2016 AMC 12A Problem 15

Question

Circles with centers

P,

Q,

and

R,

having radii

1,

2,

and

3,

respectively, lie on the same side of line

l

and are tangent to

l

at

P',

Q',

and

R',

respectively, with

Q'

between

P'

and

R'.

The circle with center

Q

is externally tangent to each of the other two circles. What is the area of

\triangle PQR?

0

\sqrt{\dfrac{2}{3}}

1

\sqrt{6}-\sqrt{2}

\sqrt{\dfrac{3}{2}}

The horizontal distance between two tangent points is

\sqrt{(r_1+r_2)^2-(r_2-r_1)^2}

Place

P,Q,R

at heights

1,2,3

and use the shoelace formula

The centers lie at heights

1,

2,

and

3

above line

l.

Since circle

Q

is externally tangent to circle

P,

we have

PQ=3,

so the horizontal distance is

P'Q'=\sqrt{3^2-1^2}=\sqrt{8}.

Since circle

Q

is tangent to circle

R,

we have

QR=5,

so

Q'R'=\sqrt{5^2-1^2}=\sqrt{24}.

Place

P=(0,1),

Q=(\sqrt8,2),

and

R=(\sqrt8+\sqrt{24},3).

By the shoelace formula, the area is

\begin{gathered} \small \dfrac12\left|\sqrt8(3-1)+(\sqrt8+\sqrt{24})(1-2)\right|\\ =\dfrac12\left(\sqrt{24}-\sqrt8\right)\\ =\sqrt6-\sqrt2. \end{gathered}

Thus, the correct answer is D.

Problem 15 in other years