General – Page 32 – High School Geometry

Mix and match cevians

We (can always) construct two cevians that concur with a median.

In a triangle, a cevian is a line segment from a vertex to the opposite side. A median is a special cevian that divides the opposite side in the ratio $1:1$

. Our aim in this post is to modify the ratios in which cevians divide the opposite sides, in such a way that concurrency is forced.

Ratios 1:2, 1:1, 1:2

In the diagram below, let cevian $AL$ divide side $BC$ such that $BL:LC=1:2$ ; let $BM$ be a median, so that $CM:MA=1:1$ ; let cevian $CN$ divide side $BA$ in the ratio $1:2$ .

We show that at the point of concurrency, median $BM$ is bisected in the ratio $1:1$ , while cevians $AL$ and $CN$ are divided in ratios $3:1$ . (Compare this with the manner in which the centroid divides the three medians of a triangle.)

Using the given ratios, PROVE that $AL,BM,CN$ are concurrent.

Just like proving that the three medians of a triangle are concurrent, this question belongs at the simplest level of simplicity. We have:

$\begin{equation*} \begin{split} \frac{BL}{LC}\times\frac{CM}{MA}\times\frac{AN}{NB}&=\frac{1}{2}\times\frac{1}{1}\times\frac{2}{1}\\ &=1 \end{split} \end{equation*}$

By the converse of Ceva’s theorem, we conclude that $AL,BM,CN$ are concurrent.

In the diagram below, the small letters $p,q,r,s,t,u$ represent the areas of the triangles in which they are located. PROVE that $u=3r$ , given that $BN:NA=1:2$ , $BL:LC=1:2$ , $CM:MA=1:1$ .

Let’s do this in four steps:

First, we have $u=t$ . Consider $\triangle AGM$ and $\triangle CGM$ . They have the same altitudes, measured from their common vertex $G$ . Since their base lengths $CM$ and $MA$ are also equal, their areas must be equal as well. This gives $u=t$ .
Next we have $s=2r$ and $p=2q$ . Consider $\triangle BGL$ and $\triangle GLC$ . They have the same altitudes, measured from their common vertex $G$ . Since $BL:LC=1:2$ , the area of $\triangle GLC$ is twice that of $\triangle BGL,$ so $s=2r$ . For the same reason we have $p=2q$ , when $\triangle AGN$ and $\triangle BGN$ are compared.
Our diagram simplifies to:
We now show that $q=r$ . Consider $\triangle ABM$ and $\triangle BMC$ . They have the same altitudes, measured from their common vertex $B$ . Since $CM:MA=1:1$ , these two triangles have the same areas, so $u+2q+q=r+2r+u$ . This gives $q=r$ .
Lastly, we show that $u=3r$ . Consider $\triangle ABL$ and $\triangle ALC$ . They have the same altitudes, measured from their common vertex $A$ . Since $BL:LC=1:2$ , the area of $\triangle ALC$ is twice the area of $\triangle ABL$ . Thus, $u+u+2r=2(2q+q+r)$ , and so $u=3q$ . Since $q=r$ from the previous step, we get $u=3r$ as desired.

Suppose that $BL:LC=1:2$ , $CM:MA=1:1$ , and $BN:NA=1:2$ in the diagram below:

PROVE that at the point of concurrency, $BM$ is bisected in the ratio $1:1$ , $AL$ is divided in the ratio $3:1$ and $CN$ is divided in the ratio $3:1$ .

Let the point of concurrency be $G$ and let the area of $\triangle BGL$ be denoted by $r$ . Then, in view of the previous example, the simplified area diagram, in terms of $r$ , is:

Consider $\triangle ABG$ and $\triangle AGM$ . They have the same areas, namely $3r$ . Since they also have the same altitudes (measured from their common vertex $A$ ), it must be the case that their base lengths are also equal. Thus $BG=GM$ .

Now consider $\triangle ACG$ and $\triangle CGL$ . They have the same altitudes, measured from their common vertex $C$ . The area of $\triangle ACG$ is $6r$ , while the area of $\triangle CGL$ is $2r$ . Thus their base lengths $AG$ and $GL$ must be in the ratio of their areas, so $AG:GL=6r:2r=3:1$ .

Similarly we have $CG:GN=3:1$ .

$\triangle ABC$ has vertices at $A(2,4)$ , $B(0,0)$ , and $C(6,0)$ . Cevians $AL,BM,CN$ divide sides $BC,CA,BA$ in the ratios $1:2,1:1,1:2$ , respectively. Determine the coordinates of the point of concurrency.

