Question

Circles $C_1,C_2,C_3$ have their centres at $(0,0),(12,0),(24,0)$ and have radii $1,2$ and $4$ respectively. Line $t_1$ is a common internal tangent to $C_1$ and $C_2$ and has a positive slope and line $t_2$ is a common internal tangent to $C_2$ and $C_3$ and has a negative slope. Given that lines $t_1$ and $t_2$ intersect at $(x,y)$ and that $x=p-q\surd r$, where $p,q$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r$.

• A. $p+q+r=26$
• B. $p+q+r=24$
• C. $p+q+r=28$
• D. $p+q+r=27$

Answer 1

The correct option is D $p+q+r=27$

Call the centers $O_1,O_2,O_3$, the points of tangency$r_1,r_2,s_1,s_2$

(with r on $t_1$ and s on $t_2$, and $s_2$ on $C_2$),

and the intersection of each common internal tangent to the X-axis r, s.

$△O_1{r}_{1}r\sim △O_2{r}_{2}r$

since both triangles have a right angle and have vertical angles,

and the same goes for $△O_2{s}_{2}s\sim △O_3{s}_{1}s$.

By proportionality, we find that $O_{1}r=4$; s

olving $△O_1{r}_{1}r$ by the Pythagorean theorem yields $r_{1}r=\sqrt{15}$.

On $C_3$, we can do the same thing to get $O_3{s}_1=4$ and $s_{1}s=4\sqrt{3}$.

The vertical altitude of each of $△O_1{r}_{1}r$

and $△O_3{s}_{1}s$ can each by found by the formula

$c\cdot h=a\cdot b$ (as both products equal twice of the area of the triangle).

Thus, the respective heights are $\frac{\sqrt{15}}{4}$ and $2\sqrt{3}$.

The horizontal distance from each altitude to the intersection

of the tangent with the x-axis can also

be determined by the Pythagorean theorem:

$\sqrt{15-\frac{15}{16}}=\frac{15}{4}$, and by 30-60-90: $6$.

From this information, the slope of each tangent can be uncovered.

The slope of $t_1 = \dfrac{\Delta y}{\Delta x} = \dfrac{d\frac{\sqrt{15}}{4}}{\dfrac{15}{4}}

= \dfrac{1}{\sqrt{15}}$

The slope of $t_2=-\frac{2\sqrt{3}}{6}=-\frac{1}{\sqrt{3}}$.

The equation of $t_1$ can be found by substituting the point

$r(4,0)$ into $y=\frac{1}{\sqrt{15}}x+b$,

so $y=\frac{1}{\sqrt{15}}x-\frac{4}{\sqrt{15}}$.

The equation of $t_2$, found by substituting point $s(16,0)$,

is $y=\frac{-1}{\sqrt{3}}x+\frac{16}{\sqrt{3}}$.

Putting these two equations together results in the desired

$\frac{1}{\sqrt{15}}x-\frac{4}{\sqrt{15}}=-\frac{1}{\sqrt{3}}x+\frac{16}{\sqrt{3}}$

$⟹x=\frac{16\sqrt{5}+4}{\sqrt{5}+1}\cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}$

$=\frac{76-12\sqrt{5}}{4}$ $=19-3\sqrt{5}$

Thus, $p+q+r=19+3+5⟹027$