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The silver ratio and a polar circle

Numerically, the silver ratio is equal to 1+\sqrt{2}, and may be denoted by \delta_s.

Next, a polar circle is a circle centred at the orthocenter of a triangle, with radius r related to the circumradius R and the side-lengths a,b,c via

(1)   \begin{equation*} \begin{split} r^2&=4R^2-\frac{1}{2}\left(a^2+b^2+c^2\right) \end{split} \end{equation*}

Now suppose that a,b,c satisfy the identity:

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

Notice that any triangle in which equation (2) holds is necessarily obtuse, and so it’s polar circle is defined, with equation (1) simplifying to:

(3)   \begin{equation*} \begin{split} r^2&=\frac{1}{2}\left(a^2+b^2-c^2\right) \end{split} \end{equation*}

Not only this, it happens that r=R precisely when \frac{a}{b} is the silver ratio.

Reduced attacks

A little bit of our private life before we proceed to something less irrelevant: We’re so glad that, as it stands, we have the upper hand in our fight against unwanted attacks:

The barrage was quite something prior to two weeks ago:

Grateful to find a barricade. It’s a different story now.

Refocused attention

Sorry we had to bother you with the above. Once in a while we’ll let you see what happens behind our scenes.

If the side-lengths a,b,c of \triangle ABC satisfy equation (2), PROVE that the radius of its polar circle satisfies equation (3).

We’re to show that if (a^2-b^2)^2=(ac)^2+(cb)^2, then the radius of the polar circle is r^2=\frac{1}{2}\left(a^2+b^2-c^2\right). Use the fact that in a non-right triangle, we have (a^2-b^2)^2=(ac)^2+(cb)^2\iff a^2+b^2=4R^2.

Suppose that (a^2-b^2)^2=(ac)^2+(cb)^2, and that the radius of the polar circle coincides with the circumradius. PROVE that \frac{a}{b} is the silver ratio.

We had r^2=\frac{1}{2}\left(a^2+b^2-c^2\right) from the preceding example. And also R^2=\frac{a^2+b^2}{4}. If we set r^2=R^2 we get a^2+b^2=2c^2 after some simplification. In addition, c^2=\frac{(a^2-b^2)^2}{a^2+b^2}. Thus: a^2+b^2=2\left(\frac{(a^2-b^2)^2}{a^2+b^2}\right). Further simplification: a^4-6a^2b^2+b^4=0\implies\left(\frac{a}{b}\right)^2=(1\pm \sqrt{2})^2. Take the positive value: \frac{a}{b}=1+\sqrt{2}, the silver ratio.

If \frac{a}{b} is the silver ratio and (a^2-b^2)^2=(ac)^2+(cb)^2, PROVE that the radius of the polar circle coincides with the circumradius.

Similar to the preceding example.

Determine the value of \angle C under the conditions \frac{a}{b}=1+\sqrt{2} and (a^2-b^2)^2=(ac)^2+(cb)^2.

We had a^2+b^2=2c^2. Since \frac{a}{b}=1+\sqrt{2}, we then obtain c^2=\sqrt{2}ab. Cosine formula: \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{2c^2-c^2}{2\left(\frac{c^2}{\sqrt{2}}\right)}=\frac{\sqrt{2}}{2}\implies C=45^{\circ}.

Radical axis

Under the conditions \frac{a}{b}=1+\sqrt{2} and (a^2-b^2)^2=(ac)^2+(cb)^2, show that the radical axis of the circumcircle and the polar circle goes through the nine-point center.

Both the circumcircle and the polar circle have the same radius, under the given conditions. Thus the radical axis is the right bisector of the line joining their centers. But then the center of one is the circumcenter, while the center of the other is the orthocenter. The midpoint of these two centers is the nine-point center.


Suppose that triangle ABC has side-lengths a,b,c, radius of polar circle r, and circumradius R. If (a^2-b^2)^2=(ac)^2+(cb)^2, then the following statements are equivalent:

  1. r=R
  2. \frac{a}{b} is the silver ratio.


