Problem:
Triangle ABC has AB=40,AC=31, and sinA=51β. This triangle is inscribed in rectangle AQRS with B on QRβ and C on RS. Find the maximum possible area of AQRS.
Solution:
Let Ξ±,Ξ², and Ξ³ denote the degree measures of β BAC,β BAQ, and β CAS, respectively, so that Ξ±+Ξ²+Ξ³=90β. Then AS=ACcosΞ³=31cosΞ³ and AQ=ABcosΞ²=40cosΞ². The area of rectangle AQRS is
ASβ
AQβ=31β
40cosΞ²cosΞ³=1240β
21β(cos(Ξ²+Ξ³)+cos(Ξ²βΞ³))=620(cos(90ββΞ±)+cos(Ξ²βΞ³))β€620(sinΞ±+1)β
The extreme value is assumed when Ξ²=Ξ³=21β(90ββsinβ151β)β39.23β giving the area 620(51β+1)=744β.
The problems and solutions on this page are the property of the MAA's American Mathematics Competitions