Bending A Triangle

Since $\triangle ABC$ and $\triangle ABP$ share the same base $AB$ and have the same perimeter, we can fix the points $A$ & $B$ as the foci and attach imaginary threads to draw an elliptical curve as shown below:

For any point $P$ on the ellipse graph and two foci points $F_{1}$ & $F_{2}$ , the sum of $P$ $F_{1}$ + $P$ $F_{2}$ is always constant. If point $P$ is not on the ellipse graph, the perimeter of $\triangle ABP$ will no longer equal to that of $\triangle ABC$ .

In this case, the total distance of $AC + CB$ (blue) = distance of $AP + PB$ (red), and we can formulate the graph as: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

Then the focal length is half of $AB = 4$ , and $b$ is the height of $\triangle ABC$ = $3$ . Then $a^2 - b^2 = 4^2; a^2 = 3^2 + 4^2; a = 5$ .

Therefore, the formula for this ellipse is $\dfrac{x^2}{5^2} + \dfrac{y^2}{3^2} = 1$ .

Since the area of $\triangle ABP$ is half of the $\triangle ABC$ , the height (also the distance from the $x$ -axis) is half of the original one; the height is $\dfrac{3}{2}$ , which is the $y$ value.

Plugging in this $y$ value, we will get $y^2 = (\dfrac{3}{2})^2 = \dfrac{9}{4}$ :

$\dfrac{x^2}{5^2} + \dfrac{\frac{9}{4}}{3^2} = 1$ ; $\ x^2 = \dfrac{75}{4}$

As a result, the distance of $CP$ = $\sqrt{\dfrac{75}{4} + (3 - \dfrac{3}{2})^2}$ = $\sqrt{\dfrac{75+9}{4}} = \sqrt{21}$ .