General – Page 22 – High School Geometry

A new point on the nine-point circle III

Look at the diagram above, where a number of things appear to be happening. The red circle that goes through points $H,U,V,X$ is our focus for now. The center $W$ of this circle is itself a point on the nine-point circle of the parent $\triangle ABC$ . We’ll examine more of the properties of $W$ in this post, especially when the side-slopes of $\triangle ABC$ form a geometric progression.

Simplified coordinates

If $\triangle ABC$ has vertices at $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ , then the coordinates of point $W$ are given by:

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

Compare (1) with its equivalent, simplified version:

(2) $\begin{equation*} x=\frac{m_2x_2-m_3x_1}{m_2-m_3},~y=\frac{m_2y_1-m_3y_2}{m_2-m_3} \end{equation*}$

Here, $m_1,m_2,m_3$ are the slopes of sides $AB,BC,CA$ , respectively. Notice that $m_1$ and the coordinates of $C(x_3,y_3)$ are explicitly “missing” in equation (2). One implication is that the calculation of $W$ requires only two sides and the slopes of those two sides. So (2) can take another form:

(3) $\begin{equation*} x=\frac{m_1x_1-m_2x_3}{m_1-m_2},~y=\frac{m_1y_3-m_2y_1}{m_1-m_2} \end{equation*}$

Or this other form:

(4) $\begin{equation*} x=\frac{m_3x_3-m_1x_2}{m_3-m_1},~y=\frac{m_3y_2-m_1y_3}{m_3-m_1} \end{equation*}$

Ample freedom. Simple concept.

Find the coordinates of point $W$

for $\triangle ABC$

with vertices located at $A(1,4)$

, $B(0,0)$

, $C(3,6)$

Set $(x_1,y_1)=(1,4)$ , $(x_2,y_2)=(0,0)$ , and $(x_3,y_3)=(3,6)$ . The slopes of sides $AB,BC$ , and $CA$ are then $m_1=4$ , $m_2=2$ , and $m_3=1$ , respectively.

Using equation (2):

$\begin{equation*} \begin{split} x&=\frac{m_2x_2-m_3x_1}{m_2-m_3},~y=\frac{m_2y_1-m_3y_2}{m_2-m_3}\\ &=\frac{2(0)-1(1)}{2-1}=-1,~y=\frac{2(4)-1(0)}{2-1}=8 \end{split} \end{equation*}$

Using equation (3):

$\begin{equation*} \begin{split} x&=\frac{m_1x_1-m_2x_3}{m_1-m_2},~y=\frac{m_1y_3-m_2y_1}{m_1-m_2}\\ &=\frac{4(1)-2(3)}{4-2}=-1,~y=\frac{4(6)-2(4)}{4-2}=8 \end{split} \end{equation*}$

Using equation (4):

$\begin{equation*} \begin{split} x&=\frac{m_3x_3-m_1x_2}{m_3-m_1},~y=\frac{m_3y_2-m_1y_3}{m_3-m_1}\\ &=\frac{4(0)-1(3)}{4-1}=-1,~y=\frac{4(6)-1(0)}{4-1}=8 \end{split} \end{equation*}$

In each case we obtain $W(-1,8)$ . It’s that point that $W$ anders here and there on the nine-point circle.

If $\triangle ABC$

has one side parallel to the $x$ -axis, PROVE that $W$

coincides with the foot of an altitude.

The converse is also true. Fix the vertices at $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ . Let the slopes of $AB,BC,CA$ be $m_1,m_2,m_3$ and suppose that $m_2=0$ . Using equation (2):

$\begin{equation*} \begin{split} x&=\frac{m_2x_2-m_3x_1}{m_2-m_3},~y=\frac{m_2y_1-m_3y_2}{m_2-m_3}\\ &=\frac{0-m_3x_1}{0-m_3}=x_1,~y=\frac{0-m_3y_2}{0-m_3}=y_2 \end{split} \end{equation*}$

Thus, $W$ is the point $(x_1,y_2=y_3)$ , which is the foot of the altitude from vertex $A$ .

If $\triangle ABC$

has two sides with opposite slopes, PROVE that $W$

is the midpoint of the third side.

