2016 AMC 12A Problem 15

Problem 15 of 25IntermediateGeometry

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

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?

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Solution

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,

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.