Note that cevian $BM$ is a median. So it is bisected $1:1$ at the point of concurrency, in view of the given ratios. This means that the midpoint of $BM$ is the point of concurrency. Since $M$ is the midpoint of $CA$ , it is the point $(4,2)$ . Then we take the midpoint of $(4,2)$ and $B(0,0)$ to get $(2,1)$ as the point of concurrency.

We can also find points $L$

and $N$

explicitly, and use the fact that cevians $AL$

and $CN$

are divided in the ratio $3:1$

at the point of concurrency.

Since $L$ divides $BC$ in the ratio $1:2$ , it is the point $\left(\frac{1\times 6+2\times 0}{1+2},\frac{1\times 0+ 2\times 0}{1+2}\right)=(2,0)$ . Then the point of concurrency divides $AL$ in the ratio $3:1$ , measured from vertex $A$ . So it is the point $\left(\frac{3\times 2+1\times 2}{3+1},\frac{3\times 0+ 1\times 4}{3+1}\right)=(2,1)$ , as before.
Since $N$ divides $BA$ in the ratio $1:2$ , it is the point $\left(\frac{2\times 0+1\times 2}{1+2},\frac{2\times 0+1\times 4}{1+2}\right)=\left(\frac{2}{3},\frac{4}{3}\right)$ . The point of concurrency divides cevian $CN$ in the ratio $3:1$ , measured from $C$ . So it is the point $\left(\frac{3\times 2/3+1\times 6}{3+1},\frac{3\times 4/3+1\times 0}{3+1}\right)=(2,1)$ , as before.

So we obtain the same point $(2,1)$ , but the last two approaches involved more steps.

Ratios m:n, 1:1, m:n

This is a slight generalization of the previous examples.

In the diagram below, the small letters $p,q,r,s,t,u$ represent the areas of the triangles in which they are located. PROVE that $u=\frac{n}{2m}\left(\frac{m+n}{m}\right)r$ , given that $BN:NA=m:n$ , $BL:LC=m:n$ , $CM:MA=1:1$ .

As before, we do this in steps:

First we have $u=t$ , because $CM:MA=1:1$ and both $\triangle AGM$ and $\triangle GMC$ have the same height.
Next we have $s=\frac{n}{m}r$ . This follows because $BL:LC=m:n$ and both $\triangle GBL$ and $\triangle GLC$ have the same altitudes, so their areas are proportional to their base lengths. Similarly, we have $p=\frac{n}{m}q$ . At this stage, our simplified diagram is:
Next we show that $q=r$ . Consider $\triangle ABM$ and $\triangle BMC$ . Since $CM:MA=1:1$ , these two triangles have the same areas, so $r+\frac{n}{m}r+u=q+\frac{n}{m}q+u$ . That is, $r+\frac{n}{m}r=q+\frac{n}{m}q$ . Re-arranging this gives $(r-q)+\frac{n}{m}(r-q)=0$ , or $(r-q)\left(1+\frac{n}{m}\right)=0$ . Since both $m$ and $n$ are positive (we’re dealing with internal division), we must have $r-q=0$ , so $r=q$ . Let’s adjust the diagram a bit:
Lastly we show that $u=\frac{n}{2m}\left(\frac{m+n}{m}\right)r$ . Consider $\triangle ABL$ and $\triangle ALC$ . They have the same altitudes, measured from their common vertex $A$ . Since $BL:LC=m:n$ , their areas follow the same ratio, meaning that $\left(r+r+\frac{n}{m}r\right):\left(\frac{n}{m}r+u+u\right)=m:n$ .
$\begin{equation*} \begin{split} \frac{2r+(n/m)r}{(n/m)r+2u}&=\frac{m}{n}\\ n\left(2r+(n/m)r\right)&=m\left((n/m)r+2u\right)\\ \frac{n}{m}\left(2r+(n/m)r\right)&=(n/m)r+2u\\ \frac{n}{m}\left(2r+(n/m)r-r\right)&=2u\\ \frac{n}{m}\left(\frac{m+n}{m}\right)r&=2u \end{split} \end{equation*}$

Thus we get $u=\frac{n}{2m}\left(\frac{m+n}{m}\right)r$ as desired.

