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 Congruent Incircles (Posted on 2017-01-01)
Let ABC be a 3-4-5 triangle with right angle C. Let D be a point on the hypotenuse. CD then partitions ABC into ACD and BCD.

Where is D located if ACD and BCD have congruent incircles?

 No Solution Yet Submitted by Brian Smith Rating: 5.0000 (1 votes)

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`  |CD|^2 = |AC|^2 + |AD|^2 - 2|AC||AD|cos(A)    y^2  =   4^2  +   x^2  - 2*4*x*(4/5)           =   x^2-32*x/5+16                 (1)The inradius of a triangle is equal to thearea divided by the semiperimeter.`
`     2*Area(ACD)        2*Area(BCD)  ---------------- = ----------------   Perimeter(ACD)     Perimeter(BCD)`
`   |AC||AD|sin(A)      |BC||BD|sin(B)  ----------------- = -----------------    |AC|+|CD|+|AD|      |BC|+|CD|+|BD|`
`   4*x*(3/5)     3*(5-x)*(4/5)  ----------- = ---------------      4+y+x         3+y+(5-x)`
`       20-7*x  y = --------                             (2)        2*x-5`
`Combining (1) and (2) give`
`   (20-7*x)^2  ------------ = x^2-32*x/5+16    (2*x-5)^2`
`               or`
`  x*(5*x^3-57*x^2+210*x-250) = 0`
`               or`
`  x*(x-5)*[5*x-16+-sqrt(6)] = 0`
`Therefore,`
`  x = 0, [16-sqrt(6)]/5, [16+sqrt(6)]/5, or 5.`
`We can disregard 0 and 5. Plugging [16+sqrt(6)]/5 into (2) gives y < 0;thus we can eliminate it also. Plugging[16-sqrt(6)]/5 into (2) gives y = sqrt(6);which is confirmed by Geometer's SketchPad.`
`Therefore,`
`  |AD| = x = [16-sqrt(6)]/5 and |CD| = y = sqrt(6),  where A is the vertex of triangle ABC opposite  the leg of length 3.`
`QED  `

 Posted by Bractals on 2017-01-01 11:00:32

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