This is a paragraph.

Category: General

Introductory; impromptu.

From reflection triangle to Euler line

If we begin with triangle ABC in which \angle A=\angle C=30^{\circ},\angle B=120^{\circ} and reflect vertex A over side BC, we obtain the orthocenter. If we reflect vertex C over side AB we obtain the orthocenter again.

If we now reflect vertex B over side AC we obtain the circumcenter.

It follows, in the case of the 30^{\circ}-30^{\circ}-120^{\circ}-triangle, that the reflection triangle is degenerate; in addition, it degenerates to a special line connecting the centroid, circumcenter and orthocenter, known as the Euler line.

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Is the 30^{\circ}-30^{\circ}-120^{\circ}-triangle the only triangle whose reflection triangle degenerates to the Euler line?

In \triangle ABC, let \angle A=\angle C=30^{\circ} and \angle B=120^{\circ}. PROVE that the reflection of vertex A over side BC coincides with the orthocenter.

To see how this holds, we’ll show that the length h_a of the altitude through A is half the distance from A to the orthocenter H. Indeed:

    \[h_a=\frac{bc}{2R}=\frac{\sqrt{3}a\times a}{2a}=\frac{\sqrt{3}a}{2}\]

and

    \[AH=\sqrt{4R^2-a^2}=\sqrt{4a^2-a^2}=\sqrt{3}a\]

This shows that the reflection of vertex A over side BC is the orthocenter H.

In \triangle ABC, let \angle A=\angle C=30^{\circ} and \angle B=120^{\circ}. PROVE that the reflection of vertex C over side AB coincides with the orthocenter.

Since a=c, we have that h_c=h_a and CH=AH. Thus, the reflection of vertex C over side AB is also the orthocenter.

In \triangle ABC, let \angle A=\angle C=30^{\circ} and \angle B=120^{\circ}. PROVE that the reflection of vertex B over side CA coincides with the circumcenter.

To see this, note that since triangle ABC is isosceles with B as the apex, the circumcenter lies on the axis of symmetry through B. We just have to show that the radius through B is twice the altitude through B. The altitude through B is:

    \[h_b=\frac{ac}{2R}=\frac{a\times a}{2a}=\frac{a}{2}\implies a=2\times h_b; \text{ that is, }R=2h_b\]

PROVE that the 30^{\circ}-30^{\circ}-120^{\circ}-triangle has a degenerate reflection triangle.

We’ve already seen this by explicitly determining the reflection triangle from the preceding three examples. Alternatively, the necessary and sufficient condition for a degenerate reflection triangle is \cos A\cos B\cos C=-\frac{3}{8}. This condition holds in the case of the 30^{\circ}-30^{\circ}-120^{\circ}-triangle:

    \[\cos 30^{\circ}\times \cos 120^{\circ}\times \cos 30^{\circ}=\frac{\sqrt{3}}{2}\times -\frac{1}{2}\times\frac{\sqrt{3}}{2}=-\frac{3}{8}\]

If a triangle has the property that the reflections of two of its vertices over the corresponding opposite sides coincide with the orthocenter, PROVE that such a triangle must be the 30^{\circ}-30^{\circ}-120^{\circ}-triangle.

Let R be the circumradius of the triangle. If the reflection of vertex A over side BC is the orthocenter, then we must have b^2+c^2=4R^2, as per the equivalent statements here. Similarly, if the reflection of vertex C over side AB is the orthocenter, then we must have a^2+b^2=4R^2. These two relations yield a=c. Since a^2+b^2=4R^2 is also equivalent to (b^2-a^2)^2=c^2(a^2+b^2), we obtain b^2=3a^2. Thus, the triangle is the 30^{\circ}-30^{\circ}-120^{\circ}-triangle.

Takeaway

In any triangle ABC, let a,b,c be the side-lengths, H the orthocenter, O the circumcenter, and R the circumradius. Then the following statements are equivalent:

  1. a=c=R
  2. \angle A=\angle C=30^{\circ}
  3. the reflection of B over AC is O
  4. the reflection of A over BC is H and the reflection of C over AB is H

Task

  • (Aufbau) In triangle ABC, let a,b,c be the side-lengths, R the circumradius, O the circumcenter, N the nine-point center, and H the orthocenter. PROVE that the following statements are equivalent:
    1. B=N
    2. a=c=R
    3. \angle A=\angle C=30^{\circ}
    4. the reflection of B over AC is O
    5. radius OA is parallel to side CB and radius OC is parallel to side AB
    6. the reflection of A over BC is H and the reflection of C over AB is H

Nine-point center equals a vertex II

Among all triangles, the only triangle whose nine-point center coincides with one of its vertices is the 30^{\circ}-30^{\circ}-120^{\circ}-triangle.

