[/quote]
,
be non-negative real numbers such that 
Prove that
be the circumcenter, symmedian point of 
.
at 
,
at 
,
at 
are concurrent.
be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly 
.
divides any of the 
'
be an integer. Prove that 
can be written as a polynomial with integer coefficients in the variables 
,
and 
.
,
are real numbers such that 
is an integral multiple of 
.
,
for all positive integers 
.
be the orthocenter of an acute-angled triangle 
.
centered at the midpoint of 
and passing through 
and 
.
,
,
and 
.
with real coefficient. It is known that 
and 
.
can't have 
distinct real roots.
be a point on side 
be a point on side 
such that lines 
and 
are parallel. Let 
be an interior point of 
.
and 
meet side 
,
and 
intersect at a point 
.
,
are collinear.
.
,
as the greatest exponent 
such that 
divides 
.
and infinitely many 
such that 
.
and 
.
is given. Let 
lies on the arc 
.
.
Y by Adventure10, Mango247
Let 
be an equilateral triangle, and let 
be a point inside this triangle. Let 
and 
be the feet of the perpendiculars from to the sides 
and 
respectively. Prove that
(a)
and
(b)![$[APF] + [BPD] + [CPE] = [APE] + [BPF] + [CPD]$](//latex.artofproblemsolving.com/f/9/c/f9c29ec1487f5db8e088a50c500cfb08fd4266f0.png)
.
The area of triangle
is denoted ![$[XYZ]$](//latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
.
See here: http://www.bmoc.maths.org/home/bmo1-2014.pdf
be an equilateral triangle, and let 
be a point inside this triangle. Let 
and 
be the feet of the perpendiculars from to the sides 
and 
respectively. Prove that(a)
and(b)
![$[APF] + [BPD] + [CPE] = [APE] + [BPF] + [CPD]$](http://latex.artofproblemsolving.com/f/9/c/f9c29ec1487f5db8e088a50c500cfb08fd4266f0.png)
.The area of triangle
is denoted ![$[XYZ]$](http://latex.artofproblemsolving.com/c/7/6/c7608a3e97e44fc5bcf0cd62a1f7dfc0a0a2e7d4.png)
.See here: http://www.bmoc.maths.org/home/bmo1-2014.pdf
are concurrent iff 
,
with 
and 
,
and 
.
and 
.
past 
and 
will be 30-60-90 triangles. Hence 
and 
,
.
or 
,
past 
,
and 
.
and 
,
.
,
and 
;
,
and 
.
;
,
in the previous equation yields 
,
.
and 
we have:
,
,
.![$[APF]=\tfrac{a}{2\sqrt{3}}(a+2b-1)$](http://latex.artofproblemsolving.com/d/b/3/db3fc542bc5cf8c3c318e94aceab72a60863141c.png)
,![$[BPD]=\tfrac{b}{2\sqrt{3}}(2-2a-b)$](http://latex.artofproblemsolving.com/6/3/3/6338f21520e2ecbc82c1e6de54aa734df9e32517.png)
,![$[CPE]=\tfrac{(3-2a-2b)}{8\sqrt{3}}(1+2a-2b)$](http://latex.artofproblemsolving.com/d/1/f/d1f1cd1b298a49c6af522380b538cba0ea15c648.png)
,![$[APE]=\tfrac{(2a+2b-1)}{8\sqrt{3}}(1+2a-2b)$](http://latex.artofproblemsolving.com/d/2/a/d2ae2a54a89b1ec3a7526dc002249e4cd6acf19a.png)
,![$[BPF]=\tfrac{(1-a)}{2\sqrt{3}}(a+2b-1)$](http://latex.artofproblemsolving.com/7/6/7/7677989faf8f5dc738cc77e41a05d5cce3ad2755.png)
,![$[CPD]=\tfrac{(1-b)}{2\sqrt{3}}(2-2a-b)$](http://latex.artofproblemsolving.com/4/1/9/419dba77faf65aca1302ab6e80be4e24f5551d42.png)
.![$[APF]+[BPD]+[CPE]=[APE]+[BPF]+[CPD]$](http://latex.artofproblemsolving.com/6/0/6/606be2923b0ead22999474f3f442cc7c39a12096.png)
,
(squares are trivial), hence the interior angles are obtuse.
with 
be the foot of the perpendicular from 
.
and![$\sum_{i=0}^{n-1}[A_i P B_{i,i+1}] = \sum_{i=0}^{n-1} [B_{i,i+1}PA_{i+1}]$](http://latex.artofproblemsolving.com/0/5/0/0500645745973a777dd2444cd3a500f4a99f47b1.png)
,
.
and 
.
