Difference between revisions of "2007 AMC 12B Problems/Problem 14"

Revision as of 23:18, 21 February 2008

Problem 14

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

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:

$\frac{s(1)}{2} + \frac{s(2)}{2} + \frac{s(3)}{2} = \frac{s^2\sqrt{3}}{4}$ where $s$ is the length of a side

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

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

@@ Line 12: / Line 12: @@
 <math>s = 4\sqrt{3} \Rightarrow \mathrm {(D)}</math>
+==See Also==
+{{AMC12 box|year=2007|ab=B|num-b=13|num-a=15}}

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Difference between revisions of "2007 AMC 12B Problems/Problem 14"

Revision as of 23:18, 21 February 2008

Problem 14

Solution

See Also