Problem:
Let AXYZB be a convex pentagon inscribed in a semicircle of diameter AB. Denote by P,Q,R,S the feet of the perpendiculars from Y onto lines AX,BX,AZ,BZ, respectively. Prove that the acute angle formed by lines PQ and RS is half the size of ∠XOZ, where O is the midpoint of segment AB.
Solution:
Solution by Titu Andreescu: Let T be the foot of the perpendicular from Y to line AB. We note the P,Q,T are the feet of the perpendiculars from Y to the sides of triangle ABX. Because Y lies on the circumcircle of triangle ABX, points P,Q,T are collinear, by Simson's theorem. Likewise, points S,R,T are collinear.
Because ∠PTS=∠PTY+∠STY, it suffices to prove that
∠PTY=∠PAY and ∠STY=∠SBY
that is, to show that quadrilaterals APYT and BSYT are cyclic, which is evident, because ∠APY=∠ATY=90∘ and ∠BTY=∠BSY=90∘.
Alternate Solution from Lenny Ng and Richard Stong: Since YQ,YR are perpendicular to BX,AZ respectively, ∠RYQ is equal to the acute angle between lines BX and AZ, which is 21​(AX+BZ)=21​(180∘−XZ) since X,Z lie on the circle with diameter AB. Also, ∠AXB=∠AZB=90∘ and so PXQY and SZRY are rectangles, whence ∠PQY=90∘−∠YXB=90∘−YB/2 and ∠YRS=90∘−∠AZY=90∘−AY/2. Finally, the angle between PQ and RS is
