Problems

We call a bisector the segment from the vertex of a triangle to the opposite side which divides in half the angle next to the starting vertex. Prove that in two congruent triangles, the corresponding bisectors are of equal length.

\(7\) identical hexagons are arranged in a pattern on the picture below. If each hexagon has an area of \(8\), what is the area of the triangle \(\triangle ABC\)?

In the triangle \(ABC\) the bisector \(BD\) coincides with the height. Prove that \(AB=BC\).

In the triangle \(ABC\) the median \(BD\) coincides with the height. Prove that \(AB=BC\).

In the triangle \(ABC\) with \(BC=12\), the median \(AE\) is perpendicular to the bisector \(BD\). Find the length of \(AB\).

On the sides of the equilateral triangle \(ABC\) three points \(D,E,F\) are chosen in such a way that the following ratios of lengths hold: \[AD:DC = CF:FB = BE:AE\] Prove that the triangle \(DEF\) is also equilateral.

Two circles with centres \(A\) and \(C\) intersect at the points \(B\) and \(D\). Prove that the segment \(AC\) is perpendicular to \(BD\). Moreover, prove that the segment \(AC\) divides \(BD\) in half.

Can you cover a \(10 \times 10\) board using only \(T\)-shaped tetrominos?

A broken calculator can only do several operations: multiply by \(2\), divide by \(2\), multiply by \(3\), divide by \(3\), multiply by \(5\), and divide by \(5\). Using this calculator any number of times, could you start with the number \(12\) and end up with \(49\)?

The numbers \(1\) through \(12\) are written on a board. You can erase any two of these numbers (call them \(a\) and \(b\)) and replace them with the number \(a+b-1\). Notice that in doing so, you remove one number from the total, so after \(11\) such operations, there will be just one number left. What could this number be?