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 January 1st, 2019, 06:20 PM #1 Newbie   Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 Detailed section of all triangles which have unexplained properties Facts for all triangles when the base of the triangle is one. Are there any explication? have three measures of three altitudes $h_0,h_1,h_2$ where sides $a,b,c$ are the constant lengths of all triangles and base $c=1$ and $h_0$ is situated on base $c$. $\frac{h_0}{h_1}=b$,$\frac{h_0}{h_2}=a$ and $h_1=\sin A$ as well $h_2=\sin B$. $\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ d = diameter $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ if I multiply by d, the diameter, it will give an answer of 1, and 1 is the base. And if I multiply the base times 4, all of 3 sides are multiple of 4 including the altitudes. d = diameter $\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ jet.jpg Last edited by skipjack; January 2nd, 2019 at 02:44 AM.
January 1st, 2019, 07:05 PM   #2
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Quote:
 Originally Posted by Larrousse And if I multiply the base times 4, all of 3 sides are multiple of 4 including the altitudes. d=diameter
For example?

January 1st, 2019, 08:19 PM   #3
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I have three measures of three altitudes $h_0,h_1,h_2$, where sides $a,b,c$ are the constant lengths of all triangles and base $c=1$ and $h_0$ is situated on base $c$.
$\frac{h_0}{h_1}=b$,$\frac{h_0}{h_2}=a$ and $h_1=\sin A$ as well $h_2=\sin B$.

$\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$

d = diameter

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ if I multiply by d, the diameter, it will give an answer of 1, and 1 is the base. And if I multiply the base times 4, all of 3 sides are multiple of 4 including the altitudes. d = diameter

$\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$

Sides 5,5,4 where 4 is the base and by dividing 5÷4=1.25, 1.25, 4÷4=1 and the base equals 1.
Attached Images
 jet.jpg (83.9 KB, 1 views)

Last edited by skipjack; January 2nd, 2019 at 02:45 AM.

January 1st, 2019, 08:43 PM   #4
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Quote:
 Originally Posted by Larrousse I have three measures of three altitudes $h_0,h_1,h_2$ where sides $a,b,c$ are the constant lengths of all triangles and base $c=1$ and $h_0$ is situated on base $c$. $\frac{h_0}{h_1}=b$,$\frac{h_0}{h_2}=a$ and $h_1=\sin A$ as well $h_2=\sin B$. $\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ d=diameter $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ if I multiply by d, the diameter, it will give an answer of 1, and 1 is the base. And if I multiply the base times 4, all of 3 sides are multiple of 4 including the altitudes. d = diameter $\frac{a}{\sin C}=\frac{b}{\sin B}=\frac{c}{\sin C}=d$ $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$ Attachments jet.jpg
You didn't answer anything but just reposted the original post.

I don't know what you mean by "base" but if you take the following equation:
$\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{d}$

and multiply all sides by "d" you get
$\displaystyle d \cdot \dfrac{\sin A}{a}= d \cdot \dfrac{\sin B}{b}= d \cdot \frac{\sin C}{c}=d \cdot \dfrac{1}{d}$

and you get
$\displaystyle \dfrac{d~\sin A}{a}= \dfrac{d~\sin B}{b}= \dfrac{d~\sin C}{c}= 1$

If you multiply both all sides by 4d you get
$\displaystyle \dfrac{4d~\sin A}{a}= \dfrac{4d~\sin B}{b}= \dfrac{4d~\sin C}{c}= 4$

There's no magic here. Geometrically speaking all you are doing is rescaling your triangle into a similar one.

-Dan

Addendum That is if I'm interpreting the problem correctly.

Last edited by skipjack; January 2nd, 2019 at 02:41 AM.

