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 3 INCIRCLES (Posted on 2014-01-25)
In ΔABC, AB = BC. If one chooses D on side AB and J on CD such that AJ ⊥ CD and the incircles of ΔACJ, ΔADJ, and ΔBCD all have radius r, then r = AJ/4.

Prove this using (i)Geometry (ii)Trignometry

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 3.0000 (1 votes)

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 Solution Comment 1 of 1
`For the right triangles AJC and AJD to have an equal inradius,the triangles must be congruent. WLOG let |AC| = AD| = 1 andx = /CAJ and /DAJ. Then the following lengths and angles are `
`easily found`
`   |AJ| = cos(x)   |CJ| = |DJ| = sin(x)   |BC| = 1/[2*cos(2x)]   |BD| = [1 - 2*cos(2x)]/[2*cos(2x)]   /ACJ = /ADJ = 90-x   /CBD = 180-4x`
`From the configuration we have`
`   /ACD < /ACB < 90`
`        or`
`   30 < x < 45`
`        or`
`   sqrt(2)/2 < cos(x) < sqrt(3)/2                    (1) `
`The inradius of a triangle is equal to its area divided by itssemiperimeter. For triangle AJC,`
`                |AJ||CJ|/2               cos(x)*sin(x)     r = ------------------------ = -----------------------          [|AJ| + |CJ| + |AC|]/2     |cos(x) + sin(x) + 1]`
`For triangle BCD,`
`           |DB||DC|*sin(/BCD)/2     r = ------------------------ =           [|DB| + |DC| + |BC|]/2`
`          cos(x)*sin(x)*[1 - 2cos(2x)]       = -------------------------------          1 - cos(2x) + 2sin(x)*cos(2x)`
`Setting these two equal to each other gives,`
`              [1 - 2cos(2x)]     -------------------------------      1 - cos(2x) + 2sin(x)*cos(2x)`
`                  1     = -----------------------                             (2)        |cos(x) + sin(x) + 1]`
`Using MathCad cos(x) = 4/5 is the only solution of (2)that satisfies (1). Therefore, sin(x) = 3/5 and`
`              (4/5)*(3/5)         1         r = --------------------- = --- = |AJ|/4          [(4/5) + (3/5) + 1]     5    `
`QED      `

Edited on January 25, 2014, 9:23 pm
 Posted by Bractals on 2014-01-25 21:21:52

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