Find $\overline{BC}$ in this triangle

Circle $H$ has an equation of $(x - 10)^2 + (y - 5)^2 = 25$ , and using implicit differentiation on this equation with $m = \frac{dy}{dx}$ we can obtain $2(x - 10) + 2(y - 5)m = 0$ . Also, a tangent point on $AB$ and a tangent point on $AC$ to this circle will both fit the equation $y = mx + 16$ .

These three equations solve to $m = -\cfrac{12}{5}$ and $m = -\cfrac{8}{15}$ , so $AB$ has an equation of $y = -\cfrac{12}{5}x + 16$ and $AC$ has an equation of $y = -\cfrac{8}{15}x + 16$ .

Setting $y = 0$ on both of these equations and solving gives $B(\cfrac{20}{3}, 0)$ and $C(30, 0)$ , so that $d = BC = 30 - \cfrac{20}{3} = \cfrac{70}{3}$ , and $3d = \boxed{70}$ .

Let the tangent AB intersect the circle at point P, tangent BC intersect at Q, and tangent AC intersect at R. Furthermore, let the origin be O. Before reading on, draw a diagram with these labelled points first!

Firstly, find AP. $AH^2 = (0 - 10)^2 + (16 - 5)^2 = 221$ , and $PH^2 = 5^2 = 25$ (radius), hence $AP = \sqrt{AH^2 - PH^2} = \sqrt{221-25} = 14$ .

Since tangents from the same point have the same length, BP = BQ = x, so OB = OQ - BQ = 10 - x, and AB = AP + PB = 14 + x. Thus by Pythagoras, $16^2 + (10 - x)^2 = (14+x)^2 \Rightarrow 256 + 100 - 20x + x^2 = 196 + 28x + x^2 \Rightarrow x = \frac{10}{3}$ .

Then using the property again, RC = QC = a, so OC = OQ + QC = 10 + a, and AC = AR + RC = 14 + a (AP = AR). Therefore $16 + (10 +a)^2 = (14 + a)^2 \Rightarrow 256 + 100 + 20a + a^2 = 196 + 28a + a^2 \Rightarrow a = 20$ .

Thus BC = BQ + QC = x + a, so $3 BC = 3 \left(20 + \frac{10}{3} \right) = \boxed{70}$ .

Let the origin be $O(0,0)$ , $\angle CAH =\angle BAH = \alpha$ and $\angle OAH = \beta$ . Then $\sin \alpha = \dfrac {HD}{AH} = \dfrac 5{\sqrt{(10-0)^2+(5-16)^2}} = \dfrac 5{\sqrt{221}} \implies \tan \alpha = \dfrac 5{14}$ . And $\tan \beta = \dfrac {10}{16-5} = \dfrac {10}{11}$ . Then we have:

$\begin{aligned} OC & = AO \cdot \tan \angle CAO = 16 \cdot \tan (\beta+\alpha) = 16 \cdot \frac {\frac {10}{11}+\frac 5{14}}{1-\frac {10}{11}\frac 5{14}} = 30 \\ OB & = AO \cdot \tan \angle BAO = 16 \cdot \tan (\beta-\alpha) = 16 \cdot \frac {\frac {10}{11}-\frac 5{14}}{1+\frac {10}{11}\frac 5{14}} = \frac {20}3 \\ d & = OC - OB = \frac {70}3 \\ \implies 3d & = \boxed{70} \end{aligned}$

First, we'll find the distance $\overline{AH}$

$\overline{AH} = \sqrt{ 10^2 + 11^2 } = \sqrt{ 221 }$

Then the two tangents $AB$ and $AC$ are separated by an angle $2 \theta$ , where

$\theta = \sin^{-1} ( \dfrac{5} { \sqrt{221} })$

from which, $\cos \theta = \sqrt{ 1 - \dfrac{25}{221} } = \dfrac{\sqrt{ 196 } }{ \sqrt{221}}$

Here is the key to this problem. The vector extending from point $A$ to a point on either of the two tangents makes an angle $\theta$ with the vector $\vec{AH}$ . Let $\vec{r} = (x, y)$ be any point on either of the two tangents, then the above statement implies that,

$( \vec{r} - \vec{A} ) \cdot \dfrac{ \vec{AH} } { | \vec{A H} |} = \cos \theta | \vec{r} - \vec{A} |$

Now, the vector $\vec{AH} = (10, -11) = [10 , -11 ]^T$ , so in matrix notation, the above dot product is written as,

$( \mathbf{r} - \mathbf{A} )^T \begin{bmatrix} 10 \\ -11 \end{bmatrix} = \sqrt{196} | \mathbf{r} - \mathbf{A} |$

Squaring,

$(\mathbf{r} - \mathbf{A})^T \begin{bmatrix} 100 && -110 \\ -110 && 121 \end{bmatrix} (\mathbf{r} - \mathbf{A} ) = 196 (\mathbf{r} - \mathbf{A} )^T(\mathbf{r} - \mathbf{A} )$

subtracting the right hand side from the left hand side, we obtain,

$(\mathbf{r} - \mathbf{A})^T \begin{bmatrix} -96 && -110 \\ -110 && -75 \end{bmatrix} (\mathbf{r} - \mathbf{A} ) = 0$

this is the equation of the two tangents.

Plugging in $y = 0$ yields the x-coordinates of points $B$ and $C$ :

$[x, -16]^T \begin{bmatrix} -96 && -110 \\ -110 && -75 \end{bmatrix} \begin{bmatrix} x \\ -16 \end{bmatrix} = 0$

Expanding this quadratic form, we obtain,

$-96 x^2 - 220 (-16)(x) - 75 (-16)^2 = 0$

which simplifies to,

$3 x^2 - 110 x + 600 = 0$

and this factors to,

$(3 x - 20 ) ( x - 30 ) = 0$

Hence, $B = (\dfrac{20}{3}, 0 )$ and $C = (30 , 0 )$

and $d = \overline{BC} = 30 - \dfrac{20}{3} = \dfrac{70}{3}$

Hence, $3 d = \boxed{70}$