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

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<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>
 
<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>
  
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==Solution==
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==Solution 1==
 
Drawing <math>\overline{PA}</math>, <math>\overline{PB}</math>, and <math>\overline{PC}</math>, <math>\triangle ABC</math> is split into three smaller triangles. The altitudes of these triangles are given in the problem as <math>PQ</math>, <math>PR</math>, and <math>PS</math>.
 
Drawing <math>\overline{PA}</math>, <math>\overline{PB}</math>, and <math>\overline{PC}</math>, <math>\triangle ABC</math> is split into three smaller triangles. The altitudes of these triangles are given in the problem as <math>PQ</math>, <math>PR</math>, and <math>PS</math>.
  
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*Note - This is called [[Viviani's Theorem]].
 
*Note - This is called [[Viviani's Theorem]].
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==Solution 2==
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The 60 degree angles suggest constructing 30-60-90 triangles. As such, let the foot of the altitude from <math>Q</math> to <math>\overline{BC}</math> be <math>D</math>. Also, since <math>\angle{PQD}=60</math>, let the foot of the altitude from <math>P</math> to <math>\overline{QD}</math> be <math>E</math>. Since <math>\triangle{QEP}</math> is 30-60-90, and <math>PQ=1</math>, <math>QE=\frac{1}{2}</math>. Also, since <math>PR=2</math>, <math>DE=2</math>. Thus, <math>QD=\frac{5}{2}</math>. Once again, since <math>\triangle{QBD}</math> is 30-60-90, <math>QB=\frac{5\sqrt{3}}{3}</math>. Similar reasoning to <math>\overline{AQ}</math> and summing the segments yields <math>\boxed{(D) 4\sqrt{3}}</math>
  
 
==See Also==
 
==See Also==

Revision as of 19:51, 26 July 2022

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

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

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 $\boxed{(D) 4\sqrt{3}}$

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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