I got a "pent"-up desire to solve this problem

by shiningsunnyday, Apr 14, 2016, 8:08 AM

[Poland 2010]

In equiangular pentagon $ABCDE$, prove that the perpendicular bisectors of $AE, BC$ and the angle bisector of $\angle{CDE}$ are concurrent.

[asy]
 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(10.22326024677836cm); 
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 = -3.1992362104443917, xmax = 7.024024036333968, ymin = -1.6419385941339073, ymax = 4.836680066688859;  /* image dimensions */
pen qqwuqq = rgb(0.,0.39215686274509803,0.); 

draw(arc((3.6659863945578213,-0.9386194167589497),0.3887171196493673,72.,180.)--(3.6659863945578213,-0.9386194167589497)--cycle, qqwuqq); 
draw(arc((4.785429822950617,2.5066731936045823),0.3887171196493673,144.,252.)--(4.785429822950617,2.5066731936045823)--cycle, qqwuqq); 
draw(arc((2.1752106212148203,4.403108451081842),0.3887171196493673,-144.,-36.)--(2.1752106212148203,4.403108451081842)--cycle, qqwuqq); 
draw(arc((-1.3830172790818867,1.8179045571810635),0.3887171196493673,-72.,36.)--(-1.3830172790818867,1.8179045571810635)--cycle, qqwuqq); 
draw(arc((-0.48736834681057295,-0.9386194167589498),0.3887171196493673,0.,108.)--(-0.48736834681057295,-0.9386194167589498)--cycle, qqwuqq); 
 /* draw figures */
draw((-1.3830172790818867,1.8179045571810635)--(2.1752106212148203,4.403108451081842)); 
draw((2.1752106212148203,4.403108451081842)--(4.785429822950617,2.5066731936045823)); 
draw((4.785429822950617,2.5066731936045823)--(3.6659863945578213,-0.9386194167589497)); 
draw((3.6659863945578213,-0.9386194167589497)--(-0.48736834681057295,-0.9386194167589498)); 
draw((-0.48736834681057295,-0.9386194167589498)--(-1.3830172790818867,1.8179045571810635)); 
draw((0.39609667106646684,3.1105065041314526)--(1.4252952535158208,1.6939361826736026)); 
draw((4.225708108754219,0.7840268884228163)--(1.4252952535158208,1.6939361826736026)); 
draw((-0.48736834681057295,-0.9386194167589498)--(1.4252952535158208,1.6939361826736026), linetype("4 4")); 
 /* dots and labels */
dot((3.6659863945578213,-0.9386194167589497),dotstyle); 
label("$C$", (3.745842993957637,-1.1495635759113771), NE * labelscalefactor); 
dot((4.785429822950617,2.5066731936045823),dotstyle); 
label("$B$", (4.886079878262447,2.5562062980792453), NE * labelscalefactor); 
dot((2.1752106212148203,4.403108451081842),dotstyle); 
label("$A$", (2.139145566073586,4.564578082934303), NE * labelscalefactor); 
dot((-1.3830172790818867,1.8179045571810635),dotstyle); 
label("$E$", (-1.59253878256034,1.8824299573536776), NE * labelscalefactor); 
dot((-0.48736834681057295,-0.9386194167589498),linewidth(3.pt) + dotstyle); 
label("$D$", (-0.6207459834369219,-1.1625208132330227), NE * labelscalefactor); 
dot((0.39609667106646684,3.1105065041314526),linewidth(3.pt) + dotstyle); 
dot((4.225708108754219,0.7840268884228163),linewidth(3.pt) + dotstyle); 
dot((1.4252952535158208,1.6939361826736026),linewidth(3.pt) + dotstyle); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 /* end of picture */
[/asy]

Proof without words
This post has been edited 5 times. Last edited by shiningsunnyday, Apr 14, 2016, 1:27 PM

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Hm, if we extend $DC$ to meet $BA$ at $F$ and we extend $DE$ to meet $BA$ at $G$, we see that the perpendicular bisectors are angle bisectors of $EGA$ and $CFD$, and we know that the angle bisectors of a triangle are concurrent.

Nice solution. :)
This post has been edited 1 time. Last edited by shiningsunnyday, Apr 17, 2016, 12:55 AM

by SantaDragon, Apr 16, 2016, 7:22 PM

The ones who are crazy enough to think they can change the world are the ones who do.

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  • The blog is locked right?

    by First, Apr 14, 2018, 6:00 PM

  • Great, amazing, inspiring blog. Good luck in life, and just know I aspire to succeed as you will in the future.

    by mgrimalo, Apr 7, 2018, 6:19 PM

  • Yesyesyes

    by shiningsunnyday, Mar 29, 2018, 5:30 PM

  • :O a new background picture

    by MathAwesome123, Mar 29, 2018, 3:39 PM

  • did you get into MIT?

    by 15Pandabears, Mar 15, 2018, 10:42 PM

  • wait what new site?

    by yegkatie, Feb 11, 2018, 1:49 AM

  • Yea, doing a bit of cleaning before migrating to new site

    by shiningsunnyday, Jan 21, 2018, 2:43 PM

  • Were there posts made in December 2017 for this blog and then deleted?

    I ask because I was purging my thunderbird inbox and I found emails indicating new blog posts of yours.

    email do not lie

    by jonlin1000, Jan 21, 2018, 12:12 AM

  • @below sorry not accepting contribs

    by shiningsunnyday, Dec 11, 2017, 11:15 AM

  • contrib plez?
    also wow this blog is very popular

    by DavidUsa, Dec 10, 2017, 7:53 PM

  • @First: lol same

    first shout of december

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  • XD this blog is hilarious

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  • @wu2481632: stop encouraging SSD to procrastinate(blog entries are fun but procrastination isn't).

    by First, Aug 7, 2017, 5:02 PM

  • 3.5 weeks without a post :o

    by Flash12, Aug 4, 2017, 8:10 AM

  • First august shout!!

    by adik7, Aug 1, 2017, 6:52 AM

416 shouts
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