Approximate Pythagorean Identity II

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 $s=BF_C$

(or $s=AF_C$

). If the side-lengths $a,b,c$

satisfy equation (1), PROVE that $(c+2s)^2=a^2+b^2$

Consider the following diagram:

Let $R$ be the circumradius and suppose that $b> a$ (if $a> b$ , take $s=AF_C$ ). 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

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

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

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

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:

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

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

Task

(Early fourties) In a non-right triangle , let be the side-lengths, the altitudes, the feet of the altitudes from the respective vertices, the circumradius, the circumcenter, the nine-point center, and the orthocenter. PROVE that the following forty 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 $|3b^2-a^2|:|b^2-3a^2|$ .