Finding the length between the two vertices of two different triangles having the same base side and same perimeter by a shorter method.

areaconic sectionseuclidean-geometrygeometrytriangles

Here is the figure to give you a better idea about the problem

My approach:-

  1. Using distance formula I found the lengths of AC,CB and AB. Using them I found out the perimeter of
    triangle ABC and it came out to be 18 square units. Hence, $$AB+BC+CA=18 units$$

  2. Then for triagle ABP, the perimeter can be written as sum of lengths of AB,AP and BP. Length of AB is
    8 units, that gives us the sum of lengths of AP and BP as 10 units. Hence,
    $$AP+BP=10 units$$

  3. Found out the area of ABC is 12 units. That means the area of ABP is 6 units.

  4. Assumed coordinates of P as (g,h). Hence the area of ABP should be, $\frac{(AB)(h)}{2}=6$. This gives
    us h=3/2.

  5. To get 'g', I wrote the distance AP and BP using distance formula,added them and equated them to 10.
    This gives us an equation where 'g' is the only variable. And I found its value.

What I am looking for:-

A shorter method to solve this problem if there is any. Because my approach took a lot of calculations and time.

Best Answer

Since $AP+BP=AC+BC,$ we see that $P$ is placed on the ellipse with focuses $(-4,0)$ and $(4,0)$, which gives $c=4$ and $b=y_{A}=3$, which gives $a=5$ and an equation of the ellipse it's: $$\frac{x^2}{25}+\frac{y^2}{9}=1.$$

Also, from the given: $y_{P}=\frac{3}{2}.$

Can you end it now?

I got $PC=\sqrt{21}.$

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