Twenty one equivalent statements – High School Geometry

In a bid to mirror the number of statements in the Invertible Matrix Theorem, to mimic right triangles, as well as to have some memory of $2021$

, we’ve assembled $21$

statements that are equivalent in certain triangles. Most of these statements are trivial, but what is vital is that right triangles and the triangles that satisfy the statements are rivals. Vital rivals.

Problem statement

Let $\triangle ABC$

be a non-right triangle with side-lengths $a,b,c$

, altitudes $h_A,h_B,h_C$

, circumradius $R$

, circumcenter $O$

, orthocenter $H$

, and nine-point center $N$

. Then the following statements are equivalent:

$AH=b$
$BH=a$
$CH=2h_C$
$\cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}$
$\cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}$
$\cos C=\frac{2ab}{a^2+b^2}$
$\cos^2 A+\cos^2 B=1$
$\sin^2 A+\sin^2 B=1$
$a\cos A+b\cos B=0$
$h_A^2+h_B^2=AB^2$
$\frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}$
$a^2+b^2=4R^2$
$A-B=\pm 90^{\circ}$
$(a^2-b^2)^2=(ac)^2+(cb)^2$
the orthic triangle is obtuse isosceles
radius $OC$ is parallel to side $AB$
the nine-point center lies on $AB$
the geometric mean theorem holds
the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
segment $CH$ is tangent to the circumcircle at point $C$
the orthocenter is a reflection of vertex $C$ over side $AB$ .

The connection between the orthic triangle and the geometric mean theorem seems cool. The restriction to non-right triangles is crucal: some of the statements hold in right triangles, but not the entire chain of statements.

Our favorites: $14\&15\&17\&18\&20\&21$ . Notice the last two.

Partial solutions

A complete proof will require about $21$ implications, and we’ve already seen some of these implications in previous iterations of our discussions: May 14, May 28, June 14, and June 28. Below we make some random selections.

Suppose that $A-B=\pm 90^{\circ}$

in $\triangle ABC$

. PROVE that the radius through $C$

is parallel to side $AB$

To be specific, let $B-A=90^{\circ}$ and set $A=\theta$ . Then $B=90^{\circ}+\theta$ and $C=90^{\circ}-2\theta$ . Consider the circumcircle shown below:

The angle which the major arc $AC$ subtends at the center of the circle is twice the angle it subtends at the circumference, and so:

$\textrm{reflex}~\angle AOC=2(90^{\circ}+\theta)\implies \textrm{obtuse}~\angle AOC=180^{\circ}-2\theta$

Since $\triangle AOC$ is isosceles, we have that

$\angle OAC=\angle OCA=\theta$

This shows that $OC$ is parallel to $AB$ .

Let $O$

be the circumcenter of $\triangle ABC$

. If radius $OC$

is parallel to side $AB$

, PROVE that $A-B=\pm 90^{\circ}$

Begin by drawing the circumcircle:

If $\angle BAC=\theta$ , then $\angle OCA=\theta$ , as marked above. Reason: radius $OC$ is parallel to side $AB$ by assumption. Next, $\triangle AOC$ is isosceles, and so $\angle OAC=\theta$ . In turn, this yields $\angle AOC=180^{\circ}-2\theta$ . The major arc $AC$ now subtends an angle of $180^{\circ}+2\theta$ at the center of the circle. This means that it subtends an angle of $90^{\circ}+\theta$ at the circumference. Thus, $\angle ABC=90^{\circ}+\theta$ . The difference between $\angle BAC$ and $\angle ABC$ is then $90^{\circ}$ .

A different orientation having angle $\theta$ placed at $B$ will also yield the same conclusion.

Let $O$

and $H$

be the circumcenter and orthocenter of $\triangle ABC$

. If $A-B=\pm 90^{\circ}$

, PROVE that $OC$

is perpendicular to $CH$

Let $B-A=90^{\circ}$ and set $A=\theta$ . Then $B=90^{\circ}+\theta$ and $C=90^{\circ}-2\theta$ . Draw the circumcircle, like so:

Since $\triangle ABC$ is obtuse, its orthocenter $H$ is situated outside the triangle as shown above. Join $OC$ and $CH$ . Extend side $AB$ to meet $CH$ at $F$ . Since $F$ now becomes the foot of the altitude from $C$ , we have that $BFH=90^{\circ}$ . Since $OC$ is parallel to $AB$ , it follows that $\angle OCH=90^{\circ}$ .

Radius is perpendicular to tangent at the point of contact. Implies: $HC$ is a tangent to the circumcircle at $C$ .