Suppose that $BL:LC=m:n$ , $CM:MA=1:1$ , and $BN:NA=m:n$ in the diagram below:

PROVE that at the point of concurrency, $BM$ is divided in the ratio $2m:n$ , $AL$ is divided in the ratio $(m+n):m$ , and $CN$ is divided in the ratio $(m+n):m$ .

Let the point of concurrency be $G$ and let the area of $\triangle BGL$ be $r$ square units. Then, in terms of $r$ , we have the following area diagram:

Consider $\triangle ABG$ and $\triangle AGM$ . They have the same altitudes, measured from their common vertex $A$ . So the ratio of their areas gives the ratio of their base lengths. In other words, $BG:GM=\left(r+\frac{n}{m}r\right):\frac{n}{2m}\left(\frac{m+n}{m}\right)r$ .

$\begin{equation*} \begin{split} BG:GM&=\left(\frac{m+n}{m}\right)r:\frac{n}{2m}\left(\frac{m+n}{m}\right)r\\ &=1:\frac{n}{2m}\\ &=2m:n \end{split} \end{equation*}$

To show that $AG:GL=(m+n):m$ , consider $\triangle AGC$ and $\triangle GCL$ . They have the same altitude, measured from their common vertex $C$ . So the ratio of their areas gives the ratio of their base lengths.

$\begin{equation*} \begin{split} AG:GL&=\left[\frac{n}{2m}\left(\frac{m+n}{m}\right)r+\frac{n}{2m}\left(\frac{m+n}{m}\right)r\right]:\frac{n}{m}r\\ &=\frac{n}{m}\left(\frac{m+n}{m}\right)r:\frac{n}{m}r\\ &=\frac{m+n}{m}:1\\ &=(m+n):m \end{split} \end{equation*}$

Similarly, $CG:GN=(m+n):m$ .

PROVE that the centroid of a triangle divides the three medians in the ratio $2:1$ .

In the case of three medians, we have $m:n=1:1$ . So, using this in the previous example:

the median $BM$ is divided in the ratio $2m:n=2(1):1=2:1;$
the median $AL$ is divided in the ratio $(m+n):m=(1+1):1=2:1;$
the median $CN$ is divided in the ratio $(m+n):m=(1+1):1=2:1.$

In $\triangle ABC$ , $CN$ is a median. In what ratio should cevians $AL$ and $BM$ divide sides $BC$ and $CA$ so that the point of concurrency and the centroid trisect median $CN$ ?

Let $G$ be the centroid, and let the point of concurrency be $P$ . Then $G$ divides median $CN$ in ratio $2:1$ , measured from vertex $C$ ; that is, $CG:GN=2:1$ . Since $CN$ is to be trisected, we want $CP:PG:GN=1:1:1$ , or $CP:PN=1:2$ . From the previous example, the median $CN$ is divided in the ratio $2m:n$ at the point of concurrency. Therefore $2m:n=1:2$ , or $n=4m$ . So cevian $AL$ should divide side $BC$ in the ratio $1:4$ and cevian $BM$ should divide side $CA$ in the ratio $4:1$ .

For any right triangle with legs parallel to the coordinate axes, PROVE that the slopes of its three medians form a geometric progression with common ratio $-2$ .

For simplicity, fix the vertices at $A(0,a)$ , $B(0,0)$ , and $C(c,0)$ ; a little modification will be needed for the general case.

We now calculate the slopes of medians $CN,BM,AL$ :

$\begin{equation*} \begin{split} \textrm{slope of median}~ CN&=-\frac{a}{2c}\\ \textrm{slope of median}~ BM&=\frac{a}{c}\\ \textrm{slope of median}~ AL&=-\frac{2a}{c} \end{split} \end{equation*}$

These form a geometric progression with common ratio of $-2$ .

(There’s something interesting here. In our first post of this year, we saw that if the slopes of the sides of a triangle form a geometric sequence with $-2$ as common ratio, then the triangle contains a vertical median and a horizontal median. Now, the above example says that if a triangle contains a vertical side and a horizontal side, then the slopes of its three medians form a geometric progression with a common ratio of $-2$ . Did you notice something? See exercise 5 at the end.)

Give an example of a triangle in which one side, one median, and another cevian have slopes forming an arithmetic progression.

Consider $\triangle ABC$ below, in which $CN$ is a median, $AL$ is a cevian:

The slopes of $AL,AC,CN$ are $-5,-3,-1$ , respectively. They form an arithmetic progression with a common difference of $2$ .

