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Can you provide a proof for the following proposition:

Proposition. Given any quadrilateral $ABCD$. Let $P,Q,R,S$ be nine-point centers of triangles $\triangle ABD$,$\triangle ABC$,$\triangle BCD$ and $\triangle ACD$ respectively. Then, the quadrilateral $ABCD$ is cyclic if and only if $P,Q,R,S$ are concyclic.

enter image description here

GeoGebra applet that demonstrates this proposition can be found here.

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    $\begingroup$ Might the Art of Problem Solving website be a good place to ask this question? $\endgroup$
    – Liam Baker
    Apr 26, 2021 at 17:18

4 Answers 4

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Regarding the "only if" part, a stronger result actually holds: if the quadrilateral $ABCD$ is cyclic, then $PQRS$ is similar to $ABCD$, and so it is cyclic, too.

This is stated, without proof, in many elementary geometry textbooks, see for instance at p. 44 of

D. G. Wells: The Penguin dictionary of curious and interesting geometry, New York, NY: Penguin Books. xiv, 285 p. (1991). ZBL0856.00005.

A proof can be found at p. 36 of the paper

F. V. Morley: Notes on the cyclic quadrilateral, Annals of Math. (2) 22, 35-42, 43 (1920). ZBL47.0566.01.

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    $\begingroup$ It is similar when $ABCD$ is cyclic, but is it always similar? $\endgroup$ Apr 26, 2021 at 5:28
  • $\begingroup$ @FedorPetrov: I do not know. It seems to me that Morley's proof only works for cyclic quadrilaterals, unless I am missing something. $\endgroup$ Apr 26, 2021 at 5:33
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    $\begingroup$ Then it is not a stronger result: it does not prove the "if" part. $\endgroup$ Apr 26, 2021 at 6:03
  • $\begingroup$ You are right: I will edit the answer adding your observation. – $\endgroup$ Apr 26, 2021 at 6:34
  • $\begingroup$ The 'if' direction is not hard to prove if 'nine-point centre' is replaced with 'centroid', since in that case $PQRS$ is always similar to $ABCD$. $\endgroup$
    – Liam Baker
    Apr 26, 2021 at 17:16
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This is not an answer but a comment. However, it will be too long and it would be awkward to break it up. It is of course natural to ask what is special about the nine point centre here. One can put the question in the following context. Given the shape of a quadrilateral $ABCD$, what is the shape of $PQRS$ (constructed as above but using any of the triangle centres protocolled in the online Encyclopedia of Triangle Centers)?

The shape of a quadrilateral is the unique pair $(p_1,q_1)$ and $(p_2,q_2)$ for which it is similar to the one with $(0,0)$, $(1,0)$ and these two points as vertices (in terms of complex numbers they are just $$ \frac{z_C-z_A}{z_B-z_A},\frac{z_D-z_A}{z_B-z_A}. $$

Many structural properties of a quadrilateral (in particular cyclicity) can be expressed as a simple equation in the $p$'s and $q$'s. For cyclicity, one equates the coordinates of the circumcentres of $ABC$ and $ABD$.

It is then a simple, if usually tedious, task to compute the shape of $PQRS$ (for a given centre function) in terms of that of $ABCD$ (easily automatised using Mathematica) and so provide far-reaching generalisations of Morley's result.

To be explicit, if $Z_1$ and $Z_2$ are $p_1+iq_1$, resp. $p_2+iq_2$, then we can easily compute the shape of $PQRS$ by computing the complex numbers which specify its vertices (using the centre function) and then forming the corresponding quotients as above.

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The proposition is almost true. Francesco Polizzi has given the "if" part. For the "only if" part, the following is true:

If $P,Q,R,S$ are concyclic then either $P=Q=R=S$ or $A,B,C,D$ are concyclic.

The first case of $P=Q=R=S$ can only happen for non-convex quadrilaterals.

The proof can be done using complex numbers, similar to what bathalf15320 has suggested.

Let $a, b, c, d, p, q, r, s$ be complex numbers representing the corresponding points of the quadrilateral and the nine-point centers. It is not too hard to compute the relationship of these numbers. For example $s = \frac{\bar{a} ( b^2 - c^2 ) + \bar{b} ( c^2 - a^2 ) + \bar{c} ( a^2 - b^2 ) }{\bar{a} ( b - c ) + \bar{b} ( c - a ) + \bar{c} ( a - b )}$, and similar for $p$, $q$, and $r$.