Let $O$

and $H$

be the circumcenter and orthocenter of $\triangle ABC$

. If segment $CH$

is a tangent to the circumcircle at $C$

, PROVE that $a^2+b^2=4R^2$

Apply the Pythagorean theorem to the right triangle $OCH$ below:

$\begin{equation*} \begin{split} OH^2&=OC^2+CH^2\\ 9R^2-a^2-b^2-c^2&=R^2+(4R^2-c^2)\\ \therefore a^2+b^2&=4R^2 \end{split} \end{equation*}$

Let $O$

and $H$

be the circumcenter and orthocenter of $\triangle ABC$

. If segment $CH$

is tangent to the circumcircle at $C$

, PROVE that $H$

is a reflection of $C$

over side $AB$

We had $a^2+b^2=4R^2$ from the preceding example. This in turn implies $BH=a$ . So $\triangle BCH$ below is isosceles with $BH=BC$ :

Extend side $AB$ to meet $CH$ at $F$ . Since $F$ now becomes the foot of the altitude from $C$ , we have that $BFH=90^{\circ}$ . Altitude $BF$ bisects the base, so $CF=FH$ . This proves that $H$ is a reflection of $C$ over side $AB$ .

In $\triangle ABC$

, suppose that $(a^2-b^2)^2=(ac)^2+(cb)^2$

. PROVE that the nine-point center $N$

lies on side $AB$

internally or externally.

We focus on the internal case and use the segment addition postulate. Note that in any triangle $ABC$ with orthocenter $H$ , circumcenter $O$ , and nine-point center $N$ we have:

$AN^2=\frac{2AH^2+2AO^2-OH^2}{4}$

We proved before that $AH=b$ when $(a^2-b^2)^2=(ac)^2+(cb)^2$ . So

$AN^2=\frac{2AH^2+2AO^2-OH^2}{4}=\frac{2b^2+2R^2-(9R^2-a^2-b^2-c^2)}{4}$

After some simplifications, we obtain

$AN^2=\frac{(a^2-3b^2)^2}{16(a^2+b^2)}\implies AN=\frac{|(a^2-3b^2)|}{4\sqrt{(a^2+b^2)}}$

Similarly:

$BN=\frac{|(b^2-3a^2)|}{4\sqrt{(a^2+b^2)}}$

Let’s examine the absolute values. There are four cases to consider.

First, we can’t have $a^2-3b^2> 0$ and $b^2-3a^2> 0$ simultaneously. Otherwise, their sum must be greater than zero as well; but their sum is $-2(a^2+b^2)< 0$ .

Next, suppose that $3b^2-a^2> 0$ and $3a^2-b^2> 0$ . Then the sum is $2(a^2+b)^2$ , and so:

$AN=\frac{3b^2-a^2}{4\sqrt{(a^2+b^2)}},~~BN=\frac{3a^2-b^2}{4\sqrt{(a^2+b^2)}}\implies AN+BN=\frac{\sqrt{a^2+b^2}}{2}$

Because $a^2+b^2=4R^2$ , this leads to $AN+BN=R$ . This is a special case. If the points $A,B,N$ aren’t co-linear, then in $\triangle ABN$ , the median through $N$ passes through the nine-point circle, and so the length of this median is a radius of the nine-point circle, namely $\frac{R}{2}$ . We now have a triangle $ABN$ in which the sum of two sides is $R$ and a median has length $\frac{R}{2}$ . This is impossible (see the exercises at the end). Indeed, the side-lengths of $\triangle ABN$ have to be of the form $\frac{R-c}{2},\frac{R+c}{2},c$ for sides $AN,BN,AB$ (or sides $BN,AN,AB$ ). If we compute the cosine of the angle at $N$ , we obtain

$\cos N=\frac{\left(\frac{R-c}{2}\right)^2+\left(\frac{R+c}{2}\right)^2-c^2}{2\times \frac{R-c}{2}\times \frac{R+c}{2}}=1\implies\angle N=0$

The third and fourth cases are the same. For example, $3b^2-a^2> 0$ and $b^2-3a^2> 0$ . Then take

$AN=\frac{3b^2-a^2}{4\sqrt{(a^2+b^2)}},~~BN=\frac{b^2-3a^2}{4\sqrt{(a^2+b^2)}}$

and obtain

$AN+BN=\frac{4(b^2-a^2)}{4\sqrt{(a^2+b^2)}}=c$

If the nine-point center $N$

lies on $AB$

internally, PROVE that $(a^2-b^2)^2=(ac)^2+(cb)^2$

Using the fact that $AN+BN=AB=c$ , we have:

$\sqrt{\frac{2AH^2+2AO^2-OH^2}{4}}+\sqrt{\frac{2BH^2+2BO^2-OH^2}{4}}=c$

After all the trouble of the previous example, we don’t want to bother you with another seemingly lengthy procedure, but note that $(a^2-b^2)^2=(ac)^2+(cb)^2$ results after simplifications.

