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:

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

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 $\triangle ABC$ .