Characterizing right triangles

Take an altitude of a triangle and a median from the same vertex. Then compute their product. If the product equals the area of the triangle, then the triangle must be a right triangle.

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The above property is exclusive to right triangles, and there are many other characterizations of right triangles, as seen in this list.

Today’s post will push at least two more entries into the list, in addition to that of July 14 \cdots.

PROVE that the area of a right triangle is equal to the product of the altitude and median from the 90^{\circ} vertex.

Consider a right triangle ABC in which \angle C=90^{\circ} as shown below:

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The length of the median to the hypotenuse is half the length of the hypotenuse: m_c=\frac{1}{2}c. Thus, the area is:

    \[\frac{AB\times h_c}{2}=\frac{c\times h_c}{2}=\frac{c}{2}h_c=m_ch_c\]

If the area of a triangle is equal to the product of the altitude and median from the same vertex, PROVE that the triangle must contain a 90^{\circ} angle.

Suppose that the area of \triangle ABC is equal to the product of the altitude h_c and median m_c from the same vertex C. Let a,b,c be the side-lengths and let R be the circumradius. Then

    \[h_c=\frac{ab}{2R},m_c=\frac{\sqrt{2a^2+2b^2-c^2}}{2}\]

The area can be given as \frac{ab\sin C}{2}, and R itself can be written as R=\frac{c}{2\sin C}, by the extended law of sines. So:

    \begin{equation*} \begin{split} \frac{ab\sin C}{2}&=\frac{ab}{2R}\times \frac{\sqrt{2a^2+2b^2-c^2}}{2}\\ \frac{abc}{4R}&=\frac{ab\sqrt{2a^2+2b^2-c^2}}{4R}\\ c&=\sqrt{2a^2+2b^2-c^2}\\ c^2&=2a^2+2b^2-c^2\\ 2c^2&=2a^2+2b^2\\ \therefore c^2&=a^2+b^2 \end{split} \end{equation*}

If the circle with diameter OH coincides with the nine-point circle of a triangle, PROVE that the triangle is a right triangle. As usual, O and H are the circumcenter and orthocenter, respectively.

Let R be the circumradius of the parent triangle ABC. If the circle with diameter OH coincides with the nine-point circle, then we must have OH=R, since the radius of the nine-point circle is \frac{R}{2}.

    \[\therefore 9R^2-a^2-b^2-c^2=R^2\implies a^2+b^2+c^2=8R^2,\]

the latter equation is one of the characterizations of right triangles given here.

In the next example we have a weaker requirement — we just ask that the nine-point circle pass through O or H.

If the nine-point circle passes through the circumcenter of a triangle, PROVE that the triangle is a right triangle.

Suppose that the nine-point circle passes through O. Let N be the nine-point center. Then NO is a radius of the nine-point circle, namely \frac{R}{2}. But then NO=\frac{OH}{2}.

    \begin{equation*} \begin{split} \frac{R}{2}&=\frac{\sqrt{9R^2-a^2-b^2-c^2}}{2}\\ \therefore 8R^2&=a^2+b^2+c^2 \end{split} \end{equation*}

Similarly, if the nine-point circle of a triangle passes through the orthocenter of the triangle, then the parent triangle must be a right triangle.

In \triangle ABC, suppose that the square of the length of the bisector
of \angle C is half the square of the harmonic mean of a and b. PROVE that \angle C=90^{\circ}.

The harmonic mean of a and b is \frac{2ab}{a+b}. The square of the length of the bisector of \angle C is given by l^2=\left(1-\left(\frac{c}{a+b}\right)^2\right). We get

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

and then finally c^2=a^2+b^2 after simplifications.

Takeaway

In \triangle ABC, let a,b,c be the side-lengths, m_c the median from vertex C, h_c the altitude from vertex C, H the orthocenter, and O the circumcenter. Then the following statements are equivalent:

  1. the area of \triangle ABC equals m_ch_c
  2. \triangle ABC is a right triangle with \angle C=90^{\circ}
  3. the nine-point circle of \triangle ABC passes through O
  4. the nine-point circle of \triangle ABC passes through H
  5. the circle with diameter OH coincides with the nine-point circle of \triangle ABC
  6. m_c is the geometric mean of the two equal segments it creates on the opposite side AB
  7. the square of the length of the bisector of angle C is half the square of the harmonic mean of a and b.

In the case of statement 6, one can replace the geometric mean with arithmetic mean or harmonic mean. Why?

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