1. ## Pentagon problem

How can you show that a regular pentagon can be divided into 2010 triangles so that one is type A and 2009 are type B?

The middle triangle is type A.
The sides triangles are type B.
Thank you!

2. Shouldn't that be "divided in 2011 (or 2009) triangles.....2010 (or 2008) are type B" ?

And what do you mean by "type B": similar triangles?

3. 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 $\angle D.$

$\Delta DFC$ is a Type A triangle.
$\Delta AFD$ 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 $DFC$
. . and we have: 1 Type A, 4 Type B.
and so on . . .

Therefore, starting with Type A triangle $ACD$,
. . perform the angle-bisection 2007 times

and we will have 1 Type A and 2009 Type B triangles.

4. Thank you for your help

5. Originally Posted by GuMing

How can you show that a regular pentagon can be divided into 2010 triangles so that one is type A and 2009 are type B?

The middle triangle is type A.
The sides triangles are type B.
Thank you!
The fact that (the years) "2010" and "2009" appear in the problem hint to us that this question may be from a maths competition.

This thread will remain closed as we conduct further investigation (we have zero tolerance for cheaters!!). If everything ends up being OK, it will be reopened.