An inverted Pythagorean identity II

Today’s focus is on the equation

(1) $\begin{equation*} \frac{1}{D_a^2}+\frac{1}{D_b^2}=\frac{1}{R^2} \end{equation*}$

that can be used to characterize right triangles.

To this end, let $ABC$ be a right triangle in which $\angle C=90^{\circ}$ , and

$O$ the circumcenter
$R$ the circumradius
$D_a$ and $D_b$ the diameters of triangles $BCO$ and $ACO$ .

Then equation (1) holds.

Preliminary calculations

Consider a right triangle $ABC$

in which $\angle C=90^{\circ}$

. Let $O$

be the circumcenter of $ABC$

, and let $O_a$

and $O_b$

denote the circumcenters of triangles $BCO$

and $ACO$

, respectively. PROVE that triangle $O_aO_bO$

is similar to the parent triangle $ABC$

A sample diagram is shown below:

Since $O_a$ is the circumcenter of $\triangle BCO$ , it lies somewhere on the right bisector of side $BC$ and so when $O_a$ is joined to $O$ , the segment $OO_a$ is then parallel to side $AC$ . Similarly, $OO_b$ is parallel to side $BC$ . Thus $\angle O_aOO_b=90^{\circ}$ . Now join $O$ to $C$ and use angle chasing to complete the similarity argument.

Consider a right triangle $ABC$

in which $\angle C=90^{\circ}$

. Let $O$

be the circumcenter of $ABC$

, and let $R_a$

and $R_b$

denote the circumradii of triangles $BCO$

and $ACO$

, respectively. PROVE that $aR_b=bR_a$

From the previous example, we saw the similarity between triangle $OO_aO_b$ and the parent triangle $ABC$ :

So:

$\frac{O_bO}{AC}=\frac{OO_a}{CB}=\frac{O_aO_b}{BA}\implies \frac{R_b}{b}=\frac{R_a}{a}=\frac{O_aO_b}{c}$

We obtain $aR_b=bR_a$ as required. Notice also that we have

(2) $\begin{equation*} \frac{R_a}{a}=\frac{R_b}{b}=\frac{\sqrt{R_a^2+R_b^2}}{\sqrt{a^2+b^2}} \end{equation*}$

Partial characterization

Partial in the sense that there’s another triangle for which the equivalence in the next two examples works. We’ll see that later.

Consider a right triangle $ABC$

in which $\angle C=90^{\circ}$

. Let $O$

be the circumcenter of $ABC$

, and let $D_a$

and $D_b$

denote the circumdiameters of triangles $BCO$

and $ACO$

, respectively. PROVE that equation (1) holds, namely: $\frac{1}{D_a^2}+\frac{1}{D_b^2}=\frac{1}{R^2}$

, where $R$

is the circumradius of the parent triangle $ABC$

Apply the extended sine law to triangle $BCO$ below, noting that $CO=R$ :

This would give:

$2R_a=\frac{R}{\sin \beta}=\frac{R}{b/c}=\frac{R}{b/2R}=\frac{2R^2}{b}$

Similarly, applying the extended sine law to triangle $ACO$ gives

$2R_b=\frac{R}{\sin \alpha}=\frac{R}{a/c}=\frac{R}{a/2R}=\frac{2R^2}{a}$

Since $D_a=2R_a$ and $D_b=2R_b$ , we now get:

$\begin{equation*} \begin{split} \frac{1}{D_a^2}+\frac{1}{D_b^2}&=\frac{b^2}{4R^4}+\frac{a^2}{4R^4}\\ &=\frac{c^2}{4R^4}\\ &=\frac{4R^2}{4R^4}\\ &=\frac{1}{R^2} \end{split} \end{equation*}$

Consider an acute triangle $ABC$

. Let $O$

be the circumcenter of $ABC$

, and let $D_a$

and $D_b$

denote the circumdiameters of triangles $BCO$

and $ACO$

, respectively. If equation (1) holds, namely: $\frac{1}{D_a^2}+\frac{1}{D_b^2}=\frac{1}{R^2}$

, where $R$

is the circumradius of the parent triangle $ABC$

, PROVE that $ABC$

is right-angled at $C$

We’ll show in a later post that the following extension of the extended sine law holds:

