A new point on the nine-point circle I

The picture which contains a mixture of the points we’ll constantly feature in most our posts this year is:

In furtherance of our previous post, we’re still discussing point $W$ , which one can easily pinpoint in the above picture. Its coordinates are given by

(1) $\begin{equation*} \begin{split} x&=\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ y&=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)} \end{split} \end{equation*}$

relative to the vertices $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ of the parent triangle $ABC$ . It lies on the nine-point circle of $\triangle ABC$ , as we saw through numerous numerical examples previously (a proper proof will be provided later). In general, $W$ is distinct from the nine traditional points through which the nine-point circle passes (the mid-point of each side, the foot of each altitude, the midpoint of the line segment from the orthocenter to each vertex), and so we’ve chosen to call it “new”.

Today we focus on when our purportedly new point $W$ coincides with the nine traditional points through which the nine point circle passes. (For a clue as to what other new points are on the nine-point circle, see this article.)

Eleven sixteenths

Remember the fraction: $\frac{11}{16}$ . And the caption.

In example 2 below, we show that the sum of the squares of the distances from the center of the nine-point circle to the three vertices of a right triangle is always $11/16$ of the square of the hypotenuse. (Though not essential to today’s target, yet it’s what we won’t want you to forget.)

In the diagram below, PROVE that $AC=AB\cos\alpha+BC\cos\beta$ , given that $BO$ is a median.

Since $BO$ is a median, the area of $\triangle ABO$ is equal to the area of $\triangle BOC$ :

$1/2\times AB\times BO\times \sin\alpha=1/2\times BO\times BC\times \sin\beta\implies\scriptstyle AB\sin\alpha=BC\cos\alpha.$

Also, the sum of the areas of triangles $ABO$ and $BOC$ gives the area of the parent triangle $ABC$ :

$\begin{equation*} \begin{split} (AB)(BO)(\sin\alpha)/2+(BC)(BO)\sin\beta/2&=(AB)(BC)/2\\ BO\Big((AB)\sin\alpha+(BC)\cos\alpha\Big)&=(AB)(BC)\\ \implies AB\sin\alpha+BC\cos\alpha&=\frac{2(AB)(BC)}{AC}\quad\textrm{since}~BO=\frac{AC}{2}\\ \implies BC\cos\alpha+BC\cos\alpha&=\frac{2(AB)(BC)}{AC}\\ \implies \cos\alpha&=\frac{AB}{AC}\\ \implies AC&=\frac{AB}{\cos\alpha} \end{split} \end{equation*}$

We’ll now simplify $AB\cos\alpha+BC\cos\beta$ using the fact that $AB\sin\alpha=BC\cos\alpha$ , obtained previously:

$\begin{equation*} \begin{split} AB\cos\alpha+BC\cos\beta&=AB\cos\alpha+\frac{AB\sin\alpha}{\cos\alpha}\cos\beta\\ &=AB\cos\alpha+\frac{AB\sin\alpha}{\cos\alpha}\sin\alpha\\ &=\frac{AB(\cos^2\alpha+\sin^2\alpha)}{\cos\alpha}\\ &=\frac{AB}{\cos\alpha}\\ &=AC \end{split} \end{equation*}$

Let $M$ be the center of the nine-point circle of a right triangle $ABC$ , where $\angle B=90^{\circ}$ . PROVE that $AM^2+BM^2+CM^2=\frac{11}{16}AC^2$ .

Since the center $M$ of the nine-point circle is the midpoint of the orthocenter ( $B$ ) and the circumcenter ( $O$ ) of $\triangle ABC$ , we have that

(2) $\begin{equation*} BM=\frac{1}{2}BO=\frac{1}{2}\left(\frac{1}{2}AC\right)\implies BM^2=\frac{1}{16}AC^2 \end{equation*}$

The cosine law, applied to $\triangle ABM$ and $\triangle BMC$ , gives:

(3) $\begin{equation*} \begin{split} AM^2&=AB^2+BM^2-2(AB)(BM)\cos\alpha\\ CM^2&=BC^2+BM^2-2(BC)(BM)\cos\beta \end{split} \end{equation*}$