As such, it has unique properties, some of which will be shown in today’s post.

Additionally, the 30^{\circ}-30^{\circ}-120^{\circ}-triangle is also unique among all isosceles triangles, and we’ll see that later.

PROVE that in the 30^{\circ}-30^{\circ}-120^{\circ}-triangle, the two equal sides are equal to the circumradius.

Let R be the circumradius and let a,b,c be the side-lengths of a given triangle ABC. By the extended law of sines we have:

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

In the event that A=C=30^{\circ}, we get R=a=c.

If one vertex of a triangle is equidistant from the midpoints of its sides, PROVE that the triangle is the 30^{\circ}-30^{\circ}-120^{\circ}-triangle.

Suppose that vertex B of triangle ABC is equidistant from the midpoints of sides AB,BC,CA. In particular, this means that the median from vertex B is equal to half the side-lengths of AB and BC.

    \begin{equation*} \begin{split} \frac{2a^2+2c^2-b^2}{4}&=\frac{a^2}{4}\\ \implies a^2+2c^2&=b^2\\ \frac{2a^2+2c^2-b^2}{4}&=\frac{c^2}{4}\\ \implies 2a^2+c^2&=b^2\\ \therefore a&=c \end{split} \end{equation*}

By the cosine formula:

    \[\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{a^2+a^2-(a^2+2a^2)}{2a^2}=-\frac{1}{2}\implies B=120^{\circ}\]

Consequently, A=C=30^{\circ}.

PROVE that in the 30^{\circ}-30^{\circ}-120^{\circ}-triangle, one vertex is equidistant from the midpoints of the three sides.

Note that the side-lengths a,b,c are in the ratio a:b:c=1:\sqrt{3}:1, and so b^2=3a^2=3c^2. Consider the length of the median from vertex B:

    \begin{equation*} \begin{split} m_b^2&=\frac{2a^2+2c^2-b^2}{4}\\ &=\frac{2a^2+2a^2-3a^2}{4}\\ &=\frac{a^2}{4}\\ \therefore m_b&=\frac{a}{2}\\ &=\frac{c}{2} \end{split} \end{equation*}

And so vertex B is equidistant from the midpoints of all three sides.

If one vertex of a triangle is equidistant from the feet of its three altitudes, PROVE that the triangle is the 30^{\circ}-30^{\circ}-120^{\circ}-triangle.

Let the feet of the altitudes be F_a,F_b,F_c, and let the altitudes themselves be h_a,h_b,h_c. Suppose that vertex B is equidistant from F_a and F_b, that is BF_a=BF_b. Then:

    \[c^2-h_a^2=h_b^2\implies c^2-\left(\frac{bc}{2R}\right)^2=\left(\frac{ac}{2R}\right)^2\implies a^2+b^2=4R^2\]

Similarly, let BF_b=BF_c:

    \[h_b^2=a^2-h_c^2\implies \left(\frac{ac}{2R}\right)^2=a^2-\left(\frac{ab}{2R}\right)^2\implies b^2+c^2=4R^2\]

And so a=c. Since a^2+b^2=4R^2 is equivalent to (b^2-a^2)^2=(ac)^2+(cb)^2, we obtain b^2=3a^2. In turn, B=120^{\circ}. The fact that a=c then gives A=C=30^{\circ}.

PROVE that in the 30^{\circ}-30^{\circ}-120^{\circ}-triangle, one vertex is equidistant from the feet of all three altitudes.

Easy converse to the preceding example.

Takeaway

In any triangle ABC, let a,b,c be the side-lengths and R the circumradius. Then the following statements are equivalent:

  1. R=a=c
  2. \angle A=\angle C=30^{\circ}
  3. B is equidistant from the midpoints of the sides
  4. B is equidistant from the feet of the three altitudes
  5. B is equidistant from the three Euler points.

Such B must be the nine-point center.

Task

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