Slopefessor Fr1day – Page 12 – High School Geometry

In the above diagram we have the nine-point circle of triangle $ABC$ going through the midpoints $M_a,M_b,M_c$ (of sides $BC,CA,AB$ respectively), and the feet of the altitudes $F_a,F_b,F_c$ (from $A,B,C$ in that order).

It will be shown that

(1) $\begin{equation*} F_cM_c^2=F_aM_a^2+F_bM_b^2 \end{equation*}$

if (and only if) the side-lengths $a,b,c$ satisfy

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

It will also be shown, in our first example, that we have the relations

(3) $\begin{equation*} \begin{split} F_cM_c&=\left(\frac{b^2-a^2}{2c}\right)\\ F_bM_b&=\left(\frac{c^2-a^2}{2b}\right)\\ F_aM_a&=\left(\frac{c^2-b^2}{2a}\right)\\ \end{split} \end{equation*}$

in any triangle (append absolute values if need be).

In the meantime, note that the right side of equation (1) evaluates to $R^2$ for a right triangle, where $R$ is the circumradius, whereas both sides of equation (1) evaluate to $R^2$ under (2). And so in the latter case, the segment $F_cM_c$ is not just a chord, but a diameter of the nine-point circle.

Derive the three equations in (3).

Let’s do this for an acute triangle. Slight modification for an obtuse triangle will be needed. By drawing an appropriate diagram we have:

$\begin{equation*} \begin{split} M_aF_a&=\frac{a}{2}-b\cos C \text{ or } c\cos B-\frac{a}{2}\\ M_bF_b&=\frac{b}{2}-a\cos C \text{ or } c\cos A-\frac{b}{2}\\ M_cF_c&=\frac{c}{2}-a\cos B \text{ or } b\cos A-\frac{c}{2}\\ \end{split} \end{equation*}$

We can re-write the expression for $M_cF_c$ entirely in terms of the side-lengths using the cosine formula:

$\begin{equation*} \begin{split} M_cF_c&=\frac{c}{2}-a\cos B\\ &=\frac{c}{2}-a\left(\frac{a^2+c^2-b^2}{2ac}\right)\\ &=\frac{b^2-a^2}{2c} \end{split} \end{equation*}$

Similarly we obtain the expressions for $M_aF_a$ and $M_bF_b$ .

Let $ABC$

be an isosceles triangle in which $AC=BC$

. PROVE that the midpoint of side $AB$

coincides with the foot of the altitude from vertex $C$

Since $AC=BC$ , we have $a=b$ by the usual notation. Using example 1 above we have

$M_cF_c=\frac{b^2-a^2}{2c}=0$

and so $M_c=F_c$ .

In a right triangle with hypotenuse of length $c$

, PROVE that the right side of equation (1) is the square of the circumradius.

We first have $c^2=a^2+b^2$ . Moreover, if $R$ is the circumradius, then $c=2R$ for a right triangle. Now:

$\begin{equation*} \begin{split} F_aM_a^2+F_bM_b^2&=\left(\frac{c^2-b^2}{2a}\right)^2+\left(\frac{c^2-a^2}{2b}\right)^2\\ &=\left(\frac{a^2}{2a}\right)^2+\left(\frac{b^2}{2b}\right)^2\\ &=\frac{a^2+b^2}{4}\\ &=\left(\frac{c}{2}\right)^2\\ &=R^2 \end{split} \end{equation*}$

If the side-lengths of a triangle satisfy equation (2), PROVE that equation (1) holds.