Add the two equations in (3) and use the fact that $AC=AB\cos\alpha+BC\cos\beta$ from example 1:

$\begin{equation*} \begin{split} AM^2+CM^2&=\scriptstyle(AB^2+BC^2)+2BM^2-2BM\Big(AB\cos\alpha+BC\cos\beta\Big)\\ &=AC^2+2\Big(\frac{1}{16}AC^2\Big)-2\Big(\frac{1}{4}AC\Big)(AC)\\ &=AC^2+\frac{1}{8}AC^2-\frac{1}{2}AC^2\\ &=\frac{5}{8}AC^2\\ \therefore AM^2+BM^2+CM^2&=\frac{5}{8}AC^2+\frac{1}{16}AC^2\\ &=\frac{11}{16}AC^2 \end{split} \end{equation*}$

Elegant solution: CLICK to see

Let $N$

be the center of the nine-point circle of any triangle $ABC$

, and let $R$

be the circum-radius. Then we have

$AN^2+BN^2+CN^2=\frac{1}{4}\left(3R^2+a^2+b^2+c^2\right),$

where $a=BC$ , $b=CA$ , and $c=AB$ , as per usual notation. In the case of a right triangle with hypotenuse having length $c$ , we have that $R=\frac{c}{2}$ and then $a^2+b^2=c^2$ . In turn:

$AN^2+BN^2+CN^2=\frac{1}{4}\left(\frac{3c^2}{4}+c^2+c^2\right)=\frac{11}{16}c^2.$

Assuming one was not aware of the relation

$AN^2+BN^2+CN^2=\frac{1}{4}\left(3R^2+a^2+b^2+c^2\right),$

there’s still another direct proof that’s simpler than the one we presented via examples 1 and 2 above. After solving a problem, always check for better solutions.

Please see exercise 3 for a non-right triangle where a modified version of the above result holds.

Equivalent statements

We now consider when the point $W$ with coordinates given in equation (1) coincides with the nine traditional points through which the nine-point circle passes. Our findings are easy to verify, so we skip the proofs.

Let $A(x_1,y_1)$

, $B(x_2,y_2)$

, $C(x_3,y_3)$

be the vertices of $\triangle ABC$

. Then the following statements are equivalent:

$W$ coincides with the midpoint of $AC$
sides $AB$ and $BC$ have opposite slopes

In particular, for a right triangle, $W$ is precisely the circumcenter if and only if the legs have slopes $\pm 1$ .

Extraneous secret: CLICK to reveal

Totally irrelevant stuff here. The proof of $(1)\implies(2)$

in the above example was completed while the poster was in the waiting area during a doctor’s appointment. For some reason the wait time was longer than expected, but the poster was “prepared” — with pencil and paper — and thereby profitably utilized the waiting period. Always go out with your pencil and paper, if possible. Unless you already live in pencilvania.

Let $A(x_1,y_1)$

, $B(x_2,y_2)$

, $C(x_3,y_3)$

be the vertices of $\triangle ABC$

. Then the following statements are equivalent:

$W$ coincides with the foot of the altitude from vertex $A$
side $BC$ is parallel to the $x$ -axis or parallel to the $y$ -axis

Something extremely pleasant happens if we combine condition $(2)$ above with condition $(2)$ below. See exercise 3 for the exquisite product.

Let $A(x_1,y_1)$

, $B(x_2,y_2)$

, $C(x_3,y_3)$

be the vertices of $\triangle ABC$

. Then the following statements are equivalent:

$W$ coincides with the midpoint of the orthocenter and the foot of the altitude from vertex $C$
sides $BC$ and $CA$ have reciprocal slopes

Exclusive six

Our next set of equivalent statements ultimately pertain only to the right triangle — the right triangle with legs parallel to the coordinate axes.