The condition for $a, b, c, d$ to be concyclic can be written as $( a - b ) ( c - d ) ( \bar{a} \bar{b} + \bar{c} \bar{d} ) + ( a - c ) ( d - b ) ( \bar{a} \bar{c} + \bar{d} \bar{b} ) + ( a - d ) ( b - c ) ( \bar{a} \bar{d} + \bar{b} \bar{c} ) = 0$.

The condition that $p, q, r, s$ are concyclic can be checked using the following sage program

def ninept_2(p0, q0, r0, p1, q1, r1):
    return p1 * (q0^2 - r0^2) + q1 * (r0^2 - p0^2) + r1 * (p0^2 - q0^2)

def ninept_1(p0, q0, r0, p1, q1, r1):
    return p1 * (q0 - r0) + q1 * (r0 - p0) + r1 * (p0 - q0)

def condiff(phi, plo, qhi, qlo):
    return expand(phi * qlo - qhi * plo)

def concross(phi, plo, qhi, qlo, rhi, rlo, shi, slo):
    return expand(phi * qhi * rlo * slo + plo * qlo * rhi * shi)

def conpart(phi, plo, qhi, qlo, rhi, rlo, shi, slo):
    return condiff(phi, plo, qhi, qlo) * condiff(rhi, rlo, shi, slo)

var("a0 b0 c0 d0 a1 b1 c1 d1")

p_hi = ninept_2(b0, c0, d0, b1, c1, d1)
q_hi = ninept_2(a0, c0, d0, a1, c1, d1)
r_hi = ninept_2(a0, b0, d0, a1, b1, d1)
s_hi = ninept_2(a0, b0, c0, a1, b1, c1)
p_lo = ninept_1(b0, c0, d0, b1, c1, d1)
q_lo = ninept_1(a0, c0, d0, a1, c1, d1)
r_lo = ninept_1(a0, b0, d0, a1, b1, d1)
s_lo = ninept_1(a0, b0, c0, a1, b1, c1)

p_hi_bar = ninept_2(b1, c1, d1, b0, c0, d0)
q_hi_bar = ninept_2(a1, c1, d1, a0, c0, d0)
r_hi_bar = ninept_2(a1, b1, d1, a0, b0, d0)
s_hi_bar = ninept_2(a1, b1, c1, a0, b0, c0)
p_lo_bar = ninept_1(b1, c1, d1, b0, c0, d0)
q_lo_bar = ninept_1(a1, c1, d1, a0, c0, d0)
r_lo_bar = ninept_1(a1, b1, d1, a0, b0, d0)
s_lo_bar = ninept_1(a1, b1, c1, a0, b0, c0)

xx0 = conpart(p_hi, p_lo, q_hi, q_lo, r_hi, r_lo, s_hi, s_lo) * 
concross(p_hi_bar, p_lo_bar, q_hi_bar, q_lo_bar, r_hi_bar, r_lo_bar, s_hi_bar, s_lo_bar)
xx1 = conpart(p_hi, p_lo, r_hi, r_lo, s_hi, s_lo, q_hi, q_lo) * 
concross(p_hi_bar, p_lo_bar, r_hi_bar, r_lo_bar, s_hi_bar, s_lo_bar, q_hi_bar, q_lo_bar)
xx2 = conpart(p_hi, p_lo, s_hi, s_lo, q_hi, q_lo, r_hi, r_lo) * 
concross(p_hi_bar, p_lo_bar, s_hi_bar, s_lo_bar, q_hi_bar, q_lo_bar, r_hi_bar, r_lo_bar)

print(factor(expand(xx0 + xx1 + xx2)))

