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Estonian Math Competitions

Geometry Difficulty 5.7 AIME, harder Prove it Estonia

The bisector of the internal angle at the vertex $B$ of a triangle $ABC$ intersects the circumcircle of the triangle $ABC$ at a point $P$ ( $P \neq B$ ). The line through $P$ perpendicular to the line $AC$ intersects the circumcircle of the triangle $ABC$ at a point $P'$ ( $P' \neq P$ ). Prove that the quadrilateral $APCP'$ is a square if and only if $\angle ABC = 90^\circ$ .

Solutions — 2

Solution 1

As $\angle ABC = \angle AP'C$ (see figure), it suffices to show that the quadrilateral $APCP'$ is a square if and only if $\angle AP'C = 90^\circ$ .

Since $\angle ABP = \angle CBP$ , we have $AP = CP$ . Thus $PP'$ is the perpendicular bisector of the side $AC$ . Hence the line $PP'$ passes through the circumcentre of the triangle $ABC$ , i.e., the chord $PP'$ is a diameter. By Thales' theorem, $\angle PAP' = \angle PCP' = 90^\circ$ .

If $\angle AP'C = 90^\circ$ then also $\angle APC = 180^\circ - 90^\circ = 90^\circ$ . Thus the quadrilateral $APCP'$ is a rectangle and, by equality of adjacent sides $AP$ and $CP$ , a square. On the other hand, if $APCP'$ is a square then obviously $\angle AP'C = 90^\circ$ . This completes the proof.

Solution 2

Since $\angle ABC = \angle AP'C$ , it suffices to show that the quadrilateral $APCP'$ is a square if and only if $\angle AP'C = 90^\circ$ .

As $\angle ABP = \angle CBP$ , we have $AP = CP$ . Thus $PP'$ bisects the line segment $AC$ .

If $\angle AP'C = 90^\circ$ then, by Thales' theorem, $AC$ is the diameter of the circumcircle of the triangle $ABC$ . Thus the chord $PP'$ bisecting $AC$ passes through the circumcentre of the triangle $ABC$ , implying that $AC$ bisects the chord $PP'$ . As $PP' \perp AC$ , the diagonals of the quadrilateral $APCP'$ are perpendicular and bisect each other. Hence the quadrilateral $APCP'$ is a rhombus and, because of $\angle AP'C = 90^\circ$ , a square. On the other hand, if $APCP'$ is a square then obviously $\angle AP'C = 90^\circ$ . This completes the proof.

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