2
$\begingroup$

This is Exercise 13 of Chapter 6 of Ideals, Varieties, and Algorithms by Cox et al.

The problem asks to translate the following geometric theorem into polynomials and using Groebner basis to test whether the conclusion $g$ follows generically from the hypothesis $h_i$.

The three altitudes of a triangle meet at a single point $H$.

Here is the picture:

enter image description here

In the picture $ABC$ is the original triangle. I will use $u_i$ as independent variables, which could be arbitrary, $x_i$ as variables that can be determined by others.

Let $$A=(0,0)\\B=(u_1,0)\\C=(u_2,u_3)\\D=(x_1,x_2)\\E=(x_3,x_4)\\F=(x_5,x_6)\\H=(x_7,x_8)$$

The hypothesis are: $$AD\perp BC: \frac{x_2}{x_1}\frac{u_3}{u_2-u_1}=-1; \implies h_1=x_2u_3+x_1(u_2-u_1)=0\\ C,D,B \text{ colinear}: \frac{u_3}{u_2-u_1}=\frac{x_2}{x_1-u_1}\implies h_2=u_3(x_1-u_1)-x_2(u_2-u_1)=0$$ The hypothesis $h_3,h_4,h_5,h_6$ of the other two sides are obtained similarly.

The other two hypothesis are $$C,F,H \text{ colinear}\\ A,D,H \text{ colinear}$$ Where I also used slopes.

The conclusion is $$B,E,H \text{ colinear}: g=x_4(x_7-u_1)-x_8(x_3-u_1)=0$$

Using the corollary from the book, I should get $\{1\}$ as the reduced Groeber basis for the ideal $\left\langle h_1,\cdots, h_8, 1-gy\right\rangle \subset \mathbb{R}(u_1,u_2,u_3)[x_1,\cdots,x_8,y]$. But I cannot get it.

My question:

Is there mistakes in my formulation? Or did I miss anything?

Thank you very much for any help!

$\endgroup$
  • $\begingroup$ is $C = (u_2, u_3)$? $\endgroup$ – muaddib Jul 28 '15 at 13:33
  • $\begingroup$ @muaddib: You are right. That was a typo. Thank you and I'll edit. $\endgroup$ – KittyL Jul 28 '15 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.