Difference between revisions of "1950 AHSME Problems/Problem 48"
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A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is: | A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is: | ||
− | <math>\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ | + | <math> \textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle}\qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity} </math> |
− | \textbf{(B)}\ \text{Greater than the altitude of the triangle} \qquad\\ | ||
− | \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ | ||
− | \textbf{(D)}\ \text{One-half the sum of the sides of the triangle} \qquad\\ | ||
− | \textbf{(E)}\ \text{Greatest when the point is the center of gravity}</math> | ||
==Solution== | ==Solution== |
Latest revision as of 01:16, 2 November 2023
Contents
Problem
A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:
Solution
Begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be with
. We will call the aforementioned point
. Call altitude from
to
. Similarly, we will name the other two altitudes
and
. We can see that
Where h is the altitude. Multiplying both sides by
and dividing both sides by
gives us
The answer is
Note
Thie result is exactly the Viviani theorem.
Video Solution
https://www.youtube.com/watch?v=l4lAvs2P_YA&t=251s
~MathProblemSolvingSkills.com
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 47 |
Followed by Problem 49 | |
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All AHSME Problems and Solutions |
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