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

(New page: ==Problem 14== Point <math>P</math> is inside equilateral <math>\triangle ABC</math>. Points <math>Q</math>, <math>R</math>, and <math>S</math> are the feet of the perpendiculars from <mat...)
 
m (spaced out answer choices)
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Point <math>P</math> is inside equilateral <math>\triangle ABC</math>. Points <math>Q</math>, <math>R</math>, and <math>S</math> are the feet of the perpendiculars from <math>P</math> to <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math>, respectively. Given that <math>PQ=1</math>, <math>PR=2</math>, and <math>PS=3</math>, what is <math>AB</math>?
 
Point <math>P</math> is inside equilateral <math>\triangle ABC</math>. Points <math>Q</math>, <math>R</math>, and <math>S</math> are the feet of the perpendiculars from <math>P</math> to <math>\overline{AB}</math>, <math>\overline{BC}</math>, and <math>\overline{CA}</math>, respectively. Given that <math>PQ=1</math>, <math>PR=2</math>, and <math>PS=3</math>, what is <math>AB</math>?
  
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<math>\mathrm {(A)} 4</math>  <math>\mathrm {(B)} 3\sqrt{3}</math>  <math>\mathrm {(C)} 6</math>  <math>\mathrm {(D)} 4\sqrt{3}</math>  <math>\mathrm {(E)} 9</math>
+
<math>\mathrm{(A)}\ 4 \qquad \mathrm{(B)}\ 3\sqrt{3} \qquad \mathrm{(C)}\ 6 \qquad \mathrm{(D)}\ 4\sqrt{3} \qquad \mathrm{(E)}\ 9</math>
  
 
==Solution==
 
==Solution==

Revision as of 23:07, 20 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

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

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