Ask Your Question
0

Intercept theorem in sage

asked 2016-05-06 18:48:30 +0200

interception47 gravatar image

updated 2024-04-18 20:08:02 +0200

FrédéricC gravatar image

Hi,

I'm working on Grobner bases and Buchberger algorithm and begginning in Sage. I would like to find a way to prove that if we have $B, C, D$ three points, then the point $C$ is in the circle of diameter $BD$ if and only if the triangle $BCD$ is an $C$-shaped right-angled triangle.

I'm thinking about using the Intercept theorem but I would like to prove it using Sage.

Thank you

edit retag flag offensive close merge delete

1 Answer

Sort by » oldest newest most voted
0

answered 2017-04-12 19:34:23 +0200

dan_fulea gravatar image

There is a first phase of translating the synthetic geometry into analytic / algebraic geometry. Let us suppose that $B,D$ are located in the points with coordinates $(-1,0)$, and respectively $(1,0)$.

Then $C=C(x,y)$ is on the circle with diameter $BD$, iff its distance to the origin $(0,0)$ is one. We get thus the condition: $$ x^2+y^2-1 =0\ . $$ The perpendicularity condition $BC\perp CD$ is written equivalently as the fact, that the slopes product is minus one: $$ \frac{y-0}{x-(-1)}\cdot \frac{y-0}{x-(-1)} = -1\ .$$ The two conditions are obviously equivalent. It is hard to find something in sage that checks that an element is in an ideal. We may of course force the situation and type:

sage: R.<x,y> = QQ[]
sage: J = R.ideal( x^2 + y^2 - 1 )
sage: mBC = (y-0) / ( x-(-1) )
sage: mCD = (y-0) / ( x-(+1) )
sage: ( mBC*mCD + 1 ).numerator() in J
True

This is very convincing, and may make us think sage did very quickly a hard job, but after...

sage: ( mBC*mCD + 1 ).numerator()
x^2 + y^2 - 1

we can only sell the code as a joke!

edit flag offensive delete link more

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

Stats

Asked: 2016-05-06 18:48:30 +0200

Seen: 614 times

Last updated: Apr 12 '17