Slopefessor Fr1day – Page 23 – High School Geometry

A new point on the nine-point circle II

Take vertex $A$ of triangle $ABC$ , together with the orthocenter $H$ of the triangle, and our new point $W$ with coordinates

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

then the resulting triangle $AHW$ has the following property: the slopes of its sides form a geometric progression. Same with vertices $B$ and $C$ as well as the corresponding triangles $BHW$ and $CHW$ (examples 6 and 10). This construction works for all triangles, except those triangles with one side parallel to the $x$ or $y$ axis.

Rigid transversals

If we draw three line segments from $W$ to each of the three vertices $A,B,C$ of any triangle $ABC$ , then the line segments $AW,BW,CW$ either all make acute angles with the $x$ -axis or all make obtuse angles with the $x$ -axis. See example 2 below.

In $\triangle ABC$ , let $m_1,m_2,m_3$ be the slopes of sides $AB,BC,CA$ , respectively. PROVE that the slopes of the line segments $AW, BW, CW$ are $\frac{m_1m_3}{-m_2}$ , $\frac{m_1m_2}{-m_3}$ , $\frac{m_2m_3}{-m_1}$ , in that order.

Let the coordinates of the vertices be $A(x_1,y_1)$ , $B(x_2,y_2)$ , and $C(x_3,y_3)$ . With $W$ as in equation (1), we can calculate the slope of $AW$ (those of $BW$ and $CW$ are similar).

$\begin{equation*} \begin{split} \Delta x&= \frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}-x_1\\ &=\frac{(x_1-x_2)(y_2-y_3)(x_3-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ \Delta y&=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}-y_1\\ &=\frac{(y_1-y_2)(x_2-x_3)(y_1-y_3)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ \therefore \textrm{slope of AW}&=\frac{\Delta y}{\Delta x}\\ &=\frac{(y_1-y_2)(x_2-x_3)(y_1-y_3)}{(x_1-x_2)(y_2-y_3)(x_3-x_1)}\\ &=\frac{y_1-y_2}{x_1-x_2}\times\frac{y_1-y_3}{x_3-x_1}\times\frac{x_2-x_3}{y_2-y_3}\\ &=m_1\times (-m_3)\times\frac{1}{m_2}\\ &=\frac{m_1m_3}{-m_2} \end{split} \end{equation*}$

The placement of the negative sign beside $m_2$ was deliberate: to make it easy to remember the formula for the slope of $AW$ as well as those of $BW$ and $CW$ . Here’s how it works: the slope of $AW$ is the product of the slopes of the two sides ( $AB$ and $AC$ ) that originate from vertex $A$ divided by the negative of the slope of the side opposite vertex $A$ (side $BC$ ).

In any $\triangle ABC$ , PROVE that the line segments $AW,BW,CW$ either all make acute angles with the positive $x$ -axis or all make obtuse angles with the positive $x$ -axis.

The slopes from $W$ to $A,B,C$ are given by:

(2) $\begin{equation*} \frac{m_1m_3}{-m_2},~\frac{m_1m_2}{-m_3},~\frac{m_2m_3}{-m_1} \end{equation*}$

Re-write the terms in (2) as

(3) $\begin{equation*} \frac{m_1m_2m_3}{-m_2^2},~\frac{m_1m_2m_3}{-m_3^2},~\frac{m_1m_2m_3}{-m_1^2} \end{equation*}$

and consider four cases:

Case I: $m_1,m_2,m_3$ are all positive. In this case, the product $m_1m_2m_3$ is also positive and so each term in equation (3) is negative.
Case II: $m_1,m_2,m_3$ are all negative. In this case, the product $m_1m_2m_3$ is also negative and so each term in equation (3) is positive.
Case III: two of $m_1,m_2,m_3$ are positive while the third is negative; specifically, let $m_1,m_2$ be positive while $m_3$ is negative. In this case, the product $m_1m_2m_3$ is negative, and so the terms in equation (3) are all positive.
Case IV: one of $m_1,m_2,m_3$ is positive while the other two are negative; specifically, let $m_1$ be positive while $m_2$ and $m_3$ are negative. In this case, the product $m_1m_2m_3$ will be positive and so the terms in equation (3) will be negative.

In $\triangle ABC$ , let $m_1,m_2,m_3$ be the slopes of sides $AB,BC,CA$ , respectively. PROVE that the line segment $AW$ is perpendicular to side $BC$ if and only if $m_1m_3=1$ .

