such that![\[(a^2 + b^2 + c^2)^2 = 3(a^3b + b^3c + c^3a) + 1. \]](http://latex.artofproblemsolving.com/9/e/4/9e400fd807a9535ede983b75dcc307d8b75a35b6.png)
by adding every 
?
on 
and 
on 
,
and 
on 
be such that 
and 
.
with 
and 
are 
and 
with 
and 
are 
and 
,
and 
are concurrent.
are different real numbers. Prove that:
.
are on the sides 
respectively such that 
.
,
?
,
.
,
is a perfect square. Find the sum of all possible values of 
.
Let 
.
Let 
.
Let 
.
and 
share a common complex root?
![\[17! = 355687ab8096000\]](http://latex.artofproblemsolving.com/a/6/d/a6d147d3866c4f1fe6dac9481e53d1354891388b.png)
for some digits 
and 
.
that is missing above.
be the polynomial with leading coefficent 1 and rational coefficents, such that ![\[p\left(\sqrt{3 + \sqrt{3 + \sqrt{3 + \ldots}}}\right) = 0,\]](http://latex.artofproblemsolving.com/0/f/4/0f472be9f2cede0221e1b4d970d65de94dba01b9.png)
and with the least degree among all such polynomials. Find 
.
such that 
is a positive integral power of a prime positive integer."
for positive integral values of 
be the largest perfect square that divides 
such that 
is divisible by 
but not 
.
Y by
A point 
is given on the altitude through point 
in the acute triangle 
with circumcenter 
, such that 
. If the line 
intersects side 
at point 
, prove that the perpendicular bisector of , the altitude through 
and the segment 
are concurrent.
is given on the altitude through point 
in the acute triangle 
with circumcenter 
, such that 
. If the line 
intersects side 
at point 
, prove that the perpendicular bisector of , the altitude through 
and the segment 
are concurrent.This post has been edited 2 times. Last edited by Pitagar, May 16, 2021, 2:32 PM
be the foot from 
.
.
,
.
.
.
.
and 
.
,
as desired.
and 
.
be the orthocenter of 
and 
(I chose this name because if 
,
)
is the intersection point. This is equivalent to showing that 
.
and the points 
, 
, 
we know that:
Now 
and 
,
Now 
and 
,
Finally 
,
.
,
Now it's easy, because 
,
and 
,
,
,
be the foots of perpendiculars from 
intersect 
at 
.
.
becomes the unit circle. Then by the extended law of sines, it is evident that 
.![\[\angle CBT = \angle B - \angle C = (90^\circ-\angle C)-(90^\circ-\angle B) = \angle OCF.\]](http://latex.artofproblemsolving.com/5/1/6/516c89d78ac42ffbd9d7044ac19bcedd4376330c.png)
By the ratio lemma, we have![\[\frac{CT}{TE}=\frac{\sin\angle BEC}{\sin\angle BCE}\cdot\frac{\sin\angle CBT}{\sin\angle TBE}=\frac{1}{\sin\angle XAM}\cdot\frac{\sin\angle OAM}{\sin C} = \frac{OA}{AM}\cdot\frac{\sin\angle OAM}{\sin\angle XAM}=\frac{OX}{XM},\]](http://latex.artofproblemsolving.com/9/6/6/966ca704b70b950ca53c05ccfed34df381b0ce94.png)
and we're done!
and after that, if you get the idea to turn the sides into trig functions using the sine law you are pretty much set as two cosines cancel and you finally get two fractions which are 1s, so I really enjoyed bashing this one actually.
on 
such that 
as then 
even though I'm not sure if 
be the foot of the perpendicular from 
and 
is altitude, equivalent to the similarity of 
and 
,
.
.
,
,
,
and 
.
,
.
be the reflection of 
be the reflection of 
to the circumcircle of 
is the reflection of 
is the 
.
yields that 
are collinear. Reflecting back in 
.
and (as 
)
.
,
and so 
.
and so 
,
.
