The Euler line as a diameter II

Let O and H be the circumcenter and orthocenter of \triangle ABC.

Last time we saw that the circle with diameter OH passes through vertex C if and only if the equation below holds:

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

Let this circle also pass through vertex A. Then the nine-point center has to be B.

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Updated equivalence

We now add the following four equivalent statements to what we had at the end of our post on August 14:

  1. a circle with diameter OH passes through A and C
  2. the nine-point center coincides with vertex B
  3. the reflection of O over AC is B
  4. the reflection of H over BC is A

Only triangle ABC having a:b:c=1:\sqrt{3}:1 satisfies these.

In \triangle ABC, if the circle with diameter OH passes through A and C, PROVE that the nine-point center is B.

Since this circle passes through vertex C, our previous post shows that equation (1) is satisfied:

    \[(a^2-b^2)^2=(ac)^2+(cb)^2\]

Further, with OH as diameter, the two triangles CHO and AHO are right triangles with \angle HAO=90^{\circ} and \angle HCO=90^{\circ}.

    \begin{equation*} \begin{split} OH^2&=OA^2+AH^2\\ 9R^2-a^2-b^2-c^2&=R^2+(4R^2-a^2)\\ \therefore b^2+c^2&=4R^2\\ \vdots&\vdots\\ OH^2&=OC^2+CH^2\\ 9R^2-a^2-b^2-c^2&=R^2+(4R^2-c^2)\\ \therefore a^2+b^2&=4R^2\\ \vdots&\vdots\\ \therefore a&=c \end{split} \end{equation*}

By a result in this post, we conclude that the nine-point center is B.

If the nine-point center of \triangle ABC coincides with vertex B, PROVE that the reflection of the circumcenter O over side AC is B.

Easy.

If the reflection of the circumcenter O over AC is B, PROVE that the reflection of the orthocenter H over BC is A.

Easy.

If the reflection of H over BC is A, PROVE that the circle with diameter OH passes through A and C.

Easy.

Usual example

Consider \triangle ABC with vertices at A(-4,0), B(0,0), and C(2,2\sqrt{3}). Its nine-point center is B, and so it satisfies all the equivalent statements listed above.

The diagram below shows the circumcircle, the nine-point circle, and the circle with diameter OH.

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Or this one:

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Takeaway

Consider \triangle ABC with side-lengths a,b,c, circumradius R, circumcenter O, orthocenter H, and nine-point center N. If equation (1) is satisfied, then the following statements are equivalent:

  1. a circle with diameter OH passes through A and C
  2. \triangle OAC and \triangle OBC are both equilateral
  3. HA is tangent to the circumcircle at A
  4. the orthic triangle is equilateral
  5. the reflection of O over AC is B
  6. the reflection of H over BC is A
  7. \triangle ACH is equilateral
  8. N coincides with B
  9. a:b:c=1:\sqrt{3}:1
  10. a=c=R

These are some of the many equivalent descriptions of the isosceles triangle ABC in which \angle A=\angle C=30^{\circ}.

Task

  • (Early 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, and O' the reflection of O over side AB. PROVE that the following fifty-two statements are equivalent:
    1. AH=b
    2. BH=a
    3. CH=2h_C
    4. h_A=AF_B
    5. h_B=BF_A
    6. AF_C=\frac{b^2}{2R}
    7. BF_C=\frac{a^2}{2R}
    8. \frac{a}{c} =\frac{h_C}{AF_B}
    9. \frac{b}{c}=\frac{h_C}{BF_A}
    10. \frac{a}{b}=\frac{BF_A}{AF_B}
    11. R=\frac{|a^2-b^2|}{2c}
    12. h_C=R\cos C
    13. \cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}
    14. \cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}
    15. \cos C=\frac{2ab}{a^2+b^2}
    16. \cos^2 A+\cos^2 B=1
    17. \sin^2 A+\sin^2 B=1
    18. a\cos A+b\cos B=0
    19. 2\cos A\cos B+\cos C=0
    20. a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C
    21. OH^2=5R^2-c^2
    22. h_A^2+h_B^2=AB^2
    23. \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}
    24. a^2+b^2=4R^2
    25. A-B=\pm 90^{\circ}
    26. (a^2-b^2)^2=(ac)^2+(cb)^2
    27. AH^2+BH^2+CH^2=8R^2-c^2
    28. a=2R\sin A,~b=2R\cos A,~c=2R\cos 2A
    29. \triangle ABH is congruent to \triangle ABC
    30. \triangle CNO is isosceles with CN=NO
    31. \triangle CNH is isosceles with CN=NH
    32. \triangle CHO is right angled at C
    33. N is the circumcenter of \triangle CHO
    34. \triangle O'OC is right-angled at O
    35. \triangle O'HC is right-angled at H
    36. quadrilateral O'OHC is a rectangle
    37. the points O',O,C,H are concyclic with OH as diameter
    38. the reflection of O over AC lies internally on AB
    39. the reflection of O over BC lies externally on AB
    40. radius OC is parallel to side AB
    41. F_A is the reflection of F_B over side AB
    42. the nine-point center lies on AB
    43. the orthic triangle is isosceles with F_AF_C=F_BF_C
    44. the geometric mean theorem holds
    45. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
    46. the orthocenter is a reflection of vertex C over side AB
    47. segment HC is tangent to the circumcircle at point C
    48. median CM has the same length as the segment HM
    49. the bisector MO of AB is tangent to the nine-point circle at M
    50. AF_ABF_B is a convex kite with diagonals AB and F_AF_B
    51. altitude CF_C is tangent to the nine-point circle at F_C
    52. segment HF_C is tangent to the nine-point circle at F_C.
      (23 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|.

