2007 AMC 12B Problem 14

Question

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?

4

3\sqrt{3}

6

4\sqrt{3}

9

Split

\triangle ABC

into

\triangle PAB,

\triangle PBC,

\triangle PCA,

using the perpendiculars as heights

Their areas sum to

\tfrac12 s(1+2+3),

which also equals

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

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.

Problem 14 in other years