The converse is also true. Fix the vertices at $A(x_1,y_1)$ , $B(x_2,y_2)$ , $C(x_3,y_3)$ . Let the slopes of $AB,BC,CA$ be $m_1,m_2,m_3$ and suppose that $m_2=-m_3$ . Using equation (2):

$\begin{equation*} \begin{split} x&=\frac{m_2x_2-m_3x_1}{m_2-m_3}\\ &=\frac{m_2x_2+m_2x_1}{m_2+m_2}\\ &=\frac{x_2+x_1}{2}\\ y&=\frac{m_2y_1-m_3y_2}{m_2-m_3}\\ &=\frac{m_2y_1+m_2y_2}{m_2+m_2}\\ &=\frac{y_1+y_2}{2} \end{split} \end{equation*}$

Thus, $W$ is the point $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ , which is the midpoint of $AB$ .

If the slopes of the legs of a right triangle are $\pm 1$

, PROVE that $W$

coincides with the circumcenter of the triangle.

This follows from example 3, since $\pm 1$ are opposites.

Special case

Most of our points behave extremely well if the side-slopes of the parent triangle form a geometric progression. Expectedly, our favorite point $W$ is not an exception.

If the side-slopes of $\triangle ABC$

form a geometric progression with common ratio $r$

, PROVE that the slopes of the line segments $BW,AW,CW$

form a geometric progression with common ratio $r^2$

Let the slopes of sides $AB,BC,CA$ be $m_1=a$ , $m_2=ar$ , $m_3=ar^2$ . From this post, the slopes from $W$ to vertices $A,B,C$ are:

(5) $\begin{equation*} \begin{split} m_{WA}&=\frac{m_1m_3}{-m_2},~m_{WB}=\frac{m_1m_2}{-m_3},~m_{WC}=\frac{m_2m_3}{-m_1}\\ m_{WA}&=\frac{a\times ar^2}{-ar}\\ &=-ar\\ &\cdots\\ m_{WB}&=\frac{a\times ar}{-ar^2}\\ &=-\frac{a}{r}\\ &\cdots\\ m_{WC}&=\frac{ar\times ar^2}{-a}\\ &=-ar^3 \end{split} \end{equation*}$

Thus, $m_{WB},m_{WA},m_{WC}$ correspond to the geometric progression $-\frac{a}{r},-ar,-ar^3$ with common ratio $r^2$ .

If the side-slopes of $AB,BC,CA$

form a geometric progression $a,ar,ar^2$

in that order, PROVE that $W$

lies on the median through vertex $A$

(hence is co-linear with the centroid, the midpoint of $BC$

, and vertex $A$

From the preceding example, the slope from $W$ to vertex $A$ is $-ar$ . From this post, the slope from vertex $A$ to the midpoint of side $BC$ is also $-ar$ . Thus, $W$ lies on the median through vertex $A$ .

If the slopes of the sides $AB,BC,CA$

of $\triangle ABC$

form a geometric progression $a,ar,ar^2$

in that order, PROVE that the area of $\triangle ABW$

is equal to the area of $\triangle ACW$

As noted in example 6 above, the point $W$ is on the median through vertex $A$ . Suppose that $W$ lies outside the triangle (usually when $r> 0$ ). Let $M$ be the midpoint of side $BC$ . Then the area of $\triangle BMW$ is equal to the area of $MCW$ (since $WM$ is a median in $\triangle WBC$ ). But then the area of $\triangle ABM$ is also equal to the area of $\triangle AMC$ (since $AM$ is a median in $\triangle ABC$ ). Together we have that the area of $\triangle ABW$ is equal to the area of $\triangle ACW$ . Same conclusion when $W$ is inside the parent triangle.

In the diagram above, point $W$ is inside the given triangle. The areas of triangles $ABW$ and $ACW$ are equal.

Sample calculations

Since $W$ lies on the median through vertex $A$ , it is natural to ask what ratio both it and the centroid $G$ divide the entire median length $AM$ .

(6) $\begin{equation*} \frac{GW}{AW}=\frac{r^2+r+1}{3r},~\frac{MW}{GW}=\frac{3(r^2+1)}{2(r^2+r+1)}},~\frac{AM}{AW}=\frac{(r-1)^2}{2r} \end{equation*}$

As you already know, the relations in (6) hold when the slopes of the parent triangle form a geometric progression. Absolute values may be necessary in some cases.