Such a triangle is necessarily non-right. Let $R$ be its circumradius. By one of the equivalent statements here, we know that equation (2) then becomes equivalent to $a^2+b^2=4R^2$ . Re-arrange equation (2) in the form $c^2=\frac{(b^2-a^2)^2}{a^2+b^2}$ and consider both sides of equation (1):

$\begin{equation*} \begin{split} \text{LS}&=F_cM_c^2\\ &=\left(\frac{b^2-a^2}{2c}\right)^2\\ &=\left(\frac{(b^2-a^2)^2}{4\times \frac{(b^2-a^2)^2}{a^2+b^2} }\right)\\ &=\frac{a^2+b^2}{4}\\ &=R^2\\ \text{RS}&=F_aM_a^2+F_bM_b^2\\ &=\left(\frac{c^2-b^2}{2a}\right)^2+\left(\frac{c^2-a^2}{2b}\right)^2\\ &=\left(\frac{\frac{(b^2-a^2)^2}{a^2+b^2}-b^2}{2a}\right)^2+\left(\frac{\frac{(b^2-a^2)^2}{a^2+b^2}-a^2}{2b}\right)^2\\ &=\frac{\Big((b^2-a^2)^2-b^2(a^2+b^2)\Big)^2}{4a^2(a^2+b^2)^2}+\frac{\Big((b^2-a^2)^2-a^2(a^2+b^2)\Big)^2}{4b^2(a^2+b^2)^2}\\ &=\left(\frac{a^2b^4(b^2-3a^2)^2+a^4b^2(a^2-3b^2)^2}{4a^2b^2(a^2+b^2)^2}\right)\\ &=\frac{a^6+3a^4b^2+3a^2b^4+b^6}{4(a^2+b^2)^2}\\ &=\frac{(a^2+b^2)^3}{4(a^2+b^2)^2}\\ &=\frac{a^2+b^2}{4}\\ &=R^2\\ \therefore \text{LS}&=\text{RS} \end{split} \end{equation*}$

PROVE that the segment $F_cM_c$

is a diameter of the nine-point circle, if equation (2) holds.

By example 4 above, we had $F_cM_c^2=R^2$ . Since the nine-point circle goes through $F_c$ and $M_c$ and has radius equal to $\frac{R}{2}$ , the fact that $F_cM_c=R$ means that the chord $F_cM_c$ is a diameter.

No other triangle has this property.

Takeaway

In triangle $ABC$ , let $E_c$ denote the Euler point of vertex $C$ , $F_c$ the foot of the altitude from vertex $C$ , and $M_c$ the midpoint of side $AB$ . Then the following statements are equivalent:

the chord $F_cM_c$ is a diameter of the nine-point circle
$E_c=F_c$ .

The equivalent conditions range from extremely simple to moderately involved.

