Three special obtuse triangles I

We’ll be considering three obtuse triangles whose interior angles are:

  • \angle A=\angle C=30^{\circ}, \angle B=120^{\circ}
  • \angle A=15^{\circ},\angle C=60^{\circ}, \angle B=105^{\circ}
  • \angle A=22.5^{\circ},\angle C=45^{\circ}, \angle B=112.5^{\circ}

What makes them special is the fact that their side-lengths a,b,c satisfy the modified Pythagorean identity

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

with the accompanying equivalent descriptions. The first triangle has already appeared on August 14 and on October 14. The second triangle is making its debut here.

Suppose that the interior angles of \triangle ABC are \angle A=15^{\circ},\angle C=60^{\circ}, \angle B=105^{\circ}. PROVE that the side-lengths a,b,c satisfy equation (1).

Since \angle B-\angle A=90^{\circ}, the conclusion follows from one of the equivalent statements here.

Suppose that the interior angles of \triangle ABC are \angle A=15^{\circ},\angle C=60^{\circ}, \angle B=105^{\circ}. PROVE that a,b satisfy a^2-4ab+b^2=0.

In example 1 above we saw that the side-lengths satisfy equation (1). As such \cos C=\frac{2ab}{a^2+b^2}, by one of the equivalent statements here. So:

    \begin{equation*} \begin{split} \cos 60^{\circ}&=\frac{2ab}{a^2+b^2}\\ \frac{1}{2}&=\frac{2ab}{a^2+b^2}\\ \therefore a^2-4ab+b^2&=0 \end{split} \end{equation*}

A direct computation of the side-lengths a and b can also be used to prove the above relationship, but that’s a longer procedure.

Suppose that the interior angles of \triangle ABC are \angle A=15^{\circ},\angle C=60^{\circ}, \angle B=105^{\circ}. PROVE that the radius R of the circumcircle of \triangle ABC is the geometric mean of a and b; that is, R^2=ab.

By example 2 above we had a^2-4ab+b^2=0. Re-write as a^2+b^2=4ab. Using one of the equivalent statements here, we know that a^2+b^2=4R^2.

    \[\implies 4R^2=4ab\implies R^2=ab\]

One can also prove this by a direct computation of the circumradius R and the fact that \sin 15^{\circ}\sin 105^{\circ}=\frac{1}{4}. Definitely a longer procedure.

Suppose that the interior angles of \triangle ABC are \angle A=15^{\circ},\angle C=60^{\circ}, \angle B=105^{\circ}. PROVE that the side-lengths a,b,c satisfy c^2=3ab, and 2c^2=3(a-b)^2.

Let’s use the cosine formula and the fact that a^2+b^2=4ab:

    \begin{equation*} \begin{split} c^2&=a^2+b^2-2ab\cos C\\ &=(a^2+b^2)-2ab\left(\frac{1}{2}\right)\\ &=(4ab)-ab\\ \therefore c^2&=3ab \end{split} \end{equation*}

To show that 2c^2=3(a-b)^2, re-write a^2+b^2=4ab as (a-b)^2=2ab and then use the fact that c^2=3ab just proved.

Find coordinates for the vertices of a triangle whose interior angles are \angle A=15^{\circ},\angle C=60^{\circ}, \angle B=105^{\circ}.

Take the tiny triangle ABC where the vertices are located at A(0,0), B\left(\frac{\sqrt{3}}{2},\frac{3+2\sqrt{3}}{2}\right), and C(0,1).

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Takeaway

Let a,b,c be the side-lengths of \triangle ABC, and let R be the radius of its circumscribed circle. If equation (1) holds, then the following statements are equivalent:

  1. R^2=ab
  2. a^2-4ab+b^2=0.

Note the first one.

Task

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