(Notice that if a third cevian is drawn from vertex $B$ and is concurrent with cevian $AL$ and median $CN$ , then it is an altitude. This is a good example of “mixing” and “matching” cevians as used in our post’s title. Mixing means that they are different; matching means that they are concurrent.)

Takeaway

The ratios we considered in this post ( $m:n,1:1,m:n$ ) were for convenience. There are other ratios that can still ensure concurrency.

Tasks

Let be the vertices of . If there are cevians which divide sides in the ratios , respectively:
- PROVE that the point of concurrency has coordinates $\left(\frac{mx_1+nx_2+mx_3}{2m+n},\frac{my_1+ny_2+my_3}{2m+n}\right)$ ;
- Deduce that the centroid of $\triangle ABC$ is the point $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$ .
$\triangle ABC$ has vertices at $A(1,4),B(0,0),C(3,6)$ . Assuming cevians $AL,BM,CN$ divide sides $BC,CA,BA$ in the ratios $m:n,1:1,m:n$ , respectively, and they concur at the point $\left(\frac{3}{2},\frac{15}{4}\right)$ , determine $m$ and $n$ .
$\triangle ABC$ has vertices at $A(2,4),B(0,0),C(6,0)$ . Determine the lengths of cevians $AL,BM,CN$ which divide sides $BC,CA,BA$ in the ratios $1:3,3:1,1:1$ , respectively.
Let $\triangle ABC$ be a right triangle in the first quadrant, with vertices $A(0,a)$ , $B(0,0)$ , and $C(c,0)$ . For $a> 1$ , PROVE that there are three cevians $AL,BM,CN$ whose slopes form a geometric progression with common ratio $-a$ .
(Compare this with Example 9.)
Let be the vertices of . Let the slopes of medians through be denoted by , respectively. Suppose that (in that order) form a geometric progression with a common ratio of . PROVE that:
- $y_2=y_3$ ;
- $x_1=x_2$ .
  (In view of Example 9, conclude that the slopes of the three medians of a triangle form a geometric progression with common ratio $-2$ if, and only if, the triangle is a right triangle with legs parallel to the coordinate axes.)
(Opposite slopes) The right triangle with vertices at , , has very nice properties:
- Find an internal point $L$ on $BC$ such that the slope of cevian $BX$ is the negative of the slope of cevian $AL$ , and PROVE that these two cevians intersect at $\left(\frac{2}{3},\frac{3}{2}\right)$ ;
- Find an internal point $N$ on $AB$ such that the slope of cevian $BZ$ is the negative of the slope of cevian $CN$ , and PROVE that these two cevians intersect at $\left(2,\frac{1}{2}\right)$ ;
- If a cevian from $B$ and a cevian from $C$ have opposite slopes, is their point of intersection always of the form $\left(a,\frac{1}{a}\right)$ for some $a$ ?
- For any point $R$ between $A$ and $M$ , PROVE that there is a point $S$ between $B$ and $C$ such that the slopes of cevians $BR$ and $AS$ are negatives of each other;
- For any point $U$ between $M$ and $C$ , PROVE that there is a point $V$ between $A$ and $B$ such that the slopes of cevians $BU$ and $CV$ are negatives of each other.
Consider the right triangle with vertices at , , :

An arbitrary point $X$ on $AC$ can be given by the coordinates $\left(k,-\frac{a}{c}k+a\right)$ , where $0< k< c$ . PROVE that:
- $\frac{ck}{c-k}< c$ if and only if $k<\frac{c}{2}$ ;
- $\frac{a}{k}(c-k)< a$ if and only if $k> \frac{c}{2}$ ;
- for $k<\frac{c}{2}$ and $L:=\left(\frac{ck}{c-k},0\right)$ , cevians $AL$ and $BX$ have opposite slopes;
- for $k> \frac{c}{2}$ and $V:=\left(0,\frac{a}{k}(c-k)\right)$ , cevians $CV$ and $BX$ have opposite slopes;
- a cevian from $B$ and a cevian from $C$ intersect at $y=\frac{a}{2}$ if and only if they have opposite slopes (and $k<\frac{c}{2}$ );
- a cevian from $B$ and a cevian from $C$ intersect at $x=\frac{c}{2}$ if and only if they have opposite slopes (and $k>\frac{c}{2}$ ).
Let be the vertices of . Suppose that the slopes of sides and are negatives of each other, and that the slopes of medians through and are also negatives of each other. PROVE that:
- $2x_1=x_2+x_3$ (so $x_1,x_2,x_3$ form an arithmetic sequence)
- $AB=AC$ (so $\triangle ABC$ is isosceles)
- the median through vertex $A$ is vertical.
PROVE that if the slopes of the equal sides of an isosceles triangle are negatives of each other, then the $x$ -coordinates form an arithmetic progression.
For an integer $r> 1$ , let $A(1,r^2),B(0,0),C(1+r,r+r^2)$ be the vertices of $\triangle ABC$ . PROVE that a cevian from vertex $A$ divides side $BC$ in the ratio $(r^2-r+1):r$ if and only if it is an altitude.