$AW$ can also be perpendicular to $AB$ or $AC$ , but requiring it to be perpendicular to $BC$ is more convenient.

First suppose that $AW$ is perpendicular to $BC$ . Since the slope of $AW$ is $\frac{m_1m_3}{-m_2}$ and the slope of $BC$ is $m_2$ , we have:

$\frac{m_1m_3}{-m_2}\times m_2=-1\implies m_1m_3=1.$

Conversely, suppose that $m_1m_3=1$ . Let’s simplify the slope of $AW$ :

$m_{AW}=\frac{m_1m_3}{-m_2}=\frac{1}{-m_2}.$

This shows that $AW$ is perpendicular to $BC$ .

(The case where $m_1m_3=1$ and $m_2=\pm 1$ is a special case. The case where $m_1m_3=1$ and $m_2=0$ is an extremely pleasant case.)

In $\triangle ABC$ , let $m_1,m_2,m_3$ be the slopes of sides $AB,BC,CA$ , respectively. PROVE that $AW$ is parallel to $BW$ if and only if $m_2=-m_3$ (in other words, the slopes of sides $BC$ and $CA$ are negatives of each other).

First suppose that $AW$ is parallel to $BW$ . Set their slopes equal:

$\frac{m_1m_3}{-m_2}=\frac{m_1m_2}{-m_3}\implies m_2^2=m_3^2\implies m_2=\pm m_3.$

However, since we can’t have $m_2=m_3$ in a triangle, we take $m_2=-m_3$ .

Conversely, suppose that $m_2=-m_3$ and re-calculate the slopes of $AW$ and $BW$ :

$m_{AW}=\frac{m_1m_3}{-m_2}=\frac{m_1m_3}{m_3}=m_1,~~m_{BW}=\frac{m_1m_2}{-m_3}=\frac{m_1m_2}{m_2}=m_1.$

What this means is that $AW$ and $BW$ are parallel to side $AB$ (in fact, $W$ is the midpoint of $AB$ under this condition).

Let $H$ be the orthocenter of $\triangle ABC$ , and let $W$ be the point with coordinates in equation (1). PROVE that the slope of $HW$ is the negative reciprocal of the product of the slopes of sides $AB,BC,CA$ .

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 its orthocenter are given by:

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

These are very much the same as the coordinates of point $W$ , except for the extra term $(y_1-y_2)(y_2-y_3)(y_3-y_1)$ in the numerator of the $x$ -coordinate and the extra term $-(x_1-x_2)(x_2-x_3)(x_3-x_1)$ in the $y$ -coordinate. It follows that the slope of the line segment $HW$ is:

(5) $\begin{equation*} m_{HW}=\frac{-(x_1-x_2)(x_2-x_3)(x_3-x_1)}{(y_1-y_2)(y_2-y_3)(y_3-y_1)}=\frac{-1}{m_1m_2m_3} \end{equation*}$

where $m_1,m_2,m_3$ are the slopes of sides $AB,BC,CA$ .

(Main goal)

For any triangle $ABC$ with non-zero side-slopes, PROVE that the three triangles $AHW$ , $BHW$ , and $CHW$ have their side-slopes in geometric progressions.

We do this for triangle $AHW$ . The same argument can be adapted for the other two triangles.

Let the slopes of sides $AB,BC,CA$ be $m_1,m_2,m_3$ . Assume all non-zero.

The slope from vertex $A$ to the orthocenter $H$ is $\frac{-1}{m_2}$ (by the definition of the altitude from vertex $A$ ). The slope of the line segment $WA$ is $\frac{m_1m_3}{-m_2}$ (by example 1), and the slope of the line segment $HW$ is $\frac{-1}{m_1m_2m_3}$ (by example 5). Since

$\frac{m_1m_3}{-m_2}\times\frac{-1}{m_1m_2m_3}=\frac{1}{m_2^2}=\left(\frac{-1}{m_2}\right)^2,$

the slopes of sides $WA,AH,HW$ form a geometric progression.

Right triangles

In $\triangle ABC$ , let $\angle C=90^{\circ}$ . PROVE that the slope of $CW$ is the reciprocal of the slope of the hypotenuse $AB$ .

