Did you know that a right triangle can be found using only one side?

[Crazy Horse 11]

I have found a way to find right triangles using only one side.

How do I show you without being deleted?

 

[dx]

You can construct many different right triangles, from a given one side. You cannot post claims or theories here that contradict accepted results, but you can probably get away with a post along the lines of "here are my thoughts, is there something wrong with the way I am thinking?"

 

[Crazy Horse 11]

You can construct many different right triangles, from a given one side. You cannot post claims or theories here that contradict accepted results, but you can probably get away with a post along the lines of "here are my thoughts, is there something wrong with the way I am thinking?"

Yes but how do you know what their values are? My formula finds the values of the triangle using only the one side.

Here is my formula: c^2 = cx + cy what is wrong with it when the line c = x + y? For example:

c = 1.8 + 3.2 = 5. This line = 5 is made up of two segments of x = 1.8 and y = 3.2.

The square root "sqrt" of c^2 is c. I can't write any more simply than this.

c^2 = cx + cy
c = sqrt (cx + cy)
5 = sqrt ((5*1.8) + (5*3.2))
5 = sqrt (9 + 16)
5 = sqrt 25
5 = 5
side c = 5
side a = sqrt (c^2 - cy) = sqrt cx = sqrt 9 = 3
side b = sqrt (c^2 - cx) = sqrt cy = sqrt 16 = 4

 

[Fredrik]

My formula finds the values of the triangle using only the one side.

Here is my formula: c^2 = cx + cy what is wrong with it when the line c = x + y? For example:

c = 1.8 + 3.2 = 5. This line = 5 is made up of two segments of x = 1.8 and y = 3.2.

The square root "sqrt" of c^2 is c. I can't write any more simply than this.

c^2 = cx + cy
c = sqrt (cx + cy)
5 = sqrt ((5*1.8) + (5*3.2))
5 = sqrt (9 + 16)
5 = sqrt 25
5 = 5
side c = 5
side a = sqrt (c^2 - cy) = sqrt cx = sqrt 9 = 3
side b = sqrt (c^2 - cx) = sqrt cy = sqrt 16 = 4

x+y=5 is the equation of a line. There's a point on that line with coordinates x=1.8, y=3.2. But so what? It's still impossible to tell what triangle you're talking about.

The first six lines of your calculation are just a complicated way to rewrite 5=1.8+3.2 as 5=5. This doesn't prove anything about any triangle.

If the hypotenuse (longest side) of a right triangle is 5, it's possible that the other two sides are 3 and 4. That much is correct. But it's also possible that they are (for example) 1 and $2\sqrt 6$, because
$$1^2+(2\sqrt{6})^2=1+24=25=5^2.$$

 

[dx]

You have chosen to divide c (the hypotenuse) as x + y, and to each of the possible ways of dividing c into two parts, you have a right triangle.

You could have chosen c = 5 = 2.5 + 2.5, then you would get a = b = sqrt(5(2.5))

 

[Fredrik]

I have found a way to find right triangles using only one side.

How do I show you without being deleted?

For starters, you don't claim to have found something that contradicts mathematics. If I had seen the thread title and the first line of post #1 before I wrote my reply, I would have deleted the thread instantly.

 

[dx]

A picture:

 

[Fredrik]

You're quite an artist dx.

I can't tell what the lower left symbol is.

Don't spend too much time on this thread. It will probably end up deleted anyway.

 

[Crazy Horse 11]

A picture:

That is correct. The point where x = 1.8 gives the height of the triangle which when using Pythagoras theorem yields a side of 2.4 see?

c=3
b=1.8
a^2 = c^2 - b^2
a^2 = 9 - 3.24
a^2 = 5.76
a = 2.4 note: the height of the triangle is 2.4 at the point x = 1.8 or y = 3.2 on the hypotenuse = 5

the point is that c = x + y at 5 = 1.8 + 3.2 is just an example of how c^2 = cx + cy works at finding the three sides of a right triangle. I could have said 5 = 2.5 + 2.5 or 5 = 1.3 + 3.7 etc...

