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 Incircles in equilateral triangles (Posted on 2014-01-18)
In an equilateral triangle ABC the lines AC'A', BA'B', CB'C' are drawn making equal angles with AB, BC, CA, respectively, forming the triangle A'B'C', and so that the radius of the incircle of triangle A'B'C' is equal to the radius of the incircle of triangle AA'B. Find A'B' in terms of AB

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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`Let x = /BAA' and apply the law of sines to triangle AA'B:         |AB|             |BA'|            |AA'|     ------------  =  ------------  =  ------------       sin(/AA'B)       sin(/BAA')       sin(/ABA')                           or        |AB|         |BA'|          |AA'|     ----------  =  --------  =  -----------      sin(120)       sin(x)       sin(60-x)                           or     |BA'|  =  2|AB|[sin(x)*sqrt(3)]/3  and     |AA'|  =  |AB|[3cos(x) - sin(x)*sqrt(3)]/3If [PQR] denotes the area of triangle PQR, s its semiperimeter, and r its inradius; then     r = [PQR]/sFor triangle A'B'C':          [|A'B'||A'C'| sin(/B'A'C')]/2     r = -------------------------------            [|A'B'|+|B'C'|+|C'A'|]/2       = |A'B'|*sqrt(3)/6       = (|AA'| - |BA'|)*sqrt(3)/6       = |AB|*[cos(x)*sqrt(3) - 3sin(x)]/6        (1)Note:     |A'B'| = [cos(x) - sin(x)*sqrt(3)]*|AB|        (2)            All we need is x.For triangle AA'B:          [|AB||AA'| sin(/BAA')]/2     r = --------------------------            [|AB|+|BA'|+|AA'|]/2          |AB|*sin(x)[3cos(x)- sin(x)*sqrt(3)]       = --------------------------------------   (3)              3 + 3cos(x) + sin(x)*sqrt(3)Combining (1) and (2) we get     sin(x)*[3 + 8cos(x)] = [1 + cos(x)]*sqrt(3)Using MathCad we get        cos(x) = [sqrt(3) + sqrt(7)]*sqrt(3)/8  and     sin(x) = [3*sqrt(3) - sqrt(7)]/8Plugging these into (2) gives     |A'B'| = [sqrt(21) - 3]/4 * |AB|           ~= 0.39593 * |AB|QED`

Edited on January 23, 2014, 4:29 am

 Posted by Bractals on 2014-01-23 05:08:06

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