Task

(Late sixties) 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 , the Euler points, the circumradius, the circumcenter, the nine-point center, the orthocenter, the reflection of over side , the reflection of over side , and the reflection of over side . PROVE that the following sixty-eight statements are equivalent:
1. $E_c=F_c$
2. $AH=b$
3. $BH=a$
4. $OO^a=b$
5. $OO^b=a$
6. $CH=2h_c$
7. $h_a=AF_b$
8. $h_b=BF_a$
9. $AF_c=\frac{b^2}{2R}$
10. $BF_c=\frac{a^2}{2R}$
11. $\frac{a}{c} =\frac{h_c}{AF_b}$
12. $\frac{b}{c}=\frac{h_c}{BF_a}$
13. $\frac{a}{b}=\frac{BF_a}{AF_b}$
14. $R=\frac{b^2-a^2}{2c}$
15. $h_c=R\cos C$
16. $\cos A=\frac{b}{\sqrt{a^2+b^2}}$
17. $\cos B=-\frac{a}{\sqrt{a^2+b^2}}$
18. $\cos C=\frac{2ab}{a^2+b^2}$
19. $\sin A=\frac{a}{\sqrt{a^2+b^2}}$
20. $\sin B=\frac{b}{\sqrt{a^2+b^2}}$
21. $\sin C=\frac{b^2-a^2}{a^2+b^2}$
22. $\cos^2 A+\cos^2 B=1$
23. $\sin^2 A+\sin^2 B=1$
24. $a\cos A+b\cos B=0$
25. $\sin A+\cos B=0$
26. $\cos A-\sin B=0$
27. $2\cos A\cos B+\cos C=0$
28. $2\sin A\sin B-\cos C=0$
29. $\cos A\cos B+\sin A\sin B=0$
30. $a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C$
31. $\sin A\sin B=\frac{ab}{b^2-a^2}\sin C$
32. $\sin^2B-\sin^2A=\sin C$
33. $\cos^2A-\cos^2B=\sin C$
34. $OH^2=5R^2-c^2$
35. $h_a^2+h_b^2=AB^2$
36. $\frac{h_a}{a}+\frac{h_b}{b}=\frac{c}{h_c}$
37. $a^2+b^2=4R^2$
38. $\left(c+2AF_c\right)^2=a^2+b^2$ or $\left(c+2BF_c\right)^2=a^2+b^2$
39. $A-B=\pm 90^{\circ}$
40. $(a^2-b^2)^2=(ac)^2+(cb)^2$
41. $AH^2+BH^2+CH^2=8R^2-c^2$
42. $b=2R\cos A$
43. $\triangle ABH$ is congruent to $\triangle ABC$
44. $\triangle OO^aO^b$ is congruent to $\triangle ABC$
45. $\triangle CNO$ is isosceles with $CN=NO$
46. $\triangle CNH$ is isosceles with $CN=NH$
47. $\triangle CHO$ is right angled at $C$
48. $N$ is the circumcenter of $\triangle CHO$
49. $\triangle O^cOC$ is right-angled at $O$
50. $\triangle O^cHC$ is right-angled at $H$
51. quadrilateral $O^cOHC$ is a rectangle
52. the points $O^c,O,C,H$ are concyclic with $OH$ as diameter
53. the reflection $O^b$ of $O$ over $AC$ lies internally on $AB$
54. the reflection $O^a$ of $O$ over $BC$ lies externally on $AB$
55. radius $OC$ is parallel to side $AB$
56. $F_a$ is the reflection of $F_b$ over side $AB$
57. the nine-point center lies on $AB$
58. the orthic triangle is isosceles with $F_aF_c=F_bF_c$
59. the geometric mean theorem holds
60. the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
61. the orthocenter is a reflection of vertex $C$ over side $AB$
62. segment $HC$ is tangent to the circumcircle at point $C$
63. median $CM_c$ has the same length as the segment $HM_c$
64. the bisector $M_cO$ of $AB$ is tangent to the nine-point circle at $M_c$
65. $AF_aBF_b$ is a convex kite with diagonals $AB$ and $F_aF_b$
66. altitude $CF_c$ is tangent to the nine-point circle at $F_c$
67. chord $F_cM_c$ is a diameter of the nine-point circle
68. segment $HF_c$ is tangent to the nine-point circle at $F_c$ .
  ( $7$ short of the target. Next target now is to extend the initial target.)
(Extra feature) If $\triangle ABC$ satisfies equation (??), PROVE that its nine-point center $N$ divides $AB$ in the ratio $|3b^2-a^2|:|b^2-3a^2|$ .

On May 14, 2021, we showed that the side-lengths $a,b,c$

( $b< a< c$

) of the Kepler triangle satisfy the identity

(1) $\begin{equation*} (ac)^2=(ab)^2+(bc)^2 \end{equation*}$

Or, in an upside-down form:

(2) $\begin{equation*} \frac{1}{b^2}=\frac{1}{a^2}+\frac{1}{c^2}\end{equation*}$

Of all right triangles, the Kepler triangle is the only one in which both the side-lengths and the reciprocals of the side-lengths satisfy Pythagorean identities.

On top of this, the Kepler triangle is also unique in that both the altitudes and the reciprocals of the altitudes satisfy Pythagorean identities.

(3) $\begin{equation*} h_b^2=h_a^2+h_c^2 \end{equation*}$

(4) $\begin{equation*} \frac{1}{h_c^2}=\frac{1}{h_a^2}+\frac{1}{h_b^2} \end{equation*}$

(Only the Kepler triangle satisfies equation (3) above, but other right triangles satisfy equation (4).)

PROVE that equations (1) and (2) are equivalent.

