Found problems: 7
Given a convex pentagon $ABCDE$ with $AB=BC, CD=DE, \angle ABC=150^o, \angle CDE=30^o, BD=2$. Find the area of $ABCDE$.
(I.Voronovich)
Given a triangle $ABC$, three equilateral triangles $AEB, BFC$, and $CGA$ are constructed in the exterior of $ABC$. Prove that:
$(a) CE = AF = BG$;
$(b) CE, AF$, and $BG$ have a common point.
I could not find a separate topic for this question and I need one. http://en.wikipedia.org/wiki/Fermat_point of course.
Let $A_1, A_2, \ldots, A_n (n \geq 4)$ be $n$ convex sets in plane. Knowing that every three convex sets have a common point. Prove that there exists a point belonging to all the sets.
Let $ A$, $ B$, $ C$, $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$, $ X$, $ Y$, $ Z$, $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$ and $ P$ be pairwise distinct points in space such that
$ A^{\prime} \in BC;\ B^{\prime}\in CA;\ C^{\prime}\in AB;\ X^{\prime}\in YZ;\ Y^{\prime}\in ZX;\ Z^{\prime}\in XY;$
$ P \in AX;\ P\in BY;\ P\in CZ;\ P\in A^{\prime}X^{\prime};\ P\in B^{\prime}Y^{\prime};\ P\in C^{\prime}Z^{\prime}$.
Prove that
$ \frac {BA^{\prime}}{A^{\prime}C}\cdot\frac {CB^{\prime}}{B^{\prime}A}\cdot\frac {AC^{\prime}}{C^{\prime}B} \equal{} \frac {YX^{\prime}}{X^{\prime}Z}\cdot\frac {ZY^{\prime}}{Y^{\prime}X}\cdot\frac {XZ^{\prime}}{Z^{\prime}Y}$.
If $ABCDEF$ is a convex cyclic hexagon, then its diagonals $AD$, $BE$, $CF$ are concurrent if and only if $\frac{AB}{BC}\cdot \frac{CD}{DE}\cdot \frac{EF}{FA}=1$.
[i]Alternative version.[/i] Let $ABCDEF$ be a hexagon inscribed in a circle. Then, the lines $AD$, $BE$, $CF$ are concurrent if and only if $AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$.
Let $ ABC$ be a triangle, and let $ X$, $ Y$, $ Z$ be three points on the segments $ BC$, $ CA$, $ AB$, respectively. Denote by $ X^{\prime}$, $ Y^{\prime}$, $ Z^{\prime}$ the reflections of these points $ X$, $ Y$, $ Z$ in the midpoints of the segments $ BC$, $ CA$, $ AB$, respectively. Prove that $ \left\vert XYZ\right\vert \equal{}\left\vert X^{\prime}Y^{\prime}Z^{\prime}\right\vert$.
$BC$ is a segment, $M$ is point on $BC$, $A$ is a point such that $A, B, C$ are non-collinear.
(i) Prove that if $M$ is the midpoint of $BC$, then $AB^2 + AC^2 = 2(AM^2 + BM^2).$
(ii) If there exists another point (except $M$) on segment $BC$ satisfying (i), find the region of point $A$ might occupy.