Let ABC be an acute angled triangle with orthocenter H.D, E, and F are the feet of perpendicular from A,B, and C, respectively, on opposite sides. Also, let R be the circumradius of `DeltaABC`. Given `AH.CH = 3 and (AH)^(2) + (BH)^(2) + (CH)^(2) = 7`
Then answer the following
Value of `HD.HF` is

Options

A. `(9)/(64R^(3))`
B. `(9)/(8R^(3))`
C. `(8)/(9R^(3))`
D. `(64)/(9R^(3))`

Answer

B

Explanation

`AH = 2R cos A`
`BH = 2R cos B`
`CH = 2R cos C`
`HD = 2R cos B cos C`
`HE = 2R cos A cos C`
`HF = 2R cos A cos B`

Now `AH.BH.CH = 3` (given)
`rArr cosA . cos B . cos C = (3)/(8R^(3))` ...(i)
Now `AH^(2) + BH^(2) + CH^(2) = 7` (given)
`rArr 4R^(2) Sigma cos^(2) A = 7`
or `Sigma cos^(2) A = (7)/(4R^(2)`
Now we know
`cos^(2)A + cos^(2)B + cos^(2) C = 1-2 cos A cos B cos C`
`(7)/(4R^(2)) = 1 -2 xx (3)/(8R^(3))`
or `4R^(3) - 7R -3 = 0`
or `(R + 1) (2R +1) (2R - 3) = 0`
or `R = (3)/(2)`
Now `HD.HE.HF`
`= (2R cos B cos C) (2R cos A cos C) (2R cos A cos B)`
`=8R^(3) cos^(2) A cos^(2) B cos^(2)C`
`=8R^(3) xx (9)/(64R^(6)) = (9)/(8R^(3))`