2007 AMC 12B Problems/Problem 14
- 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 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
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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.