Let $A(x_1,y_1)$

, $B(x_2,y_2)$

, $C(x_3,y_3)$

be the vertices of $\triangle ABC$

. Then the following statements are equivalent:

$W=H$ , where $H$ is the orthocenter
two sides of the triangle are parallel to the coordinate axes
$W$ coincides with one of the triangle’s vertices
the slopes of the three medians form a geometric progression with common ratio $r=-2$
the slopes of the three medians form an arithmetic progression with common difference equal to $-\frac{3}{2}$ times the first term ( $d=-\frac{3}{2}a$ )
one side of the triangle is parallel to the $x$ -axis and another side and the median to it have opposite slopes.

A proper confluence of concepts right there. Among them all, the one that tickles our fancy the most is the fourth one, just because it has to do with that precious thing they call geometric progressions.

PROVE that if $W$ coincides with the orthocenter, then two sides of the triangle are parallel to the coordinate axes.

Despite the two different definitions of the line segments that yield them as points of concurrency, the orthocenter and our point $W$ have similar representations. For the orthocenter, the coordinates are:

(4) $\begin{equation*} \begin{split} x&=\scriptstyle\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)+(y_1-y_2)(y_2-y_3)(y_3-y_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ y&=\scriptstyle\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)-(x_1-x_2)(x_2-x_3)(x_3-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)} \end{split} \end{equation*}$

and so if the two points coincide, then combining (1) and (4) gives

(5) $\begin{equation*} \begin{split} (y_1-y_2)(y_2-y_3)(y_3-y_1)&=0\\ (x_1-x_2)(x_2-x_3)(x_3-x_1)&=0 \end{split} \end{equation*}$

Only one factor from each equation can be zero. For example, let’s choose $y_1-y_2=0$ . Then in the second equation we can’t have $x_1-x_2=0$ . The choice is either $x_2-x_3=0$ or $x_3-x_1=0$ . Whichever one is chosen, we’ll have one side parallel to the $x$ -axis ( $y_1=y_2$ ) and another side parallel to the $y$ -axis ( $x_2=x_3$ or $x_3=x_1$ ).

PROVE that if two sides of a triangle are parallel to the coordinate axes, then $W$ coincides with one of the triangle’s vertices.

This is an application of example 4 above. Alternatively, suppose that $x_2=x_3$ and $y_1=y_3$ . Then, using equation (1) we have:

$x=x_2=x_3,~y=y_1=y_3\implies W:=(x_3,y_3).$

PROVE that if $W$ coincides with one of the vertices of a triangle, then the slopes of the three medians form a geometric progression with common ratio $r=-2$ .

Suppose that $W:=(x_1,y_1)$ . Then we have:

(6) $\begin{equation*} \begin{split} (y_2-y_3)(x_1-x_2)(x_3-x_1)&=0\\ (y_3-y_1)(y_2-y_1)(x_2-x_3)&=0 \end{split} \end{equation*}$

We can’t have $y_2-y_3=0$ from the first equation. Otherwise, none of the factors in the second equation can be zero.

Let $x_1-x_2=0$ from the first equation. Then we can’t have $x_2-x_3=0$ in the second equation, and we can’t also have $y_2-y_1=0$ from the second equation. Only permissible option is $y_3-y_1=0$ . Thus, we have $x_1=x_2$ and $y_1=y_3$ , giving a triangle with two sides parallel to the coordinate axes.

Let $x_3-x_1=0$ from the first equation. Then, in the second equation, we can’t have $x_2-x_3=0$ and we can’t have $y_3-y_1=0$ . The only option remaining is $y_2-y_1=0$ . Thus, we have $x_1=x_3$ and $y_1=y_2$ , again giving a triangle with two sides parallel to the coordinate axes.

The rest of the proof now follows from example 9 here, or example 10 here.

PROVE that if a three-term geometric progression has common ratio $r=-2$ , then it is an arithmetic progression when re-arranged.

We’re doing the easy proofs. Let the first term of the progression be $a$ . Since the common ratio $r$ is $-2$ , enumerate the geometric progression as

$a,-2a,4a$

and then interchange the first and second terms

$-2a,a,4a$

to obtain an arithmetic progression in which the common difference is $d=3a=-\frac{3}{2}(-2a)$ .