If the length of the bisector of $\angle C$

satisfies $l^2=\frac{2a^2b^2}{a^2+b^2}$

, PROVE that $(a^2-b)^2=(ac)^2+(cb)^2$

Normally, $l^2=ab\left(1-\left(\frac{c}{a+b}\right)^2\right)$ . So

$ab\left(1-\left(\frac{c}{a+b}\right)^2\right)= \frac{2a^2b^2}{a^2+b^2}\implies (a^2-b)^2=(ac)^2+(cb)^2$

If $(a^2-b)^2=(ac)^2+(cb)^2$

, PROVE that the length $l$

of the bisector of $\angle C$

satisfies $l^2=\frac{2ab}{a^2+b^2}$

First isolate $c^2$ from $(a^2-b)^2=(ac)^2+(cb)^2$ as $c^2=\frac{(a^2-b^2)^2}{a^2+b^2}$ and then use in the standard angle bisector formula:

$l^2=ab\left(1-\left(\frac{c}{a+b}\right)^2\right)=\frac{2a^2b^2}{a^2+b^2}.$

Peculiar scenario

Below is a reward of the labour in example 6.

Find a triangle for which the nine-point center coincides with a vertex.

Take $\triangle ABC$ with $\angle A=\angle C=30^{\circ},~\angle B=120^{\circ}$ . Then $b^2=3a^2$ . The nine-point center is precisely $B$ , since (according to example 6):

$BN=\frac{|(b^2-3a^2)|}{4\sqrt{(a^2+b^2)}}=0.$

Our next post will show that no other triangle has this property.

Takeaway

Let $\triangle ABC$ be a non-right triangle with side-lengths $a,b,c$ , altitudes $h_A,h_B,h_C$ , circumradius $R$ , circumcenter $O$ , orthocenter $H$ , and nine-point center $N$ . Then the following statements are equivalent:

$AH=b$
$BH=a$
$CH=2h_C$
$R=\frac{|a^2-b^2|}{2c}$
$h_C=R\cos C$
$\cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}$
$\cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}$
$\cos C=\frac{2ab}{a^2+b^2}$
$\cos^2 A+\cos^2 B=1$
$\sin^2 A+\sin^2 B=1$
$a\cos A+b\cos B=0$
$2\cos A\cos B+\cos C=0$
$OH^2=5R^2-c^2$
$h_A^2+h_B^2=AB^2$
$\frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}$
$a^2+b^2=4R^2$
$A-B=\pm 90^{\circ}$
$(a^2-b^2)^2=(ac)^2+(cb)^2$
$AH^2+BH^2+CH^2=8R^2-c^2$
radius $OC$ is parallel to side $AB$
segment $CH$ is tangent to the circumcircle at point $C$
the nine-point center lies on $AB$
the orthic triangle is obtuse isosceles
the geometric mean theorem holds
the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
the orthocenter is a reflection of vertex $C$ over side $AB$ .

No need to wait until 20 26 for twenty six equivalent statements.

Tasks

(Identical traits) Consider a right triangle with side-lengths , circumradius , circumcenter , nine-point center , and . Let be the circumcenter of and let be its circumradius. Similarly, let be the circumcenter of and let be its circumradius.
- Show that $N,O_a,O_b$ are co-linear.
- PROVE that $R^2=aR_b$ and $R^2=bR_a$ .
- Deduce that the ratio in which the bisector of $\angle C$ divides side $AB$ is the same as the circumradii ratio $\frac{R_a}{R_b}$ .
(Identical traits) Consider a non-right triangle with side-lengths , circumradius , circumcenter , and nine-point center . Let be the circumcenter of and let be its circumradius. Similarly, let be the circumcenter of and let be its circumradius. If :
- Show that $N,O_a,O_b$ are co-linear.
- PROVE that $R^2=aR_b$ and $R^2=bR_a$ .
- Deduce that the ratio in which the bisector of $\angle C$ divides side $AB$ is the same as the circumradii ratio $\frac{R_a}{R_b}$ .
  (There goes a non-right triangle that mimics a right triangle in many aspects.)
(Isosceles trapezium) Suppose that an obtuse satisfies any of the equivalent statements considered in this post. Let be a point on the circumcircle such that is a diameter.
- PROVE that $AD=BC$ .
- Deduce that quadrilateral $ABCD$ is an isosceles trapezium.
(Impossible triangle) In triangle , suppose that and that the length of the median from is , as per the special case encountered in the course of example 6.
- PROVE that the lengths of sides $AN$ and $BN$ must be $\frac{R-c}{2}$ and $\frac{R+c}{2}$ (or the other way).
- Using the triangle inequality, deduce that the three points $A,N,B$ are co-linear.
PROVE that the following two statements are equivalent for any triangle :
- $\angle C=90^{\circ}$
- the length $l$ of the bisector of $\angle C$ satisfies $l^2=\frac{2a^2b^2}{(a+b)^2}$ .