Emre HAs wrote:
I am trying plot a circle using 3 points of a triangle. Circle must be plot inside triangle. How can i do?
Can you solve this problem not thru the using ready firmulas but thru using the Given-Find block 
Werner Exinger wrote:
Unfortunately Emre did quite some crossposting. The inscribed circle wasn't the sole purpose and it wasn't meant to be 2D. See the "main" post here: http://communities.ptc.com/message/201417#201417
Thanks! But the question was "Can you solve this problem not thru the using ready firmulas but thru using the Given-Find block?"
What is better (for students) - using ready (and unclear) formulas from reference books (and from Werner's solution) or searching own solution based on one formula - Pythagorean theorem:
Sorry, my fault. I thought the intent of your post was to help Emre with his problem.
My approach is: "use the resources and don't reinvent the wheel". Students have to learn to use reference books, too. Of course its our task as teachers to pose questions which can't simply be taken out of a book but require(! by the nature of the problem and not because I forbid the use refernce books) some autonomous work from their side. The task of finding the triangle inscribed circle sure is not one of those problems, while it can be quite challenging for students finding a suitable formula. If I would insist that my students deduce a formula, I would rather insist on an exact analytical one, too, and not a numeric appraoch using pushbutton Find. Nevertheless I agree that finding the equations of constraint can be a fine "finger excercise" for students.
In my reply I had chosen on purpose a formula which is not that common and was prepared being asked about it, but Emre didn't question it. Hopefully his teacher (I remember him writing "teacher is very brilliant, he sees everything") didn't asked him about .
This is a very interesting and important educational problem.
Now students have available Reference books in Internet and available tools of symbolic and numerical solution of Math problems (Mathcad for example) - such as systems of equations.
I recommend my students to first try to resolve the problem yourself, and then use a ready-made solutions.
The best thing in teaching is to understand where it came from ready-made formula.
What solution is better:
1.
2.
I don't see anything from your animation - connection rejected. Those flash apps are rather crappy. But it doen't mather. From a pure mathematical point of view -r seems to be a solution, as you equations use r exclusively in the form r^2. But as you are using the Pythagorean theorem and this theorem doesn't deal with oriented length, you have implicitely assumed by using it, that your lengths, especially r, are positive.
For whatever its worth find attached a vectorial solution using the well known distance (point-line) formula. Still a numeric approach only, but parameterized for better flexiblity.
Keep in mind, that the order of the variables is vital here to get the INscribed circle. You may try changing e.g. nv(C-B) for nv(B-C) and you will still get a circle touching all three sides but because auf the "wrong" orientation of the normal vector it will be one of the three outer circles.
For a task like this I won't need units, but of course this approach works with units, too.
Maybe you forgot to add the units to the guess values!?
Unfortunately we are not allowed to add a "SIUnitsOf(B)" here so the block would be useable with AND without units 😞
Werner Exinger wrote:
But I am not sure what I am supposed to see here?
There is one type of decoration for rooms - mobile. It makes no sense, but look nice. Nerves calm down. My animations of numerican methods are one type of virtual mobile .
See please more interesting animation (other method for Find) in attach (for Werner) and http://communities.ptc.com/videos/3972 (for all).
I see in this animation any living creature that at first tries to break into a triangle, and then - get out of it. The creature calm down after the solution is found.
KISS?
We are searching a point with equal distance to three straight lines. What would be more simple than using distance formula? Not sure if that would apply for stupid, too
As Einstein already said: "Make everything as simple as possible, but not simpler!" 'nough said.
Werner Exinger wrote:
KISS?
We are searching a point with equal distance to three straight lines. What would be more simple than using distance formula? Not sure if that would apply for stupid, too
As Einstein already said: "Make everything as simple as possible, but not simpler!" 'nough said.
We are searching a point with equal distance to three straight lines... and then separately calculate the radius of the circle
We can find the point of intersection of two bisectors and then separately calculate the radius of the circle.
I think an universal meyhod to solve any geometric problem - we need to find the values of n unknowns. To do this, we must create and solve a system of n algebraic equations. Numeric with Mathcad for example. KISS!
Thanks, Alan!
But 3 equations and 2 unknowns - sorry, may by I do not understand someone.
And second.
Now I am helping to translate into Mathcad one book of problems in electrical engineering.
I see in this book imaginary currents, imaginary voltage, imaginary resistance, imaginary power, etc.
I ask authors: "Guys, do you know that in nature there is no imaginary currents, voltage, resistance, power, etc."
The authors answer me that they know about it. Then I ask them, "Why did you mislead students and tell them about the imaginary currents, voltage, resistance, power, etc.?" The authors answer that they do it to use mathematics of complex numbers. Then I ask them, "Did you know that there is a modern computer tools to solve these problems without using of electrical imaginary values." The authors reply that... "You can not teach an old dog to new tricks" and "An old horse (ox) not spoil the furrow." (what is the English equivalents of this Russian proverbs?) 
I use at home two type of electrical power - real and imaginary. I pay for real power real money and for imaginary one - imaginary money
Valery Ochkov wrote:
Thanks, Alan!
But 3 equations and 2 unknowns - sorry, may by I do not understand someone.
X has two components: one real, oine imaginary, so in total there are three unknowns.
. "You can not teach an old dog to new tricks" and "An old horse (ox) not spoil the furrow." (what is the English equivalents of this Russian proverbs?)
The "old dog, new tricks" is well known in English, but I'm not aware of another similar to the "old horse (ox).." one. 
Alan