(3) $\begin{equation*} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R=4R_a\cos A=4R_b\cos B=4R_c\cos C \end{equation*}$

An absolute value may be needed in one of the cosine terms if the parent triangle is obtuse. However, since we’re working with the squares for now, there’s no need for an absolute value. We have from (3) that

$R=2R_a\cos A=2R_b\cos B=2R_c\cos C$

(Or $R=D_a\cos A=D_b\cos B=D_c\cos C$ .) Assuming $\frac{1}{D_a^2}+\frac{1}{D_b^2}=\frac{1}{R^2}$ , we then have:

$\begin{equation*} \begin{split} \frac{\cos^2A}{R^2}+\frac{\cos^2B}{R^2}&=\frac{1}{R^2}\\ \implies \cos^2A+\cos^2B&=1 \end{split} \end{equation*}$

Thus, either the triangle is right-angled at $C$ , or it’s an obtuse triangle that shares several properties in common with a right-triangle.

Pertinent consideration

We emphasize that the equation $\frac{1}{D_a^2}+\frac{1}{D_b^2}=\frac{1}{R^2}$ also holds in a certain non-right triangle setting. More on this later.

Consider $\triangle ABC$

with vertices at $A(-6,0)$

, $B(0,0)$

, $C(2,4)$

. Verify that equation (1) holds for this triangle.

We’re to show that $\frac{1}{D_a^2}+\frac{1}{D_b^2}=\frac{1}{R^2}$ , where $R$ is the circumradius, $D_a$ is the diameter of the circumcircle of $\triangle BCO$ , and $D_b$ is the diameter of the circumcircle of $\triangle ACO$ .

By calculation, the circumcenter $O$ is $(-3,4)$ . Thus the circumcenter $O_a$ of $\triangle BCO$ is $\left(-\frac{1}{2},\frac{11}{4}\right)$ , while the circumcenter $O_b$ of $\triangle ACO$ is $\left(-\frac{1}{2},-1\right)$ . Notice that the nine-point center $N$ shown in the diagram is co-linear with $O_a$ and $O_b$ , just like in a right triangle. Furthermore:

$R=5$
$R_a=\frac{5\sqrt{5}}{4}\implies D_a=\frac{5\sqrt{5}}{2}$
$R_b=\frac{5\sqrt{5}}{2}\implies D_b=5\sqrt{5}$

So:

$\begin{equation*} \begin{split} \frac{1}{D_a^2}+\frac{1}{D_b^2}&=\frac{4}{125}+\frac{1}{125}\\ &=\frac{5}{125}\\ &=\frac{1}{R^2} \end{split} \end{equation*}$

Takeaway

Suppose that triangle $ABC$ has side-lengths $a,b,c$ , circumcenter $O$ , circumradius $R$ , and $R_a, R_b$ the radii of the circumcircles of $BCO$ and $ACO$ . Then the following statements are equivalent:

$ABC$ is right-angled at $C$
$\frac{R_a}{a}=\frac{R_b}{b}=\frac{\sqrt{R_a^2+R_b^2}}{\sqrt{a^2+b^2}}$ .

