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
For the right triangles AJC and AJD to have an equal inradius,
the triangles must be congruent. WLOG let |AC| = AD| = 1 and
x = /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 its
semiperimeter. 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
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Posted by Bractals
on 2014-01-25 21:21:52 |
Solution
