Entering APMOde
by shiningsunnyday, Oct 5, 2016, 1:05 PM
2013 APMO P5 wrote:
Let
be a quadrilateral inscribed in a circle
, and let
be a point on the extension of
such that
and
are tangent to
. The tangent at
intersects
at
and the line
at
. Let
be the second point of intersection between
and
. Prove that
,
,
are collinear.


















Once again, we demolish the problem with projective.
Credits to v_Enhance for the beautiful diagram:
![[asy]/* DRAGON 0.0.9.6
Homemade Script by v_Enhance. */
import olympiad;
import cse5;
size(11cm);
real lsf=0.8000;
real lisf=2011.0;
defaultpen(fontsize(10pt));
/* Initialize Objects */
pair P = (4.0, -6.0);
pair O = (4.0, 2.0);
pair D = (7.794019966423534, -0.7329512659808339);
path DP = D--P;
pair A = (0.7383826268204343, 5.350442096025524);
pair C = (2)*(foot(O,A,P))-A;
pair B = 2*foot(D,relpoint(O--P,0.5-10/lisf),relpoint(O--P,0.5+10/lisf))-D;
pair T = IntersectionPoint(Line(B,D,lisf),Line(O,midpoint(A--C),lisf));
pair Q = IntersectionPoint(DP,Line(T,C,lisf));
pair Z = IntersectionPoint(Line(A,B,lisf),Line(T,C,lisf));
pair R = IntersectionPoint(Line(A,D,lisf),Line(T,C,lisf));
pair E_prime = (foot(O,A,Q))*(2)-A;
path BD = T--D;
path AP = A--P;
pair K = IntersectionPoint(BD,AP);
/* Draw objects */
draw(DP, rgb(0.6,0.0,0.6));
draw(CirclebyPoint(O,D), rgb(0.8,0.0,0.8) + linewidth(1.2));
draw(B--P, rgb(0.6,0.0,0.6));
draw(BD, rgb(0.8,0.0,1.0) + dashed);
draw(T--A, rgb(1.0,0.6,1.0));
draw(T--R, rgb(1.0,0.6,1.0));
draw(A--R, rgb(1.0,0.6,1.0));
draw(A--Z, rgb(1.0,0.0,0.8));
draw(AP, rgb(0.8,0.0,1.0) + dashed);
draw(B--R, rgb(1.0,0.0,0.8));
draw(A--Q, rgb(0.8,0.0,0.4) + linetype("4 4"));
/* Place dots on each point */
dot(P);
dot(D);
dot(A);
dot(C);
dot(B);
dot(T);
dot(Q);
dot(Z);
dot(R);
dot(E_prime);
dot(K);
/* Label points */
label("$P$", P, lsf * dir(0));
label("$D$", D, lsf * dir(0));
label("$A$", A, lsf * dir(135));
label("$C$", C, lsf * dir(45));
label("$B$", B, lsf * dir(225));
label("$T$", T, lsf * dir(135));
label("$Q$", Q, lsf * dir(-60));
label("$Z$", Z, lsf * dir(-90));
label("$R$", R, lsf * dir(-90));
label("$E'$", E_prime, lsf * dir(70));
label("$K$", K, lsf * dir(45));
[/asy]](//latex.artofproblemsolving.com/d/a/2/da29698a481a411535910c6836714d2f4fd60fee.png)
Note that because
is a harmonic quad,
(where
is the tangent to
at
) concur at a point, say 
Define
Because
are tangents, we have
harmonic. It suffices to show
is harmonic, which would imply
We finish by observing 
Credits to v_Enhance for the beautiful diagram:
![[asy]/* DRAGON 0.0.9.6
Homemade Script by v_Enhance. */
import olympiad;
import cse5;
size(11cm);
real lsf=0.8000;
real lisf=2011.0;
defaultpen(fontsize(10pt));
/* Initialize Objects */
pair P = (4.0, -6.0);
pair O = (4.0, 2.0);
pair D = (7.794019966423534, -0.7329512659808339);
path DP = D--P;
pair A = (0.7383826268204343, 5.350442096025524);
pair C = (2)*(foot(O,A,P))-A;
pair B = 2*foot(D,relpoint(O--P,0.5-10/lisf),relpoint(O--P,0.5+10/lisf))-D;
pair T = IntersectionPoint(Line(B,D,lisf),Line(O,midpoint(A--C),lisf));
pair Q = IntersectionPoint(DP,Line(T,C,lisf));
pair Z = IntersectionPoint(Line(A,B,lisf),Line(T,C,lisf));
pair R = IntersectionPoint(Line(A,D,lisf),Line(T,C,lisf));
pair E_prime = (foot(O,A,Q))*(2)-A;
path BD = T--D;
path AP = A--P;
pair K = IntersectionPoint(BD,AP);
/* Draw objects */
draw(DP, rgb(0.6,0.0,0.6));
draw(CirclebyPoint(O,D), rgb(0.8,0.0,0.8) + linewidth(1.2));
draw(B--P, rgb(0.6,0.0,0.6));
draw(BD, rgb(0.8,0.0,1.0) + dashed);
draw(T--A, rgb(1.0,0.6,1.0));
draw(T--R, rgb(1.0,0.6,1.0));
draw(A--R, rgb(1.0,0.6,1.0));
draw(A--Z, rgb(1.0,0.0,0.8));
draw(AP, rgb(0.8,0.0,1.0) + dashed);
draw(B--R, rgb(1.0,0.0,0.8));
draw(A--Q, rgb(0.8,0.0,0.4) + linetype("4 4"));
/* Place dots on each point */
dot(P);
dot(D);
dot(A);
dot(C);
dot(B);
dot(T);
dot(Q);
dot(Z);
dot(R);
dot(E_prime);
dot(K);
/* Label points */
label("$P$", P, lsf * dir(0));
label("$D$", D, lsf * dir(0));
label("$A$", A, lsf * dir(135));
label("$C$", C, lsf * dir(45));
label("$B$", B, lsf * dir(225));
label("$T$", T, lsf * dir(135));
label("$Q$", Q, lsf * dir(-60));
label("$Z$", Z, lsf * dir(-90));
label("$R$", R, lsf * dir(-90));
label("$E'$", E_prime, lsf * dir(70));
label("$K$", K, lsf * dir(45));
[/asy]](http://latex.artofproblemsolving.com/d/a/2/da29698a481a411535910c6836714d2f4fd60fee.png)
Note that because






Define






Tidbit
Ok this tidbit is kinda boring, but obligated to be included to complete the pun in the subject.
So my APUSH teacher from last year replied to my dad's e-mail, agreeing to help me prepare for the AP exam (again) around March, saying prior to that he'll give me some resources to self-study (though I'll still prob sign up for a local test prep class or something).
So my APUSH teacher from last year replied to my dad's e-mail, agreeing to help me prepare for the AP exam (again) around March, saying prior to that he'll give me some resources to self-study (though I'll still prob sign up for a local test prep class or something).
Tidbit 2
Also darn darn darn my first call with this teacher from this Yale's scholar program (the one I talked about regarding science fair mentor) is gonna start in less than an hour. Kinda nervous. 

This post has been edited 4 times. Last edited by shiningsunnyday, Oct 12, 2016, 1:27 PM