Library / All-Soviet-Union Mathematical Olympiad /124 of 185

All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Soviet Union

Problem:

$ABCD$ is a convex quadrilateral. $A'$ is the foot of the perpendicular from $A$ to the diagonal $BD$ , $B'$ is the foot of the perpendicular from $B$ to the diagonal $AC$ , and so on. Prove that $A'B'C'D'$ is similar to $ABCD$ .

Solution

Solution:

Let the diagonals meet at $O$ . Then $CC'O$ is similar to $AA'O$ (because $CC'O = AA'O = 90^{\circ}$ , and $\angle COC'$ , $\angle AOA'$ are opposite angles), so $A'O / C'O = AO / CO$ . Similarly, $B'O / D'O = BO / DO$ . $AA'O$ is also similar to $BB'O$ , so $A'O / B'O = AO / BO$ . Thus $OA':OB':OC':OD' = OA:OB:OC:OD$ . Hence triangles $OA'B'$ and $OAB$ are similar. Likewise $OB'C'$ and $OBC$ , $OC'D'$ and $OCD$ , and $OD'A'$ and $ODA$ . Hence result.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.