2007 AMC 12B Problems/Problem 14

The following problem is from both the 2007 AMC 12B #14 and 2007 AMC 10B #17, so both problems redirect to this page.

Problem

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 $AB$ ?

$\mathrm{(A)}\ 4 \qquad \mathrm{(B)}\ 3\sqrt{3} \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 4\sqrt{3} \qquad \mathrm{(E)}\ 9$

Solution 1

Drawing $\overline{PA}$ , $\overline{PB}$ , and $\overline{PC}$ , $\triangle ABC$ is split into three smaller triangles. The altitudes of these triangles are given in the problem as $PQ$ , $PR$ , and $PS$ .

Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get:

$\frac{s}{2} + \frac{2s}{2} + \frac{3s}{2} = \frac{s^2\sqrt{3}}{4}$ where $s$ is the length of a side of the equilateral triangle

$s = \boxed{\mathrm{(D) \ } 4\sqrt{3}}$

Note - This is called Viviani's Theorem.

Solution 2

The 60 degree angles suggest constructing 30-60-90 triangles. As such, let the foot of the altitude from $Q$ to $\overline{BC}$ be $D$ . Also, since $\angle{PQD}=60$ , let the foot of the altitude from $P$ to $\overline{QD}$ be $E$ . Since $\triangle{QEP}$ is 30-60-90, and $PQ=1$ , $QE=\frac{1}{2}$ . Also, since $PR=2$ , $DE=2$ . Thus, $QD=\frac{5}{2}$ . Once again, since $\triangle{QBD}$ is 30-60-90, $QB=\frac{5\sqrt{3}}{3}$ . Similar reasoning to $\overline{AQ}$ and summing the segments yields $\boxed{(D) 4\sqrt{3}}$

Solution 3 (Viviani's Theorem)

According Viviani's Theorem, In an equilateral triangle, the sum of the perpendicular distances from a point inside the triangle will equal the altitude of the triangle. Thus, the height of the triangle is 6.

Then we start to form a 30-60-90 triangle. By that, we divide $6/\sqrt{3} = 2\sqrt{3}$ then multiply 2, and we get an answer of $\boxed{(D) 4\sqrt{3}}$

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2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2007 AMC 12B Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Solution 3 (Viviani's Theorem)

See Also