Reflecting the circumcenter II

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

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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.

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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}).

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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.

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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:

  1. O^a=A
  2. \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 ABC, let a,b,c be the side-lengths, h_A,h_B,h_C the altitudes, F_A,F_B, F_C the feet of the altitudes from the respective vertices, R the circumradius, O the circumcenter, N the nine-point center, H the orthocenter, M the midpoint of side AB, O^c the reflection of O over side AB, O^b the reflection of O over side AC, and O^a the reflection of O over side BC. 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|.