Geo 3 Prep
by djmathman, May 24, 2017, 3:14 AM
My Olympiad Geo skills are rustier than iron, holy heck
APMO 2017 #2 wrote:
Let 
be a triangle with 
. Let 
be the intersection point of the internal bisector of angle 
and the circumcircle of . Let 
be the intersection point of the perpendicular bisector of 
with the external bisector of angle 
. Prove that the midpoint of the segment 
lies on the circumcircle of triangle 
.
be a triangle with 
. Let 
be the intersection point of the internal bisector of angle 
and the circumcircle of . Let 
be the intersection point of the perpendicular bisector of 
with the external bisector of angle 
. Prove that the midpoint of the segment 
lies on the circumcircle of triangle 
.
,
,
,
,
respectively, noting that 
.
,![\[\dfrac{DE}{EM}=\dfrac{DE}{CN}=\dfrac{DC}{CZ}.\]](http://latex.artofproblemsolving.com/f/8/5/f85d3974e4fbcdb12175baa12ca8de2111157ddb.png)
Furthermore, ![\[\angle DEM=90^\circ+\angle MEB = 90^\circ + \angle ACB = (\angle ZCN+\angle DCE)+\angle ACB=\angle ZCD.\]](http://latex.artofproblemsolving.com/f/0/8/f084653f76a3424f960476b3c58ccd5747ef05db.png)
Hence 
,![\[\triangle DEC\sim\triangle DMZ\quad\implies\quad \angle DMZ=\angle DEC = 90^\circ.\]](http://latex.artofproblemsolving.com/0/2/7/027818d0f6ff208fb7d397c9c6f4f05ae7d28c06.png)
Combining this with 
establishes the desired cyclicity.![[asy]
size(250);
defaultpen(linewidth(0.8));
pair A = dir(120), B = dir(215), C = dir(325), D = dir(270), M = (A+B)/2, N = (A+C)/2, E = (B+C)/2, P = dir(90), Z = extension(A,P,origin,N);
draw(A--B--C--A--Z--C^^Z--N^^B--D--C^^unitcircle);
draw(circumcircle(D,A,Z),linetype("4 4"));
draw(D--M--E--cycle,linetype("8 8"));
label("$A$",A,dir(origin--A));
label("$B$",B,dir(origin--B));
label("$C$",C,dir(origin--C));
label("$D$",D,dir(origin--D));
dot(M^^N^^E);
label("$M$",M,dir(origin--M));
label("$Z$",Z,dir(N--Z));
label("$E$",E,NE);
label("$N$",N,dir(Z--N));
draw(rightanglemark(Z,N,C,3)^^rightanglemark(D,E,C,3));
[/asy]](http://latex.artofproblemsolving.com/d/d/f/ddf843b3473946918faa725a01fa3337c7a9b515.png)
Geo 3 Cornell 2013 Test 1 #4 wrote:
On the circumcircle 
of triangle , two points , 
are situated. 
and 
intersect 
at 
and 
, respectively. Let 
, 
be the reflections of , across the perpendicular bisector of . Prove that 
, 
intersect on .
of triangle , two points , 
are situated. 
and 
intersect 
at 
and 
, respectively. Let 
, 
be the reflections of , across the perpendicular bisector of . Prove that 
, 
intersect on .
and 
are isogonal, with respect to 
,
and 
,
and the problem is trivial. Now assume 
with 
(which we know exists since these two circles are not homothetic). I claim that 
is the concurrence point in question.
throughout this proof. Note that since 
,
as well, with 
and 
.
,![\[\dfrac{AX}{AC} = \dfrac{AB}{AD'} \quad\implies\quad AX\cdot AD' = AB\cdot AC.\]](http://latex.artofproblemsolving.com/e/0/e/e0e8d9e761a1dd9fd22c03eb9010965c2c807204.png)
Similarly, 
are isogonal and 
.
inversion 
centered at 
sends 
and 
,
.
and 
,
.