Let $m_1,m_2,m_3$ be the slopes of sides $AB,BC,CA$ . By example 1, the slope of the line segment $CW$ is $\frac{m_2m_3}{-m_1}$ . Since the given triangle is right-angled at $C$ , sides $BC$ and $CA$ are perpendicular, and so $m_2m_3=-1$ , which in turn simplifies the slope of the line segment $CW$ :

$\frac{m_2m_3}{-m_1}=\frac{1}{m_1},~m_1\neq 0.$

Alternatively, one can also deduce this from example 5.

(The three triangles $AHW,BHW,CHW$ reduces to only two for a right triangle, since its orthocenter coincides with a vertex. But we do get a special quadrilateral.)

Related trapezium

Let $\triangle ABC$ be a right triangle in which the hypotenuse has slope $\pm 1$ and the legs are not parallel to the coordinate axes. PROVE that the quadrilateral $ABCW$ is a trapezium.

Excluding the case where the legs are parallel to the coordinate axes is essential; if not, $W$ coincides with the orthocenter, which is one of the vertices of a right triangle.

Suppose we have $\angle C=90^{\circ}$ and $m_{AB}=1$ . According to example 7, the slope of $CW$ will be $1$ as well, meaning that $CW$ is parallel to $AB$ . This gives the required trapezium $ABCW$ . Similarly, if the slope of the hypotenuse $AB$ is $-1$ (that is $m_{AB}=-1$ ), then the slope of $CW$ is again $-1$ , and this gives a trapezium $ABCW$ .

(Note that the slopes of $AW$ and $BW$ are reciprocals of each other under the above condition. Together with the slope of $CW$ as $\pm 1$ , we get that the slopes of the line segments $AW,CW,BW$ form a geometric progression, even though the slopes of the sides of the parent triangle $ABC$ do not form a geometric progression.)

Calculate the coordinates of point $W$ for $\triangle ABC$ with vertices at $A(0,0)$ , $B(10,10)$ , $C(4,12)$ . Deduce that $ABCW$ is a trapezium.

Observe that $\triangle ABC$ is right-angled at $C$ . Using equation (1) with $x_1=0,y_1=0$ , $x_2=10,y_2=10$ , and $x_3=4,y_3=12$ , we get:

$\begin{equation*} \begin{split} x&=\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{0+100(-4)+48(10)}{0+10(12)+4(-10)}\\ &=1\\ y&=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{0+100(12)+48(-10)}{0+10(12)+4(-10)}\\ &=9 \end{split} \end{equation*}$

Thus, $W$ is located at $(1,9)$ in the Cartesian plane. The slope of $CW$ is $\frac{12-9}{4-1}=1$ , and the slope of $AB$ is $1$ , so the two line segments are parallel. The slope of $AW$ is $9$ , while the slope of $BC$ is $-\frac{1}{3}$ . Therefore, we obtain a trapezium.

(Notice that the slopes of $BW, CW, AW$ form a geometric progression, despite the fact the the slopes of the parent triangle $ABC$ do not form a geometric progression.)

Calculate the coordinates of $W$ for $\triangle ABC$ with vertices at $A(0,0)$ , $B(6,12)$ , and $C(5,20)$ . Deduce that $AWBC$ is a trapezium. Verify also that the slopes of the sides of triangles $AHW$ , $BHW$ , and $CHW$ form geometric progressions.

Unlike in the preceding example, the given triangle $ABC$ is not right-angled this time. So this example is saying that we can still obtain this special trapezium even when the starting triangle is not right-angled. Using equation (1) with $x_1=0,y_1=0$ , $x_2=6,y_2=12$ , and $x_3=5,y_3=20$ , we get:

$\begin{equation*} \begin{split} x&=\frac{x_1y_1(x_3-x_2)+x_2y_2(x_1-x_3)+x_3y_3(x_2-x_1)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{0+72(-5)+100(6)}{0+6(20)+5(-12)}\\ &=4\\ y&=\frac{x_1y_1(y_2-y_3)+x_2y_2(y_3-y_1)+x_3y_3(y_1-y_2)}{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}\\ &=\frac{0+72(20)+100(-12)}{0+6(20)+5(-12)}\\ &=4 \end{split} \end{equation*}$

Thus, $W$ is the point $(4,4)$ . With this, the slope of $BW$ is $\frac{12-4}{6-4}=4$ . Since the slope of $AC$ is also $4$ , it follows that $AC$ is parallel to $BW$ . The slope of $AW$ is $1$ and the slope of $BC$ is $-8$ . So we obtain a trapezium $AWBC$ .

