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
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)]/3
If [PQR] denotes the area of triangle PQR, s its
semiperimeter, and r its inradius; then
r = [PQR]/s
For 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)]/8
Plugging these into (2) gives
|A'B'| = [sqrt(21) - 3]/4 * |AB|
~= 0.39593 * |AB|
QED
Edited on January 23, 2014, 4:29 am
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Posted by Bractals
on 2014-01-23 05:08:06 |
Solution