,![\[\measuredangle AYT' \equiv\measuredangle AYB = \dfrac{\widehat{AB} + \widehat{EC}}2 = \dfrac{\widehat{AB}+\widehat{BE'}}2 = \measuredangle AD'E'\equiv \measuredangle AD'T',\]](http://latex.artofproblemsolving.com/d/f/9/df9dc1bf9a7907165357692df7dfa147b531073c.png)
where here arc measures are also directed. Similarly, 
.
and immediately got distracted with the spiral similarity configuration that emerged.Romania TST 2007, Cosmin Pohoata wrote:
Let 
be a triangle, let 
be the tangency points of the incircle 
to the sides 
, respectively 
, and let 
be the midpoint of the side 
. Let 
, let 
be the circle of diameter , and let 
be the other (than 
) intersection points of 
, respectively 
, with 
. Prove that
![\[ \frac {NX} {NY} = \frac {AC} {AB}.
\]](//latex.artofproblemsolving.com/6/6/5/665ac88bcf46454f6920a95e959d73e1ac3a9705.png)
be a triangle, let 
be the tangency points of the incircle 
to the sides 
, respectively 
, and let 
be the midpoint of the side 
. Let 
, let 
be the circle of diameter , and let 
be the other (than 
) intersection points of 
, respectively 
, with 
. Prove that![\[ \frac {NX} {NY} = \frac {AC} {AB}.
\]](http://latex.artofproblemsolving.com/6/6/5/665ac88bcf46454f6920a95e959d73e1ac3a9705.png)
,
and 
,
,
are concyclic. Thus ![\[\angle XEC = \angle XIC = 90^\circ - \frac {\angle A}2 = 180^\circ - \angle FEC,\]](http://latex.artofproblemsolving.com/c/0/e/c0ef1fb2dd3da0b15868b3efc9dbd0a0df4b5a84.png)
whence 
.![\[\dfrac{N'E}{N'F} = \dfrac{AE\sin\angle N'AE}{AF\sin\angle N'AF} = \dfrac{\sin\angle N'AE}{\sin\angle N'AF}\]](http://latex.artofproblemsolving.com/3/9/b/39b2d927b020ea622316ca4b0bd8e20f8bc3e587.png)
and ![\[\dfrac{N'E}{N'F} = \dfrac{DE\sin\frac C2}{DF\sin\frac B2} = \dfrac{2r\cos\frac C2\sin\frac C2}{2r\cos\frac B2\sin\frac B2} = \dfrac{\sin C}{\sin B} = \dfrac{\sin\angle MAC}{\sin\angle MAB}.\]](http://latex.artofproblemsolving.com/5/5/2/552b51d98e0938673afb189899b09a2f113fcbff.png)
Combining this with 
yields that 
,
,
.
and 
are isogonal with respect to either triangle (so that e.g. if 
is placed on 
,
would be isogonal), and so Steiner guarantees that ![\[\dfrac{NX}{NY}\cdot\dfrac{CD}{DB} = \left(\dfrac{IC}{IB}\right)^2.\]](http://latex.artofproblemsolving.com/b/2/2/b229e93e701f238c57eba253c35f99d8b4aa60d6.png)
Thus ![\[\dfrac{NX}{NY} = \dfrac{DB}{DC}\left(\dfrac{IC}{IB}\right)^2 = \dfrac{DB}{DC}\cdot\dfrac{\sin^2\frac B2}{\sin^2\frac C2} = \dfrac{DE\sin\frac C2}{DF\sin\frac B2} = \dfrac{AC}{AB}.\]](http://latex.artofproblemsolving.com/3/a/5/3a53d2d3f23e6ca541fa7a0dfb9d1ce7ffcc1f46.png)
This post has been edited 1 time. Last edited by djmathman, May 24, 2017, 3:15 AM
,
,
.
,
,
bisects 
,
.
,
meet at incenter 
.
;
and 
gg(g)
April 2025
November 2024