The chords $AC$ and $BD$ of the circle intersect at point $P$. The perpendiculars to $AC$ and $BD$ at points $C$ and $D$, respectively, intersect at point $Q$. Prove that the lines $AB$ and $PQ$ are perpendicular.
Problem
Source: Sharygin 2005 VIII CR 1
Tags: geometry, Chords, perpendicular
Kagebaka
18.08.2019 06:42
Let $CQ$ and $DQ$ intersect $AB$ at $E$ and $F,$ respectively. Observe that since $PQ$ is perpendicular to the line connecting the circumcenters of $\triangle CEP$ and $\triangle DFP$ (the $P$-midline of $\triangle PEF$) and $P$ is an intersection of the two circles, $PQ$ must be the radical axis of the two circles, and since $DF$ and $EC$ meet on $PQ,$ $ECDF$ must be cyclic as well. As a result, we get that $\measuredangle CPG=\measuredangle CPQ=\measuredangle CDQ=\measuredangle CEG$ so $ECPG$ is cyclic and thus $\measuredangle PCE=\measuredangle EGP=90$ as desired. $\blacksquare$
GeoMetrix
18.08.2019 11:29
pretty angle chase let $\angle DPA=\alpha$ and $\angle PQD=\beta$ now $\angle PDQ +\angle PCQ= 90+90=180$ so we have $PDQC$ is cyclic. $\angle QPD = 90-\beta$ so $\angle APF= 90 +\beta-\alpha$ now $\angle DPC =180- \alpha$ and as $PDQC$ is cyclic we have $\angle DQC=\alpha$ and so we have $\angle CQP =\angle DQC -\angle DQP= \alpha -\beta$ . Cyclicity of $PDQC$ also gives $\angle BDC \equiv\angle PDC =\angle PQC =\alpha-\beta$ and now as $ABCD$ is cyclic we conclude $\angle BDC=\angle CAB=\alpha-\beta$ and so we get $\angle PFA= 180-(\angle PAF + \angle APF)=180-(\alpha-\beta+90 +\beta-\alpha)=180-90=90$ and so we are done $\blacksquare$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -21.78872864605096, xmax = 28.25671209404774, ymin = -17.921555613911423, ymax = 14.194219349222717; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-1.18,1.35), 5.869310010554903), linewidth(2) + wrwrwr); draw((2.44,5.97)--(-6.162777920030094,4.45172922732797), linewidth(2) + wrwrwr); draw((0.2593201949161543,-4.3400929145758695)--(4.3098420790716805,-0.7261584589943815), linewidth(2) + wrwrwr); draw((-6.162777920030094,4.45172922732797)--(4.3098420790716805,-0.7261584589943815), linewidth(2) + wrwrwr); draw((0.2593201949161543,-4.3400929145758695)--(2.44,5.97), linewidth(2) + wrwrwr); draw((xmin, -1.1208066814548219*xmin-4.049445107477677)--(xmax, -1.1208066814548219*xmax-4.049445107477677), linewidth(2) + wrwrwr); /* line */ draw((xmin, 2.022566079748257*xmin-9.443098857196468)--(xmax, 2.022566079748257*xmax-9.443098857196468), linewidth(2) + wrwrwr); /* line */ draw((xmin, -17.627776936675154*xmin + 24.274552323705024)--(xmax, -17.627776936675154*xmax + 24.274552323705024), linewidth(2) + wrwrwr); /* line */ /* dots and labels */ dot((-1.18,1.35),dotstyle); label("$O$", (-1.0349763181360132,1.6828591641980655), NE * labelscalefactor); dot((2.44,5.97),dotstyle); label("$B$", (2.5772274098238666,6.3130475791284475), NE * labelscalefactor); dot((-6.162777920030094,4.45172922732797),dotstyle); label("$A$", (-6.026385105862392,4.76965144081832), NE * labelscalefactor); dot((0.2593201949161543,-4.3400929145758695),dotstyle); label("$C$", (0.3770669573392124,-3.9981521534115525), NE * labelscalefactor); dot((4.3098420790716805,-0.7261584589943815),dotstyle); label("$D$", (4.449005705221259,-0.3859484254516799), NE * labelscalefactor); dot((1.334813400079058,0.7447594550264622),linewidth(4pt) + dotstyle); label("$P$", (1.4607280757271763,0.993256634314817), NE * labelscalefactor); dot((1.7158810486270326,-5.97261605136056),linewidth(4pt) + dotstyle); label("$Q$", (1.8547866642318904,-5.705739370265311), NE * labelscalefactor); dot((1.052286094454999,5.725087777087222),linewidth(4pt) + dotstyle); label("$F$", (1.1980223500573668,5.984665422041187), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]
Delta0001
18.08.2019 12:11
Proof Without Words
GeoMetrix
18.08.2019 12:13
Ok @below. But that's quite similar to what I did
Delta0001
18.08.2019 12:57
amaanmathbuddy_2006 wrote: what did i do then pls dont double post Your proof had words Also, double posting means a user is posting two consecutive posts (with probably same meaning) Also, I don't see how your proof is same as mine
zuss77
18.08.2019 13:08
$PDQC$ - cyclic with diameter $PQ$. By Reim's $AB$ is parallel to touchline at $P$, so $AB \perp PQ$. edit typo. thanks amar
amar_04
18.08.2019 13:15
zuss77 wrote: $PDQP$ - cyclic with diameter $PQ$. By Reim's $AB$ is parallel to touchline at $P$, so $AB \perp PQ$. Should have been posted in HSM. BTW @above it should be $PDQC$.