The output of that program is -(a1*b0^2*c0 - a0^2*b1*c0 - a1*b0*c0^2 + a0*b1*c0^2 + a0^2*b0*c1 - a0*b0^2*c1 - a1*b0^2*d0 + a0^2*b1*d0 + a1*c0^2*d0 - b1*c0^2*d0 - a0^2*c1*d0 + b0^2*c1*d0 + a1*b0*d0^2 - a0*b1*d0^2 - a1*c0*d0^2 + b1*c0*d0^2 + a0*c1*d0^2 - b0*c1*d0^2 - a0^2*b0*d1 + a0*b0^2*d1 + a0^2*c0*d1 - b0^2*c0*d1 - a0*c0^2*d1 + b0*c0^2*d1)^2*(a0*a1*b1*c0 - a1*b0*b1*c0 - a0*a1*b0*c1 + a0*b0*b1*c1 + a1*b0*c0*c1 - a0*b1*c0*c1 - a0*a1*b1*d0 + a1*b0*b1*d0 + a0*a1*c1*d0 - b0*b1*c1*d0 - a1*c0*c1*d0 + b1*c0*c1*d0 + a0*a1*b0*d1 - a0*b0*b1*d1 - a0*a1*c0*d1 + b0*b1*c0*d1 + a0*c0*c1*d1 - b0*c0*c1*d1 - a1*b0*d0*d1 + a0*b1*d0*d1 + a1*c0*d0*d1 - b1*c0*d0*d1 - a0*c1*d0*d1 + b0*c1*d0*d1)*(a1^2*b1*c0 - a1*b1^2*c0 - a1^2*b0*c1 + a0*b1^2*c1 + a1*b0*c1^2 - a0*b1*c1^2 - a1^2*b1*d0 + a1*b1^2*d0 + a1^2*c1*d0 - b1^2*c1*d0 - a1*c1^2*d0 + b1*c1^2*d0 + a1^2*b0*d1 - a0*b1^2*d1 - a1^2*c0*d1 + b1^2*c0*d1 + a0*c1^2*d1 - b0*c1^2*d1 - a1*b0*d1^2 + a0*b1*d1^2 + a1*c0*d1^2 - b1*c0*d1^2 - a0*c1*d1^2 + b0*c1*d1^2)^2 where I'm using a0 to represent $a$ and a1 to represent $\bar{a}$ etc. This output is of the form $- U^2 V W^2$, where $U = \bar{W}$ and $V=0$ is the condition for $a,b,c,d$ to be concyclic.

So, if $P,Q,R,S$ are concyclic either $A,B,C,D$ are concyclic or $U = 0$. It turns out, that this condition is exactly the condition for $P=Q$ (provided $A,B,C,D$ is non-degenerate), and then automatically $P=Q=R=S$.

An example for the degenerate case would be $A(0, 0)$, $B(1, 0)$, $C(2, -1)$, $D(2, 2)$ with nine-point center for each triangle at $( \frac{5}{4}, \frac{1}{4} )$.

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Here is a proof using analytic tools, complex numbers, and formulas for the involved points in terms of them. Points in the plane will be denoted by capital letters like $A,B,C,D;N;P,Q,R,S$ and the corresponding affixes will be their lower cousins, respectively $a,b,c,d;n;p,q,r,s\in\Bbb C$, and decorations (sub- or upper indices) will be kept. First let us recall some facts.


Incircle test: Four points $a,b,c,d\in\Bbb C$ are on a circle (i.e. concyclic), iff $$ \begin{vmatrix} 1 & a & \bar a & a\bar a\\ 1 & b & \bar b & b\bar b\\ 1 & c & \bar c & c\bar c\\ 1 & d & \bar d & d\bar d \end{vmatrix} =0\ . $$ Proof: The above expression is kept invariant when simultaneously translating or rotating $a,b,c,d$, and rescaling introduces only a non-zero factor. So we may and do assume $a,b,c$ on the unit circle. Then the determinant has on the fourth column the etries $1,1,1,d\bar d$, and after subtracting the first column we obtain the factor $d\bar d-1$ for the determinant. (Times an area factor.) $\square$

Cross-ratio test: Four points $a,b,c,d\in\Bbb C$ are on a circle or on a line, iff their cross-ratio is a real number.

Proof: This is equivalent to the incircle test after clearing denominators.

Centroid $X(2)$: The centroid $G=X(2)$ of $\Delta ABC$ is given by $g=x(2)=\frac 13(a+b+c)$.

Circumcenter $X(3)$: The circumcenter $O=X(3)$ of $\Delta ABC$ is given by $$ o=x(3)= \frac { \begin{vmatrix} 1 & a & a\bar a\\ 1 & b & b\bar b\\ 1 & c & c\bar c \end{vmatrix} } { \begin{vmatrix} 1 & a & \bar a\\ 1 & b & \bar b\\ 1 & c & \bar c \end{vmatrix} } \ . $$

Nine-point center $X(5)$: The center $N=X(5)$ of the nine-point circle satisfies $3\overrightarrow{NG}=\overrightarrow{NO}$, so $3(g-n)=(o-n)$, giving $2n=3g-o$, i.e. $$ n = x(5) = \frac 12(a+b+c)-\frac 12\cdot \frac { \begin{vmatrix} 1 & a & a\bar a\\ 1 & b & b\bar b\\ 1 & c & c\bar c \end{vmatrix} } { \begin{vmatrix} 1 & a & \bar a\\ 1 & b & \bar b\\ 1 & c & \bar c \end{vmatrix} } \ . $$


Now for the given problem. Let us start with four points $A,B,C,D$ in general position.

