Difference between revisions of "2007 AMC 12B Problems/Problem 14"
(→Solution) |
|||
Line 1: | Line 1: | ||
{{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>? | ||
Revision as of 16:34, 15 February 2021
- 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
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
- Note - This is called Viviani's Theorem.
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.