Pentagon in a SemicircleThis is Problem 1 from the 2010 USAMO 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.
 |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander BogomolnyLet 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 1(A write-up in the Mathematics Magazine.) Let I be the foot of the perpendicular from Y to line AB. We note the P, Q, I 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, I are collinear, by Simsonfs theorem. Likewise, points S, R, I are collinear. We need to show that ∠XOZ = 2∠PIS or
Because ∠PIS = ∠PIY + ∠SIY, it suffices to prove that
∠PIY = ∠PAY and ∠SIY = ∠SBY;
that is, to show that quadrilaterals APYI and BSYI are cyclic, which is evident, because Solution 2Let AZ intercept BX at C, PQ and RS intercept at I. The acute angle formed by lines PQ and RS is
∠PIS = ∠PQY + ∠SRY - ∠QYR = ∠PQY + ∠SRY - ∠RCB
because angles QYR and RCB have pairwise perpendicular sides. But ∠RCB subtends arcs AX and BZ; ∠PQY = ∠PXY subtends arc AY; ∠SRY = ∠SZY subtends arc BY. Therefore, ∠PIS subtends the arc(AY) + arc(BY) - arc(AX) - arc(BZ) = arc(XZ) = ½∠XOZ. References|Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
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