Problems

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.

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

In the triangle $ABC$ the lines $AE$ and $CD$ are the bisectors of the angles $\mathrm{\angle }BAC$ and $\mathrm{\angle }BCA$, intersecting at the point $I$. In the triangle $BDE$ the lines $DG$ and $EF$ are the bisectors of the angles $\mathrm{\angle }BDE$ and $\mathrm{\angle }BED$, 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 $\mathrm{\angle }ADF=\mathrm{\angle }BDE$, $\mathrm{\angle }AFD=\mathrm{\angle }CFE$, $\mathrm{\angle }CEF=\mathrm{\angle }BED$. 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?