General – Page 24 – High School Geometry

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 $DC^2=AC^2+BC^2$ ).

Geometric sequences in all triangles

Welcome to our first post of 2021!!!

New points

Marked below are four points $N,P,W,R$ associated with any $\triangle ABC$ :

$RN$ is a diameter of the circumcircle of $\triangle ABC$ , while $WP$ is a diameter of the nine-point circle of $\triangle ABC$ . Notice that the Euler line $HO$ of the parent $\triangle ABC$ turns out to be a median in $\triangle HRN$ , just as are $RP$ and $NW$ .

Part of our goal for $2021$ is a detailed discussion of these four points, although the discussion will not follow a linear fashion. For now, we’ll first consider $W$ — in a bid to deliver on the promise made here last year (see the comment preceding example 5 there).

Let $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ be the vertices of the parent $\triangle ABC$ . Relative to these vertices, $W$ has coordinates:

Observe that $W$ is a (horizontal and vertical) translation of the orthocenter of the parent $\triangle ABC$ . Also, the “common denominator” in (1) above

$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$

is twice the area of $\triangle ABC$ ; as a result,

$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\neq 0,$

and so no $W$ orries about dividing by zero: point $W$ is $W$ ell defined. Al $W$ ays.

Nice properties

Let’s enumerate some of the properties of point $W$ :

it lies on the nine-point circle of the parent $\triangle ABC$
it is a point of concurrency of certain line segments (to be featured in the future)
if a right triangle has legs with slopes $\pm 1$ , then the point $W$ coincides with the circumcenter of the right triangle
if a right triangle has legs parallel to the coordinate axes (slopes $0,\infty$ ), then the point $W$ coincides with the orthocenter of the right triangle
if the end points of any side of a triangle are joined to $W$ , the resulting triangle has slopes in geometric progression (so long as the parent triangle has proper slopes $\implies\neq 0,\infty$ )
if any vertex of a triangle is joined to $W$ and the orthocenter, then the resulting triangle has slopes in geometric progression (so long as the parent triangle has proper slopes $\implies\neq 0,\infty$ )

You get the gist: our favourite from the above list are the last two. At least.

Numerical problems

Our next move today is to emphasize the first and second to last properties listed above through numerous, numerical instances. As such, the examples below are repetitive; please bear with us.

Consider $\triangle ABC$ with vertices at $A(4,5)$ , $B(2,0)$ , and $C(6,2)$ . PROVE that the point $W\left(\frac{7}{2},\frac{5}{4}\right)$ lies on the nine-point circle of $\triangle ABC$ .

Among the points through which the nine-point circle passes are the midpoints of the sides of the parent triangle. In the present case, the midpoints of sides $AB,BC,CA$ are $(3,5/2),(4,1),(5,7/2)$ , respectively. They are the points $X,Y,Z$ shown in the diagram below.

Behind the scenes, we’ve calculated the original triangle’s circumcenter and orthocenter; they are the points $\left(\frac{53}{16},\frac{19}{8}\right)$ and $\left(\frac{43}{8},\frac{9}{4}\right)$ , respectively.

Let $M$ the center of the nine-point circle. It is the midpoint of the circumcenter and orthocenter of the parent triangle, like so:

$\left(\frac{53/16+43/8}{2},\frac{19/8+9/4}{2}\right)=\left(\frac{139}{32},\frac{37}{16}\right).$

Lastly, we show that $MW=MX=MY=MZ$ . By definition we already have $MX=MY=MZ$ .

$\begin{equation*} \begin{split} MW^2&=\left(\frac{139}{32}-\frac{7}{2}\right)^2+\left(\frac{37}{16}-\frac{5}{4}\right)^2\\ &=\left(\frac{27}{32}\right)^2+\left(\frac{34}{32}\right)^2\\ &=\frac{1885}{1024}\\ MX^2&=\left(\frac{139}{32}-3\right)^2+\left(\frac{37}{16}-\frac{5}{2}\right)^2\\ &=\left(\frac{43}{32}\right)^2+\left(\frac{-6}{32}\right)^2\\ &=\frac{1885}{1024}. \end{split} \end{equation*}$

Since $MW=MX=MY=MZ$ , we conclude that $W$ is on the nine-point circle of $\triangle ABC$ .

