geometric locus

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

geometric locus

G H J

Reveal topic tags

geometry unsolved

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.

44 posts

#1 h Dec 11, 2006, 1:56 PM • 2 Y

Y by Adventure10, Mango247

Find the locus of points M inside an equilateral triangle ABC such that $\angle MAB+\angle MBC+\angle MCA=\frac{\pi}{2}$ .

Z K Y

The post below has been deleted. Click to close.

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

44 posts

#2 h Nov 2, 2007, 7:16 PM • 2 Y

Y by Adventure10, Mango247

Hi!
Nobody?

Z K Y

The post below has been deleted. Click to close.

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

1327 posts

#3 h Nov 2, 2007, 9:59 PM • 2 Y

Y by Adventure10 and 1 other user

I think that the altitudes $AD,$ $BE,$ $CF$ of the given equilateral triangle $\bigtriangleup ABC,$ is the wanted Locus of the point $M,$ with the property as the problem states.

It is clear that the centroid $G$ of $\bigtriangleup ABC,$ has the property as the problem states

$($ $\angle GAB + \angle GBC + \angle GCA = 30^{o} + 30^{o} + 30^{o} = 90^{o}$ $).$

We consider an arbitrary point $M,$ on $CF$ $($ $M,$ between $G,$ $F$ $)$ and it is easy to show that $\angle MAB + \angle MBC + \angle MCA = 90^{o},$ because of $\angle GAM = \angle GBM.$

$\bullet$ WE denote as $P,$ an arbitrary point between $B,$ $M$ $($ considered as an arbitrary point inwardly to $\bigtriangleup ABC,$ doesn’t lie on one of the altitudes $AD,$ $BE,$ $CF$ $)$ and we will prove that $\angle PAB + \angle PBC + \angle PCA \neq 90^{o}.$

We denote as $A',$ the reflexion of $A,$ with respect to the line segment $BM$ and then, we have $\angle PAM = \angle PA'M$ $,(1)$

We denote the point $Q\equiv AD\cap BM$ and it is easy to show that the circumcircle of the triangle $\bigtriangleup A'CM,$ passes through the point $Q,$ because of $\angle QA'M = \angle QAM = \angle QBG = \angle QCG\equiv QCM.$

So, the circumcircle of the triangle $\bigtriangleup CMP,$ doesn’t pass through the point $C$ and then, we have that $\angle MCP\neq MA'P = \angle MAP.$

We conclude now, that $\angle PAB + \angle PBC + \angle PCA \neq 90^{o}$ $($ In my schema, the point $C$ lies inwardly to the circumcircle of $\bigtriangleup CMP,$ because of the point $P,$ has been considered between $B,$ $M$ and then, in this case of $P,$ we have $\angle MCP > \angle MA'P = \angle MAP$ $\Longrightarrow$ $\angle PAB + \angle PBC + \angle PCA > 90^{o}$ $).$

Hence, we conclude that any point $M,$ inwardly to $\bigtriangleup ABC,$ with the property as the problem states, lies on one of its altitudes $AD,$ $BE,$ $CF,$ as the wanted Locus of $M$ and the solution is completed.

Kostas Vittas.

Attachments:

t=123822.pdf (6kb)

Z K Y

The post below has been deleted. Click to close.

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

316 posts

#4 h Nov 3, 2007, 12:58 AM • 2 Y

Y by Adventure10, Mango247

Did anyone try solving the equation:

$\sin{\alpha}\sin{\beta}\cos({\alpha+\beta})=\sin({\frac{\pi}{3}-\alpha})\sin({\frac{\pi}{3}-\beta})$

Z K Y

The post below has been deleted. Click to close.

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

1270 posts

April

#5 h Nov 3, 2007, 3:48 AM • 2 Y

Y by Adventure10, Mango247

The following problem can be solved easily by use trigonometric, but in fact, it has a very nice synthetic solution.

Find the locus of point $M$ inside the square $ABCD$ such that $\angle MAB + \angle MBC + \angle MCD + \angle MDA = 180^\circ$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a