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

Latest revision as of 14:33, 1 August 2024

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

~ghfhgvghj10

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.

@@ Line 1: / Line 1: @@
 {{duplicate|[[2007 AMC 12B Problems|2007 AMC 12B #14]] and [[2007 AMC 10B Problems|2007 AMC 10B #17]]}}
-==Problem 14==
+==Problem==
 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>?
 <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 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>.
 Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get:
-<cmath>\frac{s(1)}{2} + \frac{s(2)}{2} + \frac{s(3)}{2} = \frac{s^2\sqrt{3}}{4}</cmath>
+<cmath>\frac{s}{2} + \frac{2s}{2} + \frac{3s}{2} = \frac{s^2\sqrt{3}}{4}</cmath>
-where <math>s</math> is the length of a side
+where <math>s</math> is the length of a side of the equilateral triangle
 <cmath>s = \boxed{\mathrm{(D) \ } 4\sqrt{3}}</cmath>
+*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 <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>
+==Solution 3 (Viviani's Theorem)==
+According to 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 <math>6/\sqrt{3} = 2\sqrt{3}</math> then multiply 2, and we get an answer of <math>\boxed{(D) 4\sqrt{3}}</math>
+~ghfhgvghj10
 ==See Also==

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