Behind the scences, we calculated the orthocenter of triangle $ABC$ ; it is the point $H\left(36,\frac{9}{2}\right)$ . Together with $W(4,4)$ and the given vertices $A(0,0)$ , $B(6,12)$ , $C(5,20)$ , we have the following slopes:

$\scriptstyle m_{HW}=\frac{1}{64},~m_{AW}=1,~m_{BW}=4,~m_{CW}=16,~m_{HA}=\frac{1}{8},~m_{HB}=-\frac{1}{4},~m_{HC}=-\frac{1}{2}.$

The slopes of the sides of $\triangle AHW$ are then $\frac{1}{64},\frac{1}{8},1$ ; the slopes of the sides of $\triangle BHW$ are $\frac{1}{64},-\frac{1}{4},4$ ; the slopes of the sides of $\triangle CHW$ are $\frac{1}{64},-\frac{1}{2},16$ . These form geometric progressions with common ratios $8,-16,-32$ , respectively.

(Notice that the slopes of $AW, BW, CW$ form a geometric progression, despite the fact the the slopes of the parent triangle $ABC$ do not form a geometric progression.)

Takeaway

Let $W$ be the point on the nine-point circle of $\triangle ABC$ whose coordinates were given in equation (1). Let $H$ be the orthocenter of $\triangle ABC$ . Then the following statements are equivalent:

the slopes of sides $AB$ and $AC$ are reciprocals of each other
the line segment $WA$ is perpendicular to side $BC$
$W$ is the midpoint of the line segment $AH$ .

You’ll be getting used to equivalent statements by now.

Tasks

For $\triangle ABC$ with vertices at $A(-3,6)$ , $B(0,0)$ , $C(2,16)$ , calculate the slopes of line segments $AW$ , $BW$ , and $CW$ . Verify that $AW$ is perpendicular to $AB$ .
Let be such that side is parallel to the -axis. PROVE that:
- $W$ coincides with the foot of the altitude from vertex $A$
- none of the triangles $AWH,BWH,CWH$ has slopes in geometric progression.
  (This is the only exception to the main point in today’s discussion.)
In $\triangle ABC$ , let $W$ be as given in equation (1). PROVE that the slopes of $AW$ and $BW$ are reciprocals of each other if and only if the slope of side $AB$ is $\pm 1.$
PROVE if the slopes of the sides of a triangle form a geometric progression with common ratio $r$ , then the slopes of $BW,AW,CW$ form a geometric progression with common ratio $r^2$ .
In , let be the slopes of sides . PROVE that the following statements are equivalent:
- $WA$ is parallel to $BC$
- $m_2^2+m_1m_3=0$
  (Under this condition, the slopes of line segments $BW,AW,CW$ form a geometric progression.)

An application of Stewart’s theorem

Let $N$ be the nine-point center of triangle $ABC$ . We showed in our previous post that if the parent triangle is right-angled with $\angle B=90^{\circ}$ , then

(1) $\begin{equation*}AN^2+BN^2+CN^2=\frac{11}{16}AC^2. \end{equation*}$

We’ll now prove what holds more generally for any triangle:

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

$R$ being the circumradius and $a,b,c$ the usual side-lengths.

Memory recall

Let’s first be reminded of the main ingredient: Stewart’s theorem. It states that given a triangle $ABC$ with side-lengths $a,b,c$ and a cevian $AW$ like the one depicted below:

then we have:

(3) $\begin{equation*} b^2m+c^2n=a(d^2+mn) \end{equation*}$

Main results

In $\triangle ABC$ with the usual notation, PROVE that the orthocenter-circumcenter distance $HO$ satisfies $HO^2=9R^2-(a^2+b^2+c^2)$ .

It was from this amazing site that we first learnt of the above identity. We’ll add our own proof of it to an already existing pool of proofs, using Stewart’s theorem as a tool. We’ll also use the formula for median lengths, and the fact that the centroid divides a median in the ratio $2:1$ (measured from a vertex) and also divides the Euler line $HO$ in the ratio $2:1$ (measured from the orthocenter).

