Short (Trivial) Results

by AlastorMoody, Feb 21, 2019, 3:08 PM

I know these are well-known results, But I got too bored studying that silly school stuff, so just wrote this post to kill my time

Result #1: Let $P$ be any point in $\Delta ABC$ , and let $\Delta A_1B_1C_1$ be the pedal triangle of $P$ WRT $\Delta ABC$ , then the perpendicular to $B_1C_1,C_1A_1,A_1B_1$ from $A,B,C$ concur at the isogonal conjugate of $P$ WRT $\Delta ABC$

$[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(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.84218824699888, xmax = 8.054690294900963, ymin = -5.997988716404487, ymax = 5.2317136957901145; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.5,4.79)--(-8.68,-4.31)--(6.26,-4.43)--cycle, linewidth(4.4) + rvwvcq); /* draw figures */ draw((-4.5,4.79)--(-8.68,-4.31), linewidth(4.4) + rvwvcq); draw((-8.68,-4.31)--(6.26,-4.43), linewidth(4.4) + rvwvcq); draw((6.26,-4.43)--(-4.5,4.79), linewidth(4.4) + rvwvcq); draw((-2.846640129516348,-0.03308057994084305)--(-2.88136699526381,-4.356575365499889), linewidth(3.2)); draw((-2.88136699526381,-4.356575365499889)--(-6.04136891517107,1.4343882468763807), linewidth(3.2)); draw((-6.04136891517107,1.4343882468763807)--(-1.1635803419998645,1.9310976536467241), linewidth(3.2)); draw((-1.1635803419998645,1.9310976536467241)--(-2.88136699526381,-4.356575365499889), linewidth(3.2)); draw((-2.846640129516348,-0.03308057994084305)--(-1.1635803419998645,1.9310976536467241), linewidth(3.2)); draw((-2.846640129516348,-0.03308057994084305)--(-6.04136891517107,1.4343882468763807), linewidth(3.2)); draw((-3.8421626438892544,-1.6700981880050354)--(-4.5,4.79), linewidth(2) + linetype("4 4")); draw((-3.8421626438892544,-1.6700981880050354)--(-8.68,-4.31), linewidth(2) + linetype("4 4")); draw((-3.8421626438892544,-1.6700981880050354)--(6.26,-4.43), linewidth(2) + linetype("4 4")); draw(circle((-3.6733200647581747,2.378459710029578), 2.549299175358211), linewidth(1.6) + linetype("4 4") + dtsfsf); draw(circle((-5.763320064758174,-2.1715402899704217), 3.616632657052973), linewidth(1.6) + linetype("4 4") + dtsfsf); draw(circle((1.7066799352418265,-2.2315402899704213), 5.056278147882365), linewidth(1.6) + linetype("4 4") + dtsfsf); draw((-2.846640129516348,-0.03308057994084305)--(-4.5,4.79), linewidth(0.8) + linetype("4 4")); draw((-2.846640129516348,-0.03308057994084305)--(-8.68,-4.31), linewidth(0.8) + linetype("4 4")); draw((-2.846640129516348,-0.03308057994084305)--(6.26,-4.43), linewidth(0.8) + linetype("4 4")); /* dots and labels */ dot((-4.5,4.79),dotstyle); label("$A$", (-4.445130183224587,4.937418046367084), NE * labelscalefactor); dot((-8.68,-4.31),dotstyle); label("$B$", (-9.014457371634796,-4.526510469289333), NE * labelscalefactor); dot((6.26,-4.43),dotstyle); label("$C$", (6.381851866601635,-4.66591367164761), NE * labelscalefactor); dot((-2.846640129516348,-0.03308057994084305),linewidth(6pt) + dotstyle); label("$P$", (-3.0046304255223863,0.2286876555985885), NE * labelscalefactor); dot((-2.88136699526381,-4.356575365499889),dotstyle); label("$A_1$", (-2.9891411808159107,-4.712381405767037), NE * labelscalefactor); dot((-1.1635803419998645,1.9310976536467241),dotstyle); label("$B_1$", (-1.0994533266259265,2.0873970203756262), NE * labelscalefactor); dot((-6.04136891517107,1.4343882468763807),dotstyle); label("$C_1$", (-6.5051997291858,1.5297842109425148), NE * labelscalefactor); dot((-3.8421626438892544,-1.6700981880050354),linewidth(6pt) + dotstyle); label("$P^*$", (-3.701646437313774,-1.475129262113696), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Proof: It's just a matter of Ceva's theorem for concurrerncy, we'll show the isogonal part, Let $P^*$ be the concurrency point, then, $\angle PAB=\angle PB_1C_1=\angle P^*AC$ $\qquad \blacksquare$