5 = 1.3 + 3.7
25 = 6.5 + 18.5
side a = sqrt 6.5
side b = sqrt 18.5
side c = sqrt 25

The Spencer Triangle Method: c = x + y does not use angles or more than one side to find the other two sides for a right triangle. This is a totally new except that I have shown it on the internet before on other cites.

It is very discouraging to have original new math concepts deleted without much consideration here on this forum. Very simple concepts run very deep in thought. That is what a discussion is for frankly.

Because it doesn't seem to make sense to one mentor and then he wipes the board clean. What a waste of prime intelligence in a raw state. You throw away a nugget of pure gold because you didn't find a polished ring that fit your finger! That is the great downfall of this forum.

Already this forum has wiped away a formula for a rational pi because it didn't look pleasing to the eye. Too Bad you short sited ones! Why couldn't you discuss where the equation was flawed?

Because you thought that your intelligence was threatened? how cowardly of you.

Up until now only Pythagoras theorem found right triangles but now another method has been found namely c^2 = cx + cy. What is stopping you from deleting history now?.

 

[Crazy Horse 11]

In reverse a 3, 4, 5 triangle is c^2 = a^2 + b^2. This is 25 = 9 + 16. 9/c = 1.8 and 16/c = 3.2. 1.8 + 3.2 = c. What a priceless discovery if I do say so myself.

 

[Crazy Horse 11]

x+y=5 is the equation of a line. There's a point on that line with coordinates x=1.8, y=3.2. But so what? It's still impossible to tell what triangle you're talking about.

The first six lines of your calculation are just a complicated way to rewrite 5=1.8+3.2 as 5=5. This doesn't prove anything about any triangle.

If the hypotenuse (longest side) of a right triangle is 5, it's possible that the other two sides are 3 and 4. That much is correct. But it's also possible that they are (for example) 1 and $2\sqrt 6$, because
$$1^2+(2\sqrt{6})^2=1+24=25=5^2.$$

c = x + y would be 1/c = 0.2 for x and 24/c = 4.8 for y then the original line or side would be
5 = 0.2 + 4.8

 

[pwsnafu]

In reverse a 3, 4, 5 triangle is c^2 = a^2 + b^2. This is 25 = 9 + 16. 9/c = 1.8 and 16/c = 3.2. 1.8 + 3.2 = c. What a priceless discovery if I do say so myself.

This is trivial. All you have done is taken $c^2 = a^2 + b^2$ and divided both sides by c to obtain $c = \frac{a^2}{c} + \frac{b^2}{c}$. And then declared $x = \frac{a^2}{c}$ and $y = \frac{b^2}{c}$. So your formula is equivalent to Pythagoras.

 

[Crazy Horse 11]

This is trivial. All you have done is taken $c^2 = a^2 + b^2$ and divided both sides by c to obtain $c = \frac{a^2}{c} + \frac{b^2}{c}$. And then declared $x = \frac{a^2}{c}$ and $y = \frac{b^2}{c}$. So your formula is equivalent to Pythagoras.

Sorry but c = x + y equals only one line. From the knowledge of one line comes the other two sides c^2 = cx + cy. In fact Pythagoras requires two sides at least. But he found his method first in time. In fact his method is equivalent to mine. Sorry for the ego.

I can find a triangle without Pythagoras but can he find a triangle with only one side and no angles? I severely doubt it very much. This is by no means trivial. Simple yes but not trivial.

 

[mfb]

There is nothing new to your method. All you do is to choose the length of the hypotenuse (here: 5) plus one additional length in your triangle (here: 1.8). You use Pythagoras in an implicit way, as pwsnafu showed. You find three values c,a,b which satisfy c^2=a^2+b^2 by construction.

In the same way, you could directly choose another side length:
5 and 2 give sqrt(5^2-2^2)=sqrt(21) as third side.
5 and 3 give sqrt(5^2-3^2)=sqrt(16)=4 as third side.
...

