Triangle trials

We find $AD$ using a formula for the medians derived from Stewart's Theorem:

$AD=\dfrac{\sqrt{2(AB^2+AC^2)-BC^2}}{2}$

$AD=\dfrac{\sqrt{2(15^2+25^2)-18^2}}{2}$

$AD=\dfrac{\sqrt{1376}}{2}=2\sqrt{86}$

Now, since the medians cut themselves in the centroid in a ratio of $2:1$ from the vertex to the midpoint, we have:

$GD=\dfrac{AD}{3}=\dfrac{2\sqrt{86}}{3}$

Hence, $a=2$ , $b=3$ , $d=86$ and $a+b+d+1=\boxed{92}$ .

B is an obtuse angle. From A drop perpendicular AE to meet CB extended at E..
Let AE=h , AD=x, BE=n.
$\\~\\ \displaystyle{ h^2 =15^2 - n^2~~also~~=25^2 - (18+n )^2 ~~~~~~~\therefore~n=\dfrac{19}{9}.\\ \therefore~~~x^2=h^2 +(n+9)^2=15^2 - n^2+(n+9)^2} \\$
$\displaystyle{\implies ~ x^2=225+18n+81=344=(2*\sqrt{ 86 })^2. \\GD=x/3=\dfrac{2}{3}* \sqrt {86}\\ a+b+d+1=2+3+86+1=92 } \\ \boxed {92}$

By Stewart's theorem, man+dad = bmb+cnc, where m = BD, n = DC, a = BC, b = AC, c = AB, and d = AD. So

18 * 9^2+18 * d^2 = 9 * 25^2+9 * 15^2,

d = 2 * sqrt(86).

Therefore GD = d/3 = 2 * sqrt(86)/3. Hence the answer is 2+3+86+1 = 92.

Use Appolonius theorem,find the median 2 * sqrt 86,then 1/3 of it is (2 * 86)/3,so 2+3+86+1=92

Carnot Theorem applied to angle(A) = angle(CAB), leads to:

cos(A) = (|AB|^2 + |AC|^2 - |BC|^2)/(2|AB||AC|);

cos(A) = (15^2 + 25^2 - 18^2)/(2 15 25) = 263/375;

if we denote C1 the projection on axis x of a system of coordinates

(O,x,y),

O := A, x // AB, y ⊥ x in A,

|CC1| = |AC|sin(A) = 25√(1 - (263/375)^2) = √(375^2 - 263^2)/15;

the coordinates of point C are:

xc = |AC|cos(A) = 25*263/375 = 263/15;

yc = |CC1| = √(375^2 - 263^2)/15;

the coordinates of the centroid G are:

xg = (xa + xb + xc)/3 = (0 + 15 + 263/15)/3 = 488/45;

yg = (ya + yb + yc)/3 = (0 + 0 + √(375^2 - 263^2)/15)/3 = √(375^2 -

263^2)/45;

the midpoint D between B and C has coordinates:

xd = (xb + xc)/2 = (15 + 263/15)/2 = 488/30;

yd = (yb + yc)/2 = (0 + √(375^2 - 263^2)/15)/2 = √(375^2 - 263^2)/30;

|AD| = √((xd - xa)^2 + (yd - ya)^2);

|AD| = √((488/30)^2 + (375^2 - 263^2)/30^2) = 2√86;

|GD| = |AD|/3 = (2/3)√86 = (a/b)√d;

the question is the value of a + b + d + 1, so:

a + b + d + 1 = 2 + 3 + 86 + 1 = 92

The coordinates of centroid $G$ are given by:

$x_G = \cfrac{x_A+x_B+x_C}{3}, \quad \quad y_G = \cfrac{y_A+y_B+y_C}{3}$

Let $B(0,0)$ , then $C(18,0)$ and $D(9,0)$ . We have:

$x_A^2+y_A^2 = AB^2 = 15^2 = 225\quad ...(1)$

$(18-x_A)^2+y_A^2 = AC^2 = 25^2 = 625\quad ...(2)$

$(2)-(1): 342 - 36x_A = 400 \quad \Rightarrow x_A = \cfrac{-76}{36} = \cfrac{-19}{9}$

Substitute $x_A$ in equation (1):

$\left(\cfrac{-19}{9}\right)^2 + y_A^2 = 225\quad \Rightarrow y_A = \sqrt{225-\cfrac{361}{81}} = \cfrac{2}{9}\sqrt{4466}$

Therefore,

$x_G = \cfrac{\frac{-19}{9}+0+18}{3}\quad \quad y_G = \cfrac{\frac{2}{9}\sqrt{4466}+0+0}{3}$

$\quad \quad = \cfrac{143}{27}\quad \quad \quad \quad \quad \quad \quad = \cfrac{2\sqrt{4466}}{27}$

The length $GD = \sqrt{(x_G-x_D)^2+(y_G-y_D)^2}$

$\quad \quad \quad \quad \quad \quad = \sqrt{(\frac{143}{27}-9)^2+(\frac{2\sqrt{4466}}{27}-0)^2}$

$\quad \quad \quad \quad \quad \quad = \sqrt{(\frac{-100}{27})^2+(\frac{2\sqrt{4466}}{27})^2}$

$\quad \quad \quad \quad \quad \quad = \cfrac{1}{27} \sqrt{10000+17864}$

$\quad \quad \quad \quad \quad \quad = \cfrac{1}{27} \sqrt{27864} = \cfrac{18}{27} \sqrt{86} = \cfrac{2}{3} \sqrt{86}$

Therefore, $a=2,\space b=3,\space c=1,\space d=86$ .

$\Rightarrow a+b+c+d = \boxed{92}$