View topic - Prove that some circles are concurrent. • Art of Problem Solving

Page 1 of 1

[ 6 posts ]

Author	Message
freemind Posts: 337 Location: MIT or Moldova	Posted: Mar 29, 2008, 7:47 am • # 1 Let $\omega$ be the circumcircle of $ABC$ and let $D$ be a fixed point on $BC$ , $D\neq B$ , $D\neq C$ . Let $X$ be a variable point on $(BC)$ , $X\neq D$ . Let $Y$ be the second intersection point of $AX$ and $\omega$ . Prove that the circumcircle of $XYD$ passes through a fixed point. _________________ The fate of equilibrium is to end the eternity...

¬[ƒ(Gabriel)³²¹º]¼

Posts: 350
Location: Italy
Italy

Blog: View Blog

Posted: Mar 29, 2008, 8:13 am • # 2

w.l.o.g. $\beta \ge \gamma$ . Let E the second intersection between the circumcircle of XYD and $\omega$ . DE meets $\omega$ on F. So 'cause DXYE is cyclic $\angle FEY = \angle YAC + \angle BCA$ , so $\angle FEC = \angle BCA$ and then F is the symmetric of A w.r.t. the axis of BC. Then 'cause E is the intersection between FD and $\omega$ that are all fixed it's also fixed.

The QuattoMaster 6000

Posts: 1196
Location: California
India

Posted: Mar 29, 2008, 9:12 am • # 3

freemind wrote:

Let $\omega$ be the circumcircle of $ABC$ and let $D$ be a fixed point on $BC$ , $D\neq B$ , $D\neq C$ . Let $X$ be a variable point on $(BC)$ , $X\neq D$ . Let $Y$ be the second intersection point of $AX$ and $\omega$ . Prove that the circumcircle of $XYD$ passes through a fixed point.

Solution

Consider two points on $BC$ not equal to $D$ , $B$ , or $C$ . Call them $X$ and $X'$ . Let $AX\perp BC$ and $X'$ be a random point. Let $AX$ and $AX'$ meet the circle at $Y$ and $Y'$ respectively. Allow the circumcircle of $\triangle DXY$ to meet $\omega$ at $Z$ . Let $\angle Y'ZY = \alpha$ , so since $\angle DXY = 90\implies \angle DZY = 180 - \angle DXY = 90$ , which means that $\angle Y'ZD = 90 - \alpha$ . Also, $\angle Y'AY = \alpha$ and $\angle AXB = 90$ , so $\angle DX'Y' = \angle AXX' + \angle XAX' = 90 + \alpha$ . This results in $\angle DX'Y' + \angle DZY' = 90 - \alpha + 90 + \alpha=180$ , so $X'DZY'$ is cyclic, which means that the circumcircle of $X'Y'D$ passes through $Z$ . Yet, $Z$ is fixed because it is the intersection between the circumcircle of $XDY$ (which is fixed because $X$ , $Y$ , and $D$ are fixed points) and $\omega$ , which is also fixed. Hence, the circumcircle of $\triangle X'DY'$ passes through a fixed point, as desired.

andyciup

Posts: 446
Location: Aici
Romania

Posted: Mar 29, 2008, 9:24 am • # 4

Let $X_1$ be a fixed point on $BC$ and $X_{2}$ be any point on $BC$ ( $X_{1}, X_{2}$ are distinct from D). Let $AX_{i}$ cut the circumcenter of the triangle $ABC$ at $Y_{i}$ .
Let the circumcenter of the triangle $DX_{1}Y_1$ touch the circumcenter of the triangle $ABC$ at $T$ .

Then $\angle TDX_{1}= 180^{\circ}-\angle{TY_1A}=\angle{TBA}=180^{\circ}-\angle{TY_{2}A}$ , therefore the point $T$ belongs to the circumcenter of the triangle $DX_{2}Y_{2}$ , and since the points $X_{2}, Y_{2}$ are mobile, the problem is solved.

_________________
$\mathbb{A}_{n}\triangle \pi \in \mathbb{I}$
${\mathbb{CI}^{u}\cdot P(a)}^{N}$
"Cum Deus facit mundu,calculit"-Leibniz

Altheman

Posts: 6205
Location: Illinois
United States

Blog: View Blog

Posted: Mar 29, 2008, 4:42 pm • # 5

Suppose that the diagram is as given (the other situations are the same). Let $AD\cap \omega = A,F$ . Let $\Gamma$ be the circle through $D$ , $F$ that is tangent to $BC$ . Clearly $\Gamma$ is fixed. Let $\Gamma\cap \omega = E,F$ (so $E$ is fixed). Now $\angle XYE = \angle AYE = \pi - \angle AFE = \pi - \angle AFD = \pi - \angle EDX$ so $DXYE$ is cyclic. Hence the circumcircle of $\triangle DXY$ passes through a fixed point, $E$ .

Attachments:

Fixed Point.pdf [15.29 KiB]
Downloaded 127 times

_________________
-Alex
Altheman's Problem Column

kops723 Posts: 93 Location: Illinois, U.S.	Posted: Mar 29, 2008, 5:56 pm • # 6 For the record, the circumcircle of DXY passes through D, a fixed point But we know what you mean.

Display posts from previous: Sort by

Share on Facebook

Previous topic | Next topic

Community » Olympiad Section » Geometry » Geometry Proposed & Own Problems

All times are UTC - 8 hours [ DST ]

Moderators: yetti, MithsApprentice, N.T.TUAN, Peter, darij grinberg, orl, pohoatza, pbornsztein, High School Olympiad Moderators

Page 1 of 1

[ 6 posts ]

Prove that some circles are concurrent.

Login