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        riddles >> medium >> Reconstruct Triangle from Points (Geometry)
        
(Message started by: Barukh on Sep 24^th, 2007, 12:46am)

Title: Reconstruct Triangle from Points (Geometry)
Post by Barukh on Sep 24^th, 2007, 12:46am

A well-known class of geometric construction problems requires construction of a triangle from three given parts (sides, angles, cevians, etc.). I believe several problems from this class were presented at this forum.

The less known but not less fascinating is construction of a triangle from three given points. Here’s one:

Given 3 points that are feet of the altitudes of a triangle, reconstruct the triangle.

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Eigenray on Sep 24^th, 2007, 3:49am

Three weeks ago, I wouldn't have known how to do this ;)

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Barukh on Sep 24^th, 2007, 4:20am

on 09/24/07 at 03:49:00, Eigenray wrote:

Three weeks ago, I wouldn't have known how to do this ;)

I am glad you learnt something new with a little help from me... :)

BTW, how many solutions are there?

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Eigenray on Sep 24^th, 2007, 7:41am

Looks like [hide]4[/hide], but could also be [hide]1 or infinity[/hide] in degenerate cases.

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Barukh on Sep 24^th, 2007, 9:36am

on 09/24/07 at 07:41:56, Eigenray wrote:

[hide]1 or infinity[/hide] in degenerate cases.

What are the cases when you get a single solution?

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Eigenray on Sep 24^th, 2007, 10:52am

on 09/24/07 at 09:36:08, Barukh wrote:

What are the cases when you get a single solution?

When the points are collinear?

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Barukh on Sep 26^th, 2007, 1:15am

on 09/24/07 at 10:52:17, Eigenray wrote:

When the points are collinear?

Right.

Here is another one:

2. Given: one vertex, incenter, circumcenter.

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Eigenray on Sep 26^th, 2007, 3:39am

Here is a rather uninspired solution:

[hideb]Change coordinates so that the vertex is at O=(0,0), the incenter I=(1,0), and the circumcenter C=(a,b). Construct lines through O of slope http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifm. For each line, mark the point X_i on the line such that the perpendicular bisector of OX goes through C. Then we require that angle OX₁I = angle IX₁X₂. Solving for X_i in terms of m, and equating the tangents of the angles, gives m = (2a-1)/http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{4b²+4a-1}, which of course is constructible.[/hideb]

But I'm quite sure that's not what you had in mind.

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Barukh on Sep 26^th, 2007, 3:51am

on 09/26/07 at 03:39:58, Eigenray wrote:

But I'm quite sure that's not what you had in mind.

Confirmed.

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Aryabhatta on Sep 27^th, 2007, 12:59am

on 09/26/07 at 03:51:13, Barukh wrote:

Confirmed.

Another method:
[hide]

Use the fact that the distance between the incenter and circumcenter is sqrt(R(R-2r)) where R is the circum-radius and r is the in-radius.

Using this we can draw the incircle and circumcircle. The tangents to the incircle through one vertex and their intersections with the circumcircle determine the triangle.
[/hide]

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Barukh on Sep 27^th, 2007, 6:19am

Yes, exactly. Good job, Aryabhatta! :D

BTW, how would you constuct the incircle?

Title: Re: Reconstruct Triangle from Points (Geometry)
Post by Aryabhatta on Sep 27^th, 2007, 11:09am

on 09/27/07 at 06:19:46, Barukh wrote:

Yes, exactly. Good job, Aryabhatta! :D

BTW, how would you constuct the incircle?

To construct the circle, we construct the radius first. To do that we construct x = R-2r. Let y = R

Then we are given sqrt(xy) and y. We use the well known fact that the perpendicular on the hypotenuse of a right triangle is of length sqrt(xy) where x and y are the lengths of the two segments of the hypotenuse so formed because of the perpendicular on it.

Given sqrt(xy) and y, we can easily construct the whole triangle and, as a result, we can construct x.