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Difference between revisions of "2007 AMC 12B Problems/Problem 14"
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{{duplicate|[[2007 AMC 12B Problems|2007 AMC 12B #14]] and [[2007 AMC 10B Problems|2007 AMC 10B #17]]}}
 
==Problem 14==
 
==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 <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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Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get:
 
Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get:
  
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<math>\frac{s(1)}{2} + \frac{s(2)}{2} + \frac{s(3)}{2} = \frac{s^2\sqrt{3}}{4}</math> where <math>s</math> is the length of a side
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<cmath>\frac{s(1)}{2} + \frac{s(2)}{2} + \frac{s(3)}{2} = \frac{s^2\sqrt{3}}{4}</cmath>
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where <math>s</math> is the length of a side
  
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<math>s = 4\sqrt{3} \Rightarrow \mathrm {(D)}</math>
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<cmath>s = \boxed{\mathrm{(D) \ } 4\sqrt{3}}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2007|ab=B|num-b=13|num-a=15}}
 
{{AMC12 box|year=2007|ab=B|num-b=13|num-a=15}}
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{{AMC10 box|year=2007|ab=B|num-b=16|num-a=18}}

Revision as of 15:30, 5 June 2011

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

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

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

See Also

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
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