A special square (root)

The process of squaring (and taking square root) can sometimes preserve the operation of addition, as in this equation:

(1) $\begin{equation*}\left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}. \end{equation*}$

We show that if we take $(\alpha_1,\beta_1)$ as the centroid, $(\alpha_2,\beta_2)$ as the circumcenter, and $(\alpha_3,\beta_3)$ as the orthocenter of any triangle, then the equation always holds — evaluating to $\frac{1}{4}$ in every case.

Interestingly, if we append absolute values, then the equation holds twice for a right triangle — with $(\alpha_1,\beta_1)$ as the vertex containing angle $90^{\circ}$ , while $(\alpha_2,\beta_2)$ and $(\alpha_3,\beta_3)$ are the other two vertices:

(2) $\begin{equation*}\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}. \end{equation*}$

The latter equation is preferred to the former one, and is behind a little geometric application that we present.

Consider $\triangle ABC$ with vertices at $A(1,-3)=(\alpha_1,\beta_1)$ , $B(-3,-1)=(\alpha_2,\beta_2)$ , and $C(5,5)=(\alpha_3,\beta_3)$ . VERIFY that equation (2) holds in this case.

Using $(\alpha_1,\beta_1)=(1,-3)$ , $(\alpha_2,\beta_2)=(-3,-1)$ , and $(\alpha_3,\beta_3)=(5,5)$ , we have, from equation (2) that:

$\begin{equation*} \begin{split} \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(-3-1)^2+(-1--3)^2}{(5-1)^2+(5--3)^2}\\ &=\frac{(-4)^2+2^2}{4^2+8^2}\\ &=\frac{20}{80}\\ &=\frac{1}{4}\\ \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|-3-1|+|-1--3|}{|5-1|+|5--3|}\right)^2\\ &=\left(\frac{|-4|+2}{4+8}\right)^2\\ &=\left(\frac{6}{12}\right)^2\\ &=\frac{1}{4} \end{split} \end{equation*}$

CL!!!!!!!!!!!

Consider the same $\triangle ABC$ in the previous example: $A(1,-3)$ , $B(-3,-1)$ , $C(5,5)$ . Using $(\alpha_1,\beta_1)$ as the centroid, $(\alpha_2,\beta_2)$ as the circumcenter, and $(\alpha_3,\beta_3)$ as the orthocenter, VERIFY that equation (1) holds.

Since $\triangle ABC$ is a right triangle, its circumcenter is the midpoint of the hypotenuse (in this case the midpoint of side $BC$ ), and its orthocenter is at the vertex containing angle $90^{\circ}$ (vertex $A$ ). The centroid is obtained as in any other triangle, using $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$ or $\left(\frac{\alpha_1+\alpha_2+\alpha_3}{3},\frac{\beta_1+\beta_2+\beta_3}{3}\right)$ in the present case.

So we have $(\alpha_1,\beta_1)=\left(1,\frac{1}{3})$ , $(\alpha_2,\beta_2)=(1,2)$ , and $(\alpha_3,\beta_3)=(1,-3)$ . In equation (1):

$\begin{equation*} \begin{split} \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{0^2+(5/3)^2}{0^2+(-10/3)^2}\\ &=\frac{25}{100}\\ &=\frac{1}{4}\\ \left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2&=\left(\frac{0+(5/3)}{0+(-10/3)}\right)^2\\ &=\frac{1}{4} \end{split} \end{equation*}$

CL!!!

Right company:

In our next example we show that equation (2) doesn’t hold for a triangle that’s not right.

Consider a triangle with vertices $(0,0)$ , $(1,4)$ , $(3,6)$ . PROVE that equation (2) fails in this case.

