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Question

An altitude BD and angular bisector BE are drawn in ABC from the vertex B. It is known that the length of side AC=1 and the magnitude of angle BEC,ABD, ABE,BAC form an arithmetic progression. Let B be the image of point B with respect to side AC of ABC. If the length BB is equal to ab ; where a,bN, then the least possible value of a+b is

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Solution

Let BEC=α3β
ABD=αβ
ABE=α+β
BAC=α+3β


α+3β+α+β=α3β
α=7β
α+3β+αβ=π2
2α+2β=π2
14β+2β=π2
β=π24, α=7π24
B=2(α+β)=π2


In ABD,
tanπ6=BDλ ...(1)
In BDC,
tanπ3=BD1λ ...(2)
On dividing equation (1) by (2), we get
133=1λλ
λ=33λ
λ=34
BD=34
BB=32=34
Thus, a+b=3+4=7

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