Q is the midpoint of EF, circle related (In the World of Mathematics 520)

Art of Problem Solving

AoPS Online

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Q is the midpoint of EF, circle related (In the World of Mathematics 520)

G H J

Reveal topic tags

U sviti matematyky

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.

30650 posts

#1 h Apr 19, 2020, 1:56 PM

Y by

Let $O$ be the center of circle $\omega$ , let $KA$ and $KB$ be tangent lines to $\omega$ and let $Q$ be an arbitrary point on the chord $AB$ . Straight line $\ell \perp OQ$ passes through point $Q$ and intersects $KA, KB$ at points $E, F$ respectively. Prove that $Q$ is the midpoint of $EF$ .

(A. Shapoval, Kyiv)

Z K Y

The post below has been deleted. Click to close.

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

1119 posts

#2 h Apr 19, 2020, 2:18 PM

Y by

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -16.590834215250517, xmax = 12.89801095166288, ymin = -6.782012865305229, ymax = 9.321824566271607; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((0.91,0.89), 4), linewidth(0.7) + wrwrwr); draw((0.91,0.89)--(-0.48243042797081276,4.639818329367573), linewidth(0.7) + wrwrwr); draw((0.91,0.89)--(-0.4184612656165246,-2.882955163496718), linewidth(0.7) + wrwrwr); draw((-0.48243042797081276,4.639818329367573)--(-0.4184612656165246,-2.882955163496718), linewidth(0.7) + wrwrwr); draw((-0.45978502988989767,1.9767195150519568)--(0.91,0.89), linewidth(0.7) + wrwrwr); draw((2.5447979093276287,5.7639274138684105)--(0.91,0.89), linewidth(0.7) + wrwrwr); draw((-3.4643679691074247,-1.8104883837644976)--(0.91,0.89), linewidth(0.7) + wrwrwr); draw((-10.85,0.79)--(2.5447979093276287,5.7639274138684105), linewidth(0.7) + wrwrwr); draw((-10.85,0.79)--(-0.4184612656165246,-2.882955163496718), linewidth(0.7) + wrwrwr); draw((2.5447979093276287,5.7639274138684105)--(-3.4643679691074247,-1.8104883837644976), linewidth(0.7) + wrwrwr); /* dots and labels */ dot((0.91,0.89),dotstyle); label("$O$", (1.2458836820618384,0.5379132399569693), NE * labelscalefactor); dot((-10.85,0.79),dotstyle); label("$K$", (-10.734893331040768,1.0757037293231715), NE * labelscalefactor); dot((-0.48243042797081276,4.639818329367573),linewidth(4pt) + dotstyle); label("$A$", (-0.3674877860367667,4.870114404295821), NE * labelscalefactor); dot((-0.4184612656165246,-2.882955163496718),linewidth(4pt) + dotstyle); label("$B$", (-0.4869967836737004,-3.6449016773357155), NE * labelscalefactor); dot((-0.45978502988989767,1.9767195150519568),dotstyle); label("$Q$", (-1.0546645224491356,2.091530209237109), NE * labelscalefactor); dot((2.5447979093276287,5.7639274138684105),linewidth(4pt) + dotstyle); label("$E$", (2.73974615252351,5.1688868983881555), NE * labelscalefactor); dot((-3.4643679691074247,-1.8104883837644976),linewidth(4pt) + dotstyle); label("$F$", (-3.7734942186893776,-2.5394434491940774), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
$OQAE$ is cyclic $\implies \angle OEQ=\angle OAQ$
$OQFB$ is cyclic $\implies \angle OFQ=\angle OBQ$

Since $\angle OAQ=\angle OBQ\implies \angle OEQ=\angle OFQ\implies QE=QF$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a