Calendar fact: In terms of dates, both 1024 and 1885 share something in common. January 14, 1024=January 14, 1885=Wednesday.

Consider $\triangle ABC$ with vertices at $A(4,5)$ , $B(2,0)$ , and $C(6,2)$ . Let $W$ be the point $W\left(\frac{7}{2},\frac{5}{4}\right)$ . PROVE that the slopes of the sides of $\triangle ABW$ , $\triangle BCW$ , and $\triangle CAW$ form geometric progressions (in each case, the geometric mean is the slope of one side of the parent triangle $ABC$ ).

As you’ll recall, the point $W\left(\frac{7}{2},\frac{5}{4}\right)$ is on nine-point cirlce of $\triangle ABC$ .

Observe that the slopes of the sides of $\triangle ABC$ are $5/2,1/2,-3/2$ . They form an arithmetic progression, but not a geometric progression. However, our result holds regardless of any pattern of slopes in the parent triangle.

$W$ ith $W$ given as $W\left(\frac{7}{2},\frac{5}{4}\right)$ and the original vertices $A(4,5)$ , $B(2,0)$ , $C(6,2)$ , we can calculate the slopes of the sides of $\triangle ABW$ , $\triangle BCW$ , and $\triangle CAW$ :

for $\triangle ABW$ , the side-slopes are $\frac{5}{6}$ , $\frac{5}{2}$ , $\frac{15}{2}$ ; they form a geometric progression with common ratio $r=3$
for $\triangle BCW$ , the side-slopes are $\frac{3}{10}$ , $\frac{1}{2}$ , $\frac{5}{6}$ ; they form a geometric progression with common ratio $r=5/3$
for $\triangle CAW$ , the side-slopes are $\frac{3}{10}$ , $-\frac{3}{2}$ , $\frac{15}{2}$ ; they form a geometric progression with $r=-5$

Consider $\triangle ABC$ with vertices at $A(0,5)$ , $B(2,-1)$ , and $C(8,1)$ . PROVE that the point $W\left(\frac{4}{5},\frac{7}{5}\right)$ lies on the nine-point circle of $\triangle ABC.$

The midpoints of the sides of $\triangle ABC$ are $X(1,2)$ , $Y(5,0)$ , and $Z(4,3)$ , as shown below:

Since the parent $\triangle ABC$ is right-angled at $B$ , its orthocenter is $(2,-1)$ and its circumcenter is $(4,3)$ , the midpoint of the hypotenuse. Thus, the center of the nine-point circle is $(3,1):=M$ , say.

We check that $MW=MX$ :

$\begin{equation*} \begin{split} MW^2&=\left(3-\frac{4}{5}\right)^2+\left(1-\frac{7}{5}\right)^2\\ &=\left(11/5\right)^2+\left(-2/5\right)^2\\ &=5\\ MX^2&=(3-1)^2+(1-2)^2\\ &=5 \end{split} \end{equation*}$

Thus, the point $W$ is on the nine-point circle of $\triangle ABC$ .

Consider $\triangle ABC$ with vertices at $A(0,5)$ , $B(2,-1)$ , and $C(8,1)$ . Let $W$ be the point $W\left(\frac{4}{5},\frac{7}{5}\right)$ . PROVE that the slopes of the sides of $\triangle ABW$ , $\triangle BCW$ , and $\triangle CAW$ form geometric progressions.

Here’s a diagram showing $\triangle ABW$ , $\triangle BCW$ , and $\triangle CAW$ :

$W$ ith $W$ given as $W\left(\frac{4}{5},\frac{7}{5}\right)$ and the original vertices $A(0,5)$ , $B(2,-1)$ , $C(8,1)$ , we can calculate the slopes of the sides of $\triangle ABW$ , $\triangle BCW$ , and $\triangle CAW$ :

for $\triangle ABW$ , the side-slopes are $-2$ , $-3$ , $-\frac{9}{2}$ ; they form a geometric progression with common ratio $r=3/2$
for $\triangle BCW$ , the side-slopes are $-\frac{1}{18}$ , $\frac{1}{3}$ , $-2$ ; they form a geometric progression with common ratio $r=-6$
for $\triangle CAW$ , the side-slopes are $-\frac{1}{18}$ , $-\frac{1}{2}$ , $-\frac{9}{2}$ ; they form a geometric progression with $r=9$ .

