# Continuing generalization of the Simson line

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 -. 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? ** References:**

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

-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.

-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

-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

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

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

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

• "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? – Noah Schweber Mar 7 '17 at 18:35
• 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 – Oai Thanh Đào Mar 7 '17 at 23:25
• What is the line $Q$ is going on? $OQ$? – Fedor Petrov Apr 27 '17 at 8:35

## 1 Answer

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.

• 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 $Q=\infty$). – Ilya Bogdanov Oct 29 '18 at 11:29