Approximate Pythagorean Identity II

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If a non-right triangle, like the one shown above, satisfies the identity

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

then we can recast equation (1) in a form that looks more like the usual Pythagorean identity, namely:

(2)   \begin{equation*} (c+2s)^2=a^2+b^2 \end{equation*}

As s\rightarrow 0, we have an approximate Pythagorean identity

(3)   \begin{equation*} c^2\approx a^2+b^2 \end{equation*}


In a non-right \triangle ABC, let F_C be the foot of the altitude from vertex C, and let s=BF_C. If the side-lengths a,b,c satisfy equation (1), PROVE that (c+2s)^2=a^2+b^2.

Consider the following diagram:

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Let R be the circumradius and suppose that b> a. We’ll use some properties that are consequences of equation (1), including the one from our previous post:

    \[a^2=(BF_C)(2R)=s(2R),~c=\frac{b^2-a^2}{\sqrt{a^2+b^2}},~a^2+b^2=4R^2\]

Then:

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

Consider \triangle ABC with vertices at A(0,0), B(3,6), and C(0,10). If F_C is the foot of the altitude from vertex C and s=BF_C, verify that \left(c+2s\right)^2=a^2+b^2.

From the diagram below

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we get a^2=25, b^2=100, c^2=45\implies c=3\sqrt{5}, s=BF_C=\sqrt{5}. Equation (1) holds:

    \[(b^2-a^2)^2=(100-25)^2=5625=45(25+100)=c^2(a^2+b^2)\]

as well as equation (2), because:

    \begin{equation*} \begin{split} \left(c+2s\right)^2&=\left(3\sqrt{5}+2\times\sqrt{5}\right)^2\\ &=125\\ a^2+b^2&=25+100\\ &=125 \end{split} \end{equation*}

Consider \triangle ABC with vertices at A(0,0), B(3,-9), and C(5,5). If F_C is the foot of the altitude from vertex C and s=AF_C, verify that \left(c+2s\right)^2=a^2+b^2.

Note that A and B can switch places. From the diagram below

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we get a^2=200, b^2=50, c^2=90\implies c=3\sqrt{10}, s=AF_C=\sqrt{10}. Equation (1) holds:

    \[(a^2-b^2)^2=(200-50)^2=22500=90(200+50)=c^2(a^2+b^2)\]

as well as equation (2), because:

    \begin{equation*} \begin{split} \left(c+2s\right)^2&=\left(3\sqrt{10}+2\times\sqrt{10}\right)^2\\ &=250\\ a^2+b^2&=200+50\\ &=250 \end{split} \end{equation*}

Consider \triangle ABC with vertices at A(-6,0), B(0,0), and C(2,4). If F_C is the foot of the altitude from vertex C and s=BF_C, verify that (c+2s)^2=a^2+b^2.

From the diagram below

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we get a^2=20, b^2=80, c=AB=6,s=BF_C=2. Notice that equation (1) holds:

    \[(b^2-a^2)^2=(80-20)^2=3600=36(20+80)=c^2(a^2+b^2)\]

as well as equation (2), since:

    \begin{equation*} \begin{split} (c+2s)^2&=(6+2\times 2)^2\\ &=100\\ a^2+b^2&=20+80\\ &=100 \end{split} \end{equation*}

Consider \triangle ABC with vertices at A(4,4), B(0,0), and C(1,-2). If F_C is the foot of the altitude from vertex C and s=BF_C, verify that \left(c+2s\right)^2=a^2+b^2.

From the diagram below

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we get a^2=5, b^2=45, c^2=32\implies c=4\sqrt{2}, s=BF_C=\frac{1}{2}\sqrt{2}. Equation (1) holds:

    \[(b^2-a^2)^2=(45-5)^2=1600=32(50)=32(5+45)=c^2(a^2+b^2)\]

as well as equation (2), since

    \begin{equation*} \begin{split} \left(c+2s\right)^2&=\left(4\sqrt{2}+2\times\frac{1}{2}\sqrt{2}\right)^2\\ &=50\\ a^2+b^2&=5+45\\ &=50 \end{split} \end{equation*}

As in this post, notice that the title of today’s post can be abbreviated as API, an absolutely endearing term in computer programming.

Takeaway

In a non-right \triangle ABC, let a,b,c be the side-lengths, R the circumradius, O the circumcenter, H the orthocenter, and N the nine-point center. Then the following statements are equivalent:

  1. a^2+b^2=4R^2
  2. (a^2-b^2)^2=(ac)^2+(cb)^2
  3. \triangle CNO is isosceles with CN=NO
  4. \triangle CNH is isosceles with CN=NH
  5. N is the circumcenter of \triangle CHO
  6. \triangle CHO is right angled at C.

Don’t confuse CHO with the functional group of an aldehyde.

Task

  • (Early fourties) 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, and H the orthocenter. PROVE that the following fourty 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. 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. radius OC is parallel to side AB
    35. the nine-point center lies on AB
    36. the orthic triangle is isosceles with F_AF_C=F_BF_C
    37. the geometric mean theorem holds
    38. the bisector of \angle C has length l, where l^2=\frac{2a^2b^2}{a^2+b^2}
    39. the orthocenter is a reflection of vertex C over side AB
    40. segment HC is tangent to the circumcircle at point C
    41. the bisector MO of AB is tangent to the nine-point circle at M
    42. AF_ABF_B is a convex kite with diagonals AB and F_AF_B
    43. altitude CF_C is tangent to the nine-point circle at F_C
    44. segment HF_C is tangent to the nine-point circle at F_C.
      (Wow!!! The numbers increased so rapidly, from twenty one a few days ago.)
  • (Extra feature) If \triangle ABC satisfies equation (1), PROVE that its nine-point center N divides AB in the ratio |3b^2-a^2|:|b^2-3a^2|.