We will select $(\alpha_1,\beta_1)$ as any of the three vertices. This leads to three possibilities:

Case I: let $(\alpha_1,\beta_1)=(0,0)$ , $(\alpha_2,\beta_2)=(1,4)$ , and $(\alpha_3,\beta_3)=(3,6)$ .
$\begin{equation*} \begin{split} \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{1+4}{3+6}\right)^2\\ &=\frac{25}{81}\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{1^2+4^2}{3^2+6^2}\\ &=\frac{17}{45} \end{split} \end{equation*}$
Case II: let $(\alpha_1,\beta_1)=(1,4)$ , $(\alpha_2,\beta_2)=(0,0)$ , and $(\alpha_3,\beta_3)=(3,6)$ .
$\begin{equation*} \begin{split} \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|-1|+|-4|}{2+2}\right)^2\\ &=\frac{25}{16}\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(-1)^2+(-4)^2}{2^2+2^2}\\ &=\frac{17}{8} \end{split} \end{equation*}$
Case III: let $(\alpha_1,\beta_1)=(3,6)$ , $(\alpha_2,\beta_2)=(0,0)$ , and $(\alpha_3,\beta_3)=(1,4)$ .
$\begin{equation*} \begin{split} \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|-3|+|-6|}{|-2|+|-2|}\right)^2\\ &=\frac{81}{16}\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(-3)^2+(-6)^2}{(-2)^2+(-2)^2}\\ &=\frac{45}{8} \end{split} \end{equation*}$

Although equation (2) fails for this triangle, equation (1) still holds, so long as we take $(\alpha_1,\beta_1)$ as the centroid, $(\alpha_2,\beta_2)$ as the circumcenter, and $(\alpha_3,\beta_3)$ as the orthocenter.

Co-linear points

PROVE that equation (1) holds if the points $(\alpha_1,\beta_1)$ , $(\alpha_2,\beta_2)$ , and $(\alpha_3,\beta_3)$ lie on the same straight line (co-linear).

Ignoring degeneracies (like $\alpha_1=\alpha_3$ etc), we’re to prove that

$\left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}$

based on the co-linearity of the given points. [In the case where the slope of the line is $-1$ , absolute values will be necessary on the left. As a result of this, it is generally preferable to use equation (2).]

Co-linearity ensures that the slope between any two points will be the same, so we have:

$\frac{\beta_2-\beta_1}{\alpha_2-\alpha_1}=\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1},\implies \beta_2-\beta_1=\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1).$

Then

$\begin{equation*} \begin{split} \left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2&=\left(\frac{(\alpha_2-\alpha_1)+\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2\\ &=\left(\frac{\alpha_2-\alpha_1}{\alpha_3-\alpha_1}\right)^2\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(\alpha_2-\alpha_1)^2+\left[\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1)\right]^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\\ &=\frac{(\alpha_2-\alpha_1)^2}{(\alpha_3-\alpha_1)^2} \end{split} \end{equation*}$

In any triangle, let $(\alpha_1,\beta_1)$ , $(\alpha_2,\beta_2)$ , and $(\alpha_3,\beta_3)$ be the centroid, circumcenter, and orthocenter, respectively. PROVE that equation (1) is satisfied, and that both sides evaluate to $\frac{1}{4}$ .

This follows from the preceding example together with the well-known fact that these three principal triangle centers are co-linear.

To show that both sides evaluate to $1/4$ , consider the right member of equation (1):

$\left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}$

where the numerator represents the square of the circumcenter-centroid distance, and the denominator represents the square of the centroid-orthocenter distance. Since the centroid divides the circumcenter-orthocenter distance in the ratio $1:2$ , the right side always reduces to $(1/2)^2.$ And so is the left side.

Square root:

Equation (1) can be re-written as (if one wants, one can place absolute values on each term on the right):

$\sqrt{\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}}=\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}$

and so even square roots preserve addition in this situation.

Opposite slopes

In any triangle in which two sides have opposite slopes, PROVE that equation (2) is satisfied.