Consider the diagram below where $G$ is the centriod of $\triangle ABC$ , $M$ is the midpoint of $BC$ , and $AO=BO=CO=R$ (circumradius):

Notice our new point $W$ just $W$ andering along side $BC$ . More importantly, observe that $MO$ is perpendicular to $BC$ ( $M$ is the midpoint of $BC$ and $O$ is the circumcenter). Therefore, the Pythagorean theorem applied to $\triangle COM$ gives:

(4) $\begin{equation*} \begin{split} MO^2&=R^2-\left(\frac{a}{2}\right)^2\\ MO^2&=\frac{1}{4}(4R^2-a^2) \end{split} \end{equation*}$

Since $AG:GM=2:1$ and $HG:GO=2:1$ , Stewart’s theorem applied to $\triangle AMO$ gives:

(5) $\begin{equation*} \begin{split} MO^2(AG)+R^2(GM)&=AM(AG\times GM+GO^2)\\ MO^2\left(\frac{2}{3}AM\right)+R^2\left(\frac{1}{3}AM\right)&=AM\left(\frac{2}{9}AM^2+\frac{1}{9}HO^2\right)\\ 2MO^2+R^2&=\frac{2}{3}AM^2+\frac{1}{3}HO^2 \end{split} \end{equation*}$

The length of median $AM$ satisfies $AM^2=\frac{2b^2+2c^2-a^2}{4}$ . Thus, combining equations (4) and (5) gives:

$\begin{equation*} \begin{split} \frac{1}{2}\left(4R^2-a^2\right)+R^2&=\frac{2}{3}\left(\frac{2b^2+2c^2-a^2}{4}\right)+\frac{1}{3}HO^2\\ 3R^2-\frac{1}{2}a^2&=\frac{1}{6}(2b^2+2c^2-a^2)+\frac{1}{3}HO^2\\ 3R^2-\frac{1}{6}(2b^2+2c^2-a^2+3a^2)&=\frac{1}{3}HO^2\\ \therefore 9R^2-(a^2+b^2+c^2)&=HO^2 \end{split} \end{equation*}$

The above procedure works regardless of the vertex we started with. And it also works even in the peculiar case where $A,H,O$ are co-linear. It also doesn’t matter if both $H$ and $O$ are outside the parent triangle.

Let $H$ be the orthocenter of $\triangle ABC$ with circum-radius $R$ . PROVE that $AH^2+BH^2+CH^2=12R^2-(a^2+b^2+c^2)$ .

Apply Stewart’s theorem to $\triangle AHO$ in the diagram below:

$\begin{equation*} \begin{split} AH^2\Big(\frac{1}{3}HO\Big)+R^2\Big(\frac{2}{3}HO\Big)&=HO\Big(\frac{4}{9}AM^2+\frac{2}{9}HO^2\Big)\\ AH^2+2R^2&=\frac{4}{3}\Big(\frac{2b^2+2c^2-a^2}{4}\Big)+\frac{2}{3}HO^2\\ \therefore AH^2+2R^2&=\frac{2b^2+2c^2-a^2}{3}+\frac{2}{3}HO^2\\ \therefore BH^2+2R^2&=\frac{2a^2+2c^2-b^2}{3}+\frac{2}{3}HO^2\\ \therefore CH^2+2R^2&=\frac{2a^2+2b^2-c^2}{3}+\frac{2}{3}HO^2 \end{split} \end{equation*}$

Add all the three last equations:

$\begin{equation*} \begin{split} AH^2+BH^2+CH^2+6R^2&=a^2+b^2+c^2+2HO^2\\ \therefore AH^2+BH^2+CH^2&=a^2+b^2+c^2+2OH^2-6R^2\\ &=\scriptstyle a^2+b^2+c^2+2\Big(9R^2-(a^2+b^2+c^2)\Big)-6R^2\\ &=12R^2-(a^2+b^2+c^2) \end{split} \end{equation*}$

Let $N$ and $H$ be the nine-point center and orthocenter of $\triangle ABC$ . PROVE that $AN^2+BN^2+CN^2+HN^2=3R^2$ .

We only slightly modify the proof given in example 2, replacing the centroid with the nine-point center $N$ (note that $N$ is the midpoint of $HO$ ).

