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A special chord of the nine-point circle

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In the above diagram we have the nine-point circle of triangle ABC going through the midpoints M_a,M_b,M_c (of sides BC,CA,AB respectively), and the feet of the altitudes F_a,F_b,F_c (from A,B,C in that order).

It will be shown that

(1)   \begin{equation*} F_cM_c^2=F_aM_a^2+F_bM_b^2 \end{equation*}

if (and only if) the side-lengths a,b,c satisfy

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

It will also be shown, in our first example, that we have the relations

(3)   \begin{equation*} \begin{split} F_cM_c&=\left(\frac{b^2-a^2}{2c}\right)\\ F_bM_b&=\left(\frac{c^2-a^2}{2b}\right)\\ F_aM_a&=\left(\frac{c^2-b^2}{2a}\right)\\ \end{split} \end{equation*}

in any triangle (append absolute values if need be).

In the meantime, note that the right side of equation (1) evaluates to R^2 for a right triangle, where R is the circumradius, whereas both sides of equation (1) evaluate to R^2 under (2). And so in the latter case, the segment F_cM_c is not just a chord, but a diameter of the nine-point circle.

Derive the three equations in (3).

Let’s do this for an acute triangle. Slight modification for an obtuse triangle will be needed. By drawing an appropriate diagram we have:

    \begin{equation*} \begin{split} M_aF_a&=\frac{a}{2}-b\cos C \text{ or } c\cos B-\frac{a}{2}\\ M_bF_b&=\frac{b}{2}-a\cos C \text{ or } c\cos A-\frac{b}{2}\\ M_cF_c&=\frac{c}{2}-a\cos B \text{ or } b\cos A-\frac{c}{2}\\ \end{split} \end{equation*}

We can re-write the expression for M_cF_c entirely in terms of the side-lengths using the cosine formula:

    \begin{equation*} \begin{split} M_cF_c&=\frac{c}{2}-a\cos B\\ &=\frac{c}{2}-a\left(\frac{a^2+c^2-b^2}{2ac}\right)\\ &=\frac{b^2-a^2}{2c} \end{split} \end{equation*}

Similarly we obtain the expressions for M_aF_a and M_bF_b.

Let ABC be an isosceles triangle in which AC=BC. PROVE that the midpoint of side AB coincides with the foot of the altitude from vertex C.

Since AC=BC, we have a=b by the usual notation. Using example 1 above we have


and so M_c=F_c.

In a right triangle with hypotenuse of length c, PROVE that the right side of equation (1) is the square of the circumradius.

We first have c^2=a^2+b^2. Moreover, if R is the circumradius, then c=2R for a right triangle. Now:

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

If the side-lengths of a triangle satisfy equation (2), PROVE that equation (1) holds.

Such a triangle is necessarily non-right. Let R be its circumradius. By one of the equivalent statements here, we know that equation (2) then becomes equivalent to a^2+b^2=4R^2. Re-arrange equation (2) in the form c^2=\frac{(b^2-a^2)^2}{a^2+b^2} and consider both sides of equation (1):

    \begin{equation*} \begin{split} \text{LS}&=F_cM_c^2\\ &=\left(\frac{b^2-a^2}{2c}\right)^2\\ &=\left(\frac{(b^2-a^2)^2}{4\times \frac{(b^2-a^2)^2}{a^2+b^2} }\right)\\ &=\frac{a^2+b^2}{4}\\ &=R^2\\ \text{RS}&=F_aM_a^2+F_bM_b^2\\ &=\left(\frac{c^2-b^2}{2a}\right)^2+\left(\frac{c^2-a^2}{2b}\right)^2\\ &=\left(\frac{\frac{(b^2-a^2)^2}{a^2+b^2}-b^2}{2a}\right)^2+\left(\frac{\frac{(b^2-a^2)^2}{a^2+b^2}-a^2}{2b}\right)^2\\ &=\frac{\Big((b^2-a^2)^2-b^2(a^2+b^2)\Big)^2}{4a^2(a^2+b^2)^2}+\frac{\Big((b^2-a^2)^2-a^2(a^2+b^2)\Big)^2}{4b^2(a^2+b^2)^2}\\ &=\left(\frac{a^2b^4(b^2-3a^2)^2+a^4b^2(a^2-3b^2)^2}{4a^2b^2(a^2+b^2)^2}\right)\\ &=\frac{a^6+3a^4b^2+3a^2b^4+b^6}{4(a^2+b^2)^2}\\ &=\frac{(a^2+b^2)^3}{4(a^2+b^2)^2}\\ &=\frac{a^2+b^2}{4}\\ &=R^2\\ \therefore \text{LS}&=\text{RS} \end{split} \end{equation*}

PROVE that the segment F_cM_c is a diameter of the nine-point circle, if equation (2) holds.