We build $$ \begin{aligned} P &= N_{\Delta BCD}\ , & p &= \frac 12(b + c + d) -\frac 12 \begin{vmatrix} 1 & b & b\bar b\\ 1 & c & c\bar c\\ 1 & d & d\bar d\end{vmatrix} \Big/ \begin{vmatrix} 1 & b & \bar b\\ 1 & c & \bar c\\ 1 & d & \bar d\end{vmatrix} \\[3mm] Q &= N_{\Delta ACD}\ , & q &= \frac 12(a + c + d) -\frac 12\begin{vmatrix} 1 & a & a\bar a\\ 1 & c & c\bar c\\ 1 & d & d\bar d\end{vmatrix} \Big/ \begin{vmatrix} 1 & a & \bar a\\ 1 & c & \bar c\\ 1 & d & \bar d\end{vmatrix} \\[3mm] R &= N_{\Delta ABD}\ , & r &= \frac 12(a + b + d) -\frac 12\begin{vmatrix} 1 & a & a\bar a\\ 1 & b & b\bar b\\ 1 & d & d\bar d\end{vmatrix} \Big/ \begin{vmatrix} 1 & a & \bar a\\ 1 & b & \bar b\\ 1 & d & \bar d\end{vmatrix} \\[3mm] S &= N_{\Delta ABC}\ , & s &= \frac 12(a + b + c) -\frac 12\begin{vmatrix} 1 & a & a\bar a\\ 1 & b & b\bar b\\ 1 & c & c\bar c\end{vmatrix} \Big/ \begin{vmatrix} 1 & b & \bar b\\ 1 & c & \bar c\\ 1 & c & \bar c\end{vmatrix} \end{aligned} $$


The points $p,q,r,s\in\Bbb C$ are then concyclic, iff they are after a translation by $-(a+b+c+d)/2$, and a rescaling with factor $(-2)$. Let $p',q',r',s'$ be these points.

We may and do assume after translation, rotation and rescaling that $A,B,C$ are on the unit circle centered in the origin. So $1 = a\bar a=b\bar b=c\bar c$. We then consider $d':=\bar d$ as a "new variable" (only to make typing simpler) and obtain algebraic descriptions in terms of $a,b,c;d,d'$ for all involved points.

Explicitly, we have: $$ \begin{aligned} p' &= a + \begin{vmatrix} 1 & b & b\bar b\\ 1 & c & c\bar c\\ 1 & d & d\bar d\end{vmatrix} \Big/ \begin{vmatrix} 1 & b & \bar b\\ 1 & c & \bar c\\ 1 & d & \bar d\end{vmatrix} =a+\frac{bc(dd'-1)}{bcd'-b-c+d}\ , \\[3mm] q' &= b + \begin{vmatrix} 1 & a & a\bar a\\ 1 & c & c\bar c\\ 1 & d & d\bar d\end{vmatrix} \Big/ \begin{vmatrix} 1 & a & \bar a\\ 1 & c & \bar c\\ 1 & d & \bar d\end{vmatrix} =b+\frac{ac(dd'-1)}{acd'-a-c+d}\ , \\[3mm] r' &= c + \begin{vmatrix} 1 & a & a\bar a\\ 1 & b & b\bar b\\ 1 & d & d\bar d\end{vmatrix} \Big/ \begin{vmatrix} 1 & a & \bar a\\ 1 & b & \bar b\\ 1 & d & \bar d\end{vmatrix} =c+\frac{ab(dd'-1)}{abd'-a-b+d}\ , \\[3mm] s' &= d + \begin{vmatrix} 1 & a & a\bar a\\ 1 & b & b\bar b\\ 1 & c & c\bar c\end{vmatrix} \Big/ \begin{vmatrix} 1 & a & \bar a\\ 1 & b & \bar b\\ 1 & c & \bar c\end{vmatrix} =d\ . \end{aligned} $$


One direction is clear now. If $a,b,c,d$ are concyclic, then $dd'=d\bar d=1$, so $(p',q',r',s')=(a,b,c,d)$ concyclic, so $(p,q,r,s)$ concyclic. For the other direction we compute the cross ratio of the tuple $(p',q',r',s')$ in terms of the used variables $a,b,c,d;d'$.