In $\triangle AHO$ above, $AN$ is a median, so:

$AN^2=\frac{2AH^2+2R^2-HO^2}{4}.$

Similarly, if we connect the segment $HO$ to vertices $B$ and $C$ instead, we obtain (with $BN$ and $CN$ as medians)

$BN^2=\frac{2BH^2+2R^2-HO^2}{4},~CN^2=\frac{2CH^2+2R^2-HO^2}{4}.$

Add the expressions for $AN^2, BN^2, CN^2$ and simplify:

$\begin{equation*} \begin{split} AN^2+BN^2+CN^2&=\frac{1}{4}\left(6R^2+2(AH^2+BH^2+CH^2)-3HO^2\right)\\ &=\scriptstyle\frac{1}{4}\Big(6R^2+2(12R^2-(a^2+b^2+c^2))-3(9R^2-(a^2+b^2+c^2))\Big)\\ &=\frac{1}{4}(3R^2+a^2+b^2+c^2)\\ \end{split} \end{equation*}$

Since $HN=\frac{1}{2}HO$ , we have $HN^2=\frac{1}{4}HO^2=\frac{1}{4}(9R^2-(a^2+b^2+c^2))$ . In turn:

$\begin{equation*} \begin{split} AN^2+BN^2+CN^2+HN^2&=\scriptstyle\frac{1}{4}\Big(3R^2+a^2+b^2+c^2+9R^2-(a^2+b^2+c^2)\Big)\\ &=3R^2 \end{split} \end{equation*}$

PROVE that $AN^2+BN^2+CN^2=\frac{1}{4}(3R^2+a^2+b^2+c^2)$ , where $N$ is the nine-point center of $\triangle ABC$ . Hence deduce that in a right triangle $ABC$ with hypotenuse of length $c$ , the sum of the squares of the distances from the vertices to the nine-point center is $11/16$ the length of the hypotenuse.

We obtained $AN^2+BN^2+CN^2=\frac{1}{4}(3R^2+a^2+b^2+c^2)$ as an intermediate step in example 3 above.

Now if $\triangle ABC$ is such that $AB$ is the hypotenuse with length $c$ , then $a^2+b^2=c^2$ . The circumradius $R$ is half of the length of the hypotenuse, so $R=\frac{c}{2}$ . Substituting in the expression for $AN^2+BN^2+CN^2$ , we obtain:

$\begin{equation*} \begin{split} AN^2+BN^2+CN^2&=\frac{1}{4}(3R^2+a^2+b^2+c^2)\\ &=\frac{1}{4}\Big(3\left(\frac{c}{2}\right)^2+c^2+c^2\Big)\\ &=\frac{1}{4}\Big(\frac{11}{4}c^2\Big)\\ &=\frac{11}{16}c^2. \end{split} \end{equation*}$

We show in example 5 below that the above relation also holds in a non-right triangle setting.

Consider $\triangle ABC$ in which $a=b$ and $\cos C=-\frac{4}{7}$ . PROVE that $AN^2+BN^2+CN^2=\frac{11}{16}c^2$ .

(If you’re curious as to how we obtained $\cos C=-\frac{4}{7}$ , see example 7 for a sample procedure.)

A key ingredient in the proof is the fact that the circumradius $R$ can be expressed as $R=\frac{c}{2\sin C}=\frac{b}{2\sin B}=\frac{a}{2\sin A}$ as per the extended law of sines.

Since $\cos C=-\frac{4}{7}$ , the cosine law gives

$c^2=a^2+b^2-2ab\cos C=a^2+a^2-2a^2\left(-\frac{4}{7}\right)=\frac{22}{7}a^2.$

Also:

$\sin^2 C=1-\cos^2 C=1-\left(-\frac{4}{7}\right)^2=\frac{33}{49}.$

Equation (2) becomes:

$\begin{equation*} \begin{split} AN^2+BN^2+CN^2&=\frac{1}{4}\left(3R^2+a^2+b^2+c^2\right)\\ &=\frac{1}{4}\left(\frac{3c^2}{4\left(33/49\right)}+\frac{7}{22}c^2+\frac{7}{22}c^2+c^2\right)\\ &=\frac{c^2}{4}\left(\frac{49}{44}+\frac{14}{44}+\frac{14}{44}+\frac{44}{44}\right)\\ &=\frac{c^2}{4}\left(\frac{121}{44}\right)\\ &=\frac{11}{16}c^2. \end{split} \end{equation*}$

Show that the circumradius $R$ of an equilateral triangle $ABC$ can be given by $R=\frac{a}{\sqrt{3}}$ , where $a$ is the length of one of the sides.

