# Pentagon in a Semicircle

This 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.

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Copyright © 1996-2018 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.

What if applet does not run? |

### 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

∠PIS | = ½∠XOZ = ½(arc XZ) = ½(arc XY) + ½(arc YZ) |

= ∠XAY + ∠ZBY | |

= ∠PAY + ∠SBY. |

Because ∠PIS = ∠PIY + ∠SIY, it suffices to prove that

that is, to show that quadrilaterals APYI and BSYI are cyclic, which is evident, because

### Solution 2

Let AZ intercept BX at C, PQ and RS intercept at I. The acute angle formed by lines PQ and RS is

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

- JACEK FABRYKOWSKI, STEVEN R. DUNBAR,
__39th USA Mathematical Olympiad - 1st USA Junior Mathematical Olympiad__,*Math. Mag.*83 (2010) 313-319

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Copyright © 1996-2018 Alexander Bogomolny63197838 |