Problems

For two congruent triangles Prove that their corresponding heights are equal.

The sides \(AB\) and \(CD\) of the quadrilateral \(ABCD\) are equal, the points \(E\) and \(F\) are the midpoints of \(AB\) and \(CD\) correspondingly. Prove that the perpendicular bisectors of the segments \(BC\), \(AD\), and \(EF\) intersect at one point.

In the triangle \(ABC\) the heights \(AD\) and \(CE\) intersect at the point \(F\). It is known that \(CF=AF\). Prove that the triangle \(ABC\) is isosceles.

In the triangle \(ABC\) the angle \(\angle ABC = 120^{\circ}\). The segments \(AF,\, BE\), and \(CD\) are the bisectors of the corresponding angles of the triangle \(ABC\). Prove that the angle \(\angle DEF = 90^{\circ}\).

In the triangle \(ABC\) the lines \(AE\) and \(CD\) are the bisectors of the angles \(\angle BAC\) and \(\angle BCA\), intersecting at the point \(I\). In the triangle \(BDE\) the lines \(DG\) and \(EF\) are the bisectors of the angles \(\angle BDE\) and \(\angle BED\), intersecting at the point \(H\). Prove that the points \(B,\,H,\, I\) are situated on one straight line.

In the triangle \(ABC\) the points \(D,E,F\) are chosen on the sides \(AB, BC, AC\) in such a way that \(\angle ADF = \angle BDE\), \(\angle AFD = \angle CFE\), \(\angle CEF = \angle BED\). Prove that the segments \(AE, BF, CD\) are the heights of the triangle \(ABC\).

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?

There are real numbers written on each field of a \(m \times n\) chessboard. Some of them are negative, some are positive. In one move we can multiply all the numbers in one column or row by \(-1\). Is that always possible to obtain a chessboard where sums of numbers in each row and column are non-negative?