Tasks

(Geometric means) Suppose that the side-lengths of satisfy . Using the notation in this post, PROVE that:
1. $R^2=aR_b$
2. $R^2=bR_a$
(Geometric means) Suppose that the side-lengths of satisfy . Using the notation in this post, PROVE that:
1. $R^2=aR_b$
2. $R^2=bR_a$
(Growing membership) In a non-right triangle , let be the side-lengths, the altitudes, the feet of the altitudes from the respective vertices, the midpoints of sides in that order, the Euler points, the circumradius, the circumcenter, the nine-point center, the orthocenter, the reflection of over side , the reflection of over side , the reflection of over side , the symmedian point, the foot of the symmedian from vertex , and the radius of the polar circle. PROVE that the following eighty-five 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. $CO^c=OH$
14. $OM_a=\frac{b}{2}$
15. $OM_b=\frac{a}{2}$
16. $OM_c=h_c$
17. $CH=2h_c$
18. $h_a=AF_b$
19. $h_b=BF_a$
20. $AF_c=\frac{b^2}{2R}$
21. $BF_c=\frac{a^2}{2R}$
22. $\frac{a}{c} =\frac{h_c}{AF_b}$
23. $\frac{b}{c}=\frac{h_c}{BF_a}$
24. $\frac{a}{b}=\frac{BF_a}{AF_b}$
25. $R=\frac{b^2-a^2}{2c}$
26. $\frac{CK}{KK_c}=\left(\frac{a^2+b^2}{a^2-b^2}\right)^2$
27. $r^2=\frac{1}{2}\left(a^2+b^2-c^2\right)$
28. $h_c=R\cos C$
29. $\cos A=\frac{b}{\sqrt{a^2+b^2}}$
30. $\cos B=-\frac{a}{\sqrt{a^2+b^2}}$
31. $\cos C=\frac{2ab}{a^2+b^2}$
32. $\sin A=\frac{a}{\sqrt{a^2+b^2}}$
33. $\sin B=\frac{b}{\sqrt{a^2+b^2}}$
34. $\sin C=\frac{b^2-a^2}{a^2+b^2}$
35. $\cos^2 A+\cos^2 B=1$
36. $\sin^2 A+\sin^2 B=1$
37. $a\cos A+b\cos B=0$
38. $\sin A+\cos B=0$
39. $\cos A-\sin B=0$
40. $2\cos A\cos B+\cos C=0$
41. $2\sin A\sin B-\cos C=0$
42. $\cos A\cos B+\sin A\sin B=0$
43. $a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C$
44. $\sin A\sin B=\frac{ab}{b^2-a^2}\sin C$
45. $\sin^2B-\sin^2A=\sin C$
46. $\cos^2A-\cos^2B=\sin C$
47. $OH^2=5R^2-c^2$
48. $h_a^2+h_b^2=AB^2$
49. $\frac{h_a}{a}+\frac{h_b}{b}=\frac{c}{h_c}$
50. $a^2+b^2=4R^2$
51. $b=2R\cos A$
52. $A-B=\pm 90^{\circ}$
53. $(a^2-b^2)^2=(ac)^2+(cb)^2$
54. $AH^2+BH^2+CH^2=8R^2-c^2$
55. $\left(c+2AF_c\right)^2=a^2+b^2$ or $\left(c+2BF_c\right)^2=a^2+b^2$
56. $K_c$ and $F_c$ divide $AB$ harmonically
57. $E_a$ is the reflection of $M_b$ over side $AB$
58. $E_b$ is the reflection of $M_a$ over side $AB$
59. $\triangle ABH$ is congruent to $\triangle ABC$
60. $\triangle OO^aO^b$ is congruent to $\triangle ABC$
61. $\triangle CNO$ is isosceles with $CN=NO$
62. $\triangle CNH$ is isosceles with $CN=NH$
63. $\triangle CHO$ is right angled at $C$
64. $N$ is the circumcenter of $\triangle CHO$
65. $\triangle O^cOC$ is right-angled at $O$
66. $\triangle O^cHC$ is right-angled at $H$
67. quadrilateral $O^cOHC$ is a rectangle
68. the points $O^c,O,C,H$ are concyclic with $OH$ as diameter
69. the reflection $O^b$ of $O$ over $AC$ lies internally on $AB$
70. the reflection $O^a$ of $O$ over $BC$ lies externally on $AB$
71. radius $OC$ is parallel to side $AB$
72. $F_a$ is the reflection of $F_b$ over side $AB$
73. segment $F_aF_b$ is perpendicular to side $AB$
74. the nine-point center lies on $AB$
75. the orthic triangle is isosceles with $F_aF_c=F_bF_c$
76. the geometric mean theorem holds
77. the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
78. the orthocenter is a reflection of vertex $C$ over side $AB$
79. segment $HC$ is tangent to the circumcircle at point $C$
80. median $CM_c$ has the same length as the segment $HM_c$
81. the bisector $M_cO$ of $AB$ is tangent to the nine-point circle at $M_c$
82. $AF_aBF_b$ is a convex kite with diagonals $AB$ and $F_aF_b$
83. altitude $CF_c$ is tangent to the nine-point circle at $F_c$
84. chord $F_cM_c$ is a diameter of the nine-point circle
85. segment $HF_c$ is tangent to the nine-point circle at $F_c$ .
  (More statements reserved for later. We aim to stop at $100$ statements, if possible.)