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A property of the Kepler triangle

On May 14, 2021, we showed that the side-lengths a,b,c (b< a< c) of the Kepler triangle satisfy the identity

(1)   \begin{equation*} (ac)^2=(ab)^2+(bc)^2 \end{equation*}

Or, in an upside-down form:

(2)   \begin{equation*} \frac{1}{b^2}=\frac{1}{a^2}+\frac{1}{c^2}\end{equation*}

Of all right triangles, the Kepler triangle is the only one in which both the side-lengths and the reciprocals of the side-lengths satisfy Pythagorean identities.

On top of this, the Kepler triangle is also unique in that both the altitudes and the reciprocals of the altitudes satisfy Pythagorean identities.

(3)   \begin{equation*} h_b^2=h_a^2+h_c^2 \end{equation*}

(4)   \begin{equation*} \frac{1}{h_c^2}=\frac{1}{h_a^2}+\frac{1}{h_b^2} \end{equation*}

(Only the Kepler triangle satisfies equation (3) above, but other right triangles satisfy equation (4).)

PROVE that equations (1) and (2) are equivalent.

Our very first example of 2022 is a breeze. To see that (1)\implies(2), re-arrange

    \[(ac)^2=(ab)^2+(bc)^2=b^2(a^2+c^2)\implies \frac{1}{b^2}=\frac{a^2+c^2}{(ac)^2}=\frac{1}{a^2}+\frac{1}{c^2}\]

On the other hand, (2)\implies(1) because

    \[\frac{1}{b^2}=\frac{1}{a^2}+\frac{1}{c^2}=\frac{a^2+c^2}{(ac)^2}\implies (ac)^2=b^2(a^2+c^2)=(ab)^2+(bc)^2\]

PROVE that every right triangle satisfies equation (4).

Let R be the radius of the circumcircle of any triangle ABC with side-lengths a,b,c. Then the altitudes h_a,h_b,h_c from vertices A,B,C are given by


In the case of a right triangle with hypotenuse c we have c=2R and so h_a=b,h_b=a,h_c=\frac{ab}{c}.


PROVE that equation (3) implies equation (1).

Since h_a=\frac{bc}{2R},h_b=\frac{ac}{2R},~h_c=\frac{ab}{2R}, we have

    \[h_b^2=h_a^2+h_c^2\implies\left(\frac{ac}{2R}\right)^2=\left(\frac{bc}{2R}\right)^2 +\left(\frac{ab}{2R}\right)^2\]

Clear the common denominator and it becomes clear that


PROVE that equation (1) implies equation (3).

Suppose that equation (1) holds, so that (ac)^2=(ab)^2+(bc)^2. Divide through by 4R^2 and use the expressions for the altitudes given in example 2.

    \begin{equation*} \begin{split} (ac)^2&=(ab)^2+(bc)^2\\ \frac{(ac)^2}{4R^2}&=\frac{(ab)^2}{4R^2}+\frac{(bc)^2}{4R^2}\\ \therefore h_b^2&=h_a^2+h_c^2 \end{split} \end{equation*}

PROVE that a right triangle is precisely the Kepler triangle, if it satisfies equation (3).

If a right triangle is of Kepler type, then by our post on May 14, 2021 , its side-lengths satisfy equation (1). By the preceding two examples, equations (1) and (3) are equivalent. Thus, a Kepler triangle satisfies equation (3).

Conversely, suppose that a right triangle with hypotenuse c satisfies equation (3). Then c^2=a^2+b^2 and (ac)^2=(ab)^2+(bc)^2. Multiply both sides of c^2=a^2+b^2 by a^2 to get (ac)^2=a^4+(ab)^2. Now combine with (ac)^2=(ab)^2+(bc)^2 to get

    \[a^4=(bc)^2\implies a^2=bc\]

In other words, the side-lengths a,b,c form a geometric progression of the form b,a,c. Thus, the right triangle is of the Kepler type.


Let ABC be a (right) triangle whose side-lengths a,b,c satisfy c^2=a^2+b^2. Then the following statements are equivalent:

  1. the sequence b,a,c is geometric
  2. \frac{c}{b} is the golden ratio
  3. (ac)^2=(ab)^2+(bc)^2
  4. \frac{1}{b^2}=\frac{1}{a^2}+\frac{1}{c^2}
  5. \frac{1}{h_b^2}=\frac{1}{h_a^2}+\frac{1}{h_c^2}.

Such a triangle is the Kepler triangle.


  • (Mid sixties) 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 sixty-five 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{b^2-a^2}{2c}
    14. h_C=R\cos C
    15. \cos A=\frac{b}{\sqrt{a^2+b^2}}
    16. \cos B=-\frac{a}{\sqrt{a^2+b^2}}
    17. \cos C=\frac{2ab}{a^2+b^2}
    18. \sin A=\frac{a}{\sqrt{a^2+b^2}}
    19. \sin B=\frac{b}{\sqrt{a^2+b^2}}
    20. \sin C=\frac{b^2-a^2}{a^2+b^2}
    21. \cos^2 A+\cos^2 B=1
    22. \sin^2 A+\sin^2 B=1
    23. a\cos A+b\cos B=0
    24. \sin A+\cos B=0
    25. \cos A-\sin B=0
    26. 2\cos A\cos B+\cos C=0
    27. 2\sin A\sin B-\cos C=0
    28. \cos A\cos B+\sin A\sin B=0
    29. a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C
    30. \sin A\sin B=\frac{ab}{b^2-a^2}\sin C
    31. \sin^2B-\sin^2A=\sin C
    32. OH^2=5R^2-c^2
    33. h_A^2+h_B^2=AB^2
    34. \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}
    35. a^2+b^2=4R^2
    36. \left(c+2AF_C\right)^2=a^2+b^2 or \left(c+2BF_C\right)^2=a^2+b^2
    37. A-B=\pm 90^{\circ}
    38. (a^2-b^2)^2=(ac)^2+(cb)^2
    39. AH^2+BH^2+CH^2=8R^2-c^2
    40. b=2R\cos A
    41. \triangle ABH is congruent to \triangle ABC
    42. \triangle OO^aO^b is congruent to \triangle ABC
    43. \triangle CNO is isosceles with CN=NO
    44. \triangle CNH is isosceles with CN=NH
    45. \triangle CHO is right angled at C
    46. N is the circumcenter of \triangle CHO
    47. \triangle O^cOC is right-angled at O
    48. \triangle O^cHC is right-angled at H
    49. quadrilateral O^cOHC is a rectangle
    50. the points O^c,O,C,H are concyclic with OH as diameter
    51. the reflection O^b of O over AC lies internally on AB
    52. the reflection O^a of O over BC lies externally on AB
    53. radius OC is parallel to side AB
    54. F_A is the reflection of F_B over side AB
    55. the nine-point center lies on AB
    56. the orthic triangle is isosceles with F_AF_C=F_BF_C
    57. the geometric mean theorem holds
    58. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
    59. the orthocenter is a reflection of vertex C over side AB
    60. segment HC is tangent to the circumcircle at point C
    61. median CM has the same length as the segment HM
    62. the bisector MO of AB is tangent to the nine-point circle at M
    63. AF_ABF_B is a convex kite with diagonals AB and F_AF_B
    64. altitude CF_C is tangent to the nine-point circle at F_C
    65. segment HF_C is tangent to the nine-point circle at F_C.
      (10 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|.