# Thread: Calculating the Minimum Perimeter of a Triangle

1. ## 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 x-axis,
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.

2. ## 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.

3. ## Re: Calculating the Minimum Perimeter of a Triangle

Hello, gfbrd!

Did you make a sketch?

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 x-axis, and one on the line y = x.

Code:
      |
|                     *
|                   *
|               R *
|               o(y,y)
|             **  *
|           * *     *
|         *  *        * P
|       *   *           o(a,b)
|     *    *        *
|   *     *     *
| *    Q *  *
- - * - - - o - - - - - - - - - - - -
|     (x,0)
|
Point $P$ has given coordinates $P(a,b).$
It lies below the 45o-line.

Point $Q$ lies on the x-axis with coordinates $Q(x,0).$

Point $R$ lies on the line $y = x$ with coordinates $R(y,y).$

The lengths of the sides of the triangle are:

. . $PQ \;=\;\sqrt{(x-a)^2 + b^2}$

. . $QR \;=\;\sqrt{(x-y)^2 + y^2}$

. . $PR \;=\;\sqrt{(y-a)^2 + (y-b)^2}$

The perimeter of the triangle is:

. . $Z \;=\;\sqrt{(x-a)^2 + b^2} + \sqrt{(x-y)^2 + y^2} + \sqrt{(y-a)^2 + (y-b)^2}$

And that is the function we must minimize.

4. ## 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 x-axis", and so didn't include that condition.