Problem:
Triangle ABC has AB=40,AC=31, and sinA=51. This triangle is inscribed in rectangle AQRS with B on QR and C on RS. Find the maximum possible area of AQRS.
Solution:
Let α,β, and γ denote the degree measures of ∠BAC,∠BAQ, and ∠CAS, respectively, so that α+β+γ=90∘. Then AS=ACcosγ=31cosγ and AQ=ABcosβ=40cosβ. The area of rectangle AQRS is
AS⋅AQ=31⋅40cosβcosγ=1240⋅21(cos(β+γ)+cos(β−γ))=620(cos(90∘−α)+cos(β−γ))≤620(sinα+1)
The extreme value is assumed when β=γ=21(90∘−sin−151)≈39.23∘ giving the area 620(51+1)=744.
The problems on this page are the property of the MAA's American Mathematics Competitions