Reflecting the circumcenter I

In \triangle ABC, let a,b,c be the usual side-lengths and let O be the circumcenter. Let O^a,O^b,O^c denote the reflections of O over sides BC,CA,AB respectively. Then \triangle O^aO^bO^c is congruent to the parent \triangle ABC. If a,b,c satisfy the identity

(1)   \begin{equation*} (a^2-b^2)^2=(ac)^2+(cb)^2 \end{equation*}

then in addition we have that \triangle OO^aO^b is congruent to \triangle ABC.

Congruent triangles

In a non-right \triangle ABC, let O^a be the reflection of the circumcenter O over side BC. Then OO^a=b if and only if equation (1) holds.

Let’s prove one direction: OO^a=b\implies (1). Consider \triangle ABC shown below:

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We showed previously that AH=OO^a in any triangle with orthocenter H. Thus AH=b. By one of the equivalent statements here, we conclude that equation (1) holds.

In a non-right \triangle ABC, let O^b be the reflection of the circumcenter O over side AC. Then OO^b=a if and only if equation (1) holds.

Similar to example 1 above. If O^b is the reflection of the circumcenter over side AC, then we always have BH=OO^b in any triangle. This would then lead to BH=a, which by the equivalent statements here implies equation (1).

Let O be the circumcenter of \triangle ABC, and let O^a, O^b be the reflection of O over side BC and side CA, respectively. Then \triangle OO^aO^b is congruent to the parent \triangle ABC, if equation (1) holds.

This follows from the two preceding examples, together with the fact that the segment O^aO^b has length c. (Recall that \triangle O^aO^bO^c is congruent to the parent \triangle ABC in such a way that O^bO^a=c, O^aO^c=b, and O^cO^b=a.)

Cyclic trapezium

Let O^a be the reflection of the circumcenter O over side BC. If equation (1) holds, then the quadrilateral AOCO^a is an isosceles trapezium, hence cyclic.

Note that under the assumption of equation (1), we have that the reflection O^a lies on side AB. Let R be the circumradius of the parent triangle ABC. Then AO=R and also CO^a=R. Since radius OC is parallel to side AB (see here), we get an isosceles trapezium AOCO^a.

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Let O^a be the reflection of the circumcenter O over side BC, and let O^c be the reflection of O over side AB. If equation (1) holds, then the quadrilateral AO^cHO^a is an isosceles trapezium, hence cyclic. (Here, H is the orthocenter.)

Similar to example 4 above.

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Takeaway

In \triangle ABC, let a,b,c be the side-lengths, O the circumcenter, 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. Then the following statements are equivalent:

  1. OO^a=b
  2. OO^b=a
  3. (a^2-b^2)^2=(ac)^2+(cb)^2

As usual, we have more of these in the exercises.

Task

  • (Mid 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-six 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 CNO is isosceles with CN=NO
    35. \triangle CNH is isosceles with CN=NH
    36. \triangle CHO is right angled at C
    37. N is the circumcenter of \triangle CHO
    38. \triangle O^cOC is right-angled at O
    39. \triangle O^cHC is right-angled at H
    40. quadrilateral O^cOHC is a rectangle
    41. the points O^c,O,C,H are concyclic with OH as diameter
    42. the reflection O^b of O over AC lies internally on AB
    43. the reflection O^a of O over BC lies externally on AB
    44. radius OC is parallel to side AB
    45. F_A is the reflection of F_B over side AB
    46. the nine-point center lies on AB
    47. the orthic triangle is isosceles with F_AF_C=F_BF_C
    48. the geometric mean theorem holds
    49. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
    50. the orthocenter is a reflection of vertex C over side AB
    51. segment HC is tangent to the circumcircle at point C
    52. median CM has the same length as the segment HM
    53. the bisector MO of AB is tangent to the nine-point circle at M
    54. AF_ABF_B is a convex kite with diagonals AB and F_AF_B
    55. altitude CF_C is tangent to the nine-point circle at F_C
    56. segment HF_C is tangent to the nine-point circle at F_C.
      (19 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|.