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 $C$ .