Result #2: Let $P$ be any point in $\Delta ABC$ , Let $\Delta DEF$ be the cevian triangle of $P$ WRT $\Delta ABC$ , and Let $\Delta A_1B_1C_1$ be the anti-pedal triangle of $P$ WRT $\Delta ABC$ . Let $O_A,O_B,O_C$ be the centers of $\odot (PEF), \odot (PDF) , \odot (PDE)$ . Then, $O_AA_1, O_BB_1, O_CC_1$ are concurrent lines

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(15); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.350293039628266, xmax = 28.916539394258667, ymin = -17.062221181952488, ymax = 11.620937436545995; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.5,4.79)--(-8.68,-4.31)--(6.26,-4.43)--cycle, linewidth(4.4) + rvwvcq); /* draw figures */ draw((-4.5,4.79)--(-8.68,-4.31), linewidth(4.4) + rvwvcq); draw((-8.68,-4.31)--(6.26,-4.43), linewidth(4.4) + rvwvcq); draw((6.26,-4.43)--(-4.5,4.79), linewidth(4.4) + rvwvcq); draw((-2.846640129516348,-0.03308057994084305)--(-4.5,4.79), linewidth(0.8) + linetype("4 4")); draw((-2.846640129516348,-0.03308057994084305)--(-8.68,-4.31), linewidth(0.8) + linetype("4 4")); draw((-2.846640129516348,-0.03308057994084305)--(6.26,-4.43), linewidth(0.8) + linetype("4 4")); draw((-6.013161342972689,1.495797076303481)--(-2.846640129516348,-0.03308057994084305), linewidth(0.8) + linetype("4 4")); draw((-2.846640129516348,-0.03308057994084305)--(-0.704360199156609,1.537602326786611), linewidth(0.8) + linetype("4 4")); draw((-2.846640129516348,-0.03308057994084305)--(-1.3603510763491176,-4.368792360832537), linewidth(0.8) + linetype("4 4")); draw((-1.3603510763491176,-4.368792360832537)--(-0.704360199156609,1.537602326786611), linewidth(3.2)); draw((-0.704360199156609,1.537602326786611)--(-6.013161342972689,1.495797076303481), linewidth(3.2)); draw((-6.013161342972689,1.495797076303481)--(-1.3603510763491176,-4.368792360832537), linewidth(3.2)); draw((-13.172300297150874,1.8171213473550183)--(13.728756214110602,11.03884729037839), linewidth(2.8)); draw((-13.172300297150874,1.8171213473550183)--(0.3628905242396914,-16.64374527700292), linewidth(2.8)); draw((0.3628905242396914,-16.64374527700292)--(13.728756214110602,11.03884729037839), linewidth(2.8)); draw(circle((-0.11104230508613024,-1.5179202379453096), 3.112594459130544), linewidth(1.6) + linetype("4 4") + dtsfsf); draw(circle((-3.3698652219679457,2.9268412740889587), 3.0058113512551317), linewidth(1.6) + linetype("4 4") + dtsfsf); draw(circle((-6.587920289597997,-3.738204549625273), 5.265464924102208), linewidth(1.6) + linetype("4 4") + dtsfsf); draw((-6.587920289597997,-3.738204549625273)--(-3.3698652219679457,2.9268412740889587), linewidth(2.8)); draw((-3.3698652219679457,2.9268412740889587)--(-0.11104230508613024,-1.5179202379453096), linewidth(2.8)); draw((-0.11104230508613024,-1.5179202379453096)--(-6.587920289597997,-3.738204549625273), linewidth(2.8)); draw((xmin, -5.2429325360961485*xmin-14.741134740425656)--(xmax, -5.2429325360961485*xmax-14.741134740425656), linewidth(3.6) + dtsfsf); /* line */ draw((xmin, 0.7273360796637317*xmin + 1.0534275669480897)--(xmax, 0.7273360796637317*xmax + 1.0534275669480897), linewidth(3.6) + dtsfsf); /* line */ draw((xmin, -0.25533846642693253*xmin-1.5462736098345136)--(xmax, -0.25533846642693253*xmax-1.5462736098345136), linewidth(3.6) + dtsfsf); /* line */ /* dots and labels */ dot((-4.5,4.79),dotstyle); label("$A$", (-4.3559246031994565,5.17217211955944), NE * labelscalefactor); dot((-8.68,-4.31),dotstyle); label("$B$", (-9.538674643231577,-4.876824141266234), NE * labelscalefactor); dot((6.26,-4.43),dotstyle); label("$C$", (6.563457160532341,-5.035076050885537), NE * labelscalefactor); dot((-2.846640129516348,-0.03308057994084305),linewidth(6pt) + dotstyle); label("$P$", (-3.2481612358643464,0.6224297180045086), NE * labelscalefactor); dot((-1.3603510763491176,-4.368792360832537),dotstyle); label("$D$", (-1.190886410813428,-3.9668756609552482), NE * labelscalefactor); dot((-0.704360199156609,1.537602326786611),dotstyle); label("$E$", (-0.5578787723362224,1.9280079723637498), NE * labelscalefactor); dot((-6.013161342972689,1.495797076303481),dotstyle); label("$F$", (-5.859317744582819,1.8884449949589244), NE * labelscalefactor); dot((0.3628905242396914,-16.64374527700292),dotstyle); label("$A_1$", (0.9059513916423156,-16.58746545309458), NE * labelscalefactor); dot((13.728756214110602,11.03884729037839),dotstyle); label("$B_1$", (14.080422867449158,10.71098895623501), NE * labelscalefactor); dot((-13.172300297150874,1.8171213473550183),dotstyle); label("$C_1$", (-14.048854067381667,1.3345633112913675), NE * labelscalefactor); dot((-6.587920289597997,-3.738204549625273),dotstyle); label("$O_B$", (-7.679214705204786,-3.68993481912147), NE * labelscalefactor); dot((-3.3698652219679457,2.9268412740889587),dotstyle); label("$O_A$", (-3.1294723036498704,2.837956452674736), NE * labelscalefactor); dot((-0.11104230508613024,-1.5179202379453096),dotstyle); label("$O_C$", (-0.3600638853120956,-2.3447935873574033), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Proof: Trivial to see, $O_AO_B||A_1B_1$ and similarly for other sides, hence, by Desargues' Theorem, the result follows $\qquad \blacksquare$

