An upside-down Pythagorean identity

Towards the end of our post on June 28, we stated that if the side-lengths of a triangle satisfy the modified Pythagorean identity

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

then the following statements are equivalent:

  • \frac{b}{a} is the golden ratio
  • (ab)^2=(ac)^2+(cb)^2
  • \sin A\sin B=\sin C
  • \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}
  • h_C=c

Today’s post will now establish the above equivalence. (Note that the fourth condition, \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}, which motivated today’s title, was coined by the author of this article.)

If \frac{b}{a} is the golden ratio, PROVE that equation (1) becomes (ab)^2=(ac)^2+(cb)^2.

Suppose that \frac{b}{a} is the golden ratio, then \left(\frac{b}{a}\right)^2-\frac{b}{a}-1=0, and so b^2-ab-a^2=0. This then gives b^2-a^2=ab. In turn:

    \[(b^2-a^2)^2=(ac)^2+(cb)^2\implies  (ab)^2=(ac)^2+(cb)^2\]

In addition to equation (1), suppose that (ab)^2=(ac)^2+(cb)^2. PROVE that: \sin A\sin B=\sin C.

Let R be the circumradius. Then, by the extended law of sines, we have:

    \[\sin A=\frac{a}{2R},~\sin B=\frac{b}{2R},~\sin C=\frac{c}{2R}.\]

The assumption (ab)^2=(ac)^2+(cb)^2 implies ab=c\sqrt{a^2+b^2}. (Henceforth we’ll frequently use the fact that a^2+b^2=4R^2 is equivalent to equation (1).) So:

    \[\sin A\sin B=\left(\frac{ab}{4R^2}\right)=\frac{c\sqrt{a^2+b^2}}{4R^2}=\frac{c(2R)}{4R^2}=\frac{c}{2R}=\sin C\]

In addition to equation (1), suppose that \sin A\sin B=\sin C. PROVE that \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}.

Using the extended law of sines as in the previous example, the condition \sin A\sin B=\sin C gives ab=c(2R). Since 2R=\sqrt{a^2+b^2}, we obtain

    \[ab=c\sqrt{a^2+b^2}\implies \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}.\]

So-called upside-down Pythagorean identity. Ours works under an added condition, namely equation (1).

In addition to equation (1), suppose that \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}. PROVE that h_C=c, where h_C is the length of the altitude from vertex C and c=AB as per the usual notation.

Re-arrange the given condition \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2} as


and obtain ab=c(2R). The length of the altitude from vertex C is given by


Suppose that h_C=c, in addition to equation (1). PROVE that \frac{b}{a} is the golden ratio.

If h_C=c, then \frac{ab}{2R}=c\implies ab=c(2R)=c\sqrt{a^2+b^2}\implies (ab)^2=c^2(a^2+b^2). Now use this in equation (1) to get

    \[(b^2-a^2)^2=(ab)^2\implies b^2-a^2=ab\implies \left(\frac{b}{a}\right)^2-\frac{b}{a}-1=0\]

Thus, \frac{b}{a} is the golden ratio.


Let ABC be a triangle whose side-lengths a,b,c satisfy equation (1). Then the following statements are equivalent:

  1. \frac{b}{a} is the golden ratio
  2. \sin A\sin B=\sin C
  3. \tan C=\frac{1}{2}.

The last condition fixes the interior angles of the given triangle ABC: \angle A=45^{\circ}-\frac{1}{2}\tan^{-1}\left(\frac{1}{2}\right)\approx 31.7^{\circ}, \angle B=135^{\circ}-\frac{1}{2}\tan^{-1}\left(\frac{1}{2}\right)\approx 121.7^{\circ}, \angle C=\tan^{-1}\left(\frac{1}{2}\right)\approx 26.6^{\circ}.


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