $K$ on $A_1G_1$ .
Similarly, $N_1, M_1$ is the projection of $K$ on $B_1G_1, C_1G_1$ , respectively ,
so we get $KG_1$ is the diameter of $\mathcal{C}$ . ie. $L$ is the center of $\mathcal{C}$

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.

4149 posts

#7 h Dec 26, 2014, 3:09 PM • 4 Y

Y by wiseman, Arab, JasperL, Adventure10

Even more general, all points $U$ that verify $\triangle MNP \cup U \sim \triangle A'B'C' \cup U'$ orbit on a circle with center $U'$ that passes through $K.$ My proof is exactly the same what I did in my 1st post.

If $MU$ cuts $PN$ at $D$ and cuts $\odot(KBC)$ again at $M',$ we have $\angle KM'M=\angle KCM=\angle KAN$ $\Longrightarrow$ $M' \in \odot(KAD).$ Powers of $D$ WRT $\odot(KCA),\odot(KAB)$ are in constant ratio $\overline{DN}:\overline{DP}$ $\Longrightarrow$ center of $\odot(KAD)$ is the intersection $D' \equiv A'U' \cap B'C'$ that verifies $\overline{D'B'}:\overline{D'C'}=\overline{DN}:\overline{DP}.$ Now, powers of $U$ WRT $\odot(KBC)$ and $\odot(KAD)$ are in constant ratio $\overline{UM}:\overline{UD}$ $\Longrightarrow$ locus of $U$ is a circle centered at $U' \in A'D',$ because $\overline{U'A'}:\overline{U'D'}=\overline{UM}:\overline{UD},$ and passing through $K,M'.$

P.S. Telv's method also works in this general configuration.

Z K Y

The post below has been deleted. Click to close.

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

1618 posts

#8 h Apr 17, 2019, 10:28 AM • 3 Y

Y by Adventure10, Mango247, MS_asdfgzxcvb

Here is my solution for this problem
Solution
Let $F_1$ be first Fermat point of $\triangle$ $ABC$ ; $I$ be centroid of $\triangle$ $A'B'C'$ ; $M$ $\equiv$ $IA'$ $\cap$ ( $BCA'$ ), $N$ $\equiv$ $IB'$ $\cap$ ( $CAB'$ ), $P$ $\equiv$ $IC'$ $\cap$ ( $ABC'$ ) ( $M$ $\ne$ $A'$ , $N$ $\ne$ $B'$ , $P$ $\ne$ $C'$ )
Then: $M$ , $N$ , $P$ are midpoint of $\stackrel\frown{BC}$ , $\stackrel\frown{CA}$ , $\stackrel\frown{AB}$ of ( $BCA'$ ), ( $CAB'$ ), ( $ABC'$ ) in $\triangle$ $ABC$ and it's easy to see that $M$ , $N$ , $P$ are fixed
We have: ( $F_1M$ ; $F_1P$ ) $\equiv$ ( $F_1M$ ; $F_1B$ ) + ( $F_1B$ ; $F_1P$ ) $\equiv$ $\dfrac{5 \pi}{6}$ + $\dfrac{5 \pi}{6}$ $\equiv$ $\dfrac{\pi}{3}$ $\equiv$ ( $IM$ ; $IP$ ) (mod $\pi$ )
So: $F_1$ , $I$ , $M$ , $P$ lie on a circle
Similarly: $F_1$ , $N$ , $I$ , $P$ lie on a circle
Hence: $F_1$ , $N$ , $I$ , $M$ , $P$ lie on a circle or $E$ $\in$ ( $INMP$ ) which is fixed circle

Z K Y

The post below has been deleted. Click to close.

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

1345 posts

#9 h Apr 23, 2022, 1:42 AM • 3 Y

Y by FIREDRAGONMATH16, CoolJupiter, Sreepranad

One of the nice problems involving geometric constructions...!

This post has been edited 3 times. Last edited by Saucepan_man02, Jan 9, 2024, 6:13 AM
Reason: Edit

Z K Y

The post below has been deleted. Click to close.

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

8791 posts

#10 h Jul 21, 2022, 1:55 AM

Y by

The locus is also known as the Inner Napoleon Circle

“History is a set of lies agreed upon.”
― Napoleon Bonaparte

Z K Y

The post below has been deleted. Click to close.

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

1345 posts

#11 h Jan 9, 2024, 6:13 AM • 1 Y

Y by Sreepranad

Here is the proof:

Construction of the Locus:
Let $P$ be the Torricelli-Point of $\triangle ABC$ . Then, let $K, L, M$ denote the midpoints of arcs $BC, CA, AB$ of circles $(BPC), (CPA), (APB)$ respectively. Then, we claim that the locus $\L$ of all the possible centroids of $\triangle A'B'C'$ is all the points on $(KLM)$ except point $P$ .

Claim: The centroid of every such $\triangle A'B'C'$ lies on $\L$ .
Proof: Let $G$ denote the centroid of $\triangle A'B'C'$ . Then, notice that: $A'K, B'L, C'M$ meet at $G$ . Hence, $\measuredangle A'GF = \measuredangle KGL = 60^\circ$ .

Claim: For every point on $\L$ , there is an equilateral triangle $\triangle A'B'C'$ .
Proof: Let $J$ be a point on $\L$ . Then, extend lines $JK, JL, JM$ and let it intersect $(BPC), (CPA), (APB)$ at $A', B', C'$ respectively. We will prove that $A', C, B'$ are collinear, from which it follows that $\triangle A'B'C'$ is equilateral by symmetry.

Notice that, $\measuredangle GA'C = \measuredangle GB'A = 30^\circ$ and $\measuredangle A'GB' = 120^\circ$ . Hence, we must have $\measuredangle A'CB' = 180^\circ$ which implies $A', C, B$ are collinear. Therefore, $\triangle A'B'C'$ is equilateral.

Hence proved.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a