Folding paper

**Geometor** · July 17th 2008, 01:30 AM

An equilateral triangle AB'C' is folded along the height AM as shown:

After folding the angle between the planes ABM and ACM is 90deg.

Could anyone help me identify the "angle between the planes ABC and BMC", an illustration would help too.

Thanks.

**earboth** · July 17th 2008, 09:50 PM

Originally Posted by Geometor

An equilateral triangle AB'C' is folded along the height AM as shown:

After folding the angle between the planes ABM and ACM is 90deg.

Could anyone help me identify the "angle between the planes ABC and BMC", an illustration would help too.

Thanks.

I've attached a perspective sketch of the folded triangle.

1. You already know:
$|\overline{BM}|= |\overline{CM}|$
$\angle(BMC) = 90^\circ$

2. Draw the line $\overline{BC}$ . It is the hypotenuse of the isosceles right triangle BMC.
3. Construct the midpoint of BC = M'

4. The length of $\overline{MM'}$ ist calculated by

$|\overline{MM'}| = \frac12 \sqrt{2} \cdot |\overline{BM}|$

5. Since $|\overline{AB'}| = 2 \cdot |\overline{B'M}|$ the line segment $\overline{AM}$ has the length

$|\overline{AM}| = \sqrt{3} \cdot |\overline{BM}|$

6. The angle $\alpha$ is the angle between the plane BMC and the plane ABC. It can be calculated by

$\tan(\alpha) = \frac{|\overline{AM}|}{|\overline{MM'}|} = \frac{\sqrt{3} \cdot |\overline{BM}|}{\frac12 \sqrt{2} \cdot |\overline{BM}|} = \sqrt{6}$

And therefore the angle $\alpha \approx 67.79^\circ$

**Geometor** · July 18th 2008, 04:41 AM

I see thank you.
Thanks for psoting a solution however I just wanted to know which angle to work with so that would have saved you some effort

Thread: Folding paper

LinkBack

Thread Tools

Display

Folding paper

Similar Math Help Forum Discussions

Paper folding

[SOLVED] Paper Folding Questions

Help with math paper

a box of paper

Subtracting on paper

Search Tags