As always, this poster utilizes the national thanksgiving event to record his personal thanksgiving to God.

Among other things, the underlying reason dates back to a certain Thursday, June 14, 2018: beautiful, beautiful day. The poster’ll never forget that day. Also, the poster wishes that you’ll encounter a definite day you’ll always remember for good — in case you haven’t already.

All things being equal, the next iteration of the poster’s appreciation comes up on Tuesday, June 14, 2022. Until then: stay safe, do math, and give thanks.

The Euler line as a diameter

There is this orthocentroidal circle, whose definition is a circle having as diameter the line segment joining the orthocenter with the centroid.

Then there’s this modification we want to make. Consider instead a circle having as diameter the line segment connecting the orthocenter with the circumcenter (the Euler line). This circle passes through vertex C of a parent triangle ABC if, and only if, the side-lengths a,b,c satisfy

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

The reflection of the circumcenter over side AB also passes through this circle, under the above equivalence.

In \triangle ABC, let O and H be the circumcenter and orthocenter, respectively. Let O' be the reflection of O over AB. PROVE that OO'=CH. (Similarly, if O' were the reflection of O over BC, then OO'=AH; and if O' were the reflection of O over CA, then OO'=BH.)

Let R be the circumradius of the circle. In the diagram below

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M is the midpoint of AB and O' is the reflection of O over AB, and so OO'=2\times OM. Since MB=\frac{c}{2} and OB=R, we have:

    \begin{equation*} \begin{split} OM&=\sqrt{R^2-\left(\frac{c}{2}\right)^2}\\ &=\frac{\sqrt{4R^2-c^2}}{2}\\ \implies OO'&=\sqrt{4R^2-c^2}\\ &=CH \end{split} \end{equation*}

In \triangle ABC, let O and H be the circumcenter and orthocenter, respectively. Let O' be the reflection of O over AB. PROVE that HO'=CO=R.

In the diagram below, M is the midpoint of AB, and so HM is a median in triangle ABH.

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So:

    \begin{equation*} \begin{split} HM^2&=\frac{2AH^2+2BH^2-AB^2}{4}\\ &=\frac{2(4R^2-a^2)+2(4R^2-b^2)-c^2}{4}\\ &=\frac{16R^2-2a^2-2b^2-c^2}{4}\\ \end{split} \end{equation*}

Similarly, M is the midpoint of OO', and again HM is a median in triangle OHO':

    \begin{equation*} \begin{split} HM^2&=\frac{2O'H^2+2OH^2-O'O^2}{4}\\ 4HM^2&=2O'H^2+2OH^2-O'O^2\\ 4\left(\frac{16R^2-2a^2-2b^2-c^2}{4}\right)&=2O'H^2+2(9R^2-a^2-b^2-c^2)-\scriptstyle(4R^2-c^2)\\ 16R^2-2a^2-2b^2-c^2&=2O'H^2+2(9R^2-a^2-b^2-c^2)-\scriptstyle(4R^2-c^2)\\ \therefore R&=O'H \end{split} \end{equation*}

In \triangle ABC, let O and H be the circumcenter and orthocenter, respectively. Let O' be the reflection of O over AB. PROVE that the quadrilateral OO'HC is a parallelogram.