Our very first example of 2022 is a breeze. To see that (1) $\implies$ (2), re-arrange

$(ac)^2=(ab)^2+(bc)^2=b^2(a^2+c^2)\implies \frac{1}{b^2}=\frac{a^2+c^2}{(ac)^2}=\frac{1}{a^2}+\frac{1}{c^2}$

On the other hand, (2) $\implies$ (1) because

$\frac{1}{b^2}=\frac{1}{a^2}+\frac{1}{c^2}=\frac{a^2+c^2}{(ac)^2}\implies (ac)^2=b^2(a^2+c^2)=(ab)^2+(bc)^2$

PROVE that every right triangle satisfies equation (4).

Let $R$ be the radius of the circumcircle of any triangle $ABC$ with side-lengths $a,b,c$ . Then the altitudes $h_a,h_b,h_c$ from vertices $A,B,C$ are given by

$h_a=\frac{bc}{2R},h_b=\frac{ac}{2R},~h_c=\frac{ab}{2R}$

In the case of a right triangle with hypotenuse $c$ we have $c=2R$ and so $h_a=b,h_b=a,h_c=\frac{ab}{c}$ .

$\therefore\frac{1}{h_c^2}=\frac{c^2}{(ab)^2}=\frac{a^2+b^2}{(ab)^2}=\frac{1}{b^2}+\frac{1}{a^2}=\frac{1}{h_a^2}+\frac{1}{h_b^2}.$

PROVE that equation (3) implies equation (1).

Since $h_a=\frac{bc}{2R},h_b=\frac{ac}{2R},~h_c=\frac{ab}{2R}$ , we have

$h_b^2=h_a^2+h_c^2\implies\left(\frac{ac}{2R}\right)^2=\left(\frac{bc}{2R}\right)^2 +\left(\frac{ab}{2R}\right)^2$

Clear the common denominator and it becomes clear that

$(ac)^2=(ab)^2+(bc)^2.$

PROVE that equation (1) implies equation (3).

Suppose that equation (1) holds, so that $(ac)^2=(ab)^2+(bc)^2$ . Divide through by $4R^2$ and use the expressions for the altitudes given in example 2.

$\begin{equation*} \begin{split} (ac)^2&=(ab)^2+(bc)^2\\ \frac{(ac)^2}{4R^2}&=\frac{(ab)^2}{4R^2}+\frac{(bc)^2}{4R^2}\\ \therefore h_b^2&=h_a^2+h_c^2 \end{split} \end{equation*}$

PROVE that a right triangle is precisely the Kepler triangle, if it satisfies equation (3).

If a right triangle is of Kepler type, then by our post on May 14, 2021 , its side-lengths satisfy equation (1). By the preceding two examples, equations (1) and (3) are equivalent. Thus, a Kepler triangle satisfies equation (3).

Conversely, suppose that a right triangle with hypotenuse $c$ satisfies equation (3). Then $c^2=a^2+b^2$ and $(ac)^2=(ab)^2+(bc)^2$ . Multiply both sides of $c^2=a^2+b^2$ by $a^2$ to get $(ac)^2=a^4+(ab)^2$ . Now combine with $(ac)^2=(ab)^2+(bc)^2$ to get

$a^4=(bc)^2\implies a^2=bc$

In other words, the side-lengths $a,b,c$ form a geometric progression of the form $b,a,c$ . Thus, the right triangle is of the Kepler type.

Takeaway

Let $ABC$ be a (right) triangle whose side-lengths $a,b,c$ satisfy $c^2=a^2+b^2$ . Then the following statements are equivalent:

the sequence $b,a,c$ is geometric
$\frac{c}{b}$ is the golden ratio
$(ac)^2=(ab)^2+(bc)^2$
$\frac{1}{b^2}=\frac{1}{a^2}+\frac{1}{c^2}$
$\frac{1}{h_b^2}=\frac{1}{h_a^2}+\frac{1}{h_c^2}$ .

Such a triangle is the Kepler triangle.