and 
.![\[AB =\frac{x}{\sin \theta} - \frac{y}{\tan \theta}\ \text{and}\ BC=\frac{y}{\sin \theta}-\frac{x}{\tan \theta}.\]](http://latex.artofproblemsolving.com/7/8/1/781c091fd9fae9a647a932d7fb850aec38e11cd1.png)
past 
we get ![\[EA=\frac{y}{\tan (\theta-90^o)}\ \text{and}\ DE=\frac{y}{\sin (\theta-90^o)},\]](http://latex.artofproblemsolving.com/9/6/8/9685cd979fa813c0e8c4d2b40e45a28e9b0027c2.png)
and similarly 
gives ![\[CD+DE=\frac{BC}{\tan (\theta-90^o)}\ \text{and}\ EA+AB=\frac{BC}{\sin(\theta-90^o)}.\]](http://latex.artofproblemsolving.com/f/c/4/fc4490f8ff100a587754990bcb9b5105c55cbaa9.png)
Hence, solving for ![\[AB =\frac{x}{\sin \theta} - \frac{y}{\tan \theta}\ \text{and}\ BC=\frac{y}{\sin \theta}-\frac{x}{\tan \theta},\]](http://latex.artofproblemsolving.com/e/e/f/eef4f70edd043e76f997b089ded7fc43fec39784.png)
as claimed. Or denoting 
and 
,![\[AB=\frac{x-cy}{s}\ \text{and}\ BC=\frac{y-cx}{s}.\]](http://latex.artofproblemsolving.com/8/4/3/843de696a8647d7c3f3ad11178df346694a89c6d.png)
Now, let 
,
so that 
,
and all indices are taken ![\[\Sigma a_0 = \Sigma (2-a_0)\Leftrightarrow \Sigma 2a_0 = \Sigma 2\Leftrightarrow \Sigma a_0 = \Sigma 1=n.\]](http://latex.artofproblemsolving.com/3/6/b/36b86aefe62aa79882fd53efe5d42981b2440685.png)
By the Pythagorean Theorem, ![\[P B_{0,1}^2+(2-a_0)^2=a_1^2+B_{1,2} P^2\]](http://latex.artofproblemsolving.com/a/c/a/acaa4a91946605d3a732df37c0f8d4cfaa0c4905.png)
![\[\Leftrightarrow a_1^2-(2-a_0)^2=PB_{0,1}^2-B_{1,2} P^2.\]](http://latex.artofproblemsolving.com/5/1/7/5176289683c2910029bd3d6336ab9f2af5214fc3.png)
Similarly, ![\[a_1^2-(2-a_0)^2=PB_{0,1}^2-B_{1,2} P^2,\]](http://latex.artofproblemsolving.com/9/1/6/9161cd4beba3aeca214af4ed67acd596f9fdd3fb.png)
![\[a_2^2-(2-a_1)^2=PB_{1,2}^2-B_{2,3}P^2,\]](http://latex.artofproblemsolving.com/3/7/d/37d1f99621c57298a6360b07274b91d2049a5635.png)
![\[...\]](http://latex.artofproblemsolving.com/a/7/7/a77a80061f2acc5bff28985d471dadc862f24425.png)
![\[a_0^2-(2-a_{n-1})^2=PB_{n-1,0}^2-B_{0,1}P^2.\]](http://latex.artofproblemsolving.com/b/8/e/b8eef54dbd8aaf33b2b45d8f6a4f3f97c1c13845.png)
Hence, adding up and telescoping yields: ![\[\Sigma a_0^2 -\Sigma (2-a_0)^2= 0\Leftrightarrow \Sigma a_0^2-(2-a_0)^2=0\]](http://latex.artofproblemsolving.com/3/1/5/31528228b980d2192a48ebf6f452b0dc153619ef.png)
![\[\Leftrightarrow \Sigma 4(a_0-1)=0\Leftrightarrow \Sigma a_0=\Sigma 1=n,\]](http://latex.artofproblemsolving.com/6/3/b/63be46ffdaa163f524f17a0038fc32cf31ee7add.png)
as desired.![\[\frac{a_0((2-a_{n-1})-ca_0)}{2s}+\frac{a_1((2-a_0)-ca_1)}{2s}+\cdots +\frac{a_{n-1}((2-a_{n-2})-ca_{n-1})}{2s}\]](http://latex.artofproblemsolving.com/a/4/9/a49230090c4d3ac90bcc80cf8f0c976276dad232.png)
![\[=\frac{(2-a_0)(a_1-c(2-a_0))}{2s}+\frac{(2-a_1)(a_2-c(2-a_1))}{2s} +\cdots\]](http://latex.artofproblemsolving.com/4/0/a/40a4d2079f11fc6a6cecf2e2eb24ee9ba90cbe3c.png)
![\[\cdots+\frac{(2-a_{n-1})(a_0-c(2-a_{n-1}))}{2s}\]](http://latex.artofproblemsolving.com/5/3/0/530102fd7fa63a9d74159ed0d6d6ba26f83f13c7.png)
![\[\Leftrightarrow a_0((2-a_{n-1})-ca_0)+a_1((2-a_0)-ca_1)+\cdots +a_{n-1}((2-a_{n-2})-ca_{n-1})\]](http://latex.artofproblemsolving.com/5/e/1/5e1b6b33933d4987f9d72e015d605c459e720d3e.png)
![\[=(2-a_0)(a_1-c(2-a_0))+(2-a_1)(a_2-c(2-a_1))+\cdots+(2-a_{n-1})(a_0-c(2-a_{n-1}))\]](http://latex.artofproblemsolving.com/1/d/3/1d329e27e89d191d760e43e2a8e78c10e2534930.png)
![\[\Leftrightarrow a_0(2-a_{n-1})-ca_0^2+a_1(2-a_0)-ca_1^2+\cdots +a_{n-1}(2-a_{n-2})-ca_{n-1}^2\]](http://latex.artofproblemsolving.com/b/3/2/b32e986aad6e6853377aade5566c6689b41379f6.png)
![\[=a_1(2-a_0)-c(2-a_0)^2+a_2(2-a_1)-c(2-a_1)^2+\cdots +a_0(2-a_{n-1})-c(2-a_{n-1})^2\]](http://latex.artofproblemsolving.com/9/5/8/9583f2b06c4473b2e6dbcca17a6e6705f7e28639.png)
![\[\Leftrightarrow -c\Sigma a_0^2 = -c\Sigma (2-a_0)^2\Leftrightarrow \Sigma a_0^2=\Sigma (2-a_0)^2\Leftrightarrow \Sigma a_0^2-(2-a_0)^2=0\]](http://latex.artofproblemsolving.com/6/3/9/6393e4fd517df00e7033e40e1f493ac591bab03b.png)
