locus is a circle, equilateral and circle related 2019 Estonia Open Senior 2.5

The post below has been deleted. Click to close.

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

parmenides51

30629 posts

parmenides51

#1 h Oct 17, 2020, 2:43 PM • 3 Y

Y by Mango247, Mango247, Mango247

The plane has a circle $\omega$ and a point $A$ outside it. For any point $C$ , the point $B$ on the circle $\omega$ is defined such that $ABC$ is an equilateral triangle with vertices $A, B$ and $C$ listed clockwise. Prove that if point $B$ moves along the circle $\omega$ , then point $C$ also moves along a circle.

This post has been edited 2 times. Last edited by parmenides51, Oct 17, 2020, 5:18 PM
Reason: corrected wording, for any C, point B on circle

Z K Y

The post below has been deleted. Click to close.

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

vanstraelen

8945 posts

vanstraelen

#3 h Oct 17, 2020, 4:27 PM • 1 Y

Y by parmenides51

parmenides51 wrote:

The plane has a circle $\omega$ and a point $A$ outside it. For any point $B$ , the point $C$ on the circle $\omega$ is defined such that $ABC$ is an equilateral triangle with vertices $A, B$ and $C$ listed clockwise. Prove that if point $B$ moves along the circle $\omega$ , then point $C$ also moves along a circle.

the point $C$ on the circle $\omega$
if point $B$ moves along the circle $\omega$

Both points $B$ and $C$ on this circle ?

Z K Y

The post below has been deleted. Click to close.

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

natmath

8219 posts

natmath

#4 h Oct 17, 2020, 4:36 PM • 1 Y

Y by parmenides51

I'm not sure if this is right. But I think there is a typo, and it should be $B$ is on circle $\omega$ and $C$ is the point s.t. $ABC$ is an equilateral triangle. Then we must prove that for all $B$ on circle $\omega$ , the locus of points $C$ must trace out a circle.

Z K Y

The post below has been deleted. Click to close.

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

natmath

8219 posts

natmath

#5 h Oct 17, 2020, 4:38 PM

Y by

So like this, but $D$ is point $B$ and $F$ is point $C$ . Also I made $A$ , $D$ , and $F$ counter clockwise accidentally, so forgive my mistake.

Attachments:

This post has been edited 1 time. Last edited by natmath, Oct 17, 2020, 4:40 PM

Z K Y

The post below has been deleted. Click to close.

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

natmath

8219 posts

natmath

#6 h Oct 17, 2020, 4:41 PM

Y by

A proof of this is noting that $C$ is the rotation of $B$ about $A$ $60^\circ$ clockwise. So the locus of points $C$ is the locus of points $B$ (which is circle $\omega$ ) rotated about $A$ $60^\circ$ clockwise. Since circles are still circle after rotation, the locus of $C$ must also be a circle.

Sorry for triple posting

This post has been edited 1 time. Last edited by natmath, Oct 17, 2020, 4:41 PM

Z K Y

The post below has been deleted. Click to close.

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

parmenides51

30629 posts

parmenides51

#7 h Oct 17, 2020, 5:19 PM

Y by

vanstraelen wrote:

parmenides51 wrote:

The plane has a circle $\omega$ and a point $A$ outside it. For any point $B$ , the point $C$ on the circle $\omega$ is defined such that $ABC$ is an equilateral triangle with vertices $A, B$ and $C$ listed clockwise. Prove that if point $B$ moves along the circle $\omega$ , then point $C$ also moves along a circle.

the point $C$ on the circle $\omega$
if point $B$ moves along the circle $\omega$

Both points $B$ and $C$ on this circle ?

just corrected the wording, indeed it had a typo above, $B$ lies on $\omega$ and not $C$ , as natmath suggested

This post has been edited 1 time. Last edited by parmenides51, Oct 17, 2020, 5:20 PM

Z K Y

The post below has been deleted. Click to close.

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

vanstraelen

8945 posts

vanstraelen

#8 h Oct 17, 2020, 6:04 PM

Y by

@natmath
post #5
If you have used Geogebra, you can rename a point by right clicking.

Z K Y

The post below has been deleted. Click to close.

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

vanstraelen

8945 posts

vanstraelen

#9 h Oct 17, 2020, 7:33 PM

Y by

Given the circle $x^{2}+y^{2}r^{2}$ and the point $A(a,0)$ .

Take $B_{1}(r,0)$ , then $C_{1}(\frac{a+r}{2},\frac{(a-r)\sqrt{3}}{2})$ .
Take $B_{2}(-r,0)$ , then $C_{2}(\frac{a-r}{2},\frac{(a+r)\sqrt{3}}{2})$ .

Midpoint of our circle, midpoint of $C_{1}C_{2}\ :\ O(\frac{a}{2},\frac{a\sqrt{3}}{2})$ .
Radius of our circle also $r$ .

Z K Y