General – Page 13 – High School Geometry

An upside-down Pythagorean identity

Towards the end of our post on June 28, we stated that if the side-lengths of a triangle satisfy the modified Pythagorean identity

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

then the following statements are equivalent:

$\frac{b}{a}$ is the golden ratio
$(ab)^2=(ac)^2+(cb)^2$
$\sin A\sin B=\sin C$
$\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$
$h_C=c$

Today’s post will now establish the above equivalence. (Note that the fourth condition, $\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$ , which motivated today’s title, was coined by the author of this article.)

If $\frac{b}{a}$

is the golden ratio, PROVE that equation (1) becomes $(ab)^2=(ac)^2+(cb)^2$

Suppose that $\frac{b}{a}$ is the golden ratio, then $\left(\frac{b}{a}\right)^2-\frac{b}{a}-1=0$ , and so $b^2-ab-a^2=0$ . This then gives $b^2-a^2=ab$ . In turn:

$(b^2-a^2)^2=(ac)^2+(cb)^2\implies (ab)^2=(ac)^2+(cb)^2$

In addition to equation (1), suppose that $(ab)^2=(ac)^2+(cb)^2$

. PROVE that: $\sin A\sin B=\sin C$

Let $R$ be the circumradius. Then, by the extended law of sines, we have:

$\sin A=\frac{a}{2R},~\sin B=\frac{b}{2R},~\sin C=\frac{c}{2R}.$

The assumption $(ab)^2=(ac)^2+(cb)^2$ implies $ab=c\sqrt{a^2+b^2}$ . (Henceforth we’ll frequently use the fact that $a^2+b^2=4R^2$ is equivalent to equation (1).) So:

$\sin A\sin B=\left(\frac{ab}{4R^2}\right)=\frac{c\sqrt{a^2+b^2}}{4R^2}=\frac{c(2R)}{4R^2}=\frac{c}{2R}=\sin C$

In addition to equation (1), suppose that $\sin A\sin B=\sin C$

. PROVE that $\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Using the extended law of sines as in the previous example, the condition $\sin A\sin B=\sin C$ gives $ab=c(2R)$ . Since $2R=\sqrt{a^2+b^2}$ , we obtain

$ab=c\sqrt{a^2+b^2}\implies \frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}.$

So-called upside-down Pythagorean identity. Ours works under an added condition, namely equation (1).

In addition to equation (1), suppose that $\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$

. PROVE that $h_C=c$

, where $h_C$

is the length of the altitude from vertex $C$

and $c=AB$

as per the usual notation.

Re-arrange the given condition $\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}$ as

$a^2b^2=c^2(a^2+b^2)=c^2(4R^2)$

and obtain $ab=c(2R)$ . The length of the altitude from vertex $C$ is given by

$h_C=\frac{ab}{2R}=\frac{c(2R)}{2R}=c$

Suppose that $h_C=c$

, in addition to equation (1). PROVE that $\frac{b}{a}$

is the golden ratio.

If $h_C=c$ , then $\frac{ab}{2R}=c\implies ab=c(2R)=c\sqrt{a^2+b^2}\implies (ab)^2=c^2(a^2+b^2)$ . Now use this in equation (1) to get

$(b^2-a^2)^2=(ab)^2\implies b^2-a^2=ab\implies \left(\frac{b}{a}\right)^2-\frac{b}{a}-1=0$

Thus, $\frac{b}{a}$ is the golden ratio.

Takeaway

Let $ABC$ be a triangle whose side-lengths $a,b,c$ satisfy equation (1). Then the following statements are equivalent:

$\frac{b}{a}$ is the golden ratio
$\sin A\sin B=\sin C$
$\tan C=\frac{1}{2}$ .

The last condition fixes the interior angles of the given triangle $ABC$ : $\angle A=45^{\circ}-\frac{1}{2}\tan^{-1}\left(\frac{1}{2}\right)\approx 31.7^{\circ}$ , $\angle B=135^{\circ}-\frac{1}{2}\tan^{-1}\left(\frac{1}{2}\right)\approx 121.7^{\circ}$ , $\angle C=\tan^{-1}\left(\frac{1}{2}\right)\approx 26.6^{\circ}$ .

Task

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

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 $C(0,1)$ .

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:

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

Note the first one.

Task

(Late fifties) 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 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 $|3b^2-a^2|:|b^2-3a^2|$ .