The orthocenter and circumcenter coincide for an equilateral triangle, so the distance $HO=0$ . Using example 1 and $a=b=c$ , we have:

$0=9R^2-(a^2+a^2+a^2)\implies R=\sqrt{\frac{3a^2}{9}}=\frac{a}{\sqrt{3}}.$

In $\triangle ABC$ , suppose that $AH^2+BH^2+CH^2=c^2$ . PROVE that either $\cos C=0$ or $\cos C=\frac{3a^2+3b^2\pm\sqrt{9(a^2+b^2)^2-32a^2b^2}}{8ab}$ . If one has $a=b$ in addition, deduce that the triangle is equilateral.

One can easily check that if a right triangle satisfies $a^2+b^2=c^2$ , then $AH^2+BH^2+CH^2=c^2$ . So the above example is basically saying that the equation $AH^2+BH^2+CH^2=c^2$ is not exclusive to right triangles.

From example 2: $AH^2+BH^2+CH^2=12R^2-(a^2+b^2+c^2)$ . Set the left side equal to $c^2$ :

$\begin{equation*} \begin{split} c^2&=12R^2-a^2-b^2-c^2\\ 2c^2+a^2+b^2&=12R^2\\ 2c^2+a^2+b^2&=12\left(\frac{c}{2\sin C}\right)^2\\ 2c^2+a^2+b^2&=\left(\frac{3c^2}{\sin^2 C}\right)\\ \sin^2 C\Big(2c^2+a^2+b^2\Big)&=3c^2\\ \sin^2 C\Big(2(a^2+b^2-2ab\cos C)+a^2+b^2\Big)&=3(a^2+b^2-2ab\cos C)\\ (1-\cos^2 C)\Big(3a^2+3b^2-4ab\cos C\Big)&=3a^2+3b^2-6ab\cos C\\ 4ab\cos^3 C-(3a^2+3b^2)\cos^2 C+2ab\cos C&=0\\ \cos C\Big(4ab\cos^2C-(3a^2+3b^2)\cos C+2ab\Big)&=0 \end{split} \end{equation*}$

Thus, either $\cos C=0$ (giving a right triangle), or

$\cos C=\frac{3a^2+3b^2\pm\sqrt{9(a^2+b^2)^2-32a^2b^2}}{8ab},$

as required. Now let $a=b$ , then:

$\begin{equation*} \begin{split} \cos C&=\frac{6a^2\pm\sqrt{9(2a^2)^2-32a^4}}{8a^2}\\ &=\frac{6a^2\pm\sqrt{4a^4}}{8a^2}\\ &=\frac{6a^2\pm 2a^2}{8a^2}\\ &=1,\frac{1}{2} \end{split} \end{equation*}$

If $\cos C=1$ , then $\angle C=0^{\circ}$ , which is not allowed for a normal non-degenerate triangle. Therefore we must take $\cos C=\frac{1}{2}$ , which yields $\angle C=60^{\circ}$ . Since we already assumed that $a=b$ , it follows that we obtain an equilateral triangle.

Mere repetitions

What follows are particular cases of some of the preceding examples, applied to our new point $W$ whose coordinates are given by:

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

Let $N$ be the nine-point center of $\triangle ABC$ and let $W$ be the point given in equation (6). PROVE that $AN^2+BN^2+CN^2+WN^2=3R^2$ , if $\triangle ABC$ has two sides parallel to the coordinate axes.

Trivial stuff. If $\triangle ABC$ has legs parallel to the coordinate axes, then $W=H$ , based on one of the equivalent conditions in our previous post. The conclusion now follows from example 3.

If $\triangle ABC$ has two sides parallel to the coordinate axes, PROVE that $AW^2+BW^2+CW^2=12R^2-(a^2+b^2+c^2)$ , where $W$ is the point given by equation (6).

Trivial stuff, twice. Under the given condition, we have $W=H$ . Thus, the conclusion follows from example 2 above.

Let $\triangle ABC$ be such that two sides have slopes $\pm 1$ . PROVE that $HW^2=9R^2-(a^2+b^2+c^2)$ , where $W$ is the point with coordinates given by equation (6).

Trivial stuff, thrice. The given triangle is obviously a right triangle. Since the legs have slopes $\pm 1$ , point $W$ coincides with the circumcenter. Therefore we have $HW^2=9R^2-(a^2+b^2+c^2)$ by example 1 above.