So we compute with bare hands (although this is done immediately with computer aid): $$ \begin{aligned} p'-r' &= (a-c) + \frac{b(dd'-1)}{(\cdots)(\cdots)}\cdot\Big[\ c(abd'-a-b+d) - a(bcd'-b-c+d)\ \Big] \\ &= (a-c) + \frac{b(dd'-1)}{(\cdots)(\cdots)}\cdot(a-c)(b-d) \\ &= \frac{(a-c)}{(\cdots)(\cdots)}\Big[\ (abd'-a-b+d)(bcd'-b-c+d) + b(b-d)(dd'-1)\ \Big] \\ &= \frac{(a-c)}{(\cdots)(\cdots)}\Big[\ \Big( a(bd'-1) - (b-d)\Big)\Big( c(bd'-1) - (b-d)\Big) \\ &\qquad\qquad\qquad\qquad + b(b-d)(dd'-1)\ \Big] \\ &= \frac{(a-c)}{(\cdots)(\cdots)}\Big[\ ac(bd'-1)^2 -(b-d)(bd'-1)(a+c) + (b-d)^2 \\ &\qquad\qquad\qquad\qquad + b(b-d)(dd'-1) \ \Big] \\ &= \frac{(a-c)}{(\cdots)(\cdots)}\Big[\ ac(bd'-1)^2 -(b-d)(bd'-1)(a+c) \\ &\qquad\qquad\qquad\qquad + (b-d)\Big(b-d \ + \ bdd'-b\Big) \ \Big] \\ &= \frac{(a-c)(bd'-1)}{(\cdots)(\cdots)}\Big[\ ac(bd'-1) - (b-d)(a+c) + d(b-d) \ \Big] \\ &= \frac{(a-c)(bd'-1) \color{blue}{ \Big[\ abcd' - ab - bc - ca + d(a+b+c) -d^2 \ \Big]} } {(abd'-a-b+d)\color{brown}{(bcd'-b-c+d)}} \ . \\[3mm] p'-s' & = a-d + \frac{bc(dd'-1)}{bcd'-b-c+d} \\ &= \frac {\color{blue}{abcd' - ab - bc - ca + d(a+b+c) -d^2}} {\color{brown}{bcd'-b-c+d}} \\[3mm] \frac{p'-r'}{p'-s'} &= \frac{(a-c)(bd'-1)}{abd' - a - b + d} \\[3mm] &\text{ and after $a\leftrightarrow b$, inducing $p'\leftrightarrow q'$, and keeping $r',s'$} \\[3mm] \frac{q'-r'}{q'-s'} &= \frac{(b-c)(ad'-1)}{abd' - a - b + d}\ , \\[3mm] \operatorname{cr}(p', q'; r', s') &= \frac{p'-r'}{p'-s'} : \frac{q'-r'}{q'-s'} \\ &= \frac{(a-c)(bd'-1)}{(b-c)(ad'-1)} \\ &= \frac{a-c}{a-1/d'}: \frac{b-c}{b-1/d'} \\ &=\operatorname{cr}(a,b;c,1/d') \ . \end{aligned} $$ We can draw the conclusion for the missing direction. If $p,q,r,s$ concyclic, then $(p',q',r',s')$ concyclic, then we have a real value for the cross ratios $\operatorname{cr}(p,q,r,s)=\operatorname{cr}(p',q',r',s')=\operatorname{cr}(a,b;c,1/d')$, so the points $a,b,c,1/d'$ are concyclic, so the fourth point $1/d'=1/\bar d$ is on the unit circle, so $d$ is on the unit circle. $\square$


Addendum: The following simple sage code supports the above computation. We use the sage variables a, b, c, d, dd for the above variables $a,b,c,d,d'$.

var('a,b,c,d,dd')
pp = a + matrix(3, 3, [1, b, 1, 1, c, 1, 1, d, d*dd]).det() / matrix(3, 3, [1, b, 1/b, 1, c, 1/c, 1, d,  dd]).det()
qq = b + matrix(3, 3, [1, a, 1, 1, c, 1, 1, d, d*dd]).det() / matrix(3, 3, [1, a, 1/a, 1, c, 1/c, 1, d,  dd]).det()
rr = c + matrix(3, 3, [1, a, 1, 1, b, 1, 1, d, d*dd]).det() / matrix(3, 3, [1, a, 1/a, 1, b, 1/b, 1, d,  dd]).det()
ss = d + matrix(3, 3, [1, a, 1, 1, b, 1, 1, c,    1]).det() / matrix(3, 3, [1, a, 1/a, 1, b, 1/b, 1, c, 1/c]).det()

def cr(x0, x1, x2, x3):
    return (x2 - x0) / (x2 - x1) * (x3 - x1) / (x3 - x0)

print(f"The cross ratio of P', Q', R', S' is: {cr( pp, qq, rr, ss ).factor()}")

This gives:

The cross ratio of P', Q', R', S' is: (b*dd - 1)*(a - c)/((a*dd - 1)*(b - c))
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