 

[pwsnafu]

Sorry but c = x + y equals only one line. From the knowledge of one line comes the other two sides c^2 = cx + cy. In fact Pythagoras requires two sides at least. But he found his method first in time. In fact his method is equivalent to mine. Sorry for the ego.

I can find a triangle without Pythagoras but can he find a triangle with only one side and no angles? I severely doubt it very much. This is by no means trivial. Simple yes but not trivial.

Pythagoras cannot determine a triangle using only knowledge of one side.
Your method is equivalent to Pythagoras.
Therefore your method cannot determine a triangle using only knowledge of one side.

Do you why it doesn't work? Because if you only fix one side, there are an infinite number of triangles which satisfy it.

In order to solve anything, you are using knowledge of two sides. You just don't realize it.

 

[Crazy Horse 11]

Pythagoras cannot determine a triangle using only knowledge of one side.
Your method is equivalent to Pythagoras.
Therefore your method cannot determine a triangle using only knowledge of one side.

Do you why it doesn't work? Because if you only fix one side, there are an infinite number of triangles which satisfy it.

In order to solve anything, you are using knowledge of two sides. You just don't realize it.

Basically you lie mathematically speaking. How many different right triangles can you make from
c = 0.2 + 4.8? You say that you can arrange an infinite amount and I say that you lie.
I have side c = 5 and side a = 1 and side b = 2 sqrt 6.

Show me just one other right triangle from c = 0.2 + 4.8. I don't think that you can.

 

[Borek]

c = 0.2+4.8 = 5

a² + b² = 5²

$$a = \sqrt{25 - b^2}$$

Any b<5 you put into the equation has a corresponding a.

$$b = 1, a = \sqrt{24} = 2\sqrt 6\\ b = 2, a = \sqrt{20} = 2\sqrt 5\\ b = 3, a = \sqrt{16} = 4$$

Already three different solutions.

 

[Crazy Horse 11]

c = 0.2+4.8 = 5

a² + b² = 5²

$$a = \sqrt{25 - b^2}$$

Any b<5 you put into the equation has a corresponding a.

$$b = 1, a = \sqrt{24} = 2\sqrt 6\\ b = 2, a = \sqrt{20} = 2\sqrt 5\\ b = 3, a = \sqrt{16} = 4$$

Already three different solutions.

I notice only one solution for c = 0.2 + 4.8. Even your third offering is still c = 1.8 + 3.2. You prove my point that there is only one right triangle for each c = x + y.

I will now ask you this. What are the three sides of the right triangle from c = 12.44578 + 8.17695? You already know that it only describes one right triangle don't you by now?.

 

[dx]

You're quite an artist dx.

I can't tell what the lower left symbol is.

Haha.. that's a pretty bad drawing.

The symbol is x. Here's a better drawing. One can show that the possible positions of the vertex of the triangle with the right angle must lie on a circle, with c as the diameter.

Here x + y = c and xy = h²

So we have

a² = x² + h² = x² + xy = xc

which implies a = √(xc)

Similarly b = √(yc)

 

[Borek]

I notice only one solution for c = 0.2 + 4.8. Even your third offering is still c = 1.8 + 3.2. You prove my point that there is only one right triangle for each c = x + y.

I will now ask you this. What are the three sides of the right triangle from c = 12.44578 + 8.17695? You already know that it only describes one right triangle don't you by now?.

If you are splitting c into two given parts, you are no longer speaking of a universal solution using ONE SIDE. Your are solving an entirely different question - what are catheti of a right triangle in which hypotenuse is split by the triangle height into x and y. And then yes, it is a trivial problem with a single solution.

 

[pwsnafu]

I notice only one solution for c = 0.2 + 4.8. Even your third offering is still c = 1.8 + 3.2. You prove my point that there is only one right triangle for each c = x + y.

He gave you three triangles:
$(2\sqrt{6}, 1, 5)$
$(2\sqrt5,2,5)$
$(4,3,5)$

They are different triangles with hypotenuse is equal to 5.