By example 4 above, we had F_cM_c^2=R^2. Since the nine-point circle goes through F_c and M_c and has radius equal to \frac{R}{2}, the fact that F_cM_c=R means that the chord F_cM_c is a diameter.

No other triangle has this property.


In triangle ABC, let E_c denote the Euler point of vertex C, F_c the foot of the altitude from vertex C, and M_c the midpoint of side AB. Then the following statements are equivalent:

  1. the chord F_cM_c is a diameter of the nine-point circle
  2. E_c=F_c.

The equivalent conditions range from extremely simple to moderately involved.


  • (Late 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, M_a,M_b,M_c the midpoints of sides BC,CA,AB, 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, and O^a the reflection of O over side BC. PROVE that the following sixty-eight statements are equivalent:
    1. E_c=F_c
    2. AH=b
    3. BH=a
    4. OO^a=b
    5. OO^b=a
    6. CH=2h_c
    7. h_a=AF_b
    8. h_b=BF_a
    9. AF_c=\frac{b^2}{2R}
    10. BF_c=\frac{a^2}{2R}
    11. \frac{a}{c} =\frac{h_c}{AF_b}
    12. \frac{b}{c}=\frac{h_c}{BF_a}
    13. \frac{a}{b}=\frac{BF_a}{AF_b}
    14. R=\frac{b^2-a^2}{2c}
    15. h_c=R\cos C
    16. \cos A=\frac{b}{\sqrt{a^2+b^2}}
    17. \cos B=-\frac{a}{\sqrt{a^2+b^2}}
    18. \cos C=\frac{2ab}{a^2+b^2}
    19. \sin A=\frac{a}{\sqrt{a^2+b^2}}
    20. \sin B=\frac{b}{\sqrt{a^2+b^2}}
    21. \sin C=\frac{b^2-a^2}{a^2+b^2}
    22. \cos^2 A+\cos^2 B=1
    23. \sin^2 A+\sin^2 B=1
    24. a\cos A+b\cos B=0
    25. \sin A+\cos B=0
    26. \cos A-\sin B=0
    27. 2\cos A\cos B+\cos C=0
    28. 2\sin A\sin B-\cos C=0
    29. \cos A\cos B+\sin A\sin B=0
    30. a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C
    31. \sin A\sin B=\frac{ab}{b^2-a^2}\sin C
    32. \sin^2B-\sin^2A=\sin C
    33. \cos^2A-\cos^2B=\sin C
    34. OH^2=5R^2-c^2
    35. h_a^2+h_b^2=AB^2
    36. \frac{h_a}{a}+\frac{h_b}{b}=\frac{c}{h_c}
    37. a^2+b^2=4R^2
    38. \left(c+2AF_c\right)^2=a^2+b^2 or \left(c+2BF_c\right)^2=a^2+b^2
    39. A-B=\pm 90^{\circ}
    40. (a^2-b^2)^2=(ac)^2+(cb)^2
    41. AH^2+BH^2+CH^2=8R^2-c^2
    42. b=2R\cos A
    43. \triangle ABH is congruent to \triangle ABC
    44. \triangle OO^aO^b is congruent to \triangle ABC
    45. \triangle CNO is isosceles with CN=NO
    46. \triangle CNH is isosceles with CN=NH
    47. \triangle CHO is right angled at C
    48. N is the circumcenter of \triangle CHO
    49. \triangle O^cOC is right-angled at O
    50. \triangle O^cHC is right-angled at H
    51. quadrilateral O^cOHC is a rectangle
    52. the points O^c,O,C,H are concyclic with OH as diameter
    53. the reflection O^b of O over AC lies internally on AB
    54. the reflection O^a of O over BC lies externally on AB
    55. radius OC is parallel to side AB
    56. F_a is the reflection of F_b over side AB
    57. the nine-point center lies on AB
    58. the orthic triangle is isosceles with F_aF_c=F_bF_c
    59. the geometric mean theorem holds
    60. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
    61. the orthocenter is a reflection of vertex C over side AB
    62. segment HC is tangent to the circumcircle at point C
    63. median CM_c has the same length as the segment HM_c
    64. the bisector M_cO of AB is tangent to the nine-point circle at M_c
    65. AF_aBF_b is a convex kite with diagonals AB and F_aF_b
    66. altitude CF_c is tangent to the nine-point circle at F_c
    67. chord F_cM_c is a diameter of the nine-point circle
    68. segment HF_c is tangent to the nine-point circle at F_c.
      (7 short of the target. Next target now is to extend the initial 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|.