2 circles passing through parallel chords of parabola

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

2 circles passing through parallel chords of parabola

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: 2011 Belarus TST 6.1

The post below has been deleted. Click to close.

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

30630 posts

#1 h Jun 14, 2020, 9:20 AM

Y by

$AB$ and $CD$ are two parallel chords of a parabola. Circle $S_1$ passing through points $A,B$ intersects circle $S_2$ passing through $C,D$ at points $E,F$ . Prove that if $E$ belongs to the parabola, then $F$ also belongs to the parabola.

I.Voronovich

Z K Y

The post below has been deleted. Click to close.

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

1915 posts

#2 h Jun 14, 2020, 10:47 AM • 2 Y

Y by Bumblebee60, GeoMetrix

This actually works for any 2 degree plane curve (aka conics)

We show that if $E\in\mathcal{C}$ then $F\in\mathcal{C}$ , the reverse direction follows reversely. Let $\overline{BF}\cap S_2=\{P\}$ and $\overline{AE}\cap S_2=\{Q\}$ . Then by Reim's $\overline{PQ}\|\overline{CD}$ . So $\angle QEC=\angle PED=\angle PFD$ $(\bigstar)$ . Let $\overline{QE}\cap\overline{CF}=\{X\}$ and $\overline{DE}\cap\overline{PF}=\{Y\}$ . Then from $(\bigstar)$ we get $EFXY$ cyclic $\implies XY\|AB\|CD$ by Reims $\implies AB\cap CD\cap XY=\infty$ . So by Converse of Pascal on $ABFCDE$ we get that $F\in\mathcal C$ . So in this case $\{A,B,C,D,E\}\in\mathcal {P}$ (Parabola). So $F\in\mathcal{P}$ . $\blacksquare$

This post has been edited 5 times. Last edited by amar_04, Jun 14, 2020, 10:55 AM

Z K Y

The post below has been deleted. Click to close.

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

4909 posts

#3 h Dec 13, 2020, 12:15 PM

Y by

Lemma. Let points $A(a,ka^2),B(b,kb^2),C(c,kc^2)$ are points of intersection of parabola $y=kx^2$ with circle then center has coordinates $(-\frac{k^2(a+b)(b+c)(c+a)}{2},\frac{k^2(a^2+b^2+c^2+ab+bc+ca)+1}{2k})$

We have
$(x-a)^2+(y-ka^2)=(x-b)^2+(y-kb^2)=(x-c)^2+(y-kc^2)$
Then $(b-a)(2x-a-b)=(ka^2-kb^2)(2y-ka^2-kb^2)$
$2y= \frac{-2x}{k(a+b)}+\frac{1}{k}+k(a^2+b^2)$ and $2y=\frac{-2x}{k(a+c)}+\frac{1}{k}+k(a^2+c^2)$
$\frac{-2x}{k(a+b)}+\frac{1}{k}+k(a^2+b^2)=\frac{-2x}{k(a+c)}+\frac{1}{k}+k(a^2+c^2)$
$x=-\frac{k^2(c+b)(a+b)(a+c)}{2}$
$y=\frac{1}{2} (\frac{-2x}{k(a+b)}+\frac{1}{k}+k(a^2+b^2))=\frac{1}{2} ( k(a+c)(b+c)+\frac{1}{k}+k(a^2+b^2))= \frac{1}{2} ( k (a^2+b^2+c^2+ab+bc+ca)+\frac{1}{k})$
End of proof

Lemma 2. If points $A(a,ka^2),B(b,kb^2),C(c,kc^2),D(d,kd^2)$ are points of intersection of circle with parabola $y=kx^2$ then $a+b+c+d=0$

As circumcenter of $ABC$ and $ABD$ is same then $(a+b)(b+c)(c+a)=(a+b)(b+d)(d+a) \to c^2+(a+b)c=d^2+(a+c)d \to a+b+c+d=0$ End of proof.

We can replace parabola $y=kx^2+px+q$ with parabola $y=kx^2$
Let points $A,B,C,D,E$ has coordinates $(a,ka^2),(b,kb^2),(c,kc^2),(d,kd^2),(e,ke^2)$
$AB\parallel CD \to a+b=c+d$
Let point $T$ is other point of intersection of circumcircle of $ABE$ and parabola. Such point exists because $(x-x_0)^2+(kx^2-y_0)^2=r^2$ has $4$ solutions
Then $T=(-(a+b+e),k(a+b+e)^2)$
But we have $a+b=c+d$ so $T=(-(c+d+e),(c+d+e)^2)$ is point of intersection of circumcircle of $CDE$ with parabola.
So point $T$ is point of intersection of circumcircles and lies on parabola

Z K Y

The post below has been deleted. Click to close.

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

9782 posts

jayme

#4 h Dec 13, 2020, 12:25 PM • 2 Y

Y by amar_04, Mango247

Dear Mathlinkers,

it can be seen as a generalization of the Reim's theorem...when the two lines become a parabola...

Sincerely
Jean-Louis

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a