Let $\triangle ABC$

be equilateral with sides $AB,BC,CA$

having slopes $a,ar,ar^2$

. PROVE that $W$

is midway between vertex $A$

and centroid $G$

The common ratio $r$ of the slopes $a,ar,ar^2$ of an equilateral triangle satisfies the quadratic equation

(7) $\begin{equation*} r^2+4r+1=0 \end{equation*}$

Use equation (7) and the first ratio in equation (6) to obtain

$\frac{GW}{AW}=\left|\frac{r^2+r+1}{3r}\right |=\left|\frac{-3r}{3r}\right |=1$

Thus, $AW=GW$ and so $W$ is the midpoint of $A$ and $G$ . In addition, since the centroid divides median $AM$ in the ratio $1:2$ , it follows that $AW=WG=GM$ in this case.

Let $\triangle ABC$

be right isosceles with sides $AB,BC,CA$

having slopes $a,ar,ar^2$

. PROVE that $3GW=2AW$

Use the first ratio in equation (6) and the fact that the common ratio of the slopes $a,ar,ar^2$ of a right isosceles triangle satisfies the quadratic equation

(8) $\begin{equation*} r^2+3r+1=0 \end{equation*}$

and obtain

$\frac{GW}{AW}=\left|\frac{r^2+r+1}{3r}\right |=\left|\frac{-2r}{3r}\right |=\frac{2}{3}$

Thus, $2AW=3GW$ .

Suppose that the slopes of sides $AB,BC,CA$

form a geometric progression $a,ar,ar^2$

. If $r> 0$

, PROVE that the point $W$

lies outside the triangle.

We have:

$\frac{AM}{MW}=\frac{(r-1)^2}{r^2+1}< 1~~\textrm{when}~~r> 0$

Thus, the line segment $MW$ is longer than the median $AM$ . (If $r< 0$ , we can’t conclude that $W$ lies inside the triangle.)

Takeaway

In $\triangle ABC$ , let $G$ be the centroid, $M$ the midpoint of side $BC$ , and $W$ as given by equation (1). If the slopes of sides $AB,BC,CA$ form a geometric progression $a,ar,ar^2,$ then the following statements are equivalent:

$q=-19$ , where $q$ is the sum of quotients of slopes defined here
$r^2+3r+1=0$ or $r^2+7r+1=0$
$AB^2+CA^2=3BC^2$ or $AB^2+CA^2=\frac{7}{5}BC^2$
$m_{AB}^2+m_{CA}^2=\frac{7}{5}m_{BC}^2$ or $m_{AB}^2+m_{CA}^2=3m_{BC}^2$
$r=-\phi^2$ or $r=-\phi^4$ or $r=-\frac{1}{\phi^2}$ or $r=-\frac{1}{\phi^4}$ , where $\phi$ is the golden ratio
$\frac{AW}{GW}=\frac{3}{2}$ or $\frac{AW}{GW}=2$
$\frac{AM}{AW}=\frac{5}{2}$ or $\frac{AM}{AW}=\frac{9}{2}$
$\frac{AM}{WM}=\frac{5}{3}$ or $\frac{AM}{WM}=3$ .

Eight easy equivalences. A triangle that satisfies any of these (together with $ar^2=-1$ or $a^2r^3=-1$ ) will be a right isosceles triangle (or its “look-alike”).

Tasks

Let be such that is parallel to the -axis, and and have reciprocal slopes. PROVE that:
- the orthocenter $H$ is a reflection of vertex $C$ across the $x$ -axis
- $W$ is the foot of the altitude from vertex $C$
- $W$ is the midpoint of $C$ and $H$
Let be the slopes of the line segments from to the vertices of , and let be the side-slopes.
- PROVE that $m_{WA}\times m_{WB}\times m_{WC}=-\left(m_{AB}\times m_{BC}\times m_{CA}\right)$ .
- If $\triangle ABC$ is right-angled, deduce that the product $m_{WA}\times m_{WB}\times m_{WC}$ equals the slope of the hypotenuse.
PROVE that $BW^2+CW^2=(r^2+r^{-2})AW^2$ for any $\triangle ABC$ with side-slopes $a,ar,ar^2$ .
Find coordinates for the vertices of $\triangle ABC$ having the property that a point $W$ on its nine-point circle is twice as far from vertex $A$ as it is from the centroid $G$ .
Suppose that the slopes of sides $AB,BC,CA$ are $a,ar,ar^2$ , in that order. Let $W$ be as given in equation (1), and let $M$ be the midpoint of $BC$ . PROVE that $\frac{AW}{MW}=\frac{2r}{r^2+1}$ .

Concyclic with the orthocenter I

We illustrate how basic alterations to the classic altitude definition result in concurrent lines, in such a way that the points of concurrency are conclyclic with the orthocenter. Similar constructions applied to the medial triangle yield points that are concyclic with the circumcenter.