Result #3: Let $P$ be any point in $\Delta ABC$ with $M_A$ as midpoint of $BC$ , Let $P^*$ be the anti-complement of $P$ WRT $\Delta ABC$ , Let $AP^* \cap BC=X_1$ and let $X_2$ be the reflection of $X_1$ in $M_A$ , prove that, $PM_A$ bisects $AX_2$

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(15); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.947302729323027, xmax = 8.136760285225618, ymin = -5.436718762651805, ymax = 5.309958028780828; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-4.5,4.79)--(-8.68,-4.31)--(6.26,-4.43)--cycle, linewidth(4.4) + rvwvcq); /* draw figures */ draw((-4.5,4.79)--(-8.68,-4.31), linewidth(4.4) + rvwvcq); draw((-8.68,-4.31)--(6.26,-4.43), linewidth(4.4) + rvwvcq); draw((6.26,-4.43)--(-4.5,4.79), linewidth(4.4) + rvwvcq); draw((-4.761084502795285,-4.3414772329092735)--(-4.5,4.79), linewidth(3.2)); draw((xmin, -1.3431383230738154*xmin-5.995197370919317)--(xmax, -1.3431383230738154*xmax-5.995197370919317), linewidth(3.2) + linetype("4 4") + dtsfsf); /* line */ draw((-4.5,4.79)--(2.3410845027952853,-4.398522767090725), linewidth(3.2)); /* dots and labels */ dot((-4.5,4.79),dotstyle); label("$A$", (-4.447968812603726,4.9393829670072895), NE * labelscalefactor); dot((-8.68,-4.31),dotstyle); label("$B$", (-9.013453573653713,-4.5176926094534275), NE * labelscalefactor); dot((6.26,-4.43),dotstyle); label("$C$", (6.372822991183578,-4.651099631691902), NE * labelscalefactor); dot((-4.0521946466295855,-0.5525394484765619),linewidth(6pt) + dotstyle); label("$P$", (-3.988455736004539,-0.3820749200607316), NE * labelscalefactor); dot((-1.21,-4.37),dotstyle); label("$M_A$", (-1.3944303035897745,-4.962382683581675), NE * labelscalefactor); dot((1.1843892932591729,-2.8449211030468753),linewidth(6pt) + dotstyle); label("$P^*$", (1.2440641362378149,-2.6648173005857325), NE * labelscalefactor); dot((2.3410845027952853,-4.398522767090725),dotstyle); label("$X_1$", (2.1779132919071302,-5.006851690994499), NE * labelscalefactor); dot((-4.761084502795285,-4.3414772329092735),dotstyle); label("$X_2$", (-5.011242906499504,-4.858621666285083), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Proof: Notice from the definition, $AP^*\parallel PM_A$ . Now from mid point theorem on $AX_1X_2$ , we see that $PM_A$ bisects $AX_2$ . $\qquad \blacksquare$

