Reflecting the circumcenter II – High School Geometry

If we reflect the orthocenter of a triangle over the sides of the triangle, the reflections lie on the circumcircle of the parent triangle.

If we now reflect the circumcenter over the sides of the triangle, the reflections do not lie on the circumcircle, unless a certain condition is met.

Sixty degrees

In the presence of an interior angle of $60^{\circ}$ or $120^{\circ}$ , the reflection of the circumcenter lies on the circumcircle.

In any $\triangle ABC$

, let $O^a$

be the reflection of the circumcenter $O$

over side $BC$

. If $O^a$

lies on the circumcircle, then either $\angle A=60^{\circ}$

or $\angle A=120^{\circ}$

Let $R$ be the circumradius of the triangle, and let $H$ be its orthocenter. In the diagram below

suppose that $O^a$ actually lies on the circumcircle. Then $OO^a$ must equal $R$ , right? We showed previously that $AH=OO^a$ in any triangle with orthocenter $H$ . But then $AH$ always equals $\sqrt{4R^2-a^2}$ . And so:

$\begin{equation*} \begin{split} R&=\sqrt{4R^2-a^2}\\ a^2&=3R^2\\ a^2&=3\left(\frac{a}{2\sin A}\right)^2\\ 4\sin^2A&=3\\ \sin A&=\pm\frac{\sqrt{3}}{2}\\ \therefore \angle A&=60^{\circ},120^{\circ} \end{split} \end{equation*}$

The proof can also be accomplished by using angle arguments. Particularly the fact that the angle an arc subtends at the center of a circle is twice the angle the arc subtends at any other part of the circumference.

In $\triangle ABC$

, suppose that $\angle A=60^{\circ}$

(or that $\angle A=120^{\circ}$

). PROVE that the reflection of the circumcenter over side $BC$

lies on the circumcircle.

As before, let’s denote by $O^a$ the reflection of the circumcenter $O$ over side $BC$ . We’ll establish the result for the $\angle A=60^{\circ}$ case.

The angle which the minor arc through $B$ and $C$ subtends at the center is equal to $2\times 60^{\circ}$ ; that is, obtuse $\angle BOC=120^{\circ}$ , and so reflex $\angle BOC=240^{\circ}$ . Note that $BOCO^a$ is a rhombus, and so obtuse $\angle BO^aC=120^{\circ}$ .

Since reflex $\angle BOC=240^{\circ}=2\times 120^{\circ}=2\times\text{ obtuse }\angle BO^aC$ , we conclude that $O^a$ lies on the circumcircle.

Simple deduction

As a consequence of the preceding examples, we can characterize equilateral triangles via reflections of the circumcenter.

A triangle is equilateral if and only if the reflections of the circumcenter over the three sides ALL lie on the circumcircle of the triangle.

Easy.

Sample diagrams

Consider $\triangle ABC$

with vertices at $A(0,0)$

, $B(4,0)$

, and $C(1,\sqrt{3})$

Note that this is a right triangle in which $\angle A=60^{\circ}$ , $\angle B=30^{\circ}$ , and $\angle C=90^{\circ}$ . The circumcenter is the point $O(2,0)$ , and its reflection over side $BC$ is the point $O^a(3,\sqrt{3})$ .

Consider $\triangle ABC$

with vertices at $A(0,0)$

, $B(-4,0)$

, and $C(2,2\sqrt{3})$

Here, $\angle A=120^{\circ}$ , and $\angle B=\angle C=30^{\circ}$ . An extremely pleasant case.

Takeaway

In $\triangle ABC$ , let $O$ be the circumcenter, and let $O^a$ be the reflection of $O$ over side $BC$ . Then the following statements are equivalent:

$O^a=A$
$\angle A=120^{\circ},~\angle B=\angle C=30^{\circ}$ .

And a few more, if you view this post.

Task

(Late fifties) In a non-right triangle , let be the side-lengths, the altitudes, the feet of the altitudes from the respective vertices, the circumradius, the circumcenter, the nine-point center, the orthocenter, the midpoint of side , the reflection of over side , the reflection of over side , and the reflection of over side . PROVE that the following fifty-seven statements are equivalent:
1. $AH=b$
2. $BH=a$
3. $OO^a=b$
4. $OO^b=a$
5. $CH=2h_C$
6. $h_A=AF_B$
7. $h_B=BF_A$
8. $AF_C=\frac{b^2}{2R}$
9. $BF_C=\frac{a^2}{2R}$
10. $\frac{a}{c} =\frac{h_C}{AF_B}$
11. $\frac{b}{c}=\frac{h_C}{BF_A}$
12. $\frac{a}{b}=\frac{BF_A}{AF_B}$
13. $R=\frac{|a^2-b^2|}{2c}$
14. $h_C=R\cos C$
15. $\cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}$
16. $\cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}$
17. $\cos C=\frac{2ab}{a^2+b^2}$
18. $\cos^2 A+\cos^2 B=1$
19. $\sin^2 A+\sin^2 B=1$
20. $a\cos A+b\cos B=0$
21. $\sin A+\cos B=0$
22. $\cos A-\sin B=0$
23. $2\cos A\cos B+\cos C=0$
24. $a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C$
25. $OH^2=5R^2-c^2$
26. $h_A^2+h_B^2=AB^2$
27. $\frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}$
28. $a^2+b^2=4R^2$
29. $A-B=\pm 90^{\circ}$
30. $(a^2-b^2)^2=(ac)^2+(cb)^2$
31. $AH^2+BH^2+CH^2=8R^2-c^2$
32. $a=2R\sin A,~b=2R\cos A,~c=2R\cos 2A$
33. $\triangle ABH$ is congruent to $\triangle ABC$
34. $\triangle OO^aO^b$ is congruent to $\triangle ABC$
35. $\triangle CNO$ is isosceles with $CN=NO$
36. $\triangle CNH$ is isosceles with $CN=NH$
37. $\triangle CHO$ is right angled at $C$
38. $N$ is the circumcenter of $\triangle CHO$
39. $\triangle O^cOC$ is right-angled at $O$
40. $\triangle O^cHC$ is right-angled at $H$
41. quadrilateral $O^cOHC$ is a rectangle
42. the points $O^c,O,C,H$ are concyclic with $OH$ as diameter
43. the reflection $O^b$ of $O$ over $AC$ lies internally on $AB$
44. the reflection $O^a$ of $O$ over $BC$ lies externally on $AB$
45. radius $OC$ is parallel to side $AB$
46. $F_A$ is the reflection of $F_B$ over side $AB$
47. the nine-point center lies on $AB$
48. the orthic triangle is isosceles with $F_AF_C=F_BF_C$
49. the geometric mean theorem holds
50. the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
51. the orthocenter is a reflection of vertex $C$ over side $AB$
52. segment $HC$ is tangent to the circumcircle at point $C$
53. median $CM$ has the same length as the segment $HM$
54. the bisector $MO$ of $AB$ is tangent to the nine-point circle at $M$
55. $AF_ABF_B$ is a convex kite with diagonals $AB$ and $F_AF_B$
56. altitude $CF_C$ is tangent to the nine-point circle at $F_C$
57. segment $HF_C$ is tangent to the nine-point circle at $F_C$ .
  ( $18$ short of the target.)
(Extra feature) If $\triangle ABC$ satisfies equation (??), PROVE that its nine-point center $N$ divides $AB$ in the ratio $|3b^2-a^2|:|b^2-3a^2|$ .