Twenty one equivalent statements

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:

  1. AH=b
  2. BH=a
  3. CH=2h_C
  4. \cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}
  5. \cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}
  6. \cos C=\frac{2ab}{a^2+b^2}
  7. \cos^2 A+\cos^2 B=1
  8. \sin^2 A+\sin^2 B=1
  9. a\cos A+b\cos B=0
  10. h_A^2+h_B^2=AB^2
  11. \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}
  12. a^2+b^2=4R^2
  13. A-B=\pm 90^{\circ}
  14. (a^2-b^2)^2=(ac)^2+(cb)^2
  15. the orthic triangle is obtuse isosceles
  16. radius OC is parallel to side AB
  17. the nine-point center lies on AB
  18. the geometric mean theorem holds
  19. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
  20. segment CH is tangent to the circumcircle at point C
  21. 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:

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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:

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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:

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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:

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    \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:

Rendered by QuickLaTeX.com

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.

We 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:

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

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:

  1. AH=b
  2. BH=a
  3. CH=2h_C
  4. R=\frac{|a^2-b^2|}{2c}
  5. h_C=R\cos C
  6. \cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}
  7. \cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}
  8. \cos C=\frac{2ab}{a^2+b^2}
  9. \cos^2 A+\cos^2 B=1
  10. \sin^2 A+\sin^2 B=1
  11. a\cos A+b\cos B=0
  12. 2\cos A\cos B+\cos C=0
  13. OH^2=5R^2-c^2
  14. h_A^2+h_B^2=AB^2
  15. \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}
  16. a^2+b^2=4R^2
  17. A-B=\pm 90^{\circ}
  18. (a^2-b^2)^2=(ac)^2+(cb)^2
  19. AH^2+BH^2+CH^2=8R^2-c^2
  20. radius OC is parallel to side AB
  21. segment CH is tangent to the circumcircle at point C
  22. the nine-point center lies on AB
  23. the orthic triangle is obtuse isosceles
  24. the geometric mean theorem holds
  25. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
  26. the orthocenter is a reflection of vertex C over side AB.

No need to wait until 2026 for twenty six equivalent statements.

Tasks

  1. (Identical traits) Consider a right triangle ABC with side-lengths a,b,c, circumradius R, circumcenter O, nine-point center N, and \angle C=90^{\circ}. Let O_a be the circumcenter of \triangle BCO and let R_a be its circumradius. Similarly, let O_b be the circumcenter of \triangle CAO and let R_b 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}.
  2. (Identical traits) Consider a non-right triangle ABC with side-lengths a,b,c, circumradius R, circumcenter O, and nine-point center N. Let O_a be the circumcenter of \triangle BCO and let R_a be its circumradius. Similarly, let O_b be the circumcenter of \triangle CAO and let R_b be its circumradius. If (a^2-b^2)^2=(ac)^2+(cb)^2:
    • 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.)
  3. (Isosceles trapezium) Suppose that an obtuse \triangle ABC satisfies any of the equivalent statements considered in this post. Let D be a point on the circumcircle such that CD is a diameter.
    • PROVE that AD=BC.
    • Deduce that quadrilateral ABCD is an isosceles trapezium.
  4. (Impossible triangle) In triangle ABN, suppose that AN+NB=R and that the length of the median from N is \frac{R}{2}, 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.
  5. PROVE that the following two statements are equivalent for any triangle ABC:
    • \angle C=90^{\circ}
    • the length l of the bisector of \angle C satisfies l^2=\frac{2a^2b^2}{(a+b)^2}.