Takeaway

Let $a,b,c$ be the side-lengths of $\triangle ABC$ , and let $H,O,N$ be the orthocenter, circumcenter, and nine-point center, respectively.

The following statements are equivalent:

$HO^2=\frac{c^2}{4}$
$\cos C=0$ or $\cos C=\frac{9a^2+9b^2\pm\sqrt{(9a^2+9b^2)^2-320a^2b^2}}{20ab}.$

The following statements are also equivalent:

$AH^2+BH^2+CH^2=c^2$
$\cos C=0$ or $\cos C=\frac{3a^2+3b^2\pm\sqrt{(3a^2+3b^2)^2-32a^2b^2}}{8ab}.$

And the following two statements too are equivalent:

$AN^2+BN^2+CN^2=\frac{11}{16}c^2$
$\cos C=0$ or $\cos C=\frac{3a^2+3b^2\pm\sqrt{(3a^2+3b^2)^2+448a^2b^2}}{28ab}.$

Bottom line: Many non-right triangles share properties (relating to $H,O,N$ and stuff) that one would have thought to be exclusive to the right triangle.

Tasks

Let be the centroid of , and let be the circumcenter.
- PROVE that $AG^2+BG^2+CG^2=\frac{1}{3}\left(a^2+b^2+c^2\right)$ , where $a,b,c$ are the usual side-lengths.
- Deduce that $HO^2=3(3R^2-AG^2-BG^2-CG^2)$ .
Let $H$ and $N$ be the orthocenter and nine-point center of $\triangle ABC$ . PROVE that $AH^2+BH^2+CH^2-HO^2=AN^2+BN^2+CN^2+HN^2$ .
Let be the orthocenter, circumcenter and nine-point center of . PROVE that:
- $AH^2+BH^2+CH^2=AN^2+BN^2+CN^2+\frac{5}{4}HO^2$
- $AH^2+BH^2+CH^2=AN^2+BN^2+CN^2+5HN^2$ .
Let be the orthocenter of with circumradius and side-lengths .
- PROVE that $AH^2=4R^2-a^2$
- Deduce that the distance from vertex $A$ to the orthocenter is twice the distance from the circumcenter to the midpoint of side $BC$ .
This exercise shows that there are many non-right triangles with the property that the length of the Euler line is half of the length of a side of the triangle. Suppose that in one has . PROVE that:
- $\cos C=0$ or $\cos C=\frac{9a^2+9b^2\pm\sqrt{(9a^2+9b^2)^2-320a^2b^2}}{20ab}$
- if $a=b$ in addition, then $\cos C=0,1,\frac{4}{5}$ . (Thus, any isosceles triangle in which $a=b$ and $\cos C=4/5$ will satisfy $HO=c/2$ . And it’s even possible to realize this with a non-right, scalene triangle.)
Consider with vertices at , , . Verify that:
- the orthocenter is $H\left(2,\frac{2}{3}\right)$ and the circumcenter is $O\left(2,\frac{8}{3})$
- $HO=\frac{1}{2}AB.$
Find coordinates for the vertices of a non-right triangle $ABC$ with orthocenter $H$ and side-lengths $a,b,c$ such that $AH^2+BH^2+CH^2=c^2$ .
Find coordinates for the vertices of a non-right triangle $ABC$ with nine-point center $N$ and side-lengths $a,b,c$ such that $AN^2+BN^2+CN^2=\frac{11}{16}c^2$ .
Consider a quadrilateral with vertices at , , , . PROVE that:
- sides $AB$ and $CD$ have opposite slopes, and their lengths are in the ratio $1:3$
- sides $AD$ and $BC$ have opposite slopes, and their lengths are in the ratio $1:3$
- diagonals $AC$ and $BD$ have opposite slopes, and their lengths are in the ratio $3:5$
- the longer diagonal $BD$ bisects the shorter diagonal $AC$
- the shorter diagonal $AC$ divides the longer diagonal $BD$ in the ratio $1:9$
- $ABCD$ is a cyclic quadrilateral.
  (The point $D$ that partly facilitated these features is among the quartet of points we’ll introduce later this year, if time/space permits.)
Consider $\triangle ABC$ with vertices at $A\left(\frac{p^2-q^2}{p}\right)$ , $B(0,0)$ , $C(p,q)$ , $p\neq q$ . PROVE that $AH^2+BH^2+CH^2=4m_{c}^2$ , where $m_c$ is the length of the median from vertex $C$ .