Let $(\alpha_1,\beta_1)$ be the vertex from which the sides with opposite slopes originate, and let $(\alpha_2,\beta_2)$ and $(\alpha_3,\beta_3)$ be the two other vertices. We show that:

$\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.$

By assumption

$\frac{\beta_2-\beta_1}{\alpha_2-\alpha_1}=-\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1},\implies \beta_2-\beta_1=-\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1).$

The remainder of the proof is akin to that of the previous example; the only difference is the inclusion of absolute values:

$\begin{equation*} \begin{split} \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|\alpha_2-\alpha_1|+\left |-\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1)\right |}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2\\ &=\left(\frac{\alpha_2-\alpha_1}{\alpha_3-\alpha_1}\right)^2\\ \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(\alpha_2-\alpha_1)^2+\left[-\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)(\alpha_2-\alpha_1)\right]^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\\ &=\frac{(\alpha_2-\alpha_1)^2}{(\alpha_3-\alpha_1)^2} \end{split} \end{equation*}$

Reciprocal slopes

In any triangle in which two sides have reciprocal slopes, PROVE that equation (1) is satisfied.

Let $(\alpha_1,\beta_1)$ be the common vertex from which the sides with reciprocal slopes originate. Let the two other vertices be $(\alpha_2,\beta_2)$ and $(\alpha_3,\beta_3)$ . We have:

$\left(\frac{\beta_2-\beta_1}{\alpha_2-\alpha_1}\right)\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)=1,\implies \beta_2-\beta_1=\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}.$

Then

$\begin{equation*} \begin{split} \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(\alpha_2-\alpha_1)^2+\left[\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}\right]^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\\ &=\frac{(\alpha_2-\alpha_1)^2}{(\beta_3-\beta_1)^2}\\ \left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2&=\left(\frac{(\alpha_2-\alpha_1)+\left[\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}\right]}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2\\ &=\left(\frac{\alpha_2-\alpha_1}{\beta_3-\beta_1}\right)^2 \end{split} \end{equation}$

Favorite triangle:

Hardly would we talk about triangles without mentioning the ones we both cherish and relish — those whose slopes form geometric progressions. In today’s context, the ones with slopes $\frac{1}{r},1,r$ satisfy equation (1).

Right triangles

PROVE that any right triangle satisfies equation (2).

Let $(\alpha_1,\beta_1)$ , $(\alpha_2,\beta_2)$ , and $(\alpha_3,\beta_3)$ be the vertices of a right triangle, where $(\alpha_1,\beta_1)$ is the vertex containing angle $90^{\circ}$ . We show that:

$\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.$

The proof is much like that of , with a slight modification.

Since the slopes of the sides originating from vertex $(\alpha_1,\beta_1)$ are negative reciprocals of each other, we have:

$\left(\frac{\beta_2-\beta_1}{\alpha_2-\alpha_1}\right)\left(\frac{\beta_3-\beta_1}{\alpha_3-\alpha_1}\right)=-1,\implies \beta_2-\beta_1=-\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}.$

Then

$\begin{equation*} \begin{split} \frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}&=\frac{(\alpha_2-\alpha_1)^2+\left[-\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}\right]^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}\\ &=\frac{(\alpha_2-\alpha_1)^2}{(\beta_3-\beta_1)^2}\\ \left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2&=\left(\frac{|\alpha_2-\alpha_1|+\left |\left[-\frac{(\alpha_2-\alpha_1)(\alpha_3-\alpha_1)}{\beta_3-\beta_1}\right]\right |}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2\\ &=\left(\frac{\alpha_2-\alpha_1}{\beta_3-\beta_1}\right)^2 \end{split} \end{equation}$

In the end, only one of the four absolute values became absolutely valuable.

Absolute need: Here’s an example to show the difference an absolute value can make.

Let $(\alpha_1,\beta_1)=(7,5)$ , $(\alpha_2,\beta_2)=(5,7)$ , and $(\alpha_3,\beta_3)=(8,4)$ . Using these values in the equation

$\left(\frac{(\alpha_2-\alpha_1)+(\beta_2-\beta_1)}{(\alpha_3-\alpha_1)+(\beta_3-\beta_1)}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}$

the left side evaluates to the indeterminate form $(\frac{0}{0})^2$ , while the right side is $4$ . However, if we use absolute values on the left, then both sides evaluate to $4$ .

This is an exceptional case that happens whenever the slope of the line joining the points is $-1$ . For this reason we generally recommend using equation (2).

Geometric application

The following shapes

squares
isosceles right triangles
equilateral triangles with one side having slope of $1$
equilateral triangles with one side parallel to the $x$ -axis
isosceles triangles with one side parallel to the $x$ -axis

have something in common.

Let $(\alpha_2,\beta_2)$ and $(\alpha_3,\beta_3)$ be the endpoints of one diagonal of a square, and let $(\alpha_1,\beta_1)$ be any of the two remaining vertices. PROVE that $|\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|$ .

