Problem:
Quadrilateral XABY is inscribed in the semicircle ω with diameter XY. Segments AY and BX meet at P. Point Z is the foot of the perpendicular from P to line XY. Point C lies on ω such that line XC is perpendicular to line AZ. Let Q be the intersection of segments AY and XC. Prove that
XPBY+XQCY=AXAY
Solution:
First Solution: Note that ∠XAY=∠XBY=∠XCY=∠PZX=∠PZY=90∘. In right triangles BXY,AXY,AXP, we have
BY=XYcos∠BYX,AX=XYcos∠AXY,XP=cos∠AXPAX=cos∠AXPXYcos∠AXY
from which it follows that
XPBY=cos∠AXYcos∠BYXcos∠AXP
Likewise, we have
XQCY=cos∠AXYcos∠CYXcos∠AXQ
Adding the last two equations yields
XPBY+XQCY=cos∠AXYcos∠BYXcos∠AXP+cos∠CYXcos∠AXQ(3)
Because both CY and AZ are perpendicular to XC,∠CYX=∠AZX. Because ∠XAP= ∠XZP=90∘, quadrilateral AXZP is cyclic, from which it follows that ∠AZX=∠APX. Therefore, we have ∠CYX=∠AZX=∠APX=90∘−∠AXP or ∠CYX+∠AXP=\
90∘. Likewise, we can show that ∠BYX+∠AXQ=90∘. Consequently, we conclude that cos∠BYX=sin∠AXQ and sin∠CYX=cos∠AXP. Thus, by the addition and substraction formula, (4) becomes
XPBY+XQCY=cos∠AXYsin∠AXQsin∠CYX+cos∠CYXcos∠AXQ=cos∠AXYcos(∠CYX−∠AXQ)
Because ACYX is cyclic, ∠AXQ=∠AXC=∠CYA, implying that ∠CYX−∠AXQ= ∠CYX−∠CYA=∠AYX. Therefore,
XPBY+XQCY=cos∠AXYcos(∠CYX−∠AXQ)=cos∠AXYcos∠AYX=cos∠AXYsin∠AXY=tan∠AXY=AXAY
as desired.
This problem and solution were suggested by Zuming Feng.
Second Solution: Note that ∠XAY=∠XBY=∠XCY=∠PZX=∠PZY=90∘. In right triangles BXY,AXY,AXP, we have
B Y=X Y \cos (\angle B Y X), \quad A X=X Y \cos (\angle A X Y), \quad X P=\frac{A X}{\cos (\angle A X P)}=\frac{X Y \cos (\angle A X Y)}
from which it follows that
XPBY=cos(∠AXY)cos(∠BYX)cos(∠AXP)
Likewise, we have
XQCY=cos(∠AXY)cos(∠CYX)cos(∠AXQ)
Adding the last two equations yields
XPBY+XQCY=cos(∠AXY)cos(∠BYX)cos(∠AXP)+cos(∠CYX)cos(∠AXQ)(4)
Because both CY and AZ are perpendicular to XC,∠CYX=∠AZX. Because ∠XAP= ∠XZP=90∘, quadrilateral AXZP is cyclic, from which it follows that ∠AZX=∠APX. Therefore, we have ∠CYX=∠AZX=∠APX=90∘−∠AXP or ∠CYX+∠AXP=90∘. Likewise, we can show that ∠BYX+∠AXQ=90∘. Consequently, we conclude that\
cos(∠BYX)=sin(∠AXQ) and sin(∠CYX)=cos(∠AXP). Thus, by the addition and substraction formula, (4) becomes
XPBY+XQCY=cos(∠AXY)sin(∠AXQ)sin(∠CYX)+cos(∠CYX)cos(∠AXQ)=cos(∠AXY)cos(∠CYX−∠AXQ)
Because ACYX is cyclic, ∠AXQ=∠AXC=∠CYA, implying that ∠CYX−∠AXQ= ∠CYX−∠CYA=∠AYX. Therefore,
XPBY+XQCY=cos∠AXYcos(∠CYX−∠AXQ)=cos∠AXYcos∠AYX=cos∠AXYsin∠AXY=tan∠AXY=AXAY
as desired.
Rays YB and YC meet ray XA at B1 and C1 respectively. Because ∠PAB1=∠PBB1= 90∘,APBB1 is cyclic, in particular, ∠XB1Y=∠AB1B=∠APX. Because ∠PAX= ∠PZX=90∘,APZX is cyclic, in particular, ∠APX=∠AZX. Note that both AC and CY are perpendicular to XC,AZ∥CY and so ∠AZX=∠CYX=∠C1YX. Therefore, we have ∠XB1Y=∠APX=∠AZX=∠C1YX. It follows that triangles XYB1 and XC1Y are similar to each other, with XB and XC being corresponding altitudes. Hence
XPBY=XQCC1 and XPBY+XQCY=XQCC1+XQCY=XQC1Y
It remains to show that
XQC1Y=AXAY
which is true because triangles AYC1 and AXQ are similar to each other ( ∠C1AY= ∠QAX=90∘ and ∠AYC1=∠AYC=∠AXC=∠AXQ.)
This solution was suggested by Zuming Feng.
The problems on this page are the property of the MAA's American Mathematics Competitions