PROVE that if a three-term arithmetic progression has a common difference $d=-\frac{3}{2}a$ ( $a$ being the first term), then it is a geometric progression with common ratio $r=-2$ when re-arranged.

Let the first term of the arithmetic progression be $a$ . Since the common difference is $d=-\frac{3}{2}a$ , enumerate the terms as

$a,-\frac{1}{2}a,-2a$

and then re-arrange them as

$-\frac{1}{2}a,a,-2a.$

The latter is a geometric progression with common ratio $r=-2$ .

It remains to show that $(5)\implies (6)\implies (1)$ to complete the equivalence. Quite Easily Done.

Takeaway

For $\triangle ABC$ with vertices $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ , the following statements are equivalent:

$W$ coincides with the foot of the altitude from vertex $A$
side $BC$ is parallel to the $x$ -axis or is parallel to the $y$ -axis
the slopes of the three medians form an arithmetic progression with common difference $d=3(m_{AB}-m_{AC})$ , or the reciprocals of the slopes of the three medians form an arithmetic progression with common difference $d=\left(\frac{1}{m_{AB}}-\frac{1}{m_{AC}}\right)$ .

$W$ o $W$ .

Tasks

In a right triangle, PROVE that $W$ coincides with the circumcenter if and only if the slopes of the legs are $\pm 1$ .
(Likely converse) In $\triangle ABC$ , let $\angle B=90^{\circ}$ . If a point $M$ is such that $AM^2+BM^2+CM^2=\frac{11}{16}AC^2$ , is $M$ the nine-point center of $\triangle ABC$ ?
(Losing count) This exercise relates to a special triangle, representative of all triangles with one side parallel to the -axis and the two other sides having reciprocal slopes. Let , , be the coordinates of the vertices of . Assuming , PROVE that:
- the slopes of sides $AC$ and $BC$ are reciprocals of each other
- the nine-point center of $\triangle ABC$ lies on side $AB$
- the circumcenter of $\triangle ABC$ lies on a line through vertex $C$ and parallel to side $AB$
- the foot of the altitude from vertex $C$ is precisely the midpoint of the line segment joining $C$ to the orthocenter $H$
- the foot of the altitude from vertex $C$ is precisely the point $W$
- the circum-radius $r$ satisfies $r^2=\frac{1}{4}(AC^2+BC^2)$
- the slope of the Euler line cannot be $\pm 2$
- $AM^2+BM^2+CM^2=\frac{11}{16}(AC^2+BC^2)-\frac{1}{4}HC^2$ , where $M$ is the nine-point center (compare with example 2)
- $4AB^2=AC^2+BC^2-HC^2$
- the point $D\left(-\frac{b^2}{a},b\right)$ lies on the circumcircle of $\triangle ABC$
- $DC^2=AC^2+BC^2=AD^2+AC^2$ , where $D$ is the point given above
- $DC=2r$ , so $DC$ is a diameter of the circumcircle of $\triangle ABC$
- $HD^2=HC^2+CD^2$ , where $H$ is the orthocenter of $\triangle ABC$
- the orthic triangle is always isosceles
- the orthic triangle is equilateral, if $a=\pm\sqrt{3}b$
- the parent $\triangle ABC$ is isosceles when $a=\pm\sqrt{3}b$
- the circum-radius equals one of the side-lengths when $a=\pm\sqrt{3}b$ .
PROVE that the following statements are equivalent:
- $W$ coincides with the midpoint of $AC$
- sides $AB$ and $BC$ have opposite slopes
- the area of $\triangle ABC$ is $|(x_2-x_1)(y_2-y_3)|$ or $|(x_2-x_3)(y_2-y_1)|$
Consider in which , , and . If vertex is extended to point on the circumcircle in such a way that is parallel to , PROVE that:
- $CD$ is a diameter of the circumcircle
- $DC^2=DB^2+BC^2=DA^2+AC^2$
- $AC=DB$ (so we also have $DC^2=AC^2+BC^2$ ).