equilateral triangle

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

37 users

TOPICS

No tags match your search

M

No tags match your search

M

equilateral triangle

G H J

Reveal topic tags

geometric transformation

perpendicular bisector

geometry proposed

L

G H BBookmark kLocked kLocked NReply

Source: Ireland 1994

The post below has been deleted. Click to close.

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

1311 posts

#1 h Jun 29, 2009, 6:56 PM • 2 Y

Y by Adventure10, Mango247

Let $A,B,C$ be collinear points on the plane with $B$ between $A$ and $C$ . Equilateral triangles $ABD,BCE,CAF$ are constructed with $D,E$ on one side of the line $AC$ and $F$ on the other side. Prove that the centroids of the triangles are the vertices of an equilateral triangle, and that the centroid of this triangle lies on the line $AC$ .

Z K Y

The post below has been deleted. Click to close.

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

4147 posts

#2 h Jun 30, 2009, 12:21 AM • 2 Y

Y by Adventure10, Mango247

Let $(O_1)\equiv(ADB),$ $(O_2) \equiv(BCE)$ and $O\equiv (O_1) \cap (O_2).$ It is enough to see that $\angle AOC = 120^{\circ}$ $\Longrightarrow$ $O$ lies on the circumcircle $(O_3)$ of $\triangle CAF.$ $OA , OB$ and $OC$ are radical axes of $(O_1) \cup (O_3),$ $(O_1) \cup(O_2)$ and $(O_2) \cup (O_3)$ $\Longrightarrow$ $OA \perp O_1O_3,$ $OB \perp O_1O_2$ and $OC \perp O_2O_3.$ As a result, we have $\angle O_2O_1O_3 = \angle AOB = 60^{\circ}$ and $\angle O_1O_2O_3 = \angle BOC = 60^{\circ}$ $\Longrightarrow$ $\triangle O_1O_2O_3$ is equilateral.

Click to reveal hidden text

Centroid of $\triangle O_1O_2O_3$ lies on the line $AC$ $\Longleftrightarrow$ Sum of the oriented distances from $O_1,O_2,O_3$ to $AC$ equals zero. Letting $M,N,L$ be the midpoints of $AB,BC,AC$

$O_1M = \frac {_1}{^3}DM \ , \ O_2N = \frac {_1}{^3}EN \ , \ O_3L = \frac {_1}{^3}FL \ \Longrightarrow$

$O_1M + O_2N - O_3L = \frac {_1}{^3}(DM + EN - FL) = \frac {_1}{^6}\sqrt {3}(AC - AC) = 0$

$\Longrightarrow$ The centroid of $\triangle O_1O_2O_3$ lies on the line $AC,$ as desired.

Z K Y

The post below has been deleted. Click to close.

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

467 posts

#3 h Jun 30, 2009, 2:40 AM • 2 Y

Y by Adventure10, Mango247

using cos law
we acquire G1G2=G2G3=G1G3
we can also prove G1AX G2CY are both equilateral triangles
G1G2G3XY are coocyclic further, G1G2YX is an isosceles trapezoid thus the circumcenter lies on the perpendicular bisector of XG1,YG2
done

Attachments:

Z K Y

The post below has been deleted. Click to close.

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

469 posts

#4 h Jun 30, 2009, 4:55 AM • 2 Y

Y by Adventure10, Mango247

Denote $G_1,G_2,G_3$ respectively by the centroid of $\triangle DAB$ , $\triangle EBC$ , $\triangle FAC$ . Consider the rotation around $G_i$ , $i\in \{1,2\}$ with the angle $120^{\circ}$ , we have $\mathcal {R}^{120^{\circ}}_{G_1}: A\mapsto B$ . Similarly, we also have $\mathcal {R}^{120^{\circ}}_{G_2}: B\mapsto C$ $\Longrightarrow \mathcal {R}^{120^{\circ}}_{G_1}$ $\circ$ $\mathcal {R}^{120^{\circ}}_{G_2}: A\mapsto C$ . But ${ \mathcal {R}^{120^{\circ}}_{G_1}}$ $\circ$ $\mathcal {R}^{120^{\circ}}_{G_2}$ $= \mathcal {R}^{240^{\circ}}_{G}$ $= \mathcal {R}^{ - 120^{\circ}}_{G}$ . Therefore $\mathcal {R}^{ - 120^{\circ}}_{G}: A\mapsto C$ $\Longrightarrow G\equiv G_3$ . But we also have $\angle GG_1G_2 = \angle GG_2G_1 = \frac {120^{\circ}}{2} = 60^{\circ}$ $\Longrightarrow \triangle G_3G_1G_2$ is an equilateral triangle.
Now, let $\mathcal {Z}\left (B, - 120^{\circ},\frac {BE}{BA}\right ): A\mapsto E$ , $D\mapsto C$ , $\triangle BDA\mapsto \triangle BCE$ . Hence $G_1\mapsto G_2$ . Therefore, $\angle G_1BG_2 = 120^{\circ}$ . Let the circumcircle $(O)$ of $\triangle G_1BG_2$ intersects $AC$ at $G_4$ as the second point. We have $\angle G_1OG_2 = G_1BG_2 = 120^{\circ}$ , but $\angle G_4G_1G_2$ $= \angle G_2BG_4$ $\equiv$ $\angle G_2BC$ $= 60^{\circ}$ , $\angle G_4G_2G_1 = \angle {G_1BA} = 60^{\circ}$ . Then it is followed that through the rotation with center $G_4$ , angle $- 120^{\circ}$ , we have $\mathcal {R}^{ - 120^{\circ}}_{G_4}: G_1\mapsto G_2$ . Hence $G_4$ must be the centroid of $\triangle G_1G_2G_3$ . The result is lead as follow. $\square$

Z K Y

The post below has been deleted. Click to close.

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

7054 posts

#5 h Jul 2, 2009, 2:03 AM • 2 Y

Y by Adventure10, Mango247

Napoleon wrote:

Let $A$ , $B$ , $C$ be three points on the plane. Equilateral triangles $ABD$ , $BCE$ , $CAF$ are constructed outside of given triangle.

Prove that $AE = BF = CD$ , $AE\cap BF\cap CD\ne\emptyset$ , the centroids $G_c$ , $G_a$ , $G_b$ of the triangles $ABD$ , $BCE$ , $CAF$

are the vertices of an equilateral triangle and the triangles $ABC$ and $G_aG_bG_c$ have a common centroid $G$ and ... another properties.

Indication.

Particular case. When $B\in (AC)$ obtain the proposed problem in which the common centroid of the triangles $ABC$

(degenerated !) and $G_aG_bG_c$ is the point $G\in (BM)$ , where $M$ is the midpoint of $[AC]$ and $BG = 2\cdot GM$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a