6
$\begingroup$

In 2014, I found a nice result in plane geometry, the result is a generalization of the Simson line theorem, and there are nine proofs for this result were published in [1]-[7]. Continuing, I find a new generalization of the old result as follows:

Problem: Let $ABC$ be a triangle, let $P$ be a point in the circumcircle, the circumcenter is $O$. Let $Q$ be the point in the plane. The circles $(APQ), (BPQ), (CPQ)$ meet $OQ$ again at $A', B', C'$ respectively. Let $A_1, B_1, C_1$ be the the projections of $A', B', C'$ onto $BC, CA, AB$ respectively. Then $A_1, B_1, C_1$ are collinear, and the new line through a fixed point on the Nine point circle when $Q$ be moved on the given line (or $P$ be moved in the circumcircle). When $Q$ in infinity we get the old result (the result I found in 2014).

My question, could you give a proof for this problem?

enter image description here

Note: Check the result is true with applet by click here

Note: You can see some problem around this configuration in

** References:**

[1]-Nguyen Van Linh, Another synthetic proof of Dao's generalization of the Simson line theorem, Forum Geometricorum, 16 (2016) 57--61.

[2]-Nguyen Le Phuoc and Nguyen Chuong Chi (2016). 100.24 A synthetic proof of Dao's generalisation of the Simson line theorem. The Mathematical Gazette, 100, pp 341-345. doi:10.1017/mag.2016.77.

[3]-Leo Giugiuc, A proof of Dao’s generalization of the Simson line theorem, tạp chí Global Journal of Advanced Research on Classical and Modern Geometries, ISSN: 2284-5569, Vol.5, (2016), Issue 1, page 30-32

[4]-Tran Thanh Lam, Another synthetic proof of Dao's generalization of the Simson line theorem and its converse, Global Journal of Advanced Research on Classical and Modern Geometries, ISSN: 2284-5569, Vol.5, (2016), Issue 2, page 89-92

[5]-Ngo Quang Duong, A generalization of the Simson line theorem, to appear in Forum Geometricorum.

[6]-Three other proofs by Telv Cohl, Luis Gonzalez, Tran Quang Huy A Generalization of Simson Line

[7]-Another proof https://www.artofproblemsolving.com/community/c6h1075523p5181203

$\endgroup$
  • 1
    $\begingroup$ "Continuing, I find a new generalization of the old result as follows:" could you clarify whether you know that this general statement is correct? If so, what is the proof that you have? $\endgroup$ – Noah Schweber Mar 7 '17 at 18:35
  • $\begingroup$ Dear @NoahSchweber , I tried to proof in two days, but I can not give the final of calculator by hand, very long, so I must stop. The problem is true, You can check in geogebra.org/m/ezTcUPJn $\endgroup$ – Oai Thanh Đào Mar 7 '17 at 23:25
  • $\begingroup$ What is the line $Q$ is going on? $OQ$? $\endgroup$ – Fedor Petrov Apr 27 '17 at 8:35
4
$\begingroup$

Let $CC'$ meet a circle $\omega=(ABC)$ in a point $S\ne C'$. Then $\angle (CP,CS)=\angle (CP,CC')=\angle (QP,QC')=(QP,QO)$. Thus $AA'$, $BB'$ pass through the same point $S$. The following argument is not synthetic, but it explains what is this fixed point on an Euler circle and what is another point in which $A_1B_1C_1$ meets Euler circle. Thus it hopefully may help with a synthetic argument too.

Consider the complex coordinates for which $\omega=\{z:|z|=1\}$, $A,B,C$ correspond to complex numbers $a,b,c$, $OQ$ to a real line, $S$ to $s$. Then $C'$ corresponds to $c'=(c+s)/(1+cs)$ (this is a formula for central projection from $\omega$ to a real line from the point $s$, as may be checked for three points $1,-1,-s$). Next, a projection of $z$ to a line between $a,b$ is $(z-\bar{z}ab+a+b)/2$, as may be checked for points $a,b,0$. So, $C_1$ corresponds to $c_1=(c'(1-ab)+a+b)/2$. Denote $c_2=2c_1-(a+b+c)$. Note that $z\rightarrow 2z-(a+b+c)$ is a homothety which sends Euler circle of $ABC$ to $\omega$. Thus for points $a_2,b_2,c_2$ we should prove that they are collinear and the line passes through a point on $\omega$ not depending on $s$. We get $c_2=c'(1-ab)-c$ and I claim that $c_2$ lies on a line between $s$ and $-abc$. Indeed, the direction between $s$ and $-abc$ is a direction of $s+abc$. The direction between $s$ and $c_2$ is a direction of $c_2-s=-c'(ab+cs)$, that is, direction of $ab+cs$, but the ratio of $s+abc$ and $ab+cs$ is indeed real.

$\endgroup$
  • $\begingroup$ It seems that, when $A'$, $B'$, $C'$ are just intersections of $SA$, $SB$, $SC$ with a fixed line (and fixed $S\in \omega$), we come to the previous problem mentioned by the OP (which corresponds to $Q=\infty$). $\endgroup$ – Ilya Bogdanov Oct 29 '18 at 11:29

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.