Task

(Mid sixties) In a non-right triangle , let be the side-lengths, the altitudes, the feet of the altitudes from the respective vertices, the circumradius, the circumcenter, the nine-point center, the orthocenter, the midpoint of side , the reflection of over side , the reflection of over side , and the reflection of over side . PROVE that the following sixty-five 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. $R=\frac{b^2-a^2}{2c}$
14. $h_C=R\cos C$
15. $\cos A=\frac{b}{\sqrt{a^2+b^2}}$
16. $\cos B=-\frac{a}{\sqrt{a^2+b^2}}$
17. $\cos C=\frac{2ab}{a^2+b^2}$
18. $\sin A=\frac{a}{\sqrt{a^2+b^2}}$
19. $\sin B=\frac{b}{\sqrt{a^2+b^2}}$
20. $\sin C=\frac{b^2-a^2}{a^2+b^2}$
21. $\cos^2 A+\cos^2 B=1$
22. $\sin^2 A+\sin^2 B=1$
23. $a\cos A+b\cos B=0$
24. $\sin A+\cos B=0$
25. $\cos A-\sin B=0$
26. $2\cos A\cos B+\cos C=0$
27. $2\sin A\sin B-\cos C=0$
28. $\cos A\cos B+\sin A\sin B=0$
29. $a\cos A-b\cos B=\sqrt{a^2+b^2}\cos C$
30. $\sin A\sin B=\frac{ab}{b^2-a^2}\sin C$
31. $\sin^2B-\sin^2A=\sin C$
32. $OH^2=5R^2-c^2$
33. $h_A^2+h_B^2=AB^2$
34. $\frac{h_A}{a}+\frac{h_B}{b}=\frac{c}{h_C}$
35. $a^2+b^2=4R^2$
36. $\left(c+2AF_C\right)^2=a^2+b^2$ or $\left(c+2BF_C\right)^2=a^2+b^2$
37. $A-B=\pm 90^{\circ}$
38. $(a^2-b^2)^2=(ac)^2+(cb)^2$
39. $AH^2+BH^2+CH^2=8R^2-c^2$
40. $b=2R\cos A$
41. $\triangle ABH$ is congruent to $\triangle ABC$
42. $\triangle OO^aO^b$ is congruent to $\triangle ABC$
43. $\triangle CNO$ is isosceles with $CN=NO$
44. $\triangle CNH$ is isosceles with $CN=NH$
45. $\triangle CHO$ is right angled at $C$
46. $N$ is the circumcenter of $\triangle CHO$
47. $\triangle O^cOC$ is right-angled at $O$
48. $\triangle O^cHC$ is right-angled at $H$
49. quadrilateral $O^cOHC$ is a rectangle
50. the points $O^c,O,C,H$ are concyclic with $OH$ as diameter
51. the reflection $O^b$ of $O$ over $AC$ lies internally on $AB$
52. the reflection $O^a$ of $O$ over $BC$ lies externally on $AB$
53. radius $OC$ is parallel to side $AB$
54. $F_A$ is the reflection of $F_B$ over side $AB$
55. the nine-point center lies on $AB$
56. the orthic triangle is isosceles with $F_AF_C=F_BF_C$
57. the geometric mean theorem holds
58. the bisector of $\angle C$ has length $l$ , where $l^2=\frac{2a^2b^2}{a^2+b^2}$
59. the orthocenter is a reflection of vertex $C$ over side $AB$
60. segment $HC$ is tangent to the circumcircle at point $C$
61. median $CM$ has the same length as the segment $HM$
62. the bisector $MO$ of $AB$ is tangent to the nine-point circle at $M$
63. $AF_ABF_B$ is a convex kite with diagonals $AB$ and $F_AF_B$
64. altitude $CF_C$ is tangent to the nine-point circle at $F_C$
65. segment $HF_C$ is tangent to the nine-point circle at $F_C$ .
  ( $10$ short of the target.)
(Extra feature) If $\triangle ABC$ satisfies equation (??), PROVE that its nine-point center $N$ divides $AB$ in the ratio $|3b^2-a^2|:|b^2-3a^2|$ .

Author: Slopefessor Fr1day

A special chord of the nine-point circle

Takeaway

Task

A property of the Kepler triangle

Takeaway

Task