Unfortunately, it is a UK company that does not ship to the US. But that is okay because I do not have the money right now anyway.

]]>This is not surprising because when we go along the cicumference say from P to P'

we encounter twice height of triangle and once side of a triangle. I had so much of trouble writing above in LATEX.

I've inserted spaces using space \ space. Bob Bundy.

]]>tan(36)

http://www.mathisfunforum.com/viewtopic … 87#p378587

The Regular Icosahedron

http://www.mathisfunforum.com/viewtopic.php?id=22966

Note: I was having difficulty making my answer match the one given by Monox D. I-Fly. I have worked out the two versions are the same but I cannot get the final proof to post; why I don't know. So I'll try again later

Bob

]]>I shall take a look at that page.

]]>In any triangle the three angle bisectors meet at a point (the incentre), and a circle from this point will just touch the three sides of the triangle tangentially.

In triangle ABC, construct the bisectors of angle B and angle C. Call the point where they cross, O.

Construct a perpendicular OS, from O to the side AB.

On BC mark the point Q so that BS = BQ.

Consider the triangles BOS and BOQ.

BO is a common side, angle SBO = angle QBO ( bisection) and BS = BQ by construction.

So the triangles are congruent (SAS).

So angle BQO = BSO = 90. And SO = QO.

So a circle centred on O will go through S and Q, and BQ is a tangent to this circle. (BS was constructed to be a tangent.)

Construct R on AC so that CR = CQ.

By a similar argument, triangles CQO and CRO are congruent, so RO = QO. Therefore, the circle also touches AC tangentially at R.

Consider triangles ASO and ARO.

AO is common. SO = RO. angle ARO = angle ASO = 90. So the triangles are congruent (SSR).

Therefore SAO = RAO, so AO is the bisector of the angle BAC, Therefore the bisectors are concurrent.

Bob

]]>Proof.

In triangle ABC, mark the midpoints D, E, F of the sides.

Consider triangles AFE and ABC. They have two lines and the included angle in common, so they are similar, with a length scale factor of x2.

So EF is parallel to BC. Similarly for FD//AC and ED//AB.

I have already shown that the perpendicular bisectors of the sides of ABC are concurrent (meet at a point) . See http://www.mathisfunforum.com/viewtopic.php?id=22507

But these lines are the altitudes of DEF. So the altitudes of DEF are concurrent, and O is the orthocentre of DEF.

Will this be true for any triangle?

Label the sides of the triangle D, E and F. Draw lines parallel to each side going through the vertices to produce triangle ABC.

Because of the parallel lines and the common sides AFE, DEF, EDC and FBD are all congruent, so D, E and F are the midpoints of ABC. The result follows.

Bob

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