Extra points

Our intention at the beginning of the year was to focus most of our attention on four points $N,P,W,R$ associated with any triangle, depicted below:

By March, much has changed, and we’re grateful to be able to add seven more points, bringing the total, as at this paragraph, to eleven:

Actually, we have fourteen more points, so that the total, as at the time of this writing, is twenty five. However, these other fourteen points behave totally differently from the above mentioned eleven.

Extended period

Following the addition of twenty one extra points, our initial one-year time frame has been affected. Please help us do the math: “If it takes one year to discuss four points, how long will it take to discuss twenty five points, assuming it takes the same length of time for each point”?

Exchanging pleasantries

Welcome to our theory!

Extensive properties

With reference to the preceding diagram, we have the following individual and collective properties of the points:

each point is a point of concurrency of some lines (in this post we show how to obtain $W$ ; in the future, we’ll feature how to obtain the rest)
$R$ and $V$ are on the circumcircle of the parent $\triangle ABC$ (in fact, $RV$ is a diameter)
$S$ and $W$ are on the nine-point circle of the parent $\triangle ABC$ (in fact, $SW$ is a diameter)
the line segments $KS,OL,UH,VX$ always make an angle of $45^{\circ}$ with the positive $x$ -axis
the line segments $KO,SL,UV,HX$ always make an angle of $135^{\circ}$ with the positive $x$ -axis
quadrilaterals $SKOL$ and $HUVX$ are always rectangles (follows from the previous two)
the two rectangles $SKOL$ and $HUVX$ are “parallel”
the area of the bigger rectangle $HUVX$ is four times the area of the smaller rectangle $SKOL$
$W$ is the center of the cyclic quadrilateral $HUVX$
$T$ is the center of the cyclic quadrilateral $SKOL$
rectangles $SKOL$ and $HUVX$ can become squares in special cases (in particular, the extremely pleasant case of a triangle having two sides with reciprocal slopes and one side parallel to the $x$ -axis, and in the expectedly peculiar case of an equilateral triangle)
if points $R,H,V$ are connnected to form $\triangle RHV$ , then the Euler line $HO$ is just a median in this triangle
if the legs of a right triangle are parallel to the coordinate axes, then rectangle $SKOL$ becomes just the circumcenter and rectangle $HUVX$ becomes the orthocenter (this is a degenerate case, but it does generate additional characterizations of this particular right triangle)
if the slopes of the sides of the parent $\triangle ABC$ form a geometric progression, then rectangles $SKOL$ and $HUVX$ “sit” on the same line (the case of geometric progressions is the exquisite platform for the present concept)
our favourite point is the wandering point $W$ (this isn’t a property though, but there’s a basis for our bias)
And lots more! (this isn’t a property either, but as we explore these points in future posts, more will be shown).

essentially peculiar

Normal simplifications in the case of equilateral triangles. Squares are formed.

extremely pleasant

Nice setting: triangles with slopes of the form $0,m,\frac{1}{m}$

. Squares are also formed as in the case of equilateral triangles.

exquisite platform

Naturally superior to others. Nicest setting is when the slopes form a geometric progression.

equal points

No shape is formed, because the points coincide. Still, the triangle more than makes up for it with several characterizations.

elongated path

Nearly similar to the previous example. The rectangles are stretched out and degenerate to a straight line.

edge partners

Neighbours sharing edges. A special case of a right triangle with legs’ slopes $\pm 1$

, leading to several points coinciding.

Elementary proofs

Using an idea from our last post, we’ll now define the three lines that have point $W$ as a point of concurrency.

Let $A(x_1,y_1)$

, $B(x_2,y_2)$

, $C(x_3,y_3)$

be the coordinates of the vertices of $\triangle ABC$

, and let $m_1$

, $m_2$

, $m_3$

be the slopes of sides $AB,BC,CA$

in that order.

Define lines from $A,B,C$ as follows: the line from $A$ has slope $\frac{m_1m_3}{-m_2}$ , the line from $B$ has slope $\frac{m_1m_2}{-m_3}$ , and the line from $C$ has slope $\frac{m_2m_3}{-m_1}$ .

PROVE that the three lines defined above are concurrent.

Consider two such lines through $A$ and $B$ .

