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## Re: Best solution in Mathcad needed!

You're right on the circle. What I'm not sure of is that both angles remain constant moving P on a fixed circumference. However, my solution is not wrong. ## Re: Best solution in Mathcad needed!

Sorry, but what you show at the bottom right as a "proof" is no proof because you again use the wrong assumption PB=PC here.

Calculate the coordinates of P with your "solution" and then calculate the lengths and angles one by one without assuming any symmetry (which simply isn't there).

you are right.

## Re: Best solution in Mathcad needed!

Just noticed that an easy way to see that is when you try to add up the four angles you calculated - they should add up to 360 degree but they don't.

## Re: Best solution in Mathcad needed!

I apologize, you are right. Ich bitte um entschuldigung!

## Re: Best solution in Mathcad needed!

@-MFra- wrote:

I apologize, you are right.

Absolutely no reason to apologize for a mistake that can happen to anyone.
In fact I wasn't absolutely sure about my solution which was developed rather quick (and dirty) but after I checked with the angles as shown in the picture above as a consequence of our discussion, I feel pretty sure now.

Maybe you find the time to solve, too, and confirm.

## Re: Best solution in Mathcad needed!

Genau jetzt habe Ich diese solution gefunde, was glaubest du daruber? ## Re: Best solution in Mathcad needed!

Its hard and boring to debug just a pic, but as far as I see you have five unknowns to solve for but only 4 equations. This would mean that there would be an infinite number of solutions and the searched for solution would be amongst them, but is not necessarily the one Mathcad solve block comes up with.

BTW, are you sure that those 4 equations are independent from each other? If they are not, additional degrees of freedom comes into play.

Do you get different results if you change the guess values, lets say x:10 for example?

## Re: Best solution in Mathcad needed!

Just to be on the safe side I solved the problem with a completely different approach compared to my first (P as intersection point of two circles as mentioned in one of my replies above) and got the very same result as with my fist method. So I assume its the correct result. EDIT: This second approach can be significantly simplified as we already know one of the two points of intersection (B in my nomenclature). Therefore we only need to mirror B at the connecting line of the two circle centers to get P.

## Re: Best solution in Mathcad needed!

Here I added the condition that the sum of the areas of the single triangles is equal to that of the rectangle: 