2007 AMC 12B Problem 14

Problem 14 of 25IntermediateGeometry

Point

P

is inside equilateral

\triangle ABC.

Points

Q,

R,

and

S

are the feet of the perpendiculars from

P

to

\overline{AB},

\overline{BC},

and

\overline{CA},

respectively. Given that

PQ=1,

PR=2,

and

PS=3,

what is

AB?

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Solution

Let

s=AB.

Joining

P

to the vertices splits the triangle into

\triangle PAB,

\triangle PBC,

and

\triangle PCA,

with areas

\tfrac{s}{2},

s,

and

\tfrac{3s}{2}.

Their total is

3s,

which must equal the area

\tfrac{\sqrt3}{4}s^2

of the equilateral triangle. So

3s=\dfrac{\sqrt3}{4}s^2,

giving

s=\dfrac{12}{\sqrt3}=4\sqrt3.

Thus, the correct answer is D.

More practice

Concepts: area decomposition · equilateral triangle · triangle area

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.