(1) $\begin{equation*} \begin{split} y-y_1&=\frac{m_1m_3}{-m_2}(x-x_1)\\ y-y_2&=\frac{m_1m_2}{-m_3}(x-x_2)\\ \end{split} \end{equation*}$

The solution to the linear system (1) is:

$x=\frac{m_2x_2-m_3x_1}{m_2-m_3},~y=\frac{m_2y_1-m_3y_2}{m_2-m_3}.$

Let’s verify concurrency. We’ll prove that the solution above satisfies the equation of the line through $C$ :

(2) $\begin{equation*} y-y_3=\frac{m_2m_3}{-m_1}(x-x_3) \end{equation*}$

Note that

$m_1=\frac{y_1-y_2}{x_1-x_2},m_2=\frac{y_2-y_3}{x_2-x_3},m_3=\frac{y_3-y_1}{x_3-x_1}.$

Consider the left member of equation (2):

$\begin{equation*} \begin{split} y-y_3&=\left(\frac{m_2y_1-m_3y_2}{m_2-m_3}\right)-y_3\\ &=\frac{m_2m_3}{m_2-m_3}(x_1-x_2)\\ \end{split} \end{equation*}$

Now consider the right member of equation (2):

$\begin{equation*} \begin{split} \frac{m_2m_3}{-m_1}(x-x_3)&=\frac{m_2m_3}{-m_1}\left(\frac{m_2x_2-m_3x_1}{m_2-m_3}-x_3\right)\\ &=\frac{m_2m_3}{m_2-m_3}\left(\frac{y_2-y_1}{m_2-m_3}\right)\\ &=\frac{m_2m_3}{m_2-m_3}\frac{y_1-y_2}{m_1}\\ &=\frac{m_2m_3}{m_2-m_3}(x_1-x_2) \end{split} \end{equation*}$

Admittedly, we skipped a couple of steps, probably because we didn’t want to focus on the details of the algebra, but instead on how the lines were defined. Now go ahead and alter our definition to obtain other points of concurrency.

Given $\triangle ABC$ with vertices at $A(1,4)$ , $B(0,0)$ , $C(3,6)$ , the point $W(-1,8)$ is on its nine-point circle. Find the equations of the three lines through $A,B,C$ that intersect at $W$ .

The slopes of $AB, BC,CA$ are $4,2,1$ . Set

$m_1=4,m_2=2,m_3=1$

and define lines through $A,B,C$ with slopes $-2,-8,-\frac{1}{2}$ , respectively:

$\begin{equation*} \begin{split} \textrm{through}~A:~y-4&=-2(x-1)\implies y=-2x+6\\ \textrm{through}~B:~y-0&=-8(x-0)\implies y=-8x\\ \textrm{through}~C:~y-6&=-\frac{1}{2}(x-3)\implies y=-\frac{1}{2}x+\frac{15}{2}\\ \end{split} \end{equation*}$

These three equations are all satisfied by $x=-1,~y=8$ .

Given $\triangle ABC$ with vertices at $A(1,4)$ , $B(0,0)$ , $C(3,6)$ , define lines through the midpoints of $AB,BC,CA$ in such a way that the slopes are negatives of the slopes of the sides. PROVE that the three lines meet at the point $W(-1,8)$ .

This is actually an interesting property. Basically, we can obtain the point $W(-1,8)$ in two different ways:

through each vertex, draw lines with slopes of the form $\frac{m_im_j}{-m_k}$ , where $m_k$ is the slope of the side opposite the reference vertex
through each midpoint, draw lines with slopes that are negatives of the slopes of the sides on which the midpoints lie.

In the present case, the slopes of $AB, BC,CA$ are $4,2,1$ , and the midpoints are $\left(\frac{1}{2},2\right)$ , $\left(\frac{3}{2},3\right)$ , and $(2,5)$ , respectively. Through each of these midpoints we draw lines with slopes $-4,-2,-1$ :

$\begin{equation*} \begin{split} \textrm{through the midpoint of}~AB:~y-2&=-4\left(x-\frac{1}{2}\right)\implies \scriptstyle y=-4x+4\\ \textrm{through the midpoint of}~BC:~y-3&=-2\left(x-\frac{3}{2}\right)\implies \scriptstyle y=-2x+6\\ \textrm{through the midpoint of}~CA:~y-5&=-1(x-2)\implies \scriptstyle y=-x+7\\ \end{split} \end{equation*}$

These three equations are all satisfied by $x=-1,~y=8$ .

The point $W$ on the nine-point circle of $\triangle ABC$ .