Consider $\triangle ABC$ with vertices at $A(1,5)$ , $B(-3,-3)$ , and $C(7,3)$ . PROVE that the point $W\left(-\frac{11}{7},\frac{15}{7}\right)$ lies on the nine-point circle of $\triangle ABC.$

Behind the scenes, we’ve calculated the parent triangle’s circumcenter and orthocenter; they’re the points $\left(\frac{17}{7},-\frac{5}{7}\right)$ and $\left(\frac{1}{7},\frac{45}{7}\right)$ , respectively. Thus, the center of the nine-point circle is $\left(\frac{9}{7},\frac{20}{7}\right):=M$ , say.

The nine-point circle passes through the midpoints $X(-1,1)$ , $Y(2,0)$ , $Z(4,4)$ of the parent triangle.

We check that $MW=MX$ :

$\begin{equation*} \begin{split} MW^2&=\left(\frac{9}{7}+\frac{11}{7}\right)^2+\left(\frac{20}{7}-\frac{15}{7}\right)^2\\ &=\frac{425}{49}\\ MX^2&=\left(\frac{9}{7}+1\right)^2+\left(\frac{20}{7}-1\right)^2\\ &=\frac{425}{49}. \end{split} \end{equation*}$

Consider $\triangle ABC$ with vertices at $A(1,5)$ , $B(-3,-3)$ , and $C(7,3)$ . Let $W$ be the point $W\left(-\frac{11}{7},\frac{15}{7}\right)$ . PROVE that the slopes of the sides of $\triangle ABW$ , $\triangle BCW$ , and $\triangle CAW$ form geometric progressions.

The slopes of the sides of $\triangle ABW$ are $\frac{10}{9},2,\frac{18}{5};$ they form a geometric progression with common ratio $r=\frac{9}{5}$ .
The slopes of the sides of $\triangle BCW$ are $\frac{1}{10},\frac{3}{5},\frac{18}{5};$ they form a geometric progression with common ratio $r=6$ .
The slopes of the sides of $\triangle CAW$ are $\frac{1}{10},-\frac{1}{3},\frac{10}{9};$ they form a geometric progression with common ratio $r=-\frac{10}{3}$ .

Consider $\triangle ABC$ with vertices at $A(1,3)$ , $B(0,0)$ , and $C(2,4)$ . PROVE that the point $W\left(-1,6\right)$ lies on the nine-point circle of $\triangle ABC$ .

The midpoints of sides $AB,BC,CA$ ; they are $(1/2,3/2),(1,2),(3/2,7/2)$ , respectively. They are the points $X,Y,Z$ shown in the diagram below.

Behind the scenes, we’ve calculated the original triangle’s circumcenter and orthocenter; they are the points $\left(5,0\right)$ and $\left(-7,7\right)$ , respectively. It follows that the center of the nine-point circle in this case is the point $(-1,3.5)$ , denoted $M$ in the diagram above.

Now we check that $MW=MX=MY=MZ$ . Again, it suffices to check $MW=MX$ :

$\begin{equation*} \begin{split} MW^2&=(-1--1)^2+(6-3.5)^2\\ &=6.25\\ MX^2&=(0.5--1)^2+(1.5-3.5)^2\\ &=6.25 \end{split} \end{equation*}$

Thus, $W$ lies on the nine-point circle of $\triangle ABC$ .

Consider $\triangle ABC$ with vertices at $A(1,3)$ , $B(0,0)$ , and $C(2,4)$ . Let $W$ be the point $W\left(-1,6\right)$ . PROVE that the slopes of the sides of $\triangle ABW$ , $\triangle ACW$ , and $\triangle BCW$ form geometric progressions.

As you’ll recall, the point $W\left(-1,6\right)$ is on the nine-point circle of the parent triangle $ABC$ .

Observe that the slopes of the sides of $\triangle ABC$ are $1,2,3$ . They form an arithmetic progression, but not a geometric progression.

