Triangle-Ception III

If we let $x = BC$ , then we have that

$\dfrac{AD}{BD} = \dfrac{14 - x}{6} = \dfrac{AB}{BC} = \dfrac{8-x}{x}$

$14 x - x^2 = 48 - 6 x$

$x^2 - 20 x + 48 = 0$

$(x - 10)^2 = 100 - 48 = 52$

x = 10 - 2 $\sqrt{13}$ = 2.7889 (Where 6 - x > 0)

AD = 14 - x = 14 - (10 - 2 $\sqrt{13}$ ) = $4 + 2 \sqrt{13}$

Therefore, $m + n + p = 4 + 2 + 13 = 19$

Note that the triangle is not fixed as E and F are uncertain. Only the dimensions on the base line are fixed. Despite this, the proportion of ratios can be applied.

First, let's continue the pattern of similar triangles from $C$ towards point $D$ . These similar triangles cannot ever reach exactly point $D$ due to the construction of the diagram, but they will get closer and closer.

Due to the similarity of these infinitely many triangles, we have an infinite geometric progression of bases, whose sum is equal to $AD$ . Let's call the first base $AB$ . Let $AB=a$ , and $AD=x$ . We know the value of our infinite geometric series:

$x=\frac{a}{1-r}$

We also know that $x=a+6$ due to given information, so we have

$a+6=\frac{a}{1-r}$

We also can tell that the base of the second similar triangle in our progression is $ar$ , so we know from the given information that:

$a+ar=8$

Now we have a system of two equations and two variables. Solving for $a$ gives:

$0=a^{2}+4a-48$

Quadratic formula gives:

$a=\frac{-4 \pm 4\sqrt{13}}{2}$

We can ignore the minus solution, since $a>0$ due to it being a side length. We are left with:

$a=2\sqrt{13}-2$ .

$AD=x=a+6=4+2\sqrt{13}$ .

$4,2,13$ are all positive integers and $13$ is not divisible by the square of any prime, so our answer is:

$\large 2+4+13=\boxed{19}$

Also, the angle I gave you was just a red herring. It doesn't matter what the angles are in the similar triangles. For a visual proof, check out this desmos graph I made Adjust the values of slider $B$ and $C$ . The angles change, but the intersection with the x axis does not.

This is only to give a different approach. Lu Chee Ket's is a beautiful method. $Let ~~~~BC=x,~~~and ~~~~CD=y, ~~~~\dfrac{AB}{BC}=k\\ \therefore~~\color{#3D99F6}{x=\dfrac 8 {k+1},~~~~x+y=6.}\\ \angle FAB = \angle EBC~~\therefore FA || EB.\\ \therefore~~~ k = \dfrac {FA}{EB}=\dfrac{AD}{BD}=\dfrac{8+y}{6} ~~~\implies \color{#3D99F6}{ y=6k -8}.\\ 6=x+y=\dfrac 8 {k+1} + 6k-8.~~~\\ \implies 3k^2- 4k - 3=0.~~ Solving~ for~ po sitive~ k,\\k=\dfrac{4+\sqrt{16 + 36}}{2*3}=\color{#EC7300}{\dfrac{2 + \sqrt{13}}{3}. } \\ AD=14 - x =14 - \dfrac 8 {k+1} =14 - \dfrac 8 { \frac{2 + \sqrt{13}}{3} +1} \\$ $Rationalising~~AD=14 - \dfrac{ 24*( 5 - \sqrt{13} ) } {25-13}=14-10+2*\sqrt {13}=m+n*\sqrt p\\ m+n+p=\Large~~~\color{#D61F06}{19}$