  • (Geometric mean) Suppose that the side-lengths of \triangle ABC satisfy (a^2-b^2)^2=(ac)^2+(cb)^2. PROVE that the following statements are equivalent:
    1. \frac{a}{b} is the silver ratio
    2. the nine-point center coincides with the foot of the symmedian from vertex C
    3. the radius of the nine-point circle is the geometric mean of the two segments into which the nine-point center divides side AB.
  • (Growing membership) 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, M_a,M_b,M_c the midpoints of sides BC,CA,AB in that order, E_a,E_b,E_c the Euler points, R the circumradius, O the circumcenter, N the nine-point center, H the orthocenter, O^c the reflection of O over side AB, O^b the reflection of O over side AC, O^a the reflection of O over side BC, K the symmedian point, K_c the foot of the symmedian from vertex C, and r the radius of the polar circle. PROVE that the following eighty-three statements are equivalent:
    1. E_c=F_c
    2. AH=b
    3. BH=a
    4. AE_a=\frac{b}{2}
    5. BE_b=\frac{a}{2}
    6. E_aM_c=\frac{a}{2}
    7. E_bM_c=\frac{b}{2}
    8. E_aM_b=h_c
    9. E_aF_b=\frac{AH}{2}
    10. E_bF_a=\frac{BH}{2}
    11. OO^a=b
    12. OO^b=a
    13. OM_a=\frac{b}{2}
    14. OM_b=\frac{a}{2}
    15. OM_c=h_c
    16. CH=2h_c
    17. h_a=AF_b
    18. h_b=BF_a
    19. AF_c=\frac{b^2}{2R}
    20. BF_c=\frac{a^2}{2R}
    21. \frac{a}{c} =\frac{h_c}{AF_b}
    22. \frac{b}{c}=\frac{h_c}{BF_a}
    23. \frac{a}{b}=\frac{BF_a}{AF_b}
    24. R=\frac{b^2-a^2}{2c}
    25. \frac{CK}{KK_c}=\left(\frac{a^2+b^2}{a^2-b^2}\right)^2
    26. r^2=\frac{1}{2}\left(a^2+b^2-c^2\right)
    27. h_c=R\cos C
    28. \cos A=\frac{b}{\sqrt{a^2+b^2}}
    29. \cos B=-\frac{a}{\sqrt{a^2+b^2}}
    30. \cos C=\frac{2ab}{a^2+b^2}
    31. \sin A=\frac{a}{\sqrt{a^2+b^2}}
    32. \sin B=\frac{b}{\sqrt{a^2+b^2}}
    33. \sin C=\frac{b^2-a^2}{a^2+b^2}
    34. \cos^2 A+\cos^2 B=1
    35. \sin^2 A+\sin^2 B=1
    36. a\cos A+b\cos B=0
    37. \sin A+\cos B=0
    38. \cos A-\sin B=0
    39. 2\cos A\cos B+\cos C=0
    40. 2\sin A\sin B-\cos C=0
    41. \cos A\cos B+\sin A\sin B=0
    42. a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C
    43. \sin A\sin B=\frac{ab}{b^2-a^2}\sin C
    44. \sin^2B-\sin^2A=\sin C
    45. \cos^2A-\cos^2B=\sin C
    46. OH^2=5R^2-c^2
    47. h_a^2+h_b^2=AB^2
    48. \frac{h_a}{a}+\frac{h_b}{b}=\frac{c}{h_c}
    49. a^2+b^2=4R^2
    50. b=2R\cos A
    51. A-B=\pm 90^{\circ}
    52. (a^2-b^2)^2=(ac)^2+(cb)^2
    53. AH^2+BH^2+CH^2=8R^2-c^2
    54. \left(c+2AF_c\right)^2=a^2+b^2 or \left(c+2BF_c\right)^2=a^2+b^2
    55. E_a is the reflection of M_b over side AB
    56. E_b is the reflection of M_a over side AB
    57. \triangle ABH is congruent to \triangle ABC
    58. \triangle OO^aO^b is congruent to \triangle ABC
    59. \triangle CNO is isosceles with CN=NO
    60. \triangle CNH is isosceles with CN=NH
    61. \triangle CHO is right angled at C
    62. N is the circumcenter of \triangle CHO
    63. \triangle O^cOC is right-angled at O
    64. \triangle O^cHC is right-angled at H
    65. quadrilateral O^cOHC is a rectangle
    66. the points O^c,O,C,H are concyclic with OH as diameter
    67. the reflection O^b of O over AC lies internally on AB
    68. the reflection O^a of O over BC lies externally on AB
    69. radius OC is parallel to side AB
    70. F_a is the reflection of F_b over side AB
    71. segment F_aF_b is perpendicular to side AB
    72. the nine-point center lies on AB
    73. the orthic triangle is isosceles with F_aF_c=F_bF_c
    74. the geometric mean theorem holds
    75. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
    76. the orthocenter is a reflection of vertex C over side AB
    77. segment HC is tangent to the circumcircle at point C
    78. median CM_c has the same length as the segment HM_c
    79. the bisector M_cO of AB is tangent to the nine-point circle at M_c
    80. AF_aBF_b is a convex kite with diagonals AB and F_aF_b
    81. altitude CF_c is tangent to the nine-point circle at F_c
    82. chord F_cM_c is a diameter of the nine-point circle
    83. segment HF_c is tangent to the nine-point circle at F_c.
      (Five more members were recently added.)

There are milestones that one cannot afford to miss; otherwise, life may become boring, dry, insipid, lifeless. Surprisingly, the milestone may be as simple as learning a new thing, acquiring a new skill, and so on. In a nutshell: don’t miss your milestones; we didn’t miss ours.

That came on Thursday, June 14, 2018: beautiful, beautiful day. This person will never forget that day. By means of God’s merciful and meticulous redirection, there came a simple realization: learn HTML. Personally, such a simple thing is a big deal because an ingrained inclination meant such a step was outside regular imagination.

This then implies constant gratitude to God for that remarkable day. All things being equal, the next iteration of the public gratitude comes up on Friday, October 14, 2022. Until then: stay safe, do math, and give thanks.