Problem:
In △ABC,AB=BC, and BD is an altitude. Point E is on the extension of AC such that BE=10. The values of tan∠CBE,tan∠DBE, and tan∠ABE form a geometric progression, and the values of cot∠DBE,cot∠CBE,cot∠DBC form an arithmetic progression. What is the area of △ABC ?
Answer Choices:
A. 16
B. 350
C. 103
D. 85
E. 18 Solution:
Let ∠DBE=α and ∠DBC=β. Then ∠CBE=α−β and ∠ABE=α+β, so tan(α−β)tan(α+β)=tan2α. Thus
1+tanαtanβtanα−tanβ⋅1−tanαtanβtanα+tanβ=tan2α
from which it follows that
tan2α−tan2β=tan2α(1−tan2αtan2β)
Upon simplifying, tan2β(tan4α−1)=0, so tanα=1 and α=4π. Let DC=a and BD=b. Then cot∠DBC=ab. Because ∠CBE=4π−β and ∠ABE=4π+β, it follows that cot∠CBE=tan∠ABE=tan(4π+β)=1−ba1+ba=b−ab+a. Thus the numbers 1,b−ab+a, and ab form an arithmetic progression, so ab=b−ab+3a. Setting b=ka yields k2−2k−3=0, and the only positive solution is k=3. Hence b=2BE=52,a=352, and the area of △ABC is ab=350.