This post has been edited 27 times. Last edited by AlastorMoody, Mar 28, 2019, 4:40 PM

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In the first result, how do you know that $\angle PB_1C_1=\angle P^*AC?$

Edit(AlastorMoody):Let $PB_1 \cap AP^*=X_1$ and $AP^* \cap B_1C_1 =X_2$ , then $\Delta X_1B_1X_2 \sim \Delta X_1AB_1$ , therefore, $\angle PB_1C_1$ $=$ $\angle X_1B_1X_2$ $=$ $\angle X_1AB_1$ $=$ $\angle P^*AC$

This post has been edited 2 times. Last edited by AlastorMoody, Mar 5, 2019, 6:30 AM

by Kagebaka, Mar 4, 2019, 5:52 AM

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what a goat, u used to be friends with my brother
by bookstuffthanks, Jul 31, 2024, 12:05 PM
hello fellow moody!!
by crazyeyemoody907, Oct 31, 2023, 1:55 AM
@below I wish I started earlier / didn't have to do JEE and leave oly way before I could study conics and projective stuff which I really wanted to study . Huh, life really sucks when u are forced due to peer pressure to read sh_t u dont want to read
by kamatadu, Jan 3, 2023, 1:25 PM
Lots of good stuffs here.
by amar_04, Dec 30, 2022, 2:31 PM
But even if he went to jee he could continue with this.

Doing JEE(and completely leaving oly) seems like a insult to the oly math he knows
by HoRI_DA_GRe8, Feb 11, 2022, 2:11 PM
Ohhh did he go for JEE? Good for him, bad for us :sadge:. Hmmm so that is the reason why he is inactive
Btw @below finally everyone falls to the monopoly of JEE Coz IIT's are the best in India.
by BVKRB-, Feb 1, 2022, 12:57 PM
Kukuku first shout of 2022,why did this guy left this and went for trashy JEE
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When are you going to br alive again ,we miss you
by HoRI_DA_GRe8, Aug 11, 2021, 5:10 PM
kukuku first shout o 2021
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wow I completely forgot this blog
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This site would work faster if not all diagrams were displayed on the initial page. Anyway I like your problem selection taste.
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nice blog
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Hello everyone, nice blog
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pro blogo
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Nice !
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Thanks dude!
by AlastorMoody, Mar 1, 2019, 1:39 PM
Amazing work bro!!
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@below Thanks a lot
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Nice Blog, loved it.
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Nice work.
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@below Thanks buddy!
by AlastorMoody, Jan 29, 2019, 9:31 AM
Nice blog! 6th shout!
by PhysicsMonster_01, Jan 29, 2019, 9:21 AM
@below I am noob right now I hope to develop myself within a year or so
by AlastorMoody, Jan 28, 2019, 7:24 PM
Nice blog. But the properties are a bit 'basic'.
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Nice!!!!!!!!
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can I have contrib?
by Kagebaka, Dec 23, 2018, 8:24 PM
Math is fun!
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120 shouts

Magic of Hogwarts

Short (Trivial) Results

by AlastorMoody, Feb 21, 2019, 3:08 PM

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