Chase angles, not girls!?
by shiningsunnyday, Jun 3, 2016, 8:53 AM
For all y'all angle-chasers out there, the future isn't as bleak as it seems. As crazy as it sounds, I've discovered an astonishing truth -- girls like geometry.
For those who recognize the world-famous Kpop group Girls Generation (which as it sounds, speak on behalf of all the girls today), see what they have to say:
Do you see that smile? Do you see those curves, as they repetitively sing out "Gee gee gee gee" and "Oh oh oh oh?"
Blog readers: *mind-blown* Wow... I had no clue...
Now don't get ahead of yourselves. Let me start you guys off with the fundamentals. This is the first of two posts in spirit of Oly Geo. This post will be on similar triangles/angle-chasing and the next post will be more similar triangles and power of a point. So if you like geometry, let's get ripped!
Hm... what a bittersweet feeling, attempting this problem for days without solving it and suddenly having that moment of epiphany in bed at 1 AM before a Monday.
Yay first JMO problem.
Now time to finish up those end-of-chapter-1 problems for EGMO and chapter 2.
For those who recognize the world-famous Kpop group Girls Generation (which as it sounds, speak on behalf of all the girls today), see what they have to say:
Girls Generation wrote:
Blog readers: *mind-blown* Wow... I had no clue...
Now don't get ahead of yourselves. Let me start you guys off with the fundamentals. This is the first of two posts in spirit of Oly Geo. This post will be on similar triangles/angle-chasing and the next post will be more similar triangles and power of a point. So if you like geometry, let's get ripped!
Olympiad Forum wrote:
Given triangle 
where 
is 
degree and 
is 
degree let 
be a point inside such that 
and 
Find
Note: Geogebra is failing me, so you guys are gonna have to draw a diagram yourselves.
where 
is 
degree and 
is 
degree let 
be a point inside such that 
and 
Find
Note: Geogebra is failing me, so you guys are gonna have to draw a diagram yourselves.
is equilateral (what!?). This allows us to take advantage of the perpendicular bisectors created by the 
triangle. For one, line 
is a perpendicular bisector, implying 
and 
Now note 
is cyclic, which means 
after which the problem cracks. 
Beautiful!P55 of 106, Switzerland Final Round 2011 wrote:
Let ABCD be a parallelogram such that the triangle 
is acute and has orthocenter 
. The line through parallel to 
cuts 
and 
at 
and 
, respectively, while the line through parallel to cuts and 
at 
and 
, respectively. Prove that the points 
lie on the same circle.
![[asy]
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[/asy]](//latex.artofproblemsolving.com/d/a/6/da6cb9e4eb575fa8b579bcd0d1f620986b784b4a.png)
is acute and has orthocenter 
. The line through parallel to 
cuts 
and 
at 
and 
, respectively, while the line through parallel to cuts and 
at 
and 
, respectively. Prove that the points 
lie on the same circle.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/d/a/6/da6cb9e4eb575fa8b579bcd0d1f620986b784b4a.png)
hence the conclusion by converse of Power of a Point.2011 JMO 5 wrote:
Points 
lie on a circle 
and point lies outside the circle. The given points are such that 
lines 
and 
are tangent to , 
are collinear, and 
Prove that
bisects 
.
Diagram credit to user "application."
![[asy]
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lie on a circle 
and point lies outside the circle. The given points are such that 
lines 
and 
are tangent to , 
are collinear, and 
Prove that
bisects 
.Diagram credit to user "application."
![[asy]
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iff 
which follows from 
being cyclic, which we can show by 
Because 
is cyclic, 
which is true because they combine to intercept half of the whole circle.
hence 
is cyclic and conclusion follows.P54 of 106 wrote:
On the sides and of rhombus 
consider the points 
and 
such that 
Let 
and 
Prove that points 
and are collinear.
![[asy]
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and of rhombus 
consider the points 
and 
such that 
Let 
and 
Prove that points 
and are collinear.![[asy]
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dot((6.4,-2.16),dotstyle);
label("$D$", (6.412652323412216,-2.8317570354564103), NE * labelscalefactor);
dot((0.26,-2.16),dotstyle);
label("$C$", (-0.19107339432008227,-2.7636773888818507), NE * labelscalefactor);
dot((8.42172861272408,3.637604110017583),dotstyle);
label("$A$", (8.557161190510849,3.976207621999566), NE * labelscalefactor);
dot((2.2817286127240797,3.6376041100175835),linewidth(3.pt) + dotstyle);
label("$B$", (1.7151567097675913,3.976207621999566), NE * labelscalefactor);
dot((5.86,3.637604110017583),dotstyle);
label("$E$", (6.004174443964857,3.976207621999566), NE * labelscalefactor);
dot((7.243504891592546,0.25887432147862655),linewidth(3.pt) + dotstyle);
label("$F$", (7.501926668605171,0.26586688368605893), NE * labelscalefactor);
dot((5.105717700332717,11.73580811124824),linewidth(3.pt) + dotstyle);
label("$P$", (4.98297974534646,12.145765210946738), NE * labelscalefactor);
dot((10.795701717331196,-2.16),linewidth(3.pt) + dotstyle);
label("$Q$", (11.008028467194999,-2.661557919020011), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/3/0/6/30612611e2ee67f6b95f25749e140583ad570204.png)
that would imply 
and 
would be forced to go through 
hence collinearity. To do so, consider the following two pairs of similar triangles: 
keep in mind 
.
Done!Now time to finish up those end-of-chapter-1 problems for EGMO and chapter 2.
This post has been edited 1 time. Last edited by shiningsunnyday, Jun 3, 2016, 8:54 AM
May 2018
April 2018