Since a diagonal splits a square into two right triangles, we can apply equation (2):

$\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}$

noting that the right side evaluates to $1$ for our square. In turn:

$\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=1\implies \frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}=\pm 1.$

In view of the absolute values, we only take $\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}=1$ , which gives $|\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|$ .

In an equilateral triangle with one side parallel to the $x$ -axis, let $(\alpha_1,\beta_1)$ be the “apex”, while $(\alpha_2,\beta_2)$ and $(\alpha_3,\beta_3)$ are the two other vertices. PROVE that $|\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|$ .

In such an equilateral triangle, the slopes of the sides are of the form $-a,0,a$ . Since two of these are opposite values, we can use a previous example to deduce that equation (2) holds:

$\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}$

The right side evaluates to $1$ for our equilateral triangle, so:

Takeaway

Out of the two equations we considered, we recommend the second one:

$\left(\frac{|\alpha_2-\alpha_1|+|\beta_2-\beta_1|}{|\alpha_3-\alpha_1|+|\beta_3-\beta_1|}\right)^2=\frac{(\alpha_2-\alpha_1)^2+(\beta_2-\beta_1)^2}{(\alpha_3-\alpha_1)^2+(\beta_3-\beta_1)^2}.$

In practice, the equation works because of the ratio.

Tasks

In any isosceles right triangle, PROVE that the equation $|\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|$ holds, if $(\alpha_1,\beta_1)$ is the vertex containing angle $90^{\circ}$ , while $(\alpha_2,\beta_2)$ and $(\alpha_3,\beta_3)$ are the other two vertices.
Consider an equilateral triangle with vertices , , and . If the slope of side is , PROVE that:
- the slopes of sides $AB$ and $CA$ are reciprocals of each other;
- $|\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|$ .
Consider an equilateral triangle with vertices , , and . If the slope of side is , PROVE that:
- the slopes of sides $AB$ and $CA$ are reciprocals of each other
- $|\alpha_2-\alpha_1|+|\beta_2-\beta_1|=|\alpha_3-\alpha_1|+|\beta_3-\beta_1|$ .
Let be a right triangle in which is the point and . Side makes angle with the positive -axis, and the lengths of and are and , respectively. PROVE that one can describe:
- $A$ as the point $(x_1+b\sin\theta,y_1-b\cos\theta)$
- $B$ as the point $(x_1+a\cos\theta,y_1+a\sin\theta)$ .
Given a vertex , two side lengths and , and a parameter :
- PROVE that any triangle with vertices $(x_1,y_1)$ , $(x_1+a\cos\theta, y_1+a\sin\theta)$ , and $(x_1+b\sin\theta,y_1-b\cos\theta)$ is a right triangle
- Using $(x_1,y_1)=(1,-3)$ , $a=2\sqrt{5}$ , $b=4\sqrt{5}$ , and $\tan\theta=-\frac{1}{2}$ , verify that one obtains $(-3,-1)$ and $(5,5)$ as the remaining two vertices.
For $r\neq 0,\pm 1$ , PROVE that any triangle with side slopes $\frac{1}{r},1,r$ is isosceles.
Let , , , where , be the vertices of a triangle. PROVE that:
- the triangle is isosceles
- the slopes of the sides form a geometric sequence with $-2$ as common ratio
- equation (2) is satisfied in this case.
For a triangle with vertices , , , let , , and be the centroid, circumcenter, and orthocenter, respectively.
- calculate $(\alpha_1,\beta_1)$ , $(\apha_2,\beta_2)$ , and $(\alpha_3,\beta_3)$
- verify that equation (2) is satisfied.
Suppose that an equilateral triangle has one side parallel to the -axis (so that its slope is undefined). PROVE that:
- the slopes of the other two sides are opposites of each other
- equation (2) is satisfied.
Given , , and :
- PROVE that any quadrilateral with vertices $(x_1,y_1)$ , $(x_1+a\cos\theta,y_1+a\sin\theta)$ , $(x_1+a(\cos\theta+\sin\theta),y_1+a(\sin\theta-\cos\theta))$ , and $(x_1+a\sin\theta,y_1-a\cos\theta)$ is a square
- determine $a$ and $\theta$ for the square with vertices $(-2,-3)$ , $(-2+\sqrt{2},-3-\sqrt{2})$ , $(-2+2\sqrt{2},-3)$ , and $(-2+\sqrt{2},-3+\sqrt{2})$ .