# Pentagon problem

• May 5th 2010, 11:20 PM
GuMing
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!
• May 6th 2010, 06:01 AM
Wilmer
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?
• May 6th 2010, 10:27 AM
Soroban
Hello, GuMing!

Quote:

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.

• May 7th 2010, 04:19 AM
GuMing
• May 10th 2010, 08:49 AM
Chris L T521
Quote:

Originally Posted by GuMing