gamerrk1004
18.08.2019 13:16
Delta0001 wrote: Proof Without Words Delta0001 wrote: amaanmathbuddy_2006 wrote: what did i do then pls dont double post Your proof had words Also, double posting means a user is posting two consecutive posts (with probably same meaning) Also, I don't see how your proof is same as mine How can one be so much silly ! Please avoid doing that @Delta0001 ...
GeoMetrix
18.08.2019 13:16
@paramenindes has already mentioned why his problems shud be in hso
zuss77
18.08.2019 13:23
Why Proof Without Words is a problem? You need just 2 second to understand it. Compare it to #3... So, please, keep doing that @Delta0001.
parmenides51
18.08.2019 13:43
These problems come from the Correspodence Round of the Sharygin Geometry Olympiad of 2005. At the moment they are available only in Russian, and I am translating and posting each problem in each own thread in English in aops. I do not consider them that easy to be posted in the HSM forum. They are collected in aops here.
Rizsgtp
18.08.2019 19:11
GeoMetrix
18.08.2019 19:30
sorry
Delta0001
18.08.2019 20:14
amaanmathbuddy_2006 wrote: Why is everybody posting the same solution again and again Be mature man. First of all, everyone has a right to post their own solution however similar it may be to any other Secondly, the nature of problem is such that most of the solutions would probably have same progress (cyclic quads, angle chase etc) So don't object to others posting their solutions
Richangles
25.01.2020 22:37
let PQ meet AB at T.
If $AB \perp QT$ , then <QCA = <QTA=90°
» QCTA is cyclic
» <TQC = <TAC ....... (1)
But since <PCQ + <PDQ = 180°, then PDQC is cyclic. Thusly,
<TQC= <PQC= <PDC= <BDC= <BAC= <TAC
Hence (1) is true and the result follows.
Feridimo
04.04.2020 19:35
$ PQ\cap AB=T $
$ \angle PDQ+\angle PCQ=180\implies PDQC-cyclic $
$ Then $
$ \angle APT=\angle QPC=\angle QDC=90-\angle BDC=
90-\angle BAC\implies \angle APT+\angle BAC=90\implies \angle ATP=90 $
Marinchoo
03.11.2020 15:11
$PCQD$ is cyclic, because $\angle PCQ=90$ and $\angle PDQ=90$, so their sum is $180$. Now $\angle PQD=\angle PCD=\angle ACD=\angle DAB=\angle PBR$ where $R$ is the point of intersection of $PQ$ and $AB$. Since $\angle BPR=\angle QPD$ and $\angle DQP=\angle PBR$ then $\angle PRB=\angle PDQ$ so $AB\perp PQ$
th1nq3r
11.07.2022 23:03
$\bullet$ Denote by $R = AB \cap PQ$;
$\bullet$ $\angle PCD = \angle ABP = \angle RBP$;
$\bullet$ $\angle DCQ = \angle DPQ = \angle PBR$;
$\bullet$ $\angle PCD + \angle DCQ = 90$.
$\implies \angle RBP + \angle PBR = 90 \implies \angle PRB = 90$, as desired.
Minkowsi47
12.04.2023 07:05
2 line solution $$ QDPC is cyclic quadrilateral [\angle DQP =\angle DCP =\angle ABP ] and [\angle DPQ =\angle KPB ] $$It means $BKDQ$ Cyclic hence we are done.
cj13609517288
14.04.2023 01:10
Let $Y$ be the antipode of $B$ with respect to the circle. Then \[2\angle APQ=2\angle CDY=\overarc{YAC}=\overarc{YAB}+\overarc{BC}=180^{\circ}+2\angle CAB,\]so $\angle APQ-\angle CAB=90^{\circ}$ which immediately finishes.