 January 10th, 2019, 06:24 AM #5 Newbie   Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 According to Pythagorean theorem, these are unexplained properties for all triangles. $$\cos A + (b - \cos A)=b$$ $$\cos B + (a - \cos B)=a$$ $$(\sin B)^2+(a -\cos B)^2=b^2$$ $$(\sin A)^2+(b -\cos A)^2=a^2$$ $$(\sin B)^2+(\cos B)^2=c^2$$ $$(\sin A)^2+(\cos A)^2=c^2$$ $$(b\times\cos C)=(a -\cos B)$$ $$(a\times\cos C)=(b -\cos A)$$ $$(a\times\cos B)+(b \times\cos A)=c$$ Last edited by skipjack; January 10th, 2019 at 11:56 AM.
 January 10th, 2019, 11:53 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,628 Thanks: 2077 None of those are "unexplained". Some of them are true only when c = 1. Thanks from topsquark
 January 11th, 2019, 12:23 PM #7 Newbie   Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 These are new to me and it describes the schematic shown above, but using a simplified version of the triangles when base equals 1, the lengths of altitudes or sides are measures of angles. $(c\times\sin B)^2+(a\times\cos C)^2=b^2$ $(c\times\sin A)^2+(b\times\cos C)^2=a^2$ $(c\times\sin B)^2+(c\times\cos B)^2=c^2$ $(c\times\sin A)^2+(c\times\cos A)^2=c^2$ $(b\times\cos C)+(c\times\cos B)=a$ $(a\times\cos C)+(c\times\cos A)=b$ $(a\times\cos B)+(b \times\cos A)=c$ $(c\times\cos B)+(b \times\cos C)=a$ $(c\times\cos A)+(a\times\cos C)=b$ and to find one of the altitudes: $h_1=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$ Last edited by skipjack; January 11th, 2019 at 04:16 PM.
January 11th, 2019, 04:15 PM   #8
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Quote:
 Originally Posted by Larrousse These are new to me and it describes the schematic shown above, but using a simplified version of the triangles when base equals 1, the lengths of altitudes or sides are measures of angles. $(c\times\sin B)^2+(a\times\cos C)^2=b^2$ $(c\times\sin A)^2+(b\times\cos C)^2=a^2$ $(c\times\sin B)^2+(c\times\cos B)^2=c^2$ $(c\times\sin A)^2+(c\times\cos A)^2=c^2$ $(b\times\cos C)+(c\times\cos B)=a$ $(a\times\cos C)+(c\times\cos A)=b$ $(a\times\cos B)+(b \times\cos A)=c$ $(c\times\cos B)+(b \times\cos C)=a$ $(c\times\cos A)+(a\times\cos C)=b$ and to find one of the altitudes: $h_1=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$
There's a very good reason why the first can't be proven: It isn't correct. Apply this to the 20-40-120 triangle with a = 1, b = 1.8793, and c = 2.5321. This doesn't work for the first "identity" and I'm going to presume the rest don't either. I suspect the problem is that these hold for some specific type of triangle (not an arbitrary one).

-Dan

Last edited by skipjack; January 11th, 2019 at 04:18 PM.

 January 11th, 2019, 04:39 PM #9 Global Moderator   Joined: Dec 2006 Posts: 20,628 Thanks: 2077 The first two were evidently mistyped and should have been as follows: $(c\times\sin B)^2+(b\times\cos C)^2=b^2$ $(c\times\sin A)^2+(a\times\cos C)^2=a^2$ Thanks from topsquark
 January 11th, 2019, 05:08 PM #10 Newbie   Joined: Jan 2018 From: Ontario Posts: 11 Thanks: 0 It was my mistake. The first 2 lines were found to be with errors and I just corrected them and now it is verified and as long as sides $a+b>c. a+c>b. b+c>a$ I don't know why one triangle is different than others if they hold the same properties. The only difference is that one triangle has its three angles and and its six properties $\sin A,\sin B,\sin C,\cos A,\cos B,\cos C$. $f^1)$ $(c\times\sin B)^2+(b\times\cos C)^2=b^2$ $f^2)$ $(c\times\sin A)^2+(a\times\cos C)^2=a^2$ $(\sin B)^2+(\cos B)^2=c^2$ $(\sin A)^2+(\cos A)^2=c^2$ $(b\times\cos C)+(c\times\cos B)=a$ $(a\times\cos C)+(c\times\cos A)=b$ $(a\times\cos B)+(b \times\cos A)=c$ $(c\times\cos B)+(b \times\cos C)=a$ $(c\times\cos A)+(a\times\cos C)=b$ and to find one of the altitude: $h_1=\frac{\sqrt{2(a^2 b^2+a^2 c^2+b^2 c^2)-a^4-b^4-c^4}}{2a}$ Thanks. Last edited by skipjack; January 11th, 2019 at 05:45 PM.

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