2007 AMC 10B Problem 17

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\sqrt3

6

4\sqrt3

9

Connect

P

to the three vertices to split

\triangle ABC

into three smaller triangles

Their areas are

\dfrac{s}{2},s,\dfrac{3s}{2}

; set the sum equal to

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

Let the side length be

s.

The perpendiculars from

P

are the heights of triangles

APB,

BPC,

and

CPA,

so their areas are

\dfrac{s}{2},

s,

and

\dfrac{3s}{2}.

Their sum equals the area of

\triangle ABC,

which is also

\dfrac{\sqrt3}{4}s^2.

Hence

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

The positive solution is

s=4\sqrt3.

Thus, the correct answer is D.

Problem 17 in other years