Calculating the Minimum Perimeter of a Triangle
Hey there sorry if this is in the wrong thread, but I need some help with this problem I'm stuck on.
Give a point (a,b) with 0 < b < a
determine the minimum perimeter of a triangle with one vertex at (a,b), on the xaxis,
and one on the line y = x.
I hope someone can help/guide me through this problem, so I can understand how to do this problem thanks.
Re: Calculating the Minimum Perimeter of a Triangle
No triangle exists that minimizes that condition.
However, the infinum of the perimeters of all such triangles = twice the distance from (a,b) to the line y = x.
That's just an application of the triangle inequality after picking the first two points (one is on the line, the other is (a,b)) of any triangle.
Re: Calculating the Minimum Perimeter of a Triangle
Hello, gfbrd!
Did you make a sketch?
Quote:
Give a point (a,b) with 0 < b < a
determine the minimum perimeter of a triangle
with one vertex at (a,b), one on the xaxis, and one on the line y = x.
Code:

 *
 *
 R *
 o(y,y)
 ** *
 * * *
 * * * P
 * * o(a,b)
 * * *
 * * *
 * Q * *
  *    o            
 (x,0)

Point $\displaystyle P$ has given coordinates $\displaystyle P(a,b).$
It lies below the 45^{o}line.
Point $\displaystyle Q$ lies on the xaxis with coordinates $\displaystyle Q(x,0).$
Point $\displaystyle R$ lies on the line $\displaystyle y = x$ with coordinates $\displaystyle R(y,y).$
The lengths of the sides of the triangle are:
. . $\displaystyle PQ \;=\;\sqrt{(xa)^2 + b^2}$
. . $\displaystyle QR \;=\;\sqrt{(xy)^2 + y^2}$
. . $\displaystyle PR \;=\;\sqrt{(ya)^2 + (yb)^2}$
The perimeter of the triangle is:
. . $\displaystyle Z \;=\;\sqrt{(xa)^2 + b^2} + \sqrt{(xy)^2 + y^2} + \sqrt{(ya)^2 + (yb)^2}$
And that is the function we must minimize.
Re: Calculating the Minimum Perimeter of a Triangle
Apologies  ignore my previous post  it's wrong. I simply read right over the phrase "on the xaxis", and so didn't include that condition.