Eventful pi

Our final example is a big digression from the preceding discussion. For your pi day celebration.

Let $a,b,c$ be the side-lengths, and let $N$ be the nine-point center, of non-right $\triangle ABC$ . If $a=b$ and $AN^2+BN^2+CN^2=\frac{11}{16}c^2$ , PROVE that $\left(\frac{c}{a}\right)^2=\frac{22}{7}$ .

And so a different version of “ $\pi$ ” for your pi day celebration.

Let $R$ be the circumradius. Since

$AN^2+BN^2+CN^2=\frac{1}{4}(3R^2+a^2+b^2+c^2)\quad\textrm{and}\quad R=\frac{c}{2\sin C},$

we have $\frac{11}{16}c^2=\frac{1}{4}(3R^2+a^2+b^2+c^2).$ Further simplifications:

$\begin{equation*} \begin{split} 11c^2&=\scriptstyle 12R^2+4a^2+4b^2+4c^2\\ 7c^2-4a^2-4b^2&=12\left(\frac{c}{2\sin C}\right)^2\\ \sin^2C\left(7c^2-4a^2-4b^2\right)&=3c^2\\ (1-\cos^2 C)\Big(7(a^2+b^2-2ab\cos C)-4a^2-4b^2\Big)&=\scriptstyle 3(a^2+b^2-2ab\cos C)\\ \cos C\Big(14ab\cos^2 C-(3a^2+3b^2)\cos C-8ab\Big)&=0\\ 0,\frac{(3a^2+3b^2)\pm\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab}&=\cos C\\ \end{split} \end{equation*}$

Since $\triangle ABC$ is not a right triangle (right triangles automatically satisfy the relation $AN^2+BN^2+CN^2=\frac{11}{16}c^2$ if $c$ is taken as the hypotenuse), we ignore the $\cos C=0$ value. Then put $a=b$ and consider the negative discriminant (the positive part would give $\cos C\geq 1$ , unacceptable):

$\begin{equation*} \begin{split} \cos C&=\frac{(3a^2+3b^2)-\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab}\\ &=\frac{6a^2-\sqrt{484a^2}}{28a^2}\\ \cos C&=-\frac{4}{7} \end{split} \end{equation*}$

By the cosine law again:

$\begin{equation*} \begin{split} c^2&=a^2+b^2-2ab\cos C\\ &=2a^2-2a^2\left(-\frac{4}{7}\right)\\ \therefore c^2&=\frac{22}{7}a^2\\ \left(\frac{c}{a}\right)^2&=\frac{22}{7}. \end{split} \end{equation*}$

There $\pi$ goes.

If we didn’t require $a=b$

above, we get some other interesting possibilities. For example, if we want $\cos C=-\frac{1}{2}$

, then $\frac{a}{b}=\phi^2$

or $\left(-\frac{1}{\phi}\right)^2$

, where $\phi$

is the golden ratio.

Takeaway

The following statements are equivalent for any $\triangle ABC$ :

$W=H$ , 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.

You’re already familiar with the above six equivalent statements. By the time we finish discussing our new points, the number of characterizations will be about twenty, if not more. Be looking forward to that time.

Tasks

Given with vertices , , , and as the slopes of sides , respectively, PROVE that:
- $\frac{m_2x_2-m_3x_1}{m_2-m_3}=\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}$
- $\frac{m_2y_1-m_3y_2}{m_2-m_3}=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}$
PROVE that the line segments from $W$ to the three vertices and the three midpoints either all make acute angles with the positive $x$ -axis, or all make obtuse angles with the positive $x$ -axis.
Let , and let . PROVE that:
- $\cos C^{+}\geq 1$ (hence there’s no solution except $C^{+}=0$ )
- $-1\leq \cos C^{-}\leq 1$ .
Let as before.
- If $\cos C^{-}=-\frac{1}{2}$ , PROVE that $a^2-3ab+b^2=0$
- Under the above condition, deduce that $\frac{a}{b}=\phi^2$ , or $\frac{a}{b}=\left(-\frac{1}{\phi}\right)^2$ , where $\phi$ is the golden ratio.
Let as before.
- If $\cos C^{-}=-\frac{1}{7}$ , PROVE that $a^2-18ab+b^2=0$
- Under the above condition, deduce that $\frac{a}{b}=(2\phi+1)^2$ , or $\frac{a}{b}=\left(3-2\phi\right)^2$ , where $\phi$ is the golden ratio.