The slopes of the sides of $\triangle ABW$ are $-\frac{3}{2},3,-6;$ they form a geometric progression with common ratio $r=-2$ (remember the name: a peach).
The slopes of the sides of $\triangle BCW$ are $-\frac{2}{3},2,-6;$ they form a geometric progression with common ratio $r=-3$ .
The slopes of the sides of $\triangle CAW$ are $-\frac{2}{3},1,-\frac{3}{2};$ they form a geometric progression with common ratio $r=-\frac{3}{2}$ .

Consider $\triangle ABC$ with vertices at $A(0,0)$ , $B(3,12)$ , and $C(6,6)$ . Find coordinates for a point $W$ for which the slopes of the sides of $\triangle ABW$ , $\triangle BCW$ , and $\triangle CAW$ form three separate geometric progressions.

Use equation (1) with $(x_1,y_1)=(0,0)$ , $(x_2,y_2)=(3,12)$ , and $(x_3,y_3)=(6,6)$ :

$\begin{equation*} \begin{split} x&=\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{0+3(12)(0-6)+6(6)(3-0)}{0+3(6-0)+6(0-12)}\\ &=\frac{-108}{-54}\\ &=2\\ y&=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{-216}{-54}\\ &=4 \end{split} \end{equation*}$

The desired point is $W(2,4)$ .

$W$ ith the point $W(2,4)$ and the original coordinates $A(0,0)$ , $B(3,12)$ , $C(6,6)$ , the slopes of the sides of each of $\triangle ABW$ , $\triangle ACW$ , and $\triangle BCW$ can be calculated:

the slopes of the sides of $\triangle ABW$ are $2,4,8$ ; they form a geometric progression with common ratio $r=2$
the slopes of the sides of $\triangle ACW$ are $\frac{1}{2},-2,8$ ; they form a geometric progression with common ratio $r=-4$
the slopes of the sides of $\triangle BCW$ are $\frac{1}{2},1,2$ ; they form a geometric progression with common ratio $r=2$ .

There’s an additional point $W$ orth noting here. If one calculates the slopes from $W(2,4)$ to each of the vertices $A(0,0)$ , $B(3,12)$ , $C(6,6)$ , one obtains

$\frac{1}{2},2,8$

a geometric progression with common ratio $r=4$ . This only happens in some specific situations (see the exercises).

Given $\triangle ABC$ with vertices at $A(0,0)$ , $B(3,12)$ , $C(6,6)$ , verify that the point $W(2,4)$ found in the previous example lies on the nine point circle of $\triangle ABC$ .

The midpoints of the sides of $\triangle ABC$ are $X(1.5,6)$ , $Y(4.5,9)$ , $Z(3,3)$ ; they’re shown (in the big circle) below as red bullet points:

Behind the scenes, we’ve calculated the circumcenter and the orthocenter of $\triangle ABC$ ; they are the points $O\left(-\frac{1}{2},\frac{13}{2}\right)$ and $H(10,5)$ , respectively. Thus, the center of the nine-point circle is $\left(\frac{19}{4},\frac{23}{4}\right):=M$ , say.

We now check that $MX=MW$ :

$\begin{equation*} \begin{split} MX^2&=\left(\frac{19}{4}-\frac{3}{2}\right)^2+\left(\frac{23}{4}-6\right)^2\\ &=\frac{170}{16}\\ MW^2&=\left(\frac{19}{4}-2\right)^2+\left(\frac{23}{4}-4\right)^2\\ &=\frac{170}{16}. \end{split} \end{equation*}$

Boom.

Takeaway

Triangles with slopes in geometric progressions are very nice objects, at least because of the “algebra” that their theory brings along. However, even if the slopes of the sides of a triangle do not form a geometric progression (sighs), there’s a $W$ ay around it:

if the parent triangle has a side parallel to any of the axes, relax — already done last March.
if the parent triangle $ABC$ has no side parallel to the coordinate axes, call $W$ with coordinates

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

and obtain triangles $ABW$ , $ACW$ , $BCW$ whose side-slopes form three separate geometric progressions. (Alternatively, one can also use $W$ and the orthocenter $H$ . Then the slopes of the sides of triangles $AWH$ , $BWH$ , $CWH$ form three separate geometric progressions.)

