A square sheet of paper ABCD is folded with D falling on E along BC, A falling on F and EF intersecting AB at G. A circle is inscribed in triangle GBE with radius R .
Determine |FG| in terms of R.
Let
a = |FG|
b = |EG|
e = |BG|
f = |AG|
g = |BE|
s = the semiperimeter of triangle GBE,
t = the length of the square's side, and
x = the measure of /ADE.
s = (b + e + g)/2 (1)
t = |AB| = |BC| = |CD| = |DA| (2)
Applying the Pythagorean Theorem to triangle GBE,
b^2 = e^2 + g^2 (3)
The inradius of right triangle GBE is equal
to its semiperimeter minus its hypotenuse,
R = s - b (4)
The line segment FE is the reflection of line
segment AD about the crease of the fold (the
perpendicular bisector of line segment DE),
t = |AD| = |FE| = |FG| + |EG| = a + b (5)
ADEF is therefore an isosceles trapezoid,
/FED = /ADE = x and /AFE = /FAD = 180-x (6)
Equations (6) imply
/CED = x and /FAG = 90-x (7)
Applying the law of sines to triangle AFG,
a f
----------- = -----------
sin(/FAG) sin(/AFG)
t - b t - e
----------- = ------------
sin(90-x) sin(180-x)
t - b t - e
----------- = ------------ (8)
cos(x) sin(x)
Therefore,
t - e |CD|
------- = tan(x) = tan(/CED) = ------
t - b |CE|
t
= ------- (9)
t - g
Combining equations (1),(3)-(5), and (9) gives
|FG| = R
QED
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Posted by Bractals
on 2014-04-27 12:54:16 |
Solution