IF you are actually claiming that $c = 1.8 + 3.2$ and $c = 1.7 + 3.3$ are different, then you are using two pieces of information: the value of x and the value of y.

 

[Crazy Horse 11]

If you are splitting c into two given parts, you are no longer speaking of a universal solution using ONE SIDE. Your are solving an entirely different question - what are catheti of a right triangle in which hypotenuse is split by the triangle height into x and y. And then yes, it is a trivial problem with a single solution.

You are wrong the triangle height of 2.4 cannot be split into x = 1.8 and y = 3.2 which creates the three sides of the right triangle with a height of 2.4. The line of the hypotenuse is split by x or y and the result is the difference of the hypotenuse not the height of the triangle.

Previously your three solutions were based on Pythagoras not c^2 = cx + cy which offers only one solution. Pythagoras theorem is then trivial if my formula is equivalent to it.

Pythagoras requires two sides but my method requires only one side. Take any portion of the hypotenuse and the three sides of the particular right triangle can be quickly calculated from it. Provo!

 

[pwsnafu]

Previously your three solutions were based on Pythagoras not c^2 = cx + cy which offers only one solution. Pythagoras theorem is then trivial if my formula is equivalent to it.

If your solution only offers only one solution, but Pythagoras offers all solutions then Pythagoras is better!

Let's try a different example: find me all triangles with hypotenuse equal to $4\sqrt{6}$.

Take any portion of the hypotenuse

So you have used two pieces of information: "the length of the hypotenuse" and "the length of the portion".

 

[Crazy Horse 11]

He gave you three triangles:
$(2\sqrt{6}, 1, 5)$
$(2\sqrt5,2,5)$
$(4,3,5)$

They are different triangles with hypotenuse is equal to 5.

IF you are actually claiming that $c = 1.8 + 3.2$ and $c = 1.7 + 3.3$ are different, then you are using two pieces of information: the value of x and the value of y.

Yes that is correct but the two pieces of information make up 1 side of the triangle which is all that I have. Whoever said that I couldn't use two pieces of information? But I definitely only have one side to work from and that is a fact.

 

[Crazy Horse 11]

If your solution only offers only one solution, but Pythagoras offers all solutions then Pythagoras is better!

Let's try a different example: find me all triangles with hypotenuse equal to $4\sqrt{6}$.



So you have used two pieces of information: "the length of the hypotenuse" and "the length of the portion".

That is the point. With Pythagoras method you have no choice but to consider all the right triangles from a given hypotenuse. My method calculates individual triangles given the hypotenuse. Which one do you want? c = 1.8 + 3.2 or c = 2.5 + 2.5? Or would you want a triangle from a different c? Pythagoras can't come close to an offer like that.

 

[Crazy Horse 11]

There is nothing new to your method. All you do is to choose the length of the hypotenuse (here: 5) plus one additional length in your triangle (here: 1.8). You use Pythagoras in an implicit way, as pwsnafu showed. You find three values c,a,b which satisfy c^2=a^2+b^2 by construction.

In the same way, you could directly choose another side length:
5 and 2 give sqrt(5^2-2^2)=sqrt(21) as third side.
5 and 3 give sqrt(5^2-3^2)=sqrt(16)=4 as third side.
...

If there is nothing new about my method then why do you not see "c^2 = cx + cy" in the textbooks? Or anything pertaining to c = x + y? Exactly! Because it is new. 5 * (1.8+3.2) = 5^2. So the sqrt of 5x = one side of the triangle. This is completely new.

 

[pwsnafu]

Pythagoras can't come close to an offer like that.

Except in post #19, dx demonstrated how to use Pythagoras to do it.

Edit: I was wondering where I had seen dx's drawing when he first posted, but now I remember. It was a question on my year 9 maths mid-year exam. Strange how you remember these things.

 

[mfb]

If there is nothing new about my method then why do you not see "c^2 = cx + cy" in the textbooks?

It is a more complicated way to write the usual equations.

You can replace "5" with "545349-545344". Nobody does, as it does not add anything new.