This follows from the two preceding examples, since we now have CH=OO' and CO=HO'.

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(Main goal)

In \triangle ABC, let O and H be the circumcenter and orthocenter, respectively. Let O' be the reflection of O over AB. PROVE that the quadrilateral OO'HC is a rectangle, if the side-lengths a,b,c of \triangle ABC satisfy equation (1).

If equation (1) is satisfied, then from one of the equivalent statements here, the segment HC is perpendicuclar to the radius OC.

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Since the quadrilateral O'OCH is a parallelogram, it follows that it is a rectangle. Consequently, the points O',O,C,H are concyclic. And the circle through these four points shares the same center with the nine-point center of the parent triangle.

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In \triangle ABC, let O and H be the circumcenter and orthocenter, respectively. Let O' be the reflection of O over AB. If the points O,O',H,C are concyclic with OH as diameter, PROVE that equation (1) holds.

Since OH is a diameter, the triangle OHC is right-angled at C. By one of the equivalent statements here, we conclude that equation (1) holds.

Takeaway

In a non-right \triangle ABC, let a,b,c be the side-lengths, H the orthocenter, O the circumcenter, and O' the reflection of the circumcenter over side AB. Then the following statements are equivalent:

  1. (a^2-b^2)^2=(ac)^2+(cb)^2
  2. \triangle O'OC is right-angled at O
  3. \triangle O'HC is right-angled at H
  4. quadrilateral O'OHC is a rectangle
  5. points O',O,C,H are concyclic with OH as diameter.

Task

  • (Half century) 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, and O' the reflection of O over side AB. PROVE that the following fifty statements are equivalent:
    1. AH=b
    2. BH=a
    3. CH=2h_C
    4. h_A=AF_B
    5. h_B=BF_A
    6. AF_C=\frac{b^2}{2R}
    7. BF_C=\frac{a^2}{2R}
    8. \frac{a}{c} =\frac{h_C}{AF_B}
    9. \frac{b}{c}=\frac{h_C}{BF_A}
    10. \frac{a}{b}=\frac{BF_A}{AF_B}
    11. R=\frac{|a^2-b^2|}{2c}
    12. h_C=R\cos C
    13. \cos A=\frac{b(b^2-a^2)}{c(a^2+b^2)}
    14. \cos B=\frac{a(a^2-b^2)}{c(a^2+b^2)}
    15. \cos C=\frac{2ab}{a^2+b^2}
    16. \cos^2 A+\cos^2 B=1
    17. \sin^2 A+\sin^2 B=1
    18. a\cos A+b\cos B=0
    19. 2\cos A\cos B+\cos C=0
    20. a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C
    21. OH^2=5R^2-c^2
    22. h_A^2+h_B^2=AB^2
    23. \frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}
    24. a^2+b^2=4R^2
    25. A-B=\pm 90^{\circ}
    26. (a^2-b^2)^2=(ac)^2+(cb)^2
    27. AH^2+BH^2+CH^2=8R^2-c^2
    28. a=2R\sin A,~b=2R\cos A,~c=2R\cos 2A
    29. \triangle ABH is congruent to \triangle ABC
    30. \triangle CNO is isosceles with CN=NO
    31. \triangle CNH is isosceles with CN=NH
    32. \triangle CHO is right angled at C
    33. N is the circumcenter of \triangle CHO
    34. \triangle O'OC is right-angled at O
    35. \triangle O'HC is right-angled at H
    36. quadrilateral O'OHC is a rectangle
    37. the points O',O,C,H are concyclic with OH as diameter
    38. radius OC is parallel to side AB
    39. F_A is the reflection of F_B over side AB
    40. the nine-point center lies on AB
    41. the orthic triangle is isosceles with F_AF_C=F_BF_C
    42. the geometric mean theorem holds
    43. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
    44. the orthocenter is a reflection of vertex C over side AB
    45. segment HC is tangent to the circumcircle at point C
    46. median CM has the same length as the segment HM
    47. the bisector MO of AB is tangent to the nine-point circle at M
    48. AF_ABF_B is a convex kite with diagonals AB and F_AF_B
    49. altitude CF_C is tangent to the nine-point circle at F_C
    50. segment HF_C is tangent to the nine-point circle at F_C.
      (Quite plenty, but can we reach 75?)
  • (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|.