Hello, GuMing!
Code:
A
o
* : *
* *:* *
* ::: *
E o B *:::* B o B
* : A : *
* *:::::* *
* ::::::: *
o * * * o
D C
Show that a regular pentagon can be divided into 2010 triangles:
. . where 1 is type A and 2009 are type B.
Type A triangle has angles: .72°, 72°, 36°.
Type B triangle has angles: .36°, 36°, 108°.
In the above diagram, the pentagon is divided into: . 1 Type A, 2 Type B.
Consider the Type A triangle:
Code:
A
*
/ \
/36°\
/ \
/ 2 \ F
/ 108° *
/ *72°\
/ * \
/36°* 1 \
/ * 36° 72° \
D o - - - - - - - - - o C
Bisect $\displaystyle \angle D.$
$\displaystyle \Delta DFC$ is a Type A triangle.
$\displaystyle \Delta AFD$ is a Type B triangle.
That is, a Type A triangle can be divided into: 1 Type A and 1 Type B.
We have divided the pentagon into: 1 Type A, 3 Type B.
Repeat the process with Type A triangle $\displaystyle DFC$
. . and we have: 1 Type A, 4 Type B.
and so on . . .
Therefore, starting with Type A triangle $\displaystyle ACD$,
. . perform the angle-bisection 2007 times
and we will have 1 Type A and 2009 Type B triangles.