0xDE ([info]11011110) wrote,
@ 2008-05-16 13:05:00
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Entry tags:carnival, geometry

Carnival triangles
Here's a cute little geometric factoid that has something to do with one of the posts over at the 33rd Carnival of Mathematics. I'll leave it as a puzzle which post it belongs to...

Let ABC be any triangle in the Euclidean plane, and AD be any line. Form points A', B', and C' as the perpendicular projections of A onto BC, B onto AD, and C onto AD respectively. Then triangles ABC and A'B'C' are similar.



(Hint: in the post I have in mind, ABC is isosceles and A' is the midpoint of BC.)

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[info]samjshah.com
2008-05-18 12:55 am UTC (link)
Way elegant.
-sam shah.

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[info]11011110
2008-05-18 06:34 am UTC (link)
Thanks!

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(Anonymous)
2008-05-20 09:22 pm UTC (link)
Maybe I'm missing something here, its been a while since I took geometry but....

If D = A', then AD = AA'

Doesn't that mean A' B' & C' all fall on the line BC? How can that even constitute a triangle? Much less one that is similar to ABC.

Am I reading this incorrectly?

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[info]11011110
2008-05-20 09:25 pm UTC (link)
If D = A', then you get a degenerate case in which all three of A', B', and C' land on a single point.

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Reasons?
(Anonymous)
2008-05-21 08:38 pm UTC (link)
This is a neat fact, and it can be proven using a simple four line elementary proof. Very nice.

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Re: Reasons?
[info]11011110
2008-05-21 09:20 pm UTC (link)
Thanks. Here's the proof I had in mind (simplified a little now that I write it down): because of the right angles, A' and C' lie on the circle with diameter AC, so by Euclid III.21 angles ACB = ACA' and A'C'B' = A'C'A are equal. A symmetric argument shows that angle ABC = A'B'C'. With two equal angles, the triangles are similar.

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Re: Reasons?
(Anonymous)
2008-05-22 03:05 am UTC (link)
Exactly, clean and straightforward. However it left me thinking that it is not very insightful in terms of what is going on. Is there a way to interpret this as an inversion or some other natural geometric operation (either based on the original algebraic motivation or otherwise)?

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Re: Reasons?
(Anonymous)
2008-05-22 03:03 pm UTC (link)
Ok, I know it's bad form to answer your own question but here's a constructive, more intuive proof.

Suppose we are tasked with building a triangle similar to ABC with B'C' corresponding to the scaled BC.

The scaling factor is |cos(Θ)| where Θ is the angle formed by the lines BC and AD. One way to complete the construction of the similar triangle is to scaled the height AA' and place it appropriately on B'C' as follows:

Let P' be the projection of A' on AD. Since the angle between AA' and BC, and the angle between A'P' and AD are both 90, then the angle between AA' and AP' = Θ = angle between BC and AD. This means that the line A'P' is a scaled down projection of AA' by a factor of |cos(Θ)|.

Now where should we place the scaled height A'P' on AD? In the projected image of the pedal point A' on AD of course!

The triangles are then similar by construction.

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Re: Reasons?
(Anonymous)
2009-03-04 05:44 pm UTC (link)
A' and C' are on a circle of *radius* *AC'*, not *diameter* *AC*. (Otherwise, C' would not be the projection of C except in a few specific cases.)

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Re: Reasons?
[info]11011110
2009-03-04 07:27 pm UTC (link)
No, I meant diameter. Whenever XYZ is a right triangle (with the right angle at Y), Y lies on a circle with diameter XZ. I don't see any particular reason for AA' and AC' to be equally long, and in fact in the drawing AA' is about 6% longer than AC'.

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Carnival triangles
[info]brantac
2008-08-11 04:00 pm UTC (link)
If you draw AD' at right angles to AD, and form B" abd C" in a similar way to B' and C', you can draw a third triangle A'B"C" such that area A'B'C' + area A'B"C" = area ABC, by Pythagoras.

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