3 "Cyclic" Problems
by shiningsunnyday, Apr 7, 2016, 1:40 PM
So apparently it's only been 3 days of school after break and for some reason my math skills have already deteriorated. 
Thanks a lot my English teacher for making my brain suffer by staying up all night finishing busy work... Thankfully tomorrow is parent teacher conferences so I get tomorrow, Friday off. What better way to spend a cozy Thurs night solving problems? 
In parallelogram
with 
show that the circle passing through the projections of 
onto 
and 
respectively, passes through the center of the parallelogram.
Let be a cyclic quadrilateral. Let 
be the point not he ray 
such that 
and let 
be the point on the ray 
such that 
Prove that the line 
cuts 
at its midpoint.
Finally, here's another "cyclic" problem I'm struggling on:
Let
be positive reals such that 
Prove that ![$$\frac{a+b}{\sqrt{ab+c}}+\frac{b+c}{\sqrt{bc+a}}+\frac{c+a}{\sqrt{ca+b}} \ge 3 \sqrt[6]{abc}.$$](//latex.artofproblemsolving.com/7/d/f/7df5f219932ad37c674b4bf2dbc825e8cf212699.png)
EDIT: The two geometry problems are now solved. Yay.
EDIT 2: The solution to the inequality is now posted.
Thanks a lot my English teacher for making my brain suffer by staying up all night finishing busy work... Thankfully tomorrow is parent teacher conferences so I get tomorrow, Friday off. What better way to spend a cozy Thurs night solving problems? In parallelogram
with 
show that the circle passing through the projections of 
onto 
and 
respectively, passes through the center of the parallelogram.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/3/3/4/3344553d962e602d5ce13102cdc082c38a40ce46.png)
is cyclic, which will solve the problem. We use the fact that 
is the midpoint of ![[asy]
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[/asy]](http://latex.artofproblemsolving.com/b/5/a/b5ae8e607a8b1313c62f82fddffefb1dba753e71.png)
![[asy]
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Looks like we have to prove that this is equal to 
Note that direct chasing is quite futile and gets confusing real quick. So if only we can get some cyclic quadrilaterals to help us out...
and 
are two cyclic quadrilaterals staring straight at us. Howe can we use this fact? We can, by doing the following:
Done!Let
be a cyclic quadrilateral. Let 
be the point not he ray 
such that 
and let 
be the point on the ray 
such that 
Prove that the line 
cuts 
at its midpoint.![[asy]
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[/asy]](http://latex.artofproblemsolving.com/1/1/f/11f2ba7e1244c0961362aa64601a9dea963900b5.png)
and 
,
I wrote 
as 
partly cause I knew this problem had to be finished off by cyclic quadrilaterals.
upon setting 
.
So we would like to prove that:
But 
and 
So we're done!![[asy]
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[/asy]](http://latex.artofproblemsolving.com/e/1/d/e1d692c382bc450d58feb19f00cdebf49c41604a.png)
be the 
rotation of 
.
is the midline of triangle 
,
will be the midpoint of 
But 
We will show that triangles 
and 
are in fact congruent. Note that 
are equal by the given. 
Finally, 
by the given, so they're indeed congruent. So we're done.
Finally, here's another "cyclic" problem I'm struggling on:
Let
be positive reals such that 
Prove that ![$$\frac{a+b}{\sqrt{ab+c}}+\frac{b+c}{\sqrt{bc+a}}+\frac{c+a}{\sqrt{ca+b}} \ge 3 \sqrt[6]{abc}.$$](http://latex.artofproblemsolving.com/7/d/f/7df5f219932ad37c674b4bf2dbc825e8cf212699.png)
![$$\sum_{cyclic} \frac{a+b}{\sqrt{ab+c}} = \sum_{cyclic} \frac{(a+b)\sqrt{c}}{\sqrt{abc+c^2}} \ge \sum_{cyclic} \frac{(a+b)\sqrt{c}}{\sqrt{ab+bc+ca+c^2}} = \frac{(a+b)\sqrt{c}}{\sqrt{(a+c)(b+c)}} \ge 3 \sqrt[3]{\sqrt{abc}},$$](http://latex.artofproblemsolving.com/6/7/d/67de1b1fd878261aedde7480a9eb295cf7aeff0f.png)
as desired.
to produce 
can be factored into 
Inequalities are tricky...EDIT: The two geometry problems are now solved. Yay.
EDIT 2: The solution to the inequality is now posted.
This post has been edited 5 times. Last edited by shiningsunnyday, Apr 9, 2016, 3:38 AM
May 2018
April 2018