be the orthogonal projection of 
on 
and 
are the midpoints of 
and 
respectively. Let 
and 
are points on the minor arcs 
and 
of the circumcircle of triangle 
.
and 
are tangent if and only if 
.
be 
given points in the plane, and let 
be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points 
a non-negative real number 
,
,
.
such that 
and 
are connected by an edge for every 
.
is not fixed and the first selected vertex always has to be 
.![\[
\frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
\]](http://latex.artofproblemsolving.com/1/7/9/1795db8e4a509dd465f6ff462093ae75b04de2b4.png)
otherwise, Alice wins. Depending on 
,
for which Bob has a winning strategy.
such that![\[
f(x^2 \cdot f(x) + f(y)) = f(f(x^3)) + y
\]](http://latex.artofproblemsolving.com/8/6/5/865b10bbc5ba7721078b1c60eaeede2abfeeb1a2.png)
.
.
When does equality hold?
Y by Adventure10, Mango247
On the sides of the triangle 
, three similar isosceles triangles 
, 
, and 
are constructed. The first two are constructed externally to the triangle , but the third is placed in the same half-plane determined by the line 
as the triangle . Prove that 
is a parallelogram.
, three similar isosceles triangles 
, 
, and 
are constructed. The first two are constructed externally to the triangle , but the third is placed in the same half-plane determined by the line 
as the triangle . Prove that 
is a parallelogram.
that sends 
must send 
since 
is similar to 
.
is similar to 
.
is similar to triangle to 
,
and 
.
.
we get
and the result follows.
![\[\begin{gathered}
\left. \begin{gathered}
\frac{{PB}}{{AB}} = \frac{{RB}}{{CB}} \hfill \\
P\hat BR = A\hat BC \hfill \\
\end{gathered} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} tr.PBR \approx tr.ACB\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} {{\hat P}_2} = {{\hat A}_1} \hfill \\
\bullet \mathop {}\limits^{\mathop {}\limits^{} } similarly\mathop {}\limits^{} \mathop {}\limits^{} {{\hat Q}_2} = {{\hat A}_1}\mathop {}\limits^{} \mathop {}\limits^{} So\mathop {}\limits^{} :\mathop {}\limits^{} {{\hat P}_1} = {{\hat Q}_1}\mathop {}\limits_{\mathop {}\nolimits_{} } \hfill \\
\left. \begin{gathered}
{{\hat P}_1} + {{\hat P}_2} = {{\hat A}_1} + 2\varphi \mathop {}\limits_{} \hfill \\
{A_1} + 2\varphi + {{\hat R}_1} + {{\hat R}_2} + \theta = {360^o} \hfill \\
\end{gathered} \right\}\mathop {}\limits_{} \Rightarrow \mathop {}\limits^{} \theta = {{\hat A}_1} + 2\varphi \mathop {}\limits_{\mathop {}\limits_{\mathop {}\limits_{\mathop {}\limits_{} } } } \hfill \\
\left. \begin{gathered}
{{\hat P}_1} = {{\hat Q}_1} \hfill \\
\theta = {{\hat A}_1} + 2\varphi \hfill \\
\end{gathered} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} APRQ\mathop {}\limits^{} \# \hfill \\
\end{gathered} \]](http://latex.artofproblemsolving.com/7/9/5/795563537324b38d6f14a59a89d96d73588d96d9.png)
lie on the unit circle without loss of generality. Then,
and similarly for 
for some fixed real constant 
by the similarity condition. So, stripping away terms, it suffices to show that
after canceling the 
