Problem:
In β³ABC,AB=BC, and BD is an altitude. Point E is on the extension of AC such that BE=10. The values of tanβ CBE,tanβ DBE, and tanβ ABE form a geometric progression, and the values of cotβ DBE,cotβ CBE,cotβ DBC form an arithmetic progression. What is the area of β³ABC ?
Answer Choices:
A. 16
B. 350β
C. 103β
D. 85β
E. 18 Solution:
Let β DBE=Ξ± and β DBC=Ξ². Then β CBE=Ξ±βΞ² and β ABE=Ξ±+Ξ², so tan(Ξ±βΞ²)tan(Ξ±+Ξ²)=tan2Ξ±. Thus
Upon simplifying, tan2Ξ²(tan4Ξ±β1)=0, so tanΞ±=1 and Ξ±=4Οβ. Let DC=a and BD=b. Then cotβ DBC=abβ. Because β CBE=4ΟββΞ² and β ABE=4Οβ+Ξ², it follows that cotβ CBE=tanβ ABE=tan(4Οβ+Ξ²)=1βbaβ1+baββ=bβab+aβ. Thus the numbers 1,bβab+aβ, and abβ form an arithmetic progression, so abβ=bβab+3aβ. Setting b=ka yields k2β2kβ3=0, and the only positive solution is k=3. Hence b=2βBEβ=52β,a=352ββ, and the area of β³ABC is ab=350ββ.