With Pythagoras method you have no choice but to consider all the right triangles from a given hypotenuse.

You don't have to, but you have the freedom to do so.
Your "method" is just one possible way to pick some specific triangle from the set of all triangles (with that hypotenuse). You can do the same by choosing a, or h, or anything else in the triangle.

 

[Crazy Horse 11]

Except in post #19, dx demonstrated how to use Pythagoras to do it.

Edit: I was wondering where I had seen dx's drawing when he first posted, but now I remember. It was a question on my year 9 maths mid-year exam. Strange how you remember these things.

You can use Pythagoras for double checking. example sqrt 5x = 3. sqrt 5y = 4.

c^2 = a^2 + b^2
5^2 - 3^2 = b^2
16 = b^2
4 = b so yes Pythagoras confirms that the triangle with sides 3, 4, and 5 make a right triangle but what are x and y?

sqrt 5x = 3
5x = 9
x = 9/5
x = 1.8

sqrt 5y = 4
5y = 16
y = 16/5
y= 3.2 simply algebra finds x and y not Pythagoras theorem.

c = x + y
c = 1.8 + 3.2
c = 5 its square! without Pythagoras.

And for the record my circles formed from my rational pi equation inscribe within my squares perfectly.

 

[pwsnafu]

You can use Pythagoras for double checking.

Your expression is equivalent to Pythagoras. You even said so yourself. So really, you're just doing the same thing twice.

but what are x and y?

Why should we care?

simply algebra finds x and y not Pythagoras theorem.

Again, your formula is equivalent to Pythagoras, so you are using it indirectly.

And for the record my circles formed from my rational pi equation inscribe within my squares perfectly.

I have no idea what you mean. Please use clear English.

Edit: Anyway, to summarise

Did you know that a right triangle can be found using only one side?

If we are given the length of one side and the length of a line segment from that side, we can determine a unique right angle triangle. Yes we knew that. It is basic geometry and is equivalent to Pythagoras.

 

[Crazy Horse 11]

It is a more complicated way to write the usual equations.

You can replace "5" with "545349-545344". Nobody does, as it does not add anything new.

You don't have to, but you have the freedom to do so.
Your "method" is just one possible way to pick some specific triangle from the set of all triangles (with that hypotenuse). You can do the same by choosing a, or h, or anything else in the triangle.

A line makes up one side of the triangle what are the other two sides? boo who I don't have enough information?

okay x = 3 and it is 1/5th of the line what are the other two sides? boo who I don't have enough information?

but yes you do!
c = 5x
c = 5 * 3
c = 15
y = 15 - 3
y = 12
c = x + y
15 = 3 + 12
c^2 = cx + cy
side a = sqrt cx = sqrt 45
side b = sqrt cy = sqrt 180
side c = 15

double checking with Pythagoras:
c^2 = a^2 + b^2
225 = 45 + 180 yes the triangle that I just designed is a right triangle using my own methods. But using my methods at least we now know what the right triangle looks like.

What is your answer to the question? Personally I feel that you still don't have enough information if all you are limited to Pythagoras theorem or the law of sine's!

 

[micromass]

I don't get this. If you are given one side, then there are infinitely many right triangles which have that as a side. So why would you expect to get only one solution? Why would we care about a formula that gives only one solution?

 

[Crazy Horse 11]

What you take for granted is that you already have a right triangle to begin with for your exercises.

What you cannot do is calculate a triangle from a single side. Thank-you for allowing me to show you how I can do that using my own methods at that.

 

[Borek]

What you cannot do is calculate a triangle from a single side. Thank-you for allowing me to show you how I can do that using my own methods at that.

OK, so if you are right, you will have no problem solving my problem.

I have a triangle with hypotenuse length equal to 10. Give me the other two sides of the triangle I have on mind.

 

[Crazy Horse 11]

I don't get this. If you are given one side, then there are infinitely many right triangles which have that as a side. So why would you expect to get only one solution? Why would we care about a formula that gives only one solution?

Those are two good questions to ask of yourself from for the answer.