Tasks

Throughout, any mention of $W$ refers to the point with coordinates given by equation (1) or (2). Unless otherwise specified, assume where necessary that an arbitrary triangle $ABC$ has vertices at $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_3)$ .

(Separate character) Verify that the point $W$ is not any of the nine traditional points through which the nine-point circle passes. Under normal circumstances.
(Special case) Let $\triangle ABC$ be such that $\angle B=90^{\circ}$ . If the slopes of the legs $AB$ and $BC$ are $\pm 1$ , PROVE that $W$ coincides with the mid-point of $AC$ (that is, the circumcenter of $\triangle ABC$ . Also see exercise 8 below).
(Special case) If $y_2=y_3$ , PROVE that $W$ is precisely the foot of the altitude from vertex $A$ .
(Shortcut) In , let be the slopes of sides . PROVE that:
- $W$ has coordinates $\left(x_3-\frac{y_2-y_3}{ar(r^2-1)},\frac{r^2y_2-y_3}{r^2-1}\right).$
- the slopes from $W$ to the vertices form a geometric progression $-\frac{a}{r},~-ar,-ar^2$
  (The best setting for $W$ is $W$ hen the slopes of the sides of the parent triangle form a geometric progression.)
(Shortcut) In , let be the slopes of sides , where . PROVE that:
- $W$ has coordinates $\left(x_3-\frac{a(y_2-y_3)}{2d(a+d)},y_2+\frac{a(y_2-y_3)}{2d}\right).$
- the slopes from $W$ to the vertices are $-\frac{a(a+2d)}{a+d},-\frac{a(a+d)}{a+2d},-\frac{(a+d)(a+2d)}{a}$
- the slopes from $W$ to the vertices form a geometric progression if $d=-\frac{3a}{4}$ or $d=-\frac{3a}{2}$ or $d=(-2\pm\sqrt{2})a$
  (In each case, the common ratio is always positive — in fact, a “square”.)
- the slopes from $W$ to the vertices form an arithmetic progression if $d=\left(\frac{-3\pm\sqrt{3}}{2}\right)a$ .
  (Incidentally, the common difference of the resulting arithmetic progression is equal to $\pm d$ , depending on how it’s arranged.)
(Six colleagues) Let be the orthocenter of and let be as given in equation (1). PROVE that the six statements below are equivalent:
- $W=H$
- $W$ coincides with one of the vertices of $\triangle ABC$
- $\triangle ABC$ has two sides parallel to the coordinate axes
- 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 $d=-\frac{3}{2}a$ ( $a:=$ first term)
- one side is parallel to the $x$ -axis and another side and the median to it have opposite slopes (e.g. $m_{AB}=0$ and $m_{B}=-m_{CA}$ ).
(HOW come?) In , let the orthocenter and circumcenter be and , respectively, and let be as given in (1). If the slopes of sides are in that order, PROVE that:
- the circumcenter shares the same $y$ -coordinate as vertex $C$
- the slopes of the sides of $\triangle HOW$ form an arithmetic progression
  (One of the reasons we chose $W$ as a label for one of our four points $N,P,W,R$ — rather than $N,P,Q,R$ maybe — was because we had this triangle $HOW$ in view.)
Are the following statements equivalent?
- $W$ is the midpoint of $CA$ ; that is, $W:=\left(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\right)$
- either $\triangle ABC$ is isosceles with side $CA$ parallel to the $x$ -axis and $AB=BC$ , or two sides of $\triangle ABC$ have opposite slopes
- the area of $\triangle ABC$ is $|(x_2-x_3)(y_2-y_1)|$ or $|(x_2-x_1)(y_2-y_3)|$ .
  (This exercise shows that in some situations, $W$ behaves like the circumcenter of a right triangle which occurs at the midpoint of the hypotenuse.)
Consider a right triangle in which none of the sides is parallel to the coordinate axes. PROVE that the slope from $W$ to the $90^{\circ}$ vertex is the reciprocal of the slope of the hypotenuse.
PROVE that $W$ is a both a horizontal and a vertical translation of the orthocenter $H$ of the parent $\triangle ABC$ .