Problem:
Let AB be a diameter of circle Ο. Extend AB through A to C. Point T lies on Ο so that line CT is tangent to Ο. Point P is the foot of the perpendicular from A to line CT. Suppose AB=18, and let m denote the maximum possible length of segment BP. Find m2.
Solution:
Let O be the center of the circle, and let Q be the foot of the perpendicular from B to line CT. Segment BQ meets the circle again (other than at B ) at D. Since AB is a diameter, β ADB=90β and ADQP is a rectangle. Then
BP2β=PQ2+BQ2=AD2+BQ2=AB2βBD2+BQ2=AB2+(BQβBD)(BQ+BD)=AB2+DQ(BQ+BD)β
Note that ABQP is a trapezoid with OT as its midline. Hence AB= 2OT=AP+BQ=DQ+BQ. Consequently
BQ+BDβ=BQ+BQβDQ=2BQβDQ=2(BQ+DQ)β3DQ=2ABβ3DQβ
Combining the above shows that
BP2=AB2+DQ(2ABβ3DQ)=AB2+124β
(3DQ)(2ABβ3DQ)β.
For real numbers x and y,4xyβ€(x+y)2. Hence
BP2β€AB2+12(3DQ+2ABβ3DQ)2β=34AB2β=31296β, which equals 432β.
This maximum may be obtained by setting 3DQ=2ABβ3DQ, or 3PA= 3DQ=AB=2OT. It follows that COACβ=OTPAβ=32β, implying that CA=AB.
OR
Let β ABQ=Ξ±. Let R be on BQβ such that ORβ₯BQβ. Let r be the radius of the circle. To maximize BP2, observe that BP2=PQ2+BQ2, and from β³BRO it follows that OR=rsinΞ± and BR=rcosΞ±. Thus PQ=2rsinΞ±, and BQ=r+rcosΞ±. Then
PB2β=(2rsinΞ±)2+(r+rcosΞ±)2=r2+r2(4sin2Ξ±+1+2cosΞ±+cos2Ξ±)=r2(5β3cos2Ξ±+2cosΞ±)=β3r2(β35β+cos2Ξ±β32βcosΞ±)=β3r2((cosΞ±β31β)2β916β).β
The maximum for PB2 is thus attained when cosΞ±=31β and has the value 92(16/3)=432β.
Query: Segment BP reaches its maximum length when lines AD,BP, and OT are concurrent. Why?
The problems on this page are the property of the MAA's American Mathematics Competitions
