Intersection points of two circles

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

Intersection points of two circles

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 239 2013 J3

The post below has been deleted. Click to close.

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

198 posts

#1 h Aug 7, 2020, 10:16 AM

Y by

The altitudes $AA_1$ and $CC_1$ of an acute-angled triangle $ABC$ intersect at point $H$ . A straight line passing through $H$ parallel to line $A_1C_1$ intersects the circumscribed circles of triangles $AHC_1$ and $CHA_1$ at points $X$ and $Y$ , respectively. Prove that points $X$ and $Y$ are equidistant from the midpoint of segment $BH$ .

Z K Y

The post below has been deleted. Click to close.

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

1452 posts

#2 h Aug 7, 2020, 12:49 PM

Y by

SinaQane wrote:

The altitudes $AA_1$ and $CC_1$ of an acute-angled triangle $ABC$ intersect at point $H$ . A straight line passing through $H$ parallel to line $A_1C_1$ intersects the circumscribed circles of triangles $AHC_1$ and $CHA_1$ at points $X$ and $Y$ , respectively. Prove that points $X$ and $Y$ are equidistant from the midpoint of segment $BH$ .

Here is my solution:
Let $D$ be the midpoint of $BH$ . It's easy to see that $D$ is the center of $(BA_{1}C_{1})$ . We also have $\triangle{BAC}$ ~ $\triangle{BA_{1}C_{1}}$ .
Hence, there exists a similarity with center $B$ whichs turns $A$ into $A_{1}$ , $C$ into $C_{1}$ , $D$ into $O$ . Thus, $\angle{DA_{1}C_{1}} = \angle{OCA} = \angle{C_{1}CA_{1}} = \angle{HYA_{1}}$ . We only need to prove that $D, A_{1},Y$ are collinear. Which is true if and only if $\angle{BA_{1}D} = \angle{CA_{1}Y} \Leftrightarrow \angle{DBA_{1}} = \angle{YHC} \Leftrightarrow \angle{A_{1}C_{1}H} = \angle{YHC}$ which is obviously true since $HY / /A_{1}C_{1}$ . So $\triangle{DC_{1}A_{1}}$ ~ $\triangle{DXY}$ . Because $DA_{1} = DC_{1}$ , we deduce that $DX=DY$ .
End of the solution.

Attachments:

This post has been edited 1 time. Last edited by Akatsuki1010, Aug 7, 2020, 12:54 PM

Z K Y

The post below has been deleted. Click to close.

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

1452 posts

#3 h Aug 7, 2020, 12:52 PM

Y by

SinaQane wrote:

The altitudes $AA_1$ and $CC_1$ of an acute-angled triangle $ABC$ intersect at point $H$ . A straight line passing through $H$ parallel to line $A_1C_1$ intersects the circumscribed circles of triangles $AHC_1$ and $CHA_1$ at points $X$ and $Y$ , respectively. Prove that points $X$ and $Y$ are equidistant from the midpoint of segment $BH$ .

Can I know about the source of this problem? What is "239 2013 J3"?

Z K Y

The post below has been deleted. Click to close.

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

198 posts

#5 h Aug 7, 2020, 1:43 PM

Y by

Akatsuki1010 wrote:

Can I know about the source of this problem? What is "239 2013 J3"?

It's an open olympiad held by a school named 239 in Saint Petersburg. Its full name is "Open Math Olympiad of 239 Presidential Physics and Mathematics Lyceum in Saint-Petersburg".
You can see its problems here.

This post has been edited 1 time. Last edited by SinaQane, Aug 7, 2020, 3:33 PM

Z K Y

The post below has been deleted. Click to close.

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

5001 posts

#6 h Oct 1, 2023, 2:54 AM

Y by

Let $M$ be the midpoint of $\overline{BH}$ and $B_1$ be the foot of $\overline{BH}$ on $\overline{AC}$ , so $AB_1C_1H$ and $CB_1A_1H$ are cyclic. It suffices to show that $M,C_1,X$ and $M,A_1,Y$ are collinear, since we know $MC_1=MA_1$ and can thus finish with a homothety.

To that end, define $X'=\overline{MC_1} \cap (AB_1C_1H) \neq C_1$ . Then because $M$ lies on the perpendicular bisector of $\overline{C_1H}$ , $B_1HC_1X$ is an isosceles trapezoid, hence $\measuredangle C_1HX'=\measuredangle B_1XH=\measuredangle B_1AH=\measuredangle HC_1A_1$ , hence $\overline{X'H} \parallel \overline{A_1C_1}$ . If $Y'$ is defined similarly, we also have $\overline{Y'H} \parallel \overline{A_1C_1}$ , hence $X',H,Y'$ are collinear with $\overline{X'HY'} \parallel \overline{A_1C_1}$ , i.e. $X'=X,Y'=Y$ , as desired. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

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

715 posts

#7 h May 20, 2024, 7:55 PM

Y by

Let $M,N$ be the midpoints of $AC,BH$ respectively. $AX\perp \overline{XHY}\perp CY$ Also $MN\perp A_1C_1$ since $A_1C_1$ the radical axis of $(AC)$ and $(BH)$ . And $M$ is on the perpendicular bisector of $XY$ since $MA=MC$ and $AX\parallel CY$ . Thus $N$ is also on the perpendicular bisector of $XY$ which gives that $NX=NY$ as desired. $\blacksquare$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a