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]]}} | {{duplicate|[[2007 AMC 12B Problems|2007 AMC 12B #14]] and [[2007 AMC 10B Problems|2007 AMC 10B #17]]}} | ||
− | ==Problem | + | ==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>? | 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> | <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>. | 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: | Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get: | ||
− | <cmath>\frac{s | + | <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> | <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== | ==See Also== | ||
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{{AMC10 box|year=2007|ab=B|num-b=16|num-a=18}} | {{AMC10 box|year=2007|ab=B|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:33, 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 is inside equilateral
. Points
,
, and
are the feet of the perpendiculars from
to
,
, and
, respectively. Given that
,
, and
, what is
?
Solution 1
Drawing ,
, and
,
is split into three smaller triangles. The altitudes of these triangles are given in the problem as
,
, and
.
Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get:
where
is the length of a side of the equilateral triangle
- 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 to
be
. Also, since
, let the foot of the altitude from
to
be
. Since
is 30-60-90, and
,
. Also, since
,
. Thus,
. Once again, since
is 30-60-90,
. Similar reasoning to
and summing the segments yields
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 then